Research ArticleOn Equalities Involving Integrals of the Logarithm ofthe Riemann 120589-Function with Exponential Weight WhichAre Equivalent to the Riemann Hypothesis

Sergey K Sekatskii1 Stefano Beltraminelli2 and Danilo Merlini2

1Laboratoire de Physique de laMatiere Vivante IPSB BSP 408 Ecole Polytechnique Federale de Lausanne 1015 Lausanne Switzerland2Research Center for Mathematics and Physics (CERFIM) PO Box 1132 6600 Locarno Switzerland

Correspondence should be addressed to Sergey K Sekatskii sergueisekatskiepflch

Received 30 April 2015 Revised 4 September 2015 Accepted 6 September 2015

Academic Editor Remi Leandre

Copyright copy 2015 Sergey K Sekatskii et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

Integral equalities involving integrals of the logarithm of the Riemann 120589-functionwith exponential weight functions are introducedand it is shown that an infinite number of them are equivalent to the Riemann hypothesis Some of these equalities are testednumerically The possible contribution of the Riemann function zeroes nonlying on the critical line is rigorously estimated andshown to be extremely small in particular smaller than nine milliards of decimals for the maximal possible weight functionexp(minus2120587119905) We also show how certain Fourier transforms of the logarithm of the Riemann zeta-function taken along the real(demi)axis are expressible via elementary functions plus logarithm of the gamma-function and definite integrals thereof as well ascertain sums over trivial and nontrivial Riemann function zeroes

1 Introduction

In recent papers [1 2] we analyzed certain contour integralsinvolving the logarithm of the Riemann zeta-function andhave established an infinite number of equalities of thetype int119887+119894infin

119887minus119894infin119892(119911) ln(120589(119911))119889119911 = 119891(119887) which were proven to

be equivalent to the Riemann hypothesis (RH 120589(119911) is theRiemann zeta-function see eg [3] for definitions anddiscussion of the general properties of this function) Inparticular it was shown that all earlier known equalities ofthis type that is those of Wang [4] Volchkov [5] Balazardet al [6] and one of us [7] are certain particular cases of ourgeneral approach elaborated in [1]

In this paper we establish new integral equalities equiva-lent to RH We use exponential weight functions and in ouropinion the resulting equations are especially interesting Inparticular we were able to rigorously estimate the possiblecontribution of the Riemann function zeroes nonlying onthe critical line which were shown to be extremely smallfor example smalle8r than nine milliards of decimals for the

ldquomaximal possiblerdquo in a sense (see below) weight function119892(119911) = exp(minus2120587119911)

2 Integral Equalities with Exponential WeightFunction Equivalent to the RiemannHypothesis

The main tool for our work here is the following generalizedLittlewood theorem about contour integrals involving loga-rithm of an analytical function

Theorem 1 (the generalized Littlewood theorem) Let 119862denote the rectangle bounded by the lines 119909 = 1198831 119909 = 1198832 119910 =1198841 119910 = 1198842 where 1198831 lt 1198832 1198841 lt 1198842 and let 119891(119911) be analyticand nonzero on 119862 and meromorphic inside it let also 119892(119911) beanalytic on 119862 and meromorphic inside it Let 119865(119911) = ln(119891(119911))be the logarithm defined as follows one starts with a particulardetermination on 119909 = 1198832 and obtains the value at other pointsby continuous variation along 119910 = const from ln(1198832 + 119894119910) Ifhowever this path would cross a zero or pole of 119891(119911) one takes

Hindawi Publishing CorporationInternational Journal of AnalysisVolume 2015 Article ID 980728 8 pageshttpdxdoiorg1011552015980728

2 International Journal of Analysis

119865(119911) to be 119865(119911 plusmn 1198940) accordingly as one approaches the pathfrom above or below Let also 119865(119911) = ln(119891(119911)) be the logarithmdefined by continuous variation along any smooth curve fullylying inside the contour which avoids all poles and zeroes of119891(119911) and starts from the same particular determination on119909 = 1198832 Suppose also that the poles and zeroes of the functions119891(119911) 119892(119911) do not coincide

Then int119862119865(119911)119892(119911)119889119911 = 2120587119894(sum

120588119892119903119890119904(119892(120588119892) sdot 119865(120588119892)) minus

sum1205880119891

int1198830

120588+1198941198840

120588

1198831+1198941198840

120588

119892(119911)119889119911 + sum120588119901119900119897

119891

int119883119901119900119897

120588+119894119884119901119900119897

120588

1198831+119894119884119901119900119897

120588

119892(119911)119889119911) where the sumis over all 120588119892 which are poles of the function 119892(119911) lying inside119862 all 1205880

119891= 1198830

120588+ 1198941198840

120588which are zeroes of the function

119891(119911) counted taking into account their multiplicities (ie thecorresponding term is multiplied by 119898 for a zero of the order119898) and which lie inside 119862 and all 120588119901119900119897

119891= 119883119901119900119897

120588+ 119894119884119901119900119897

120588which

are poles of the function 119891(119911) counted taking into accounttheir multiplicities and which lie inside 119862 The assumption isthat all relevant integrals in the right-hand side of the equalityexist

Proof Our proof closely follows the well-known proof of theLittlewood theorem (or lemma) given for example in [8 p133] Consider first the function119891(119911) = 119911minus119886where 119886 = 120572+119894120573is a point of the rectangle Let 1198621015840 be the contour obtained bydescribing 119862 in the positive direction from (1198832 1198841) as far as(1198831 120573) then the straight line 119910 = 120573 as far as 120572 minus 120576 + 119894120573 thena circle of radius 120576 about 119911 = 119886 and then returning along119910 = 120573 and the rest of 119862 to the starting point see Figure 1The only poles of 119865(119911)119892(119911) in 1198621015840 are those of the function119892(119911) so that int

1198621015840119865(119911)119892(119911)119889119911 = 2120587119894sum

120588119892res(119892(120588119892) sdot 119865(120588119892)) The

integral round the small circle tends to zero with the radiusthus we have int

119862119865(119911)119892(119911)119889119911 = 2120587119894sum

120588119892res(119892(120588119892) sdot 119865(120588119892)) minus

int120572+119894120573

1198831+119894120573119892(119911)(1198651(119911) minus 1198652(119911))119889119911 where 1198651 and 1198652 are the values

of 119865 on the two paths joining 1198831 + 119894120573 to 120572 + 119894120573 Hence weobtain 1198652 from 1198651 by passing in the negative direction rounda simple zero of 119891(119911) at 119911 = 119886 we have 1198652(119911) = 1198651(119911) minus 2120587119894and correspondingly int

119862119865(119911)119892(119911)119889119911 = 2120587119894(sum

120588119892res(119892(120588119892) sdot

119865(120588119892)) minus int120572+119894120573

1198831+119894120573119892(119911)119889119911) where we introduce a notation 119865(119911)

to distinguish this function from 119865(119911) The general case noweasily follows by addition of terms corresponding to thevarious poles and zeroes of 119891(119911)

From the proof of Theorem 1 illustrated in Figure 1 itfollows that the function 119865(119911) = ln(119891(119911)) occurring inits formulation is not a continuous function along the leftborder of the contour namely along the line connecting thepoints 1198831 + 1198941198841 and 1198831 + 1198941198842 its value jumps on ∓2120587119897 whenwe pass a point1198831 + 119894119884 such that there is an 119897th order zero orpole of the function119891(119911) lying inside the contour (and not onthe integration line) and having an ordinate 119884 This functionis only piecewise continuous This circumstance is presentof course also in the original Littlewood theorem given for119892(119911) = 1 (and is well recognized in its formulation see [8])The value of ln |119891(119911)| is a continuous function along the leftborder of the contour provided of course that there are no

Starting pointof contour integration

Value of the function

Value of the function

X1 X2

Y1

Y2

(F(z) minus 2120587i)g(z)

F(z)g(z)

120572 + i120573

Figure 1 Illustrating the proof of the generalized Littlewoodtheorem

poles or zeroes of the function119891(119911) lying exactly on this sameborder

Now let us consider a rectangular contour 119862 with thevertices 119887 119887 + 119894119883 119887 +119883+ 119894119883 119887 +119883 with real 119887 gt minus2 (with thischoice we avoid the trivial Riemann zeroes) and real 119883 rarr+infin introduce the function 119892(119911) = 119894 exp 119894(119886(119911 minus 119887)) where 119886is real positive and apply the above theorem to the contourintegral int

119862119892(119911) ln(120589(119911))119889119911 Along the line (119887 119887 + 119894119883) we have

an integral int1198830119890minus119886119905lnlowast(120589(119887+119894119905))119889119905 and along the line (119887 119887+119883)

we have an integral 119894 int1198830(cos(119886119909) + 119894 sin(119886119909)) ln(120589(119887 + 119909))119889119909

Due to the known asymptotic behaviour of the Riemannzeta-function [3] in particular ln(120589(119911)) = 119874(2minus119911) for largepositive 119911 these integrals converge while integrals over lines(119883119883 + 119894119883) and (119887 + 119894119883119883 + 119894119883) vanish in the limit of large119883

If 119887 lt 1 on the border of the contour we have the poleof the Riemann function at 119911 = 1 and in the interior ofthe contour we also might have a number of zeroes of theRiemann function We definitely have an infinite numberof zeroes if 119887 lt 12 The order 119897119896 zero of the Riemannfunction 120588 = 120590119896 + 119894119905119896 for 119887 lt 120590119896 according to Theo-rem 1 contributes minus2120587119894119897119896 int

120590119896minus119887

0119890minus119886119905119896(119894 cos(119886119909) minus sin(119886119909))119889119909

= (2120587119897119896119890minus119886119905119896119886)[sin(119886(120590119896 minus 119887)) + 119894(1 minus cos(119886(120590119896 minus 119887)))] to

the contour integral value so finally the aforementioned

International Journal of Analysis 3

generalized Littlewood theorem results in the followingequality

int

infin

0

119890minus119886119905lnlowast (120589 (119887 + 119894119905)) 119889119905 minus int

infin

0

sin (119886119909) ln (120589 (119887 + 119909)) 119889119909

+ 119894 int

infin

0

cos (119886119909) ln (120589 (119887 + 119909)) 119889119909 = 2120587119886

sdot sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896[sin (119886 (120590119896 minus 119887))

+ 119894 (1 minus cos (119886 (120590119896 minus 119887)))] minus120587

119886(sin (119886 (1 minus 119887)) + 119894 (1

minus cos (119886 (1 minus 119887))))

(1)

In the last term here we took into account that the pole 119911 = 1lies exactly onto the contour border whence the coefficient is120587119886 instead of 2120587119886

Now we should remind the reader that the functionlnlowast(120589(119887 + 119894119905)) is not continuous see above (and this is thereason why we put an asterisk sign here) If we want touse a continuous logarithm function instead (which we willdenote simply ln(120589(119887 + 119894119905))) the following straightforwardmodification is to be done If real functions 119891(119909) 119892(119909)are continuous on the segment [119860 119861] and function 119891lowast(119909)coincides with 119891(119909) on [119860 119862] and is equal to 119891(119909) + ℎon (119862 119861] (ℎ is a constant 119860 le 119862 le 119861) we haveint119861

119860119892(119909)119891

lowast(119909)119889119909 = int

119861

119860119892(119909)119891(119909)119889119909 + ℎ int

119861

119862119892(119909)119889119909 and this

is trivially generalized for complex numbers and improperintegrals having counting number of discontinuities For ourparticular case we have considering one 2120587119894119897119896 discontinuityat 119905 = 119905119896 int

infin

0119890minus119886119905arglowast(120589(119887 + 119894119905))119889119905 = intinfin

0119890minus119886119905 arg(120589(119887 + 119894119905))119889119905 +

(2120587119897119896119886)119890minus119886119905119896 and the aforementioned generalization gives

int

infin

0

119890minus119886119905arglowast (120589 (119887 + 119894119905)) 119889119905

= int

infin

0

119890minus119886119905arg (120589 (119887 + 119894119905)) 119889119905 + sum

120588119905119896gt0120590119896gt119887

2120587119897119896

119886119890minus119886119905119896

(2)

Using (2) we get instead of (1)

int

infin

0

119890minus119886119905 ln (120589 (119887 + 119894119905)) 119889119905 minus int

infin

0

sin (119886119909) ln (120589 (119887 + 119909)) 119889119909

+ 119894 int

infin

0

cos (119886119909) ln (120589 (119887 + 119909)) 119889119909 = 2120587119886

sdot sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896[sin (119886 (120590119896 minus 119887))

minus 119894 cos (119886 (120590119896 minus 119887))] minus120587

119886(sin (119886 (1 minus 119887)) + 119894 (1

minus cos (119886 (1 minus 119887))))

(3)

where separation of real and imaginary parts readily givesequalities

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905 minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909

=2120587

119886sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 sin (119886 (120590119896 minus 119887))

minus120587

119886sin (119886 (1 minus 119887))

(4)

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

= minus2120587

119886sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896cos (119886 (120590119896 minus 119887))

minus120587

119886(1 minus cos (119886 (1 minus 119887)))

(5)

Here the argument is a continuous function along the contour(of course provided there are no Riemann zeroes lyingexactly on it) and the following initial values of the argumentshould be taken since we have the simple pole at 119911 = 1 onthe contour we select for real 119909 gt 1 arg(120589(119909)) = 0 and for1 gt 119887 gt minus2 arg(120589(119887 + 119894120576)) = minus120587 where real positive 120576 rarr 0

Equations (4) and (5) are our starting point and their firstapplication is the following theorem

Theorem 2 Equalities

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909 + 120587119886sin (119886 (1 minus 119887))

= 0

(6)

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

+120587

119886(1 minus cos (119886 (1 minus 119887))) = 0

(7)

where 119887 119886 are real positive numbers such that 1 gt 119887 ge 12119886(1 minus 119887) le 120587 for (6) and 119886(1 minus 119887) le 1205872 for (7) hold truefor some 119887 if and only if there are no Riemann function zeroeswith 120590 gt 119887 For 119887 = 12 these equalities are equivalent to theRiemann hypothesis

Proof For all possible Riemann 120589-function zeroes with Re120588 =120590119896 gt 119887 and hence lying inside the contour from our choice of119887 and the known properties of the Riemann function zeroeswe have 0 lt 120590119896minus119887 lt 1minus119887 le 12Then the imposed conditions119886(1minus119887) le 120587 and 119886(1minus119887) le 1205872 guaranty the same signs of all

4 International Journal of Analysis

possible contributions (2120587119886)sum120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 sin(119886(120590119896minus119887))

and minus(2120587119886)sum120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 cos(119886(120590119896minus119887)) from (4) and (5)

and also prevent them from being equal to zeroIn (6) it is interesting for 119887 = 12 to take 119886 = 2120587 thus

eliminating any problem with the integration over real axis(pole of the Riemann function coincides with zero of sine)

int

infin

0

119890minus2120587119905 ln

1003816100381610038161003816100381610038161003816120589 (1

2+ 119894119905)

1003816100381610038161003816100381610038161003816119889119905

minus int

infin

0

sin (2120587119909) ln1003816100381610038161003816100381610038161003816120589 (1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 = 0

(6a)

Similarly for (7) the choice 119886 = 120587 is useful

int

infin

0

119890minus120587119905 arg(120589 (1

2+ 119894119905)) 119889119905

+ int

infin

0

cos (120587119909) ln1003816100381610038161003816100381610038161003816120589 (1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 + 1 = 0

(7a)

Equalities (6a) (7a) are equivalent to the Riemannhypothesis They have been tested numerically using thestandard procedure Integrate in Mathematica Because thefirst integrand here decays very fast we can limit the inte-gration interval of the variable 119905 to [0 50] We need alsoto reduce the working precision (set to 50) and maximalrecursion (set to 70) For the second integral in (6a) and (7a)we set the integration interval for the variable 119909 to [0 200]and set the working precision and maximal recursion to50 After 15 hours of CPU time we obtained for (6a) theresult of 88044282 times 10minus32 and for (7a) after more than8 hours 114767406 times 10minus31 So in both cases we have acorrespondence for at least the first 30 figures

Next two similar theorems can be proved when we selectleft contour border line (119887 119887 + 119894infin) lying to the left to thecritical line

Theorem 2a Equality

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909 + 120587119886sin (119886 (1 minus 119887))

= 0

(8)

where 119886 119887 are real positive numbers such that 14 ge 119887 gt 0and 119886 = 120587119899(12 minus 119887) where 119899 is a positive integer such that119899 le 12119887 minus 1 holds true for some 119887 if and only if there are noRiemann function zeroes with 120590 le 119887

Proof In conditions of the theorem all contributions ofthe Riemann function zeroes lying on the critical line(2120587119886)sum

120588119905119896gt0120590119896gt119887119897119896119890minus119886119905119896 sin(119886(12 minus 119887)) to the contour inte-

gral value are equal to zero If there are a pair of Riemannfunction zeroes 120588 = 120590119896+119894119905119896 120588 = 1minus120590119896+119894119905119896 with 12 gt 120590119896 gt 119887both of them lie inside the contour and their contributionsexactly compensate each other thus contributing nothing tothe contour integral value If there is a zero with 120590119896 le 119887 it lies

outside the contour or on its border and hence contributesnothingwhile its partner zero 1minus120590119896+119894119905119896 lies inside the contourand contributes (2120587119886)119897119896 sin(119886(1 minus 120590119896 minus 119887)) In the conditionsof the theorem contributions of all such zeroes are nonzeroand have the same sign

Quite similarly we haveTheorem 2b whose proof one-to-one follows that of Theorem 2a

Theorem 2b Equality

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

+120587

119886(1 minus cos (119886 (1 minus 119887))) = 0

(9)

where 119886 119887 are real positive numbers such that 14 ge 119887 gt 0 and119886 = 120587(2119899+1)(1minus2119887) where 119899 is a positive integer or zero suchthat 119899 le 14119887 minus 1 holds true for some 119887 if and only if there areno Riemann function zeroes with 120590 le 119887

Unfortunately similar theorems cannot be formulated for12 gt 119887 gt 14 because no one value of 119886 can ensure thesame sign of all Riemann zeroes nonlying on the critical linecontributions for such a case

3 Rigorous Bound forthe Contribution of Riemann ZeroesNot Lying on the Critical Line

Exponential weight function appearing in the integralsconsidered in the previous section makes the problem ofestimation of themaximal possible contribution of remainingRiemann function zeroes nonlying on the critical line a rathersimple one For example for the real part we know ((4) for119887 = 12) that

int

infin

0

119890minus119886119905 ln

1003816100381610038161003816100381610038161003816120589 (1

2+ 119894119905)

1003816100381610038161003816100381610038161003816119889119905

minus int

infin

0

sin (119886119909) ln1003816100381610038161003816100381610038161003816120589 (1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 +

120587

119886sin(119886

2)

=2120587

119886sum

120588119905119896gt0

119897119896119890minus119886119905119896 sin 119886 (120590119896 minus

1

2)

(10)

where the sum is over all zeroes nonlying on the criticalline and having 120590 gt 12 How large the rhs here can beFirst we have that (2120587119886)| sum

120588119905119896gt0119897119896119890minus119886119905119896 sin 119886(120590119896 minus 12)| le

(2120587119886)sum120588119905119896gt0

119897119896119890minus119886119905119896 This last sum is given by Stieltjes inte-

gral 119868 = (2120587119886) intinfin0119890minus119886119905119889119873(119905) where119873(119905) is a discontinuous

counting function which counts the number of Riemannzeroes having the real part 120590119896 gt 12 and 119879 ge 119905119896 gt 0 takinginto account the order of zeroes The integration by partsreadily gives 119868 = (21205871198862) intinfin

0119890minus119886119905119873(119905)119889119905 About the function

119873(119905) we know that it is equal to zero when 119905 lt 119879 = 33 sdot 109[9] Further when 119905 gt 119879 it cannot exceed the function

International Journal of Analysis 5

(119905) minus (119879) where (119905) is a usual zero counting functionwhich counts zeroes in the strip 1 gt 120590119896 gt 0 119879 ge 119905119896 gt 0 (119873(119905)is equal to (119905) minus (119879) if all zeroes with 119905 gt 119879 do not lieon the critical line)Thus the contribution at question cannotexceed

119868 =120587

1198862int

infin

119879

119890minus119886119905( (119905) minus (119879)) 119889119905 (11)

Now for the function (119905) we use the known formula

(119905) =119905

2120587(ln 1199052120587minus 1) +

7

8+ 119876 (119905) (12)

with the estimation of the reminder given by Backlund in 1916[10]

|119876 (119905)| lt 0137 ln 119905 + 0443 ln ln 119905 + 4350 (13)

Then integration in (11) with (12) and (13) is trivial and weobtain that

|119868| lt1

21198863(2120587 sdot 0137 ln119879 + 1

119886ln119879 + 2120587 sdot 435) 119890minus119886119879

+2120587 sdot 0443

21198863

int

infin

119879

119890minus119886119905

ln2119905119889119905 + (

2120587 sdot 0137

21198863

+1

21198865)

sdot int

infin

119879

119890minus119886119905

119905119889119905

(14)

Of course intinfin119879(119890minus119886119905ln2119905)119889119905 lt (1119886)(119890

minus119886119879ln2119879)

intinfin

119879(119890minus119886119905119905)119889119905 lt (1119886)(119890

minus119886119879119879) and thus

|119868| lt1

21198863(0861 ln119879 + 1

119886ln119879 + 27332 + 2784

119886 ln2119879

+0861

119886119879+1

1198863119879) 119890minus119886119879

(15)

Thus for say 119886 = 1 we obtain |119868| lt 341119890minus33sdot10

9

=

341101433sdot10

9

and taking the maximal possible value of 119886119886 = 2120587 we get |119868| lt 0101119890minus2074sdot10

10

= 01011009sdot10

10

that isa precision with at least 143 milliard decimals in the first caseand of 9 milliards of decimals in the second case

To summarize we may say that combining a rigorousanalytical treatment with the known numerical results on theRiemann function zeros on the critical line we have found anequation of the form 119860 = 119861 + 120576 where 120576 = 0 is equivalent tothe RHWewere able to set a bound to 120576 of the order 1 over thenine milliardth power of 10 in this tiny factor lies the truth ofthe RH In a numerical experiment we verified the equationup to 30 decimals

Remark 3 There is a more recent calculation of Gourdonwhere it is reported that the first 119896 = 1013 Riemann zeroesare located on the critical line but we were unable to get anexact value of 119879 from the corresponding reference [11] See

also quite recent Platt paper [12] and references cited thereinfor a short review of the problem

It is also worthwhile to note that if we put the questionwhat is the attainable precision of equalities pertinent tocheck up whether there are no Riemann function zeroes with120590 lt 119887 lt 12 such a precision can be much larger because forsuch case in the conditions of Theorems 2 2a and 2b muchlarger values of 119886 can be taken for example for 119887 = 116 weare able to use the weight function 119890minus8120587119905

4 On Some Fourier Transforms ofLogarithm of the Riemann Zeta-FunctionTaken along the Real Axis and TheirRelation with the Riemann Hypothesis

We now discuss what is the situation with 119887 lt 0 that is ifwe move the left border of the contour further to the leftLet for a moment still 119887 gt minus2 First we select 119886 in such amanner that sin(119886(12 minus 119887)) = plusmn1 119886 le 120587 In doing so whenspeaking about integral equalities involving intinfin

0119890minus119886119905 ln |120589(119887 +

119894119905)|119889119905 we set all contributions of the Riemann function zeroeslying on the critical line to the ldquomaximal by modulusrdquo valueplusmn(2120587119886)sum

120588119905119896gt0120590119896=12119897119896119890minus119886119905119896 If we somehow know all ordi-

nates 119905119896 of the Riemann function zeroes 120588119896 = 120590119896+119894119905119896 we are ina position to calculate the value Σ119886RH = (2120587119886)sum120588119905119896gt0 119897119896119890

minus119886119905119896

and then we would be able to formulate a criterion equivalentto the Riemann hypothesis of the type ldquoRH holds true ifand only if the integral equality intinfin

0119890minus119886119905 ln |120589(119887 + 119894119905)|119889119905 minus

intinfin

0sin(119886119909) ln |120589(119887 + 119909)|119889119909 + (120587119886) sin(119886(1 minus 119887)) = Σ119886RH

holdsrdquo (evident changes taking into account an appearanceof trivial zeroes on the contour border are to be made if119887 lt minus2 see below) This is clear because from the formulaexpressing the contribution of zeroes to the contour integralvalue Σ119886 = (2120587119886)sum120588119905119896gt0120590119896gt119887 119897119896119890

minus119886119905119896 sin(119886(120590119896 minus 119887)) and ourchoice of 119886 119887 it immediately follows that for zeroes with120590119896 = 12 all contributions have the same sign and themoduleof contribution is smaller than it would be for zeroes with thesame ordinates but with 120590119896 = 12 | sin(119886(120590119896 minus 119887)) + sin(119886(1 minus120590119896 minus 119887))| lt |2 sin(119886(12 minus 119887))| = 2 Similar situation takesplace for an integral involving an argument of the Riemannfunction if we select cos(119886(12 minus 119887)) = plusmn1 119886 le 120587

Unfortunately we do not see how the constant Σ119886RH canbe efficiently calculated

The other way around we can select 119886 in such amanner that the condition cos(119886(12 minus 119887)) = 0 holdsThen for all such values of 119886 we obtain unconditionally(ie independent on the RH) true equalities all termsminus(2120587119886)sum

120588119905119896gt0120590119896gt119887119897119896119890minus119886119905119896 cos(119886(120590119896 minus 119887)) in (5) vanish they

are either equal to zero if 120590119896 = 12 or mutually compensateeach other for a pair of zeroes having 120590119896 = 12 plusmn 120572119896 = 12Similar unconditional integral equalities can be obtainedstarting from (4) for integrals involving logarithm of themodule of the Riemann function if we select sin(119886(12minus119887)) =0 For completeness we present these results as the followingsimple theorem

6 International Journal of Analysis

Theorem 4 For any real positive 119886 and any real 119887 such thatminus2 lt 119887 le 0 one has an integral equality

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

= minus120587

119886(1 minus cos (119886 (1 minus 119887)))

(16)

if cos(119886(12 minus 119887)) = 0 and integral equality

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909

= minus120587

119886sin (119886 (1 minus 119887))

(17)

if sin(119886(12 minus 119887)) = 0

If we move the left border of the contour further to theleft then contrary to the case 119887 gt minus2 on the lower border ofthe contour we have a number of trivial zeroes at the pointsminus2 minus4 minus6 and (4) is to be modified as

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909

=2120587

119886sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 sin (119886 (120590119896 minus 119887))

minus120587

119886sin (119886 (1 minus 119887)) + 120587

119886

119896lt[minus1198872]

sum

119896=1

sin (119886 (minus2119896 minus 119887))

(18)

Correspondingly (5) now reads

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+2120587

119886sum

120588119905119896gt0

119897119896119890minus119886119905119896 cos (119886 (120590119896 minus 119887))

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

+120587

119886(1 minus cos (119886 (1 minus 119887)))

minus120587

119886

119896lt[minus1198872]

sum

119896=1

(1 minus cos (119886 (minus2119896 minus 119887))) = 0

(19)

In this last equality we start from the following initial valueof an argument function arg(120589(119887 + 119894120576)) = minus120587 + [minus1198872] sdot120587 where real positive 120576 rarr 0 (As usual [minus1198872] means

thewhole part of positiveminus1198872number) Evident correspond-ing modifications should be done inTheorem 4

One of such unconditionally true equalities namely thatobtained from (5) putting 119886 = 2120587 and 119887 = minus72 thatis intinfin0119890minus2120587119905 arg(120589(minus72 + 119894119905))119889119905 + intinfin

0cos(2120587119909) ln |120589(minus72 +

119909)|119889119909 = 0 has also been checked up numerically (Note thathere 119887 lt minus2) In this case we calculated the first integralwith the same parameters as discussed above For the secondintegral we extended the integrationrsquos interval to [0 250] andthe maximal recursion to 100 The result 874533283 times 10minus30is similar to two previously mentioned cases Note that herearg(120589(minus72 + 119894120576)) = 0 and the contribution of the pole exactlycompensates that of the first trivial zero hence in fact the twointegrals when summing compensate each other up to 30digits as found in the computations

Interesting new possibilities appear if we take certainvalue of 119886 and introduce a sequence 119887119894 rarr minusinfin in such amanner that the condition cos(119886(12 minus 119887119894)) = 1 holds forany 119894 Below we illustrate such approach for a particular case119886 = 2120587 119887119899 = minus2119899 minus 12 (119899 is a positive integer) We have thefollowing particular case of (19)

int

infin

0

119890minus2120587119905 arg(120589 (minus2119899 minus 1

2+ 119894119905)) 119889119905 + Σ2120587

+ int

infin

0

cos (2120587119909) ln1003816100381610038161003816100381610038161003816120589 (minus2119899 minus

1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 +

1

2

minus119899

2= 0

(20)

where Σ2120587 = sum120588119905119896gt0 119897119896119890minus2120587119905119896 cos(2120587(120590119896+12)) is independent

on 119899Let us now analyze both integrals of (20) starting from

intinfin

0119890minus2120587119905 arg(120589(minus2119899 minus 12 + 119894119905))119889119905 All estimations below are

done with the 119874(1) precision because 119874(1119899) and smallerterms tend to zero for large 119899 To begin with we shouldinclude the contribution of an initial value of an argument ofthe Riemann function which is equal to arg(120589(119887 + 119894120576)) = (119899 minus1) sdot 120587 to (20) This contributes (119899 minus 1)2 to the integral valueNow let us change 120589(minus2119899minus12+119894119905) function to 120589(2119899+32minus119894119905)function using functional equation [3]

120589 (119904) =1

120587(2120587)119904 sin120587119904

2Γ (1 minus 119904) 120589 (1 minus 119904) (21)

Factor (2120587)minus2119899minus12+119894119905 in (21) contributes ln(2120587)41205872 tothe integral value For sine factor taking 119899 odd wehave sin((minus14 minus 119899 + 1198941199052)120587) = sin((14 minus 1198941199052)120587) =(radic22)(minus119894 sinh(1205871199052) + cosh(1205871199052)) Its argument isminusarctan(tanh(1205871199052)) hence we need to handle an integralminusintinfin

0119890minus2120587119905arctan(tanh(1205871199052))119889119905 Integrating by parts

we see that this contribution is equal to minus12120587 sdot (1205872) intinfin

0(eminus2120587t cosh(120587119905))119889119905 = minus(12) intinfin

0(119890minus2120587119905(119890120587119905+ 119890minus120587119905))119889119905

with the variable transform 119910 = 119890120587119905 this is easily calculated tobe equal to minus12120587 + 18

Integral involving an argument of Riemann 120589-functiontends to zero when 119899 rarr infin so next we evaluate the

International Journal of Analysis 7

contribution of the gamma-functionWe use Stirling formula[13]

ln Γ (32+ 2119899 minus 119894119905) = (1 + 2119899 minus 119894119905) ln(3

2+ 2119899 minus 119894119905)

minus (3

2+ 2119899 minus 119894119905) + 05 ln 2120587

+ 119892 (119909)

(22)

where 119892(119909) = intinfin0(12 minus 1119905 + 1(119890

119905minus 1))(119890

minus119905119909119905)119889119905 [13] and

have

Imintinfin

0

119890minus2120587119905(1 + 2119899 minus 119894119905) ln(3

2+ 2119899 minus 119894119905) 119889119905

= Im 1

2120587int

infin

0

119890minus2120587119905(minus119894 ln(3

2+ 2119899 minus 119894119905)

minus 1198941 + 2119899 minus 119894119905

32 + 2119899 minus 119894119905) 119889119905 = minus

1

41205872ln(32+ 2119899)

minus1

41205872+ 119874(

1

119899)

(23)

Further minus Imintinfin0119890minus2120587119905(32 + 2119899 minus 119894119905)119889119905 = 14120587

2 Factor05 ln 2120587 is real and contributes nothing It is not difficult tosee that integral involving 119892(119911) tends to zero with 119899 tendingto infinity

Now we need to analyze second integral in (20)that is the integral intinfin

0cos(2120587119909) ln |120589(minus12 minus 2119899 + 119909)|119889119909

With a variable change 119910 = minus12 minus 2119899 + 119909 it isequal to minusintinfin

minus12minus2119899cos(2120587119910) ln |120589(119910)|119889119910 which we divide into

minusint12

minus12minus2119899minusintinfin

12 For the first integral we use the variable

transform 119909 = 1 minus 119910 to write that it is equal tominusint32+2119899

12cos(2120587119909) ln |120589(1 minus 119909)|119889119909 and then use the same

functional equation (21) again so we need to estimate

1198682 = minusint

32+2119899

12

cos (2120587119909)

sdot ln10038161003816100381610038161003816100381610038161003816

1

120587(2120587)1minus119909 sin120587 (1 minus 119909)

2Γ (119909) 120589 (119909)

10038161003816100381610038161003816100381610038161003816

119889119909

(24)

First int32+211989912

cos(2120587119909) ln |(1120587)(2120587)1minus119909|119889119909 = 0 For anintegral involving logarithm of gamma-functionwe again use

the Stirling formula ln Γ(119909) = (119909minus12) ln(119909)minus119909 + 05 ln 2120587 +119892(119909) [13] and get

minus int

32+2119899

12

cos (2120587119909) (119909 minus 12) ln (119909) 119889119909

=1

2120587int

32+2119899

12

sin (2120587119909) (ln119909 + 1 minus 12119909) 119889119909

=1

2120587int

32+2119899

12

sin (2120587119909) (ln119909 minus 12119909) 119889119909

= minus1

2120587int

32+2119899

12

sin (2120587119909) 12119909119889119909

minus1

41205872cos (2120587119909) ln119909

1003816100381610038161003816100381610038161003816

32+2119899

12

+1

41205872int

32+2119899

12

cos (2120587119909) 1119909119889119909

=1

41205872(ln(3

2+ 2119899) + 2119899)

+1

41205872int

32+119899

12

1

119909cos (2120587119909) 119889119909

minus1

2120587int

32+2119899

12

sin (2120587119909) 12119909119889119909

(25)

To finish the consideration of the contribution of thelogarithm of gamma-function we need to add the contri-bution of 119892(119909) which is minusintinfin

12cos(2120587119909)119889119909 intinfin

0(12 minus 1119905 +

1(119890119905minus 1))(119890

minus119905119909119905)119889119905 Using cos(2120587119909) = (12)(1198902120587119909119894 + 119890minus2120587119909119894)

we can perform integration over 119909 minusintinfin12

cos(2120587119909)119890minus119905119909119889119909 =119890minus1199052(119905(1199052+ 41205872)) and thus the contribution at question is

intinfin

0(12 minus 1119905 + 1(119890

119905minus 1))(119890

minus1199052(1199052+ 41205872))119889119905

Finally we need to analyze the contribution of thesine factor in (24) which is minusint32+2119899

12cos(2120587119909) ln |sin(120587(1 minus

119909)2)|119889119909 = minusint32+211989912

cos(2120587119909) ln |cos(1205871199092)|119889119909 Integrationis to be applied with caution here due to the presence ofthe module To avoid errors it is reasonable to considerseparately integrals int1

12 int2119899+322119899+1

= int32

1 and int2

1 the latter is

an integral along the period Integrating by parts we have

minus int

1

12

cos (2120587119909) ln1003816100381610038161003816100381610038161003816cos 120587119909

2

1003816100381610038161003816100381610038161003816119889119909

= minusint

1

12

cos (2120587119909) ln(cos 1205871199092) 119889119909

= minus1

4int

1

12

sin (2120587119909) tan 1205871199092119889119909

(26)

8 International Journal of Analysis

and further

minus1

4int

1

12

sin (2120587119909) tan 1205871199092119889119909

= minusint

1

12

sin21205871199092

cos (120587119909) 119889119909

=1

2int

1

12

(cos (120587119909) minus 1) cos (120587119909) 119889119909

=1

2int

1

12

(1 + cos (2120587119909)

2minus cos (120587119909)) 119889119909

=1

8minus1

2120587sin (120587119909)

1003816100381610038161003816100381610038161003816

1

12

=1

8+1

2120587

(27)

Similarly minusint32

1cos(2120587119909) ln | cos(1205871199092)|119889119909 =

minusint32

1cos(2120587119909) ln(minus cos(1205871199092))119889119909 = 18 + 12120587 while

integral over the period is equal to 14 (use the sameapproach as above or entry number 43843 of [14] withevident variable changes) so the total contribution of thesine factor in (24) to (20) is equal to 1198992 + 14 + 1120587

Collecting everything together we see the cancelling of119874(119899) and 119874(ln(119899)) terms Thus considering the limit 119899 rarrinfin we have established the following relation expressingcertain particular value of Fourier transform of the logarithmof the Riemann zeta-function

int

infin

12

cos (2120587119909) ln |120589 (119909)| 119889119909

=1

2Σ2120587 +

7

16+1

4120587+ln (4120587)81205872minus119888119894 (120587)

81205872+119904119894 (120587)

8120587

+1

2int

infin

0

(1

2+1

119890119905minus 1minus1

119905)119890minus1199052

1199052+ 41205872119889119905

(28)

Here we used standard notation 119904119894(119909) = minusintinfin119909(sin 119902119902)119889119902

119888119894(119909) = minus intinfin

119909(cos 119902119902)119889119902 [14] Of course similar equalities

can be established for many other such Fourier transformsRepeating what has been said above about the ldquomax-

imalityrdquo of the sum Σ2120587RH = sum120588119905119896gt0

119897119896119890minus2120587119905119896 we again

could formulate a statement of the type ldquoRH holds true ifand only if the integral equality (28) with Σ2120587 = Σ2120587RHholdsrdquo Numerical calculation taking into account first 100zeroes of the Riemann function gives the value Σ2120587 =26904854 sdot 10

minus39 Equality (28) has been tested numer-ically and shown to be true without the term Σ2120587 bothsides coincide up to 39 figures after decimal and are equalto 05620211964552684935093570395449423344 Then ofcourse they start to differ

5 Conclusions

In this paper we have established a number of new criteriainvolving the integrals of the logarithm of the Riemann 120589-function and equivalent to the Riemann hypothesis thistime with the exponential weight functions Exponential

weight functions lead to rather simple expressions for thecontribution of the Riemann function zeroes not lying onthe critical line to the contour integral value This enabledobtaining a rigorous estimation of the possible error whichwas shown to be extremely small

We also show how certain Fourier transforms of thelogarithm of the Riemann zeta-function taken along thereal (demi)axis are expressible via elementary functionsplus logarithm of the gamma-function and definite integralsthereof as well as certain sums over trivial and nontrivialRiemann function zeroes In our opinion further study ofsimilar Fourier transforms is interesting and might be usefulin the Riemann researches

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] S K Sekatskii S Beltraminelli and D Merlini ldquoOn equalitiesinvolving integrals of the logarithm of the Riemann-functionand equivalent to the Riemann hypothesisrdquo Ukrainian Mathe-matical Journal vol 64 no 2 pp 218ndash228 2012

[2] S Beltraminelli D Merlini and S Sekatskii ldquoA hidden symme-try related to the Riemann hypothesis with the primes into thecritical striprdquoAsian-European Journal ofMathematics vol 3 no4 pp 555ndash563 2010

[3] E C Titchmarsh and E R Heath-Brown The Theory of theRiemann Zeta-Function Clarendon Press Oxford UK 1988

[4] F T Wang ldquoA note on the Riemann zeta-functionrdquo Bulletin ofthe American Mathematical Society vol 52 pp 319ndash321 1946

[5] V V Volchkov ldquoOn an equality equivalent to the Riemannhypothesisrdquo Ukrainian Mathematical Journal vol 47 no 3 pp491ndash493 1995

[6] M Balazard E Saias and M Yor ldquoNotes sur la fonction 120577 deRiemann 2rdquo Advances in Mathematics vol 143 no 2 pp 284ndash287 1999

[7] D Merlini ldquoThe Riemann magneton of the primesrdquo Chaos andComplexity Letters vol 2 no 1 pp 93ndash98 2006

[8] E C Titchmarsh The Theory of Functions Oxford UniversityPress Oxford UK 1939

[9] C Ramare and Y Saouter ldquoShort effective intervals containingprimesrdquo Journal ofNumberTheory vol 98 no 1 pp 10ndash33 2003

[10] R J Backlund ldquoUber die Nullstellen der Riemannschen Zeta-funktionrdquo Acta Mathematica vol 41 no 1 pp 345ndash375 1916

[11] X Gourdon ldquoThe 1013 first zeroes of the Riemann ZetaFunction and zeroes computation at very large heightrdquo 2004httpnumberscomputationfreefrConstantsMiscellaneouszetazeros1e13-1e24pdf

[12] D J Platt ldquoComputing 120587(119909) analyticallyrdquo Mathematics ofComputation vol 84 no 293 pp 1521ndash1535 2015

[13] A Erdelyi EdHigher Transcendental Functions Based in PartOn Notes Left by Harry Bateman vol 1 McGraw-Hill NewYork NY USA 1953

[14] I S Gradshtein and I M Ryzhik Tables of Integrals Series andProducts Academic New York NY USA 1990

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Differential EquationsInternational Journal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 International Journal of Analysis

119865(119911) to be 119865(119911 plusmn 1198940) accordingly as one approaches the pathfrom above or below Let also 119865(119911) = ln(119891(119911)) be the logarithmdefined by continuous variation along any smooth curve fullylying inside the contour which avoids all poles and zeroes of119891(119911) and starts from the same particular determination on119909 = 1198832 Suppose also that the poles and zeroes of the functions119891(119911) 119892(119911) do not coincide

Then int119862119865(119911)119892(119911)119889119911 = 2120587119894(sum

120588119892119903119890119904(119892(120588119892) sdot 119865(120588119892)) minus

sum1205880119891

int1198830

120588+1198941198840

120588

1198831+1198941198840

120588

119892(119911)119889119911 + sum120588119901119900119897

119891

int119883119901119900119897

120588+119894119884119901119900119897

120588

1198831+119894119884119901119900119897

120588

119892(119911)119889119911) where the sumis over all 120588119892 which are poles of the function 119892(119911) lying inside119862 all 1205880

119891= 1198830

120588+ 1198941198840

120588which are zeroes of the function

119891(119911) counted taking into account their multiplicities (ie thecorresponding term is multiplied by 119898 for a zero of the order119898) and which lie inside 119862 and all 120588119901119900119897

119891= 119883119901119900119897

120588+ 119894119884119901119900119897

120588which

are poles of the function 119891(119911) counted taking into accounttheir multiplicities and which lie inside 119862 The assumption isthat all relevant integrals in the right-hand side of the equalityexist

Proof Our proof closely follows the well-known proof of theLittlewood theorem (or lemma) given for example in [8 p133] Consider first the function119891(119911) = 119911minus119886where 119886 = 120572+119894120573is a point of the rectangle Let 1198621015840 be the contour obtained bydescribing 119862 in the positive direction from (1198832 1198841) as far as(1198831 120573) then the straight line 119910 = 120573 as far as 120572 minus 120576 + 119894120573 thena circle of radius 120576 about 119911 = 119886 and then returning along119910 = 120573 and the rest of 119862 to the starting point see Figure 1The only poles of 119865(119911)119892(119911) in 1198621015840 are those of the function119892(119911) so that int

1198621015840119865(119911)119892(119911)119889119911 = 2120587119894sum

120588119892res(119892(120588119892) sdot 119865(120588119892)) The

integral round the small circle tends to zero with the radiusthus we have int

119862119865(119911)119892(119911)119889119911 = 2120587119894sum

120588119892res(119892(120588119892) sdot 119865(120588119892)) minus

int120572+119894120573

1198831+119894120573119892(119911)(1198651(119911) minus 1198652(119911))119889119911 where 1198651 and 1198652 are the values

of 119865 on the two paths joining 1198831 + 119894120573 to 120572 + 119894120573 Hence weobtain 1198652 from 1198651 by passing in the negative direction rounda simple zero of 119891(119911) at 119911 = 119886 we have 1198652(119911) = 1198651(119911) minus 2120587119894and correspondingly int

119862119865(119911)119892(119911)119889119911 = 2120587119894(sum

120588119892res(119892(120588119892) sdot

119865(120588119892)) minus int120572+119894120573

1198831+119894120573119892(119911)119889119911) where we introduce a notation 119865(119911)

to distinguish this function from 119865(119911) The general case noweasily follows by addition of terms corresponding to thevarious poles and zeroes of 119891(119911)

From the proof of Theorem 1 illustrated in Figure 1 itfollows that the function 119865(119911) = ln(119891(119911)) occurring inits formulation is not a continuous function along the leftborder of the contour namely along the line connecting thepoints 1198831 + 1198941198841 and 1198831 + 1198941198842 its value jumps on ∓2120587119897 whenwe pass a point1198831 + 119894119884 such that there is an 119897th order zero orpole of the function119891(119911) lying inside the contour (and not onthe integration line) and having an ordinate 119884 This functionis only piecewise continuous This circumstance is presentof course also in the original Littlewood theorem given for119892(119911) = 1 (and is well recognized in its formulation see [8])The value of ln |119891(119911)| is a continuous function along the leftborder of the contour provided of course that there are no

Starting pointof contour integration

Value of the function

Value of the function

X1 X2

Y1

Y2

(F(z) minus 2120587i)g(z)

F(z)g(z)

120572 + i120573

Figure 1 Illustrating the proof of the generalized Littlewoodtheorem

poles or zeroes of the function119891(119911) lying exactly on this sameborder

Now let us consider a rectangular contour 119862 with thevertices 119887 119887 + 119894119883 119887 +119883+ 119894119883 119887 +119883 with real 119887 gt minus2 (with thischoice we avoid the trivial Riemann zeroes) and real 119883 rarr+infin introduce the function 119892(119911) = 119894 exp 119894(119886(119911 minus 119887)) where 119886is real positive and apply the above theorem to the contourintegral int

119862119892(119911) ln(120589(119911))119889119911 Along the line (119887 119887 + 119894119883) we have

an integral int1198830119890minus119886119905lnlowast(120589(119887+119894119905))119889119905 and along the line (119887 119887+119883)

we have an integral 119894 int1198830(cos(119886119909) + 119894 sin(119886119909)) ln(120589(119887 + 119909))119889119909

Due to the known asymptotic behaviour of the Riemannzeta-function [3] in particular ln(120589(119911)) = 119874(2minus119911) for largepositive 119911 these integrals converge while integrals over lines(119883119883 + 119894119883) and (119887 + 119894119883119883 + 119894119883) vanish in the limit of large119883

If 119887 lt 1 on the border of the contour we have the poleof the Riemann function at 119911 = 1 and in the interior ofthe contour we also might have a number of zeroes of theRiemann function We definitely have an infinite numberof zeroes if 119887 lt 12 The order 119897119896 zero of the Riemannfunction 120588 = 120590119896 + 119894119905119896 for 119887 lt 120590119896 according to Theo-rem 1 contributes minus2120587119894119897119896 int

120590119896minus119887

0119890minus119886119905119896(119894 cos(119886119909) minus sin(119886119909))119889119909

= (2120587119897119896119890minus119886119905119896119886)[sin(119886(120590119896 minus 119887)) + 119894(1 minus cos(119886(120590119896 minus 119887)))] to

the contour integral value so finally the aforementioned

International Journal of Analysis 3

generalized Littlewood theorem results in the followingequality

int

infin

0

119890minus119886119905lnlowast (120589 (119887 + 119894119905)) 119889119905 minus int

infin

0

sin (119886119909) ln (120589 (119887 + 119909)) 119889119909

+ 119894 int

infin

0

cos (119886119909) ln (120589 (119887 + 119909)) 119889119909 = 2120587119886

sdot sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896[sin (119886 (120590119896 minus 119887))

+ 119894 (1 minus cos (119886 (120590119896 minus 119887)))] minus120587

119886(sin (119886 (1 minus 119887)) + 119894 (1

minus cos (119886 (1 minus 119887))))

(1)

In the last term here we took into account that the pole 119911 = 1lies exactly onto the contour border whence the coefficient is120587119886 instead of 2120587119886

Now we should remind the reader that the functionlnlowast(120589(119887 + 119894119905)) is not continuous see above (and this is thereason why we put an asterisk sign here) If we want touse a continuous logarithm function instead (which we willdenote simply ln(120589(119887 + 119894119905))) the following straightforwardmodification is to be done If real functions 119891(119909) 119892(119909)are continuous on the segment [119860 119861] and function 119891lowast(119909)coincides with 119891(119909) on [119860 119862] and is equal to 119891(119909) + ℎon (119862 119861] (ℎ is a constant 119860 le 119862 le 119861) we haveint119861

119860119892(119909)119891

lowast(119909)119889119909 = int

119861

119860119892(119909)119891(119909)119889119909 + ℎ int

119861

119862119892(119909)119889119909 and this

is trivially generalized for complex numbers and improperintegrals having counting number of discontinuities For ourparticular case we have considering one 2120587119894119897119896 discontinuityat 119905 = 119905119896 int

infin

0119890minus119886119905arglowast(120589(119887 + 119894119905))119889119905 = intinfin

0119890minus119886119905 arg(120589(119887 + 119894119905))119889119905 +

(2120587119897119896119886)119890minus119886119905119896 and the aforementioned generalization gives

int

infin

0

119890minus119886119905arglowast (120589 (119887 + 119894119905)) 119889119905

= int

infin

0

119890minus119886119905arg (120589 (119887 + 119894119905)) 119889119905 + sum

120588119905119896gt0120590119896gt119887

2120587119897119896

119886119890minus119886119905119896

(2)

Using (2) we get instead of (1)

int

infin

0

119890minus119886119905 ln (120589 (119887 + 119894119905)) 119889119905 minus int

infin

0

sin (119886119909) ln (120589 (119887 + 119909)) 119889119909

+ 119894 int

infin

0

cos (119886119909) ln (120589 (119887 + 119909)) 119889119909 = 2120587119886

sdot sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896[sin (119886 (120590119896 minus 119887))

minus 119894 cos (119886 (120590119896 minus 119887))] minus120587

119886(sin (119886 (1 minus 119887)) + 119894 (1

minus cos (119886 (1 minus 119887))))

(3)

where separation of real and imaginary parts readily givesequalities

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905 minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909

=2120587

119886sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 sin (119886 (120590119896 minus 119887))

minus120587

119886sin (119886 (1 minus 119887))

(4)

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

= minus2120587

119886sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896cos (119886 (120590119896 minus 119887))

minus120587

119886(1 minus cos (119886 (1 minus 119887)))

(5)

Here the argument is a continuous function along the contour(of course provided there are no Riemann zeroes lyingexactly on it) and the following initial values of the argumentshould be taken since we have the simple pole at 119911 = 1 onthe contour we select for real 119909 gt 1 arg(120589(119909)) = 0 and for1 gt 119887 gt minus2 arg(120589(119887 + 119894120576)) = minus120587 where real positive 120576 rarr 0

Equations (4) and (5) are our starting point and their firstapplication is the following theorem

Theorem 2 Equalities

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909 + 120587119886sin (119886 (1 minus 119887))

= 0

(6)

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

+120587

119886(1 minus cos (119886 (1 minus 119887))) = 0

(7)

where 119887 119886 are real positive numbers such that 1 gt 119887 ge 12119886(1 minus 119887) le 120587 for (6) and 119886(1 minus 119887) le 1205872 for (7) hold truefor some 119887 if and only if there are no Riemann function zeroeswith 120590 gt 119887 For 119887 = 12 these equalities are equivalent to theRiemann hypothesis

Proof For all possible Riemann 120589-function zeroes with Re120588 =120590119896 gt 119887 and hence lying inside the contour from our choice of119887 and the known properties of the Riemann function zeroeswe have 0 lt 120590119896minus119887 lt 1minus119887 le 12Then the imposed conditions119886(1minus119887) le 120587 and 119886(1minus119887) le 1205872 guaranty the same signs of all

4 International Journal of Analysis

possible contributions (2120587119886)sum120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 sin(119886(120590119896minus119887))

and minus(2120587119886)sum120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 cos(119886(120590119896minus119887)) from (4) and (5)

and also prevent them from being equal to zeroIn (6) it is interesting for 119887 = 12 to take 119886 = 2120587 thus

eliminating any problem with the integration over real axis(pole of the Riemann function coincides with zero of sine)

int

infin

0

119890minus2120587119905 ln

1003816100381610038161003816100381610038161003816120589 (1

2+ 119894119905)

1003816100381610038161003816100381610038161003816119889119905

minus int

infin

0

sin (2120587119909) ln1003816100381610038161003816100381610038161003816120589 (1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 = 0

(6a)

Similarly for (7) the choice 119886 = 120587 is useful

int

infin

0

119890minus120587119905 arg(120589 (1

2+ 119894119905)) 119889119905

+ int

infin

0

cos (120587119909) ln1003816100381610038161003816100381610038161003816120589 (1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 + 1 = 0

(7a)

Equalities (6a) (7a) are equivalent to the Riemannhypothesis They have been tested numerically using thestandard procedure Integrate in Mathematica Because thefirst integrand here decays very fast we can limit the inte-gration interval of the variable 119905 to [0 50] We need alsoto reduce the working precision (set to 50) and maximalrecursion (set to 70) For the second integral in (6a) and (7a)we set the integration interval for the variable 119909 to [0 200]and set the working precision and maximal recursion to50 After 15 hours of CPU time we obtained for (6a) theresult of 88044282 times 10minus32 and for (7a) after more than8 hours 114767406 times 10minus31 So in both cases we have acorrespondence for at least the first 30 figures

Next two similar theorems can be proved when we selectleft contour border line (119887 119887 + 119894infin) lying to the left to thecritical line

Theorem 2a Equality

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909 + 120587119886sin (119886 (1 minus 119887))

= 0

(8)

where 119886 119887 are real positive numbers such that 14 ge 119887 gt 0and 119886 = 120587119899(12 minus 119887) where 119899 is a positive integer such that119899 le 12119887 minus 1 holds true for some 119887 if and only if there are noRiemann function zeroes with 120590 le 119887

Proof In conditions of the theorem all contributions ofthe Riemann function zeroes lying on the critical line(2120587119886)sum

120588119905119896gt0120590119896gt119887119897119896119890minus119886119905119896 sin(119886(12 minus 119887)) to the contour inte-

gral value are equal to zero If there are a pair of Riemannfunction zeroes 120588 = 120590119896+119894119905119896 120588 = 1minus120590119896+119894119905119896 with 12 gt 120590119896 gt 119887both of them lie inside the contour and their contributionsexactly compensate each other thus contributing nothing tothe contour integral value If there is a zero with 120590119896 le 119887 it lies

outside the contour or on its border and hence contributesnothingwhile its partner zero 1minus120590119896+119894119905119896 lies inside the contourand contributes (2120587119886)119897119896 sin(119886(1 minus 120590119896 minus 119887)) In the conditionsof the theorem contributions of all such zeroes are nonzeroand have the same sign

Quite similarly we haveTheorem 2b whose proof one-to-one follows that of Theorem 2a

Theorem 2b Equality

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

+120587

119886(1 minus cos (119886 (1 minus 119887))) = 0

(9)

where 119886 119887 are real positive numbers such that 14 ge 119887 gt 0 and119886 = 120587(2119899+1)(1minus2119887) where 119899 is a positive integer or zero suchthat 119899 le 14119887 minus 1 holds true for some 119887 if and only if there areno Riemann function zeroes with 120590 le 119887

Unfortunately similar theorems cannot be formulated for12 gt 119887 gt 14 because no one value of 119886 can ensure thesame sign of all Riemann zeroes nonlying on the critical linecontributions for such a case

3 Rigorous Bound forthe Contribution of Riemann ZeroesNot Lying on the Critical Line

Exponential weight function appearing in the integralsconsidered in the previous section makes the problem ofestimation of themaximal possible contribution of remainingRiemann function zeroes nonlying on the critical line a rathersimple one For example for the real part we know ((4) for119887 = 12) that

int

infin

0

119890minus119886119905 ln

1003816100381610038161003816100381610038161003816120589 (1

2+ 119894119905)

1003816100381610038161003816100381610038161003816119889119905

minus int

infin

0

sin (119886119909) ln1003816100381610038161003816100381610038161003816120589 (1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 +

120587

119886sin(119886

2)

=2120587

119886sum

120588119905119896gt0

119897119896119890minus119886119905119896 sin 119886 (120590119896 minus

1

2)

(10)

where the sum is over all zeroes nonlying on the criticalline and having 120590 gt 12 How large the rhs here can beFirst we have that (2120587119886)| sum

120588119905119896gt0119897119896119890minus119886119905119896 sin 119886(120590119896 minus 12)| le

(2120587119886)sum120588119905119896gt0

119897119896119890minus119886119905119896 This last sum is given by Stieltjes inte-

gral 119868 = (2120587119886) intinfin0119890minus119886119905119889119873(119905) where119873(119905) is a discontinuous

counting function which counts the number of Riemannzeroes having the real part 120590119896 gt 12 and 119879 ge 119905119896 gt 0 takinginto account the order of zeroes The integration by partsreadily gives 119868 = (21205871198862) intinfin

0119890minus119886119905119873(119905)119889119905 About the function

119873(119905) we know that it is equal to zero when 119905 lt 119879 = 33 sdot 109[9] Further when 119905 gt 119879 it cannot exceed the function

International Journal of Analysis 5

(119905) minus (119879) where (119905) is a usual zero counting functionwhich counts zeroes in the strip 1 gt 120590119896 gt 0 119879 ge 119905119896 gt 0 (119873(119905)is equal to (119905) minus (119879) if all zeroes with 119905 gt 119879 do not lieon the critical line)Thus the contribution at question cannotexceed

119868 =120587

1198862int

infin

119879

119890minus119886119905( (119905) minus (119879)) 119889119905 (11)

Now for the function (119905) we use the known formula

(119905) =119905

2120587(ln 1199052120587minus 1) +

7

8+ 119876 (119905) (12)

with the estimation of the reminder given by Backlund in 1916[10]

|119876 (119905)| lt 0137 ln 119905 + 0443 ln ln 119905 + 4350 (13)

Then integration in (11) with (12) and (13) is trivial and weobtain that

|119868| lt1

21198863(2120587 sdot 0137 ln119879 + 1

119886ln119879 + 2120587 sdot 435) 119890minus119886119879

+2120587 sdot 0443

21198863

int

infin

119879

119890minus119886119905

ln2119905119889119905 + (

2120587 sdot 0137

21198863

+1

21198865)

sdot int

infin

119879

119890minus119886119905

119905119889119905

(14)

Of course intinfin119879(119890minus119886119905ln2119905)119889119905 lt (1119886)(119890

minus119886119879ln2119879)

intinfin

119879(119890minus119886119905119905)119889119905 lt (1119886)(119890

minus119886119879119879) and thus

|119868| lt1

21198863(0861 ln119879 + 1

119886ln119879 + 27332 + 2784

119886 ln2119879

+0861

119886119879+1

1198863119879) 119890minus119886119879

(15)

Thus for say 119886 = 1 we obtain |119868| lt 341119890minus33sdot10

9

=

341101433sdot10

9

and taking the maximal possible value of 119886119886 = 2120587 we get |119868| lt 0101119890minus2074sdot10

10

= 01011009sdot10

10

that isa precision with at least 143 milliard decimals in the first caseand of 9 milliards of decimals in the second case

To summarize we may say that combining a rigorousanalytical treatment with the known numerical results on theRiemann function zeros on the critical line we have found anequation of the form 119860 = 119861 + 120576 where 120576 = 0 is equivalent tothe RHWewere able to set a bound to 120576 of the order 1 over thenine milliardth power of 10 in this tiny factor lies the truth ofthe RH In a numerical experiment we verified the equationup to 30 decimals

Remark 3 There is a more recent calculation of Gourdonwhere it is reported that the first 119896 = 1013 Riemann zeroesare located on the critical line but we were unable to get anexact value of 119879 from the corresponding reference [11] See

also quite recent Platt paper [12] and references cited thereinfor a short review of the problem

It is also worthwhile to note that if we put the questionwhat is the attainable precision of equalities pertinent tocheck up whether there are no Riemann function zeroes with120590 lt 119887 lt 12 such a precision can be much larger because forsuch case in the conditions of Theorems 2 2a and 2b muchlarger values of 119886 can be taken for example for 119887 = 116 weare able to use the weight function 119890minus8120587119905

4 On Some Fourier Transforms ofLogarithm of the Riemann Zeta-FunctionTaken along the Real Axis and TheirRelation with the Riemann Hypothesis

We now discuss what is the situation with 119887 lt 0 that is ifwe move the left border of the contour further to the leftLet for a moment still 119887 gt minus2 First we select 119886 in such amanner that sin(119886(12 minus 119887)) = plusmn1 119886 le 120587 In doing so whenspeaking about integral equalities involving intinfin

0119890minus119886119905 ln |120589(119887 +

119894119905)|119889119905 we set all contributions of the Riemann function zeroeslying on the critical line to the ldquomaximal by modulusrdquo valueplusmn(2120587119886)sum

120588119905119896gt0120590119896=12119897119896119890minus119886119905119896 If we somehow know all ordi-

nates 119905119896 of the Riemann function zeroes 120588119896 = 120590119896+119894119905119896 we are ina position to calculate the value Σ119886RH = (2120587119886)sum120588119905119896gt0 119897119896119890

minus119886119905119896

and then we would be able to formulate a criterion equivalentto the Riemann hypothesis of the type ldquoRH holds true ifand only if the integral equality intinfin

0119890minus119886119905 ln |120589(119887 + 119894119905)|119889119905 minus

intinfin

0sin(119886119909) ln |120589(119887 + 119909)|119889119909 + (120587119886) sin(119886(1 minus 119887)) = Σ119886RH

holdsrdquo (evident changes taking into account an appearanceof trivial zeroes on the contour border are to be made if119887 lt minus2 see below) This is clear because from the formulaexpressing the contribution of zeroes to the contour integralvalue Σ119886 = (2120587119886)sum120588119905119896gt0120590119896gt119887 119897119896119890

minus119886119905119896 sin(119886(120590119896 minus 119887)) and ourchoice of 119886 119887 it immediately follows that for zeroes with120590119896 = 12 all contributions have the same sign and themoduleof contribution is smaller than it would be for zeroes with thesame ordinates but with 120590119896 = 12 | sin(119886(120590119896 minus 119887)) + sin(119886(1 minus120590119896 minus 119887))| lt |2 sin(119886(12 minus 119887))| = 2 Similar situation takesplace for an integral involving an argument of the Riemannfunction if we select cos(119886(12 minus 119887)) = plusmn1 119886 le 120587

Unfortunately we do not see how the constant Σ119886RH canbe efficiently calculated

The other way around we can select 119886 in such amanner that the condition cos(119886(12 minus 119887)) = 0 holdsThen for all such values of 119886 we obtain unconditionally(ie independent on the RH) true equalities all termsminus(2120587119886)sum

120588119905119896gt0120590119896gt119887119897119896119890minus119886119905119896 cos(119886(120590119896 minus 119887)) in (5) vanish they

are either equal to zero if 120590119896 = 12 or mutually compensateeach other for a pair of zeroes having 120590119896 = 12 plusmn 120572119896 = 12Similar unconditional integral equalities can be obtainedstarting from (4) for integrals involving logarithm of themodule of the Riemann function if we select sin(119886(12minus119887)) =0 For completeness we present these results as the followingsimple theorem

6 International Journal of Analysis

Theorem 4 For any real positive 119886 and any real 119887 such thatminus2 lt 119887 le 0 one has an integral equality

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

= minus120587

119886(1 minus cos (119886 (1 minus 119887)))

(16)

if cos(119886(12 minus 119887)) = 0 and integral equality

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909

= minus120587

119886sin (119886 (1 minus 119887))

(17)

if sin(119886(12 minus 119887)) = 0

If we move the left border of the contour further to theleft then contrary to the case 119887 gt minus2 on the lower border ofthe contour we have a number of trivial zeroes at the pointsminus2 minus4 minus6 and (4) is to be modified as

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909

=2120587

119886sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 sin (119886 (120590119896 minus 119887))

minus120587

119886sin (119886 (1 minus 119887)) + 120587

119886

119896lt[minus1198872]

sum

119896=1

sin (119886 (minus2119896 minus 119887))

(18)

Correspondingly (5) now reads

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+2120587

119886sum

120588119905119896gt0

119897119896119890minus119886119905119896 cos (119886 (120590119896 minus 119887))

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

+120587

119886(1 minus cos (119886 (1 minus 119887)))

minus120587

119886

119896lt[minus1198872]

sum

119896=1

(1 minus cos (119886 (minus2119896 minus 119887))) = 0

(19)

In this last equality we start from the following initial valueof an argument function arg(120589(119887 + 119894120576)) = minus120587 + [minus1198872] sdot120587 where real positive 120576 rarr 0 (As usual [minus1198872] means

thewhole part of positiveminus1198872number) Evident correspond-ing modifications should be done inTheorem 4

One of such unconditionally true equalities namely thatobtained from (5) putting 119886 = 2120587 and 119887 = minus72 thatis intinfin0119890minus2120587119905 arg(120589(minus72 + 119894119905))119889119905 + intinfin

0cos(2120587119909) ln |120589(minus72 +

119909)|119889119909 = 0 has also been checked up numerically (Note thathere 119887 lt minus2) In this case we calculated the first integralwith the same parameters as discussed above For the secondintegral we extended the integrationrsquos interval to [0 250] andthe maximal recursion to 100 The result 874533283 times 10minus30is similar to two previously mentioned cases Note that herearg(120589(minus72 + 119894120576)) = 0 and the contribution of the pole exactlycompensates that of the first trivial zero hence in fact the twointegrals when summing compensate each other up to 30digits as found in the computations

Interesting new possibilities appear if we take certainvalue of 119886 and introduce a sequence 119887119894 rarr minusinfin in such amanner that the condition cos(119886(12 minus 119887119894)) = 1 holds forany 119894 Below we illustrate such approach for a particular case119886 = 2120587 119887119899 = minus2119899 minus 12 (119899 is a positive integer) We have thefollowing particular case of (19)

int

infin

0

119890minus2120587119905 arg(120589 (minus2119899 minus 1

2+ 119894119905)) 119889119905 + Σ2120587

+ int

infin

0

cos (2120587119909) ln1003816100381610038161003816100381610038161003816120589 (minus2119899 minus

1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 +

1

2

minus119899

2= 0

(20)

where Σ2120587 = sum120588119905119896gt0 119897119896119890minus2120587119905119896 cos(2120587(120590119896+12)) is independent

on 119899Let us now analyze both integrals of (20) starting from

intinfin

0119890minus2120587119905 arg(120589(minus2119899 minus 12 + 119894119905))119889119905 All estimations below are

done with the 119874(1) precision because 119874(1119899) and smallerterms tend to zero for large 119899 To begin with we shouldinclude the contribution of an initial value of an argument ofthe Riemann function which is equal to arg(120589(119887 + 119894120576)) = (119899 minus1) sdot 120587 to (20) This contributes (119899 minus 1)2 to the integral valueNow let us change 120589(minus2119899minus12+119894119905) function to 120589(2119899+32minus119894119905)function using functional equation [3]

120589 (119904) =1

120587(2120587)119904 sin120587119904

2Γ (1 minus 119904) 120589 (1 minus 119904) (21)

Factor (2120587)minus2119899minus12+119894119905 in (21) contributes ln(2120587)41205872 tothe integral value For sine factor taking 119899 odd wehave sin((minus14 minus 119899 + 1198941199052)120587) = sin((14 minus 1198941199052)120587) =(radic22)(minus119894 sinh(1205871199052) + cosh(1205871199052)) Its argument isminusarctan(tanh(1205871199052)) hence we need to handle an integralminusintinfin

0119890minus2120587119905arctan(tanh(1205871199052))119889119905 Integrating by parts

we see that this contribution is equal to minus12120587 sdot (1205872) intinfin

0(eminus2120587t cosh(120587119905))119889119905 = minus(12) intinfin

0(119890minus2120587119905(119890120587119905+ 119890minus120587119905))119889119905

with the variable transform 119910 = 119890120587119905 this is easily calculated tobe equal to minus12120587 + 18

Integral involving an argument of Riemann 120589-functiontends to zero when 119899 rarr infin so next we evaluate the

International Journal of Analysis 7

contribution of the gamma-functionWe use Stirling formula[13]

ln Γ (32+ 2119899 minus 119894119905) = (1 + 2119899 minus 119894119905) ln(3

2+ 2119899 minus 119894119905)

minus (3

2+ 2119899 minus 119894119905) + 05 ln 2120587

+ 119892 (119909)

(22)

where 119892(119909) = intinfin0(12 minus 1119905 + 1(119890

119905minus 1))(119890

minus119905119909119905)119889119905 [13] and

have

Imintinfin

0

119890minus2120587119905(1 + 2119899 minus 119894119905) ln(3

2+ 2119899 minus 119894119905) 119889119905

= Im 1

2120587int

infin

0

119890minus2120587119905(minus119894 ln(3

2+ 2119899 minus 119894119905)

minus 1198941 + 2119899 minus 119894119905

32 + 2119899 minus 119894119905) 119889119905 = minus

1

41205872ln(32+ 2119899)

minus1

41205872+ 119874(

1

119899)

(23)

Further minus Imintinfin0119890minus2120587119905(32 + 2119899 minus 119894119905)119889119905 = 14120587

2 Factor05 ln 2120587 is real and contributes nothing It is not difficult tosee that integral involving 119892(119911) tends to zero with 119899 tendingto infinity

Now we need to analyze second integral in (20)that is the integral intinfin

0cos(2120587119909) ln |120589(minus12 minus 2119899 + 119909)|119889119909

With a variable change 119910 = minus12 minus 2119899 + 119909 it isequal to minusintinfin

minus12minus2119899cos(2120587119910) ln |120589(119910)|119889119910 which we divide into

minusint12

minus12minus2119899minusintinfin

12 For the first integral we use the variable

transform 119909 = 1 minus 119910 to write that it is equal tominusint32+2119899

12cos(2120587119909) ln |120589(1 minus 119909)|119889119909 and then use the same

functional equation (21) again so we need to estimate

1198682 = minusint

32+2119899

12

cos (2120587119909)

sdot ln10038161003816100381610038161003816100381610038161003816

1

120587(2120587)1minus119909 sin120587 (1 minus 119909)

2Γ (119909) 120589 (119909)

10038161003816100381610038161003816100381610038161003816

119889119909

(24)

First int32+211989912

cos(2120587119909) ln |(1120587)(2120587)1minus119909|119889119909 = 0 For anintegral involving logarithm of gamma-functionwe again use

the Stirling formula ln Γ(119909) = (119909minus12) ln(119909)minus119909 + 05 ln 2120587 +119892(119909) [13] and get

minus int

32+2119899

12

cos (2120587119909) (119909 minus 12) ln (119909) 119889119909

=1

2120587int

32+2119899

12

sin (2120587119909) (ln119909 + 1 minus 12119909) 119889119909

=1

2120587int

32+2119899

12

sin (2120587119909) (ln119909 minus 12119909) 119889119909

= minus1

2120587int

32+2119899

12

sin (2120587119909) 12119909119889119909

minus1

41205872cos (2120587119909) ln119909

1003816100381610038161003816100381610038161003816

32+2119899

12

+1

41205872int

32+2119899

12

cos (2120587119909) 1119909119889119909

=1

41205872(ln(3

2+ 2119899) + 2119899)

+1

41205872int

32+119899

12

1

119909cos (2120587119909) 119889119909

minus1

2120587int

32+2119899

12

sin (2120587119909) 12119909119889119909

(25)

To finish the consideration of the contribution of thelogarithm of gamma-function we need to add the contri-bution of 119892(119909) which is minusintinfin

12cos(2120587119909)119889119909 intinfin

0(12 minus 1119905 +

1(119890119905minus 1))(119890

minus119905119909119905)119889119905 Using cos(2120587119909) = (12)(1198902120587119909119894 + 119890minus2120587119909119894)

we can perform integration over 119909 minusintinfin12

cos(2120587119909)119890minus119905119909119889119909 =119890minus1199052(119905(1199052+ 41205872)) and thus the contribution at question is

intinfin

0(12 minus 1119905 + 1(119890

119905minus 1))(119890

minus1199052(1199052+ 41205872))119889119905

Finally we need to analyze the contribution of thesine factor in (24) which is minusint32+2119899

12cos(2120587119909) ln |sin(120587(1 minus

119909)2)|119889119909 = minusint32+211989912

cos(2120587119909) ln |cos(1205871199092)|119889119909 Integrationis to be applied with caution here due to the presence ofthe module To avoid errors it is reasonable to considerseparately integrals int1

12 int2119899+322119899+1

= int32

1 and int2

1 the latter is

an integral along the period Integrating by parts we have

minus int

1

12

cos (2120587119909) ln1003816100381610038161003816100381610038161003816cos 120587119909

2

1003816100381610038161003816100381610038161003816119889119909

= minusint

1

12

cos (2120587119909) ln(cos 1205871199092) 119889119909

= minus1

4int

1

12

sin (2120587119909) tan 1205871199092119889119909

(26)

8 International Journal of Analysis

and further

minus1

4int

1

12

sin (2120587119909) tan 1205871199092119889119909

= minusint

1

12

sin21205871199092

cos (120587119909) 119889119909

=1

2int

1

12

(cos (120587119909) minus 1) cos (120587119909) 119889119909

=1

2int

1

12

(1 + cos (2120587119909)

2minus cos (120587119909)) 119889119909

=1

8minus1

2120587sin (120587119909)

1003816100381610038161003816100381610038161003816

1

12

=1

8+1

2120587

(27)

Similarly minusint32

1cos(2120587119909) ln | cos(1205871199092)|119889119909 =

minusint32

1cos(2120587119909) ln(minus cos(1205871199092))119889119909 = 18 + 12120587 while

integral over the period is equal to 14 (use the sameapproach as above or entry number 43843 of [14] withevident variable changes) so the total contribution of thesine factor in (24) to (20) is equal to 1198992 + 14 + 1120587

Collecting everything together we see the cancelling of119874(119899) and 119874(ln(119899)) terms Thus considering the limit 119899 rarrinfin we have established the following relation expressingcertain particular value of Fourier transform of the logarithmof the Riemann zeta-function

int

infin

12

cos (2120587119909) ln |120589 (119909)| 119889119909

=1

2Σ2120587 +

7

16+1

4120587+ln (4120587)81205872minus119888119894 (120587)

81205872+119904119894 (120587)

8120587

+1

2int

infin

0

(1

2+1

119890119905minus 1minus1

119905)119890minus1199052

1199052+ 41205872119889119905

(28)

Here we used standard notation 119904119894(119909) = minusintinfin119909(sin 119902119902)119889119902

119888119894(119909) = minus intinfin

119909(cos 119902119902)119889119902 [14] Of course similar equalities

can be established for many other such Fourier transformsRepeating what has been said above about the ldquomax-

imalityrdquo of the sum Σ2120587RH = sum120588119905119896gt0

119897119896119890minus2120587119905119896 we again

could formulate a statement of the type ldquoRH holds true ifand only if the integral equality (28) with Σ2120587 = Σ2120587RHholdsrdquo Numerical calculation taking into account first 100zeroes of the Riemann function gives the value Σ2120587 =26904854 sdot 10

minus39 Equality (28) has been tested numer-ically and shown to be true without the term Σ2120587 bothsides coincide up to 39 figures after decimal and are equalto 05620211964552684935093570395449423344 Then ofcourse they start to differ

5 Conclusions

In this paper we have established a number of new criteriainvolving the integrals of the logarithm of the Riemann 120589-function and equivalent to the Riemann hypothesis thistime with the exponential weight functions Exponential

weight functions lead to rather simple expressions for thecontribution of the Riemann function zeroes not lying onthe critical line to the contour integral value This enabledobtaining a rigorous estimation of the possible error whichwas shown to be extremely small

We also show how certain Fourier transforms of thelogarithm of the Riemann zeta-function taken along thereal (demi)axis are expressible via elementary functionsplus logarithm of the gamma-function and definite integralsthereof as well as certain sums over trivial and nontrivialRiemann function zeroes In our opinion further study ofsimilar Fourier transforms is interesting and might be usefulin the Riemann researches

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] S K Sekatskii S Beltraminelli and D Merlini ldquoOn equalitiesinvolving integrals of the logarithm of the Riemann-functionand equivalent to the Riemann hypothesisrdquo Ukrainian Mathe-matical Journal vol 64 no 2 pp 218ndash228 2012

[2] S Beltraminelli D Merlini and S Sekatskii ldquoA hidden symme-try related to the Riemann hypothesis with the primes into thecritical striprdquoAsian-European Journal ofMathematics vol 3 no4 pp 555ndash563 2010

[3] E C Titchmarsh and E R Heath-Brown The Theory of theRiemann Zeta-Function Clarendon Press Oxford UK 1988

[4] F T Wang ldquoA note on the Riemann zeta-functionrdquo Bulletin ofthe American Mathematical Society vol 52 pp 319ndash321 1946

[5] V V Volchkov ldquoOn an equality equivalent to the Riemannhypothesisrdquo Ukrainian Mathematical Journal vol 47 no 3 pp491ndash493 1995

[6] M Balazard E Saias and M Yor ldquoNotes sur la fonction 120577 deRiemann 2rdquo Advances in Mathematics vol 143 no 2 pp 284ndash287 1999

[7] D Merlini ldquoThe Riemann magneton of the primesrdquo Chaos andComplexity Letters vol 2 no 1 pp 93ndash98 2006

[8] E C Titchmarsh The Theory of Functions Oxford UniversityPress Oxford UK 1939

[9] C Ramare and Y Saouter ldquoShort effective intervals containingprimesrdquo Journal ofNumberTheory vol 98 no 1 pp 10ndash33 2003

[10] R J Backlund ldquoUber die Nullstellen der Riemannschen Zeta-funktionrdquo Acta Mathematica vol 41 no 1 pp 345ndash375 1916

[11] X Gourdon ldquoThe 1013 first zeroes of the Riemann ZetaFunction and zeroes computation at very large heightrdquo 2004httpnumberscomputationfreefrConstantsMiscellaneouszetazeros1e13-1e24pdf

[12] D J Platt ldquoComputing 120587(119909) analyticallyrdquo Mathematics ofComputation vol 84 no 293 pp 1521ndash1535 2015

[13] A Erdelyi EdHigher Transcendental Functions Based in PartOn Notes Left by Harry Bateman vol 1 McGraw-Hill NewYork NY USA 1953

[14] I S Gradshtein and I M Ryzhik Tables of Integrals Series andProducts Academic New York NY USA 1990

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Differential EquationsInternational Journal of

Volume 2014

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

International Journal of Analysis 3

generalized Littlewood theorem results in the followingequality

int

infin

0

119890minus119886119905lnlowast (120589 (119887 + 119894119905)) 119889119905 minus int

infin

0

sin (119886119909) ln (120589 (119887 + 119909)) 119889119909

+ 119894 int

infin

0

cos (119886119909) ln (120589 (119887 + 119909)) 119889119909 = 2120587119886

sdot sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896[sin (119886 (120590119896 minus 119887))

+ 119894 (1 minus cos (119886 (120590119896 minus 119887)))] minus120587

119886(sin (119886 (1 minus 119887)) + 119894 (1

minus cos (119886 (1 minus 119887))))

(1)

In the last term here we took into account that the pole 119911 = 1lies exactly onto the contour border whence the coefficient is120587119886 instead of 2120587119886

Now we should remind the reader that the functionlnlowast(120589(119887 + 119894119905)) is not continuous see above (and this is thereason why we put an asterisk sign here) If we want touse a continuous logarithm function instead (which we willdenote simply ln(120589(119887 + 119894119905))) the following straightforwardmodification is to be done If real functions 119891(119909) 119892(119909)are continuous on the segment [119860 119861] and function 119891lowast(119909)coincides with 119891(119909) on [119860 119862] and is equal to 119891(119909) + ℎon (119862 119861] (ℎ is a constant 119860 le 119862 le 119861) we haveint119861

119860119892(119909)119891

lowast(119909)119889119909 = int

119861

119860119892(119909)119891(119909)119889119909 + ℎ int

119861

119862119892(119909)119889119909 and this

is trivially generalized for complex numbers and improperintegrals having counting number of discontinuities For ourparticular case we have considering one 2120587119894119897119896 discontinuityat 119905 = 119905119896 int

infin

0119890minus119886119905arglowast(120589(119887 + 119894119905))119889119905 = intinfin

0119890minus119886119905 arg(120589(119887 + 119894119905))119889119905 +

(2120587119897119896119886)119890minus119886119905119896 and the aforementioned generalization gives

int

infin

0

119890minus119886119905arglowast (120589 (119887 + 119894119905)) 119889119905

= int

infin

0

119890minus119886119905arg (120589 (119887 + 119894119905)) 119889119905 + sum

120588119905119896gt0120590119896gt119887

2120587119897119896

119886119890minus119886119905119896

(2)

Using (2) we get instead of (1)

int

infin

0

119890minus119886119905 ln (120589 (119887 + 119894119905)) 119889119905 minus int

infin

0

sin (119886119909) ln (120589 (119887 + 119909)) 119889119909

+ 119894 int

infin

0

cos (119886119909) ln (120589 (119887 + 119909)) 119889119909 = 2120587119886

sdot sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896[sin (119886 (120590119896 minus 119887))

minus 119894 cos (119886 (120590119896 minus 119887))] minus120587

119886(sin (119886 (1 minus 119887)) + 119894 (1

minus cos (119886 (1 minus 119887))))

(3)

where separation of real and imaginary parts readily givesequalities

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905 minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909

=2120587

119886sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 sin (119886 (120590119896 minus 119887))

minus120587

119886sin (119886 (1 minus 119887))

(4)

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

= minus2120587

119886sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896cos (119886 (120590119896 minus 119887))

minus120587

119886(1 minus cos (119886 (1 minus 119887)))

(5)

Here the argument is a continuous function along the contour(of course provided there are no Riemann zeroes lyingexactly on it) and the following initial values of the argumentshould be taken since we have the simple pole at 119911 = 1 onthe contour we select for real 119909 gt 1 arg(120589(119909)) = 0 and for1 gt 119887 gt minus2 arg(120589(119887 + 119894120576)) = minus120587 where real positive 120576 rarr 0

Equations (4) and (5) are our starting point and their firstapplication is the following theorem

Theorem 2 Equalities

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909 + 120587119886sin (119886 (1 minus 119887))

= 0

(6)

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

+120587

119886(1 minus cos (119886 (1 minus 119887))) = 0

(7)

where 119887 119886 are real positive numbers such that 1 gt 119887 ge 12119886(1 minus 119887) le 120587 for (6) and 119886(1 minus 119887) le 1205872 for (7) hold truefor some 119887 if and only if there are no Riemann function zeroeswith 120590 gt 119887 For 119887 = 12 these equalities are equivalent to theRiemann hypothesis

Proof For all possible Riemann 120589-function zeroes with Re120588 =120590119896 gt 119887 and hence lying inside the contour from our choice of119887 and the known properties of the Riemann function zeroeswe have 0 lt 120590119896minus119887 lt 1minus119887 le 12Then the imposed conditions119886(1minus119887) le 120587 and 119886(1minus119887) le 1205872 guaranty the same signs of all

4 International Journal of Analysis

possible contributions (2120587119886)sum120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 sin(119886(120590119896minus119887))

and minus(2120587119886)sum120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 cos(119886(120590119896minus119887)) from (4) and (5)

and also prevent them from being equal to zeroIn (6) it is interesting for 119887 = 12 to take 119886 = 2120587 thus

eliminating any problem with the integration over real axis(pole of the Riemann function coincides with zero of sine)

int

infin

0

119890minus2120587119905 ln

1003816100381610038161003816100381610038161003816120589 (1

2+ 119894119905)

1003816100381610038161003816100381610038161003816119889119905

minus int

infin

0

sin (2120587119909) ln1003816100381610038161003816100381610038161003816120589 (1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 = 0

(6a)

Similarly for (7) the choice 119886 = 120587 is useful

int

infin

0

119890minus120587119905 arg(120589 (1

2+ 119894119905)) 119889119905

+ int

infin

0

cos (120587119909) ln1003816100381610038161003816100381610038161003816120589 (1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 + 1 = 0

(7a)

Equalities (6a) (7a) are equivalent to the Riemannhypothesis They have been tested numerically using thestandard procedure Integrate in Mathematica Because thefirst integrand here decays very fast we can limit the inte-gration interval of the variable 119905 to [0 50] We need alsoto reduce the working precision (set to 50) and maximalrecursion (set to 70) For the second integral in (6a) and (7a)we set the integration interval for the variable 119909 to [0 200]and set the working precision and maximal recursion to50 After 15 hours of CPU time we obtained for (6a) theresult of 88044282 times 10minus32 and for (7a) after more than8 hours 114767406 times 10minus31 So in both cases we have acorrespondence for at least the first 30 figures

Next two similar theorems can be proved when we selectleft contour border line (119887 119887 + 119894infin) lying to the left to thecritical line

Theorem 2a Equality

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909 + 120587119886sin (119886 (1 minus 119887))

= 0

(8)

where 119886 119887 are real positive numbers such that 14 ge 119887 gt 0and 119886 = 120587119899(12 minus 119887) where 119899 is a positive integer such that119899 le 12119887 minus 1 holds true for some 119887 if and only if there are noRiemann function zeroes with 120590 le 119887

Proof In conditions of the theorem all contributions ofthe Riemann function zeroes lying on the critical line(2120587119886)sum

120588119905119896gt0120590119896gt119887119897119896119890minus119886119905119896 sin(119886(12 minus 119887)) to the contour inte-

gral value are equal to zero If there are a pair of Riemannfunction zeroes 120588 = 120590119896+119894119905119896 120588 = 1minus120590119896+119894119905119896 with 12 gt 120590119896 gt 119887both of them lie inside the contour and their contributionsexactly compensate each other thus contributing nothing tothe contour integral value If there is a zero with 120590119896 le 119887 it lies

outside the contour or on its border and hence contributesnothingwhile its partner zero 1minus120590119896+119894119905119896 lies inside the contourand contributes (2120587119886)119897119896 sin(119886(1 minus 120590119896 minus 119887)) In the conditionsof the theorem contributions of all such zeroes are nonzeroand have the same sign

Quite similarly we haveTheorem 2b whose proof one-to-one follows that of Theorem 2a

Theorem 2b Equality

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

+120587

119886(1 minus cos (119886 (1 minus 119887))) = 0

(9)

where 119886 119887 are real positive numbers such that 14 ge 119887 gt 0 and119886 = 120587(2119899+1)(1minus2119887) where 119899 is a positive integer or zero suchthat 119899 le 14119887 minus 1 holds true for some 119887 if and only if there areno Riemann function zeroes with 120590 le 119887

Unfortunately similar theorems cannot be formulated for12 gt 119887 gt 14 because no one value of 119886 can ensure thesame sign of all Riemann zeroes nonlying on the critical linecontributions for such a case

3 Rigorous Bound forthe Contribution of Riemann ZeroesNot Lying on the Critical Line

Exponential weight function appearing in the integralsconsidered in the previous section makes the problem ofestimation of themaximal possible contribution of remainingRiemann function zeroes nonlying on the critical line a rathersimple one For example for the real part we know ((4) for119887 = 12) that

int

infin

0

119890minus119886119905 ln

1003816100381610038161003816100381610038161003816120589 (1

2+ 119894119905)

1003816100381610038161003816100381610038161003816119889119905

minus int

infin

0

sin (119886119909) ln1003816100381610038161003816100381610038161003816120589 (1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 +

120587

119886sin(119886

2)

=2120587

119886sum

120588119905119896gt0

119897119896119890minus119886119905119896 sin 119886 (120590119896 minus

1

2)

(10)

where the sum is over all zeroes nonlying on the criticalline and having 120590 gt 12 How large the rhs here can beFirst we have that (2120587119886)| sum

120588119905119896gt0119897119896119890minus119886119905119896 sin 119886(120590119896 minus 12)| le

(2120587119886)sum120588119905119896gt0

119897119896119890minus119886119905119896 This last sum is given by Stieltjes inte-

gral 119868 = (2120587119886) intinfin0119890minus119886119905119889119873(119905) where119873(119905) is a discontinuous

counting function which counts the number of Riemannzeroes having the real part 120590119896 gt 12 and 119879 ge 119905119896 gt 0 takinginto account the order of zeroes The integration by partsreadily gives 119868 = (21205871198862) intinfin

0119890minus119886119905119873(119905)119889119905 About the function

119873(119905) we know that it is equal to zero when 119905 lt 119879 = 33 sdot 109[9] Further when 119905 gt 119879 it cannot exceed the function

International Journal of Analysis 5

(119905) minus (119879) where (119905) is a usual zero counting functionwhich counts zeroes in the strip 1 gt 120590119896 gt 0 119879 ge 119905119896 gt 0 (119873(119905)is equal to (119905) minus (119879) if all zeroes with 119905 gt 119879 do not lieon the critical line)Thus the contribution at question cannotexceed

119868 =120587

1198862int

infin

119879

119890minus119886119905( (119905) minus (119879)) 119889119905 (11)

Now for the function (119905) we use the known formula

(119905) =119905

2120587(ln 1199052120587minus 1) +

7

8+ 119876 (119905) (12)

with the estimation of the reminder given by Backlund in 1916[10]

|119876 (119905)| lt 0137 ln 119905 + 0443 ln ln 119905 + 4350 (13)

Then integration in (11) with (12) and (13) is trivial and weobtain that

|119868| lt1

21198863(2120587 sdot 0137 ln119879 + 1

119886ln119879 + 2120587 sdot 435) 119890minus119886119879

+2120587 sdot 0443

21198863

int

infin

119879

119890minus119886119905

ln2119905119889119905 + (

2120587 sdot 0137

21198863

+1

21198865)

sdot int

infin

119879

119890minus119886119905

119905119889119905

(14)

Of course intinfin119879(119890minus119886119905ln2119905)119889119905 lt (1119886)(119890

minus119886119879ln2119879)

intinfin

119879(119890minus119886119905119905)119889119905 lt (1119886)(119890

minus119886119879119879) and thus

|119868| lt1

21198863(0861 ln119879 + 1

119886ln119879 + 27332 + 2784

119886 ln2119879

+0861

119886119879+1

1198863119879) 119890minus119886119879

(15)

Thus for say 119886 = 1 we obtain |119868| lt 341119890minus33sdot10

9

=

341101433sdot10

9

and taking the maximal possible value of 119886119886 = 2120587 we get |119868| lt 0101119890minus2074sdot10

10

= 01011009sdot10

10

that isa precision with at least 143 milliard decimals in the first caseand of 9 milliards of decimals in the second case

To summarize we may say that combining a rigorousanalytical treatment with the known numerical results on theRiemann function zeros on the critical line we have found anequation of the form 119860 = 119861 + 120576 where 120576 = 0 is equivalent tothe RHWewere able to set a bound to 120576 of the order 1 over thenine milliardth power of 10 in this tiny factor lies the truth ofthe RH In a numerical experiment we verified the equationup to 30 decimals

Remark 3 There is a more recent calculation of Gourdonwhere it is reported that the first 119896 = 1013 Riemann zeroesare located on the critical line but we were unable to get anexact value of 119879 from the corresponding reference [11] See

also quite recent Platt paper [12] and references cited thereinfor a short review of the problem

It is also worthwhile to note that if we put the questionwhat is the attainable precision of equalities pertinent tocheck up whether there are no Riemann function zeroes with120590 lt 119887 lt 12 such a precision can be much larger because forsuch case in the conditions of Theorems 2 2a and 2b muchlarger values of 119886 can be taken for example for 119887 = 116 weare able to use the weight function 119890minus8120587119905

4 On Some Fourier Transforms ofLogarithm of the Riemann Zeta-FunctionTaken along the Real Axis and TheirRelation with the Riemann Hypothesis

We now discuss what is the situation with 119887 lt 0 that is ifwe move the left border of the contour further to the leftLet for a moment still 119887 gt minus2 First we select 119886 in such amanner that sin(119886(12 minus 119887)) = plusmn1 119886 le 120587 In doing so whenspeaking about integral equalities involving intinfin

0119890minus119886119905 ln |120589(119887 +

119894119905)|119889119905 we set all contributions of the Riemann function zeroeslying on the critical line to the ldquomaximal by modulusrdquo valueplusmn(2120587119886)sum

120588119905119896gt0120590119896=12119897119896119890minus119886119905119896 If we somehow know all ordi-

nates 119905119896 of the Riemann function zeroes 120588119896 = 120590119896+119894119905119896 we are ina position to calculate the value Σ119886RH = (2120587119886)sum120588119905119896gt0 119897119896119890

minus119886119905119896

and then we would be able to formulate a criterion equivalentto the Riemann hypothesis of the type ldquoRH holds true ifand only if the integral equality intinfin

0119890minus119886119905 ln |120589(119887 + 119894119905)|119889119905 minus

intinfin

0sin(119886119909) ln |120589(119887 + 119909)|119889119909 + (120587119886) sin(119886(1 minus 119887)) = Σ119886RH

holdsrdquo (evident changes taking into account an appearanceof trivial zeroes on the contour border are to be made if119887 lt minus2 see below) This is clear because from the formulaexpressing the contribution of zeroes to the contour integralvalue Σ119886 = (2120587119886)sum120588119905119896gt0120590119896gt119887 119897119896119890

minus119886119905119896 sin(119886(120590119896 minus 119887)) and ourchoice of 119886 119887 it immediately follows that for zeroes with120590119896 = 12 all contributions have the same sign and themoduleof contribution is smaller than it would be for zeroes with thesame ordinates but with 120590119896 = 12 | sin(119886(120590119896 minus 119887)) + sin(119886(1 minus120590119896 minus 119887))| lt |2 sin(119886(12 minus 119887))| = 2 Similar situation takesplace for an integral involving an argument of the Riemannfunction if we select cos(119886(12 minus 119887)) = plusmn1 119886 le 120587

Unfortunately we do not see how the constant Σ119886RH canbe efficiently calculated

The other way around we can select 119886 in such amanner that the condition cos(119886(12 minus 119887)) = 0 holdsThen for all such values of 119886 we obtain unconditionally(ie independent on the RH) true equalities all termsminus(2120587119886)sum

120588119905119896gt0120590119896gt119887119897119896119890minus119886119905119896 cos(119886(120590119896 minus 119887)) in (5) vanish they

are either equal to zero if 120590119896 = 12 or mutually compensateeach other for a pair of zeroes having 120590119896 = 12 plusmn 120572119896 = 12Similar unconditional integral equalities can be obtainedstarting from (4) for integrals involving logarithm of themodule of the Riemann function if we select sin(119886(12minus119887)) =0 For completeness we present these results as the followingsimple theorem

6 International Journal of Analysis

Theorem 4 For any real positive 119886 and any real 119887 such thatminus2 lt 119887 le 0 one has an integral equality

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

= minus120587

119886(1 minus cos (119886 (1 minus 119887)))

(16)

if cos(119886(12 minus 119887)) = 0 and integral equality

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909

= minus120587

119886sin (119886 (1 minus 119887))

(17)

if sin(119886(12 minus 119887)) = 0

If we move the left border of the contour further to theleft then contrary to the case 119887 gt minus2 on the lower border ofthe contour we have a number of trivial zeroes at the pointsminus2 minus4 minus6 and (4) is to be modified as

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909

=2120587

119886sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 sin (119886 (120590119896 minus 119887))

minus120587

119886sin (119886 (1 minus 119887)) + 120587

119886

119896lt[minus1198872]

sum

119896=1

sin (119886 (minus2119896 minus 119887))

(18)

Correspondingly (5) now reads

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+2120587

119886sum

120588119905119896gt0

119897119896119890minus119886119905119896 cos (119886 (120590119896 minus 119887))

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

+120587

119886(1 minus cos (119886 (1 minus 119887)))

minus120587

119886

119896lt[minus1198872]

sum

119896=1

(1 minus cos (119886 (minus2119896 minus 119887))) = 0

(19)

In this last equality we start from the following initial valueof an argument function arg(120589(119887 + 119894120576)) = minus120587 + [minus1198872] sdot120587 where real positive 120576 rarr 0 (As usual [minus1198872] means

thewhole part of positiveminus1198872number) Evident correspond-ing modifications should be done inTheorem 4

One of such unconditionally true equalities namely thatobtained from (5) putting 119886 = 2120587 and 119887 = minus72 thatis intinfin0119890minus2120587119905 arg(120589(minus72 + 119894119905))119889119905 + intinfin

0cos(2120587119909) ln |120589(minus72 +

119909)|119889119909 = 0 has also been checked up numerically (Note thathere 119887 lt minus2) In this case we calculated the first integralwith the same parameters as discussed above For the secondintegral we extended the integrationrsquos interval to [0 250] andthe maximal recursion to 100 The result 874533283 times 10minus30is similar to two previously mentioned cases Note that herearg(120589(minus72 + 119894120576)) = 0 and the contribution of the pole exactlycompensates that of the first trivial zero hence in fact the twointegrals when summing compensate each other up to 30digits as found in the computations

Interesting new possibilities appear if we take certainvalue of 119886 and introduce a sequence 119887119894 rarr minusinfin in such amanner that the condition cos(119886(12 minus 119887119894)) = 1 holds forany 119894 Below we illustrate such approach for a particular case119886 = 2120587 119887119899 = minus2119899 minus 12 (119899 is a positive integer) We have thefollowing particular case of (19)

int

infin

0

119890minus2120587119905 arg(120589 (minus2119899 minus 1

2+ 119894119905)) 119889119905 + Σ2120587

+ int

infin

0

cos (2120587119909) ln1003816100381610038161003816100381610038161003816120589 (minus2119899 minus

1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 +

1

2

minus119899

2= 0

(20)

where Σ2120587 = sum120588119905119896gt0 119897119896119890minus2120587119905119896 cos(2120587(120590119896+12)) is independent

on 119899Let us now analyze both integrals of (20) starting from

intinfin

0119890minus2120587119905 arg(120589(minus2119899 minus 12 + 119894119905))119889119905 All estimations below are

done with the 119874(1) precision because 119874(1119899) and smallerterms tend to zero for large 119899 To begin with we shouldinclude the contribution of an initial value of an argument ofthe Riemann function which is equal to arg(120589(119887 + 119894120576)) = (119899 minus1) sdot 120587 to (20) This contributes (119899 minus 1)2 to the integral valueNow let us change 120589(minus2119899minus12+119894119905) function to 120589(2119899+32minus119894119905)function using functional equation [3]

120589 (119904) =1

120587(2120587)119904 sin120587119904

2Γ (1 minus 119904) 120589 (1 minus 119904) (21)

Factor (2120587)minus2119899minus12+119894119905 in (21) contributes ln(2120587)41205872 tothe integral value For sine factor taking 119899 odd wehave sin((minus14 minus 119899 + 1198941199052)120587) = sin((14 minus 1198941199052)120587) =(radic22)(minus119894 sinh(1205871199052) + cosh(1205871199052)) Its argument isminusarctan(tanh(1205871199052)) hence we need to handle an integralminusintinfin

0119890minus2120587119905arctan(tanh(1205871199052))119889119905 Integrating by parts

we see that this contribution is equal to minus12120587 sdot (1205872) intinfin

0(eminus2120587t cosh(120587119905))119889119905 = minus(12) intinfin

0(119890minus2120587119905(119890120587119905+ 119890minus120587119905))119889119905

with the variable transform 119910 = 119890120587119905 this is easily calculated tobe equal to minus12120587 + 18

Integral involving an argument of Riemann 120589-functiontends to zero when 119899 rarr infin so next we evaluate the

International Journal of Analysis 7

contribution of the gamma-functionWe use Stirling formula[13]

ln Γ (32+ 2119899 minus 119894119905) = (1 + 2119899 minus 119894119905) ln(3

2+ 2119899 minus 119894119905)

minus (3

2+ 2119899 minus 119894119905) + 05 ln 2120587

+ 119892 (119909)

(22)

where 119892(119909) = intinfin0(12 minus 1119905 + 1(119890

119905minus 1))(119890

minus119905119909119905)119889119905 [13] and

have

Imintinfin

0

119890minus2120587119905(1 + 2119899 minus 119894119905) ln(3

2+ 2119899 minus 119894119905) 119889119905

= Im 1

2120587int

infin

0

119890minus2120587119905(minus119894 ln(3

2+ 2119899 minus 119894119905)

minus 1198941 + 2119899 minus 119894119905

32 + 2119899 minus 119894119905) 119889119905 = minus

1

41205872ln(32+ 2119899)

minus1

41205872+ 119874(

1

119899)

(23)

Further minus Imintinfin0119890minus2120587119905(32 + 2119899 minus 119894119905)119889119905 = 14120587

2 Factor05 ln 2120587 is real and contributes nothing It is not difficult tosee that integral involving 119892(119911) tends to zero with 119899 tendingto infinity

Now we need to analyze second integral in (20)that is the integral intinfin

0cos(2120587119909) ln |120589(minus12 minus 2119899 + 119909)|119889119909

With a variable change 119910 = minus12 minus 2119899 + 119909 it isequal to minusintinfin

minus12minus2119899cos(2120587119910) ln |120589(119910)|119889119910 which we divide into

minusint12

minus12minus2119899minusintinfin

12 For the first integral we use the variable

transform 119909 = 1 minus 119910 to write that it is equal tominusint32+2119899

12cos(2120587119909) ln |120589(1 minus 119909)|119889119909 and then use the same

functional equation (21) again so we need to estimate

1198682 = minusint

32+2119899

12

cos (2120587119909)

sdot ln10038161003816100381610038161003816100381610038161003816

1

120587(2120587)1minus119909 sin120587 (1 minus 119909)

2Γ (119909) 120589 (119909)

10038161003816100381610038161003816100381610038161003816

119889119909

(24)

First int32+211989912

cos(2120587119909) ln |(1120587)(2120587)1minus119909|119889119909 = 0 For anintegral involving logarithm of gamma-functionwe again use

the Stirling formula ln Γ(119909) = (119909minus12) ln(119909)minus119909 + 05 ln 2120587 +119892(119909) [13] and get

minus int

32+2119899

12

cos (2120587119909) (119909 minus 12) ln (119909) 119889119909

=1

2120587int

32+2119899

12

sin (2120587119909) (ln119909 + 1 minus 12119909) 119889119909

=1

2120587int

32+2119899

12

sin (2120587119909) (ln119909 minus 12119909) 119889119909

= minus1

2120587int

32+2119899

12

sin (2120587119909) 12119909119889119909

minus1

41205872cos (2120587119909) ln119909

1003816100381610038161003816100381610038161003816

32+2119899

12

+1

41205872int

32+2119899

12

cos (2120587119909) 1119909119889119909

=1

41205872(ln(3

2+ 2119899) + 2119899)

+1

41205872int

32+119899

12

1

119909cos (2120587119909) 119889119909

minus1

2120587int

32+2119899

12

sin (2120587119909) 12119909119889119909

(25)

To finish the consideration of the contribution of thelogarithm of gamma-function we need to add the contri-bution of 119892(119909) which is minusintinfin

12cos(2120587119909)119889119909 intinfin

0(12 minus 1119905 +

1(119890119905minus 1))(119890

minus119905119909119905)119889119905 Using cos(2120587119909) = (12)(1198902120587119909119894 + 119890minus2120587119909119894)

we can perform integration over 119909 minusintinfin12

cos(2120587119909)119890minus119905119909119889119909 =119890minus1199052(119905(1199052+ 41205872)) and thus the contribution at question is

intinfin

0(12 minus 1119905 + 1(119890

119905minus 1))(119890

minus1199052(1199052+ 41205872))119889119905

Finally we need to analyze the contribution of thesine factor in (24) which is minusint32+2119899

12cos(2120587119909) ln |sin(120587(1 minus

119909)2)|119889119909 = minusint32+211989912

cos(2120587119909) ln |cos(1205871199092)|119889119909 Integrationis to be applied with caution here due to the presence ofthe module To avoid errors it is reasonable to considerseparately integrals int1

12 int2119899+322119899+1

= int32

1 and int2

1 the latter is

an integral along the period Integrating by parts we have

minus int

1

12

cos (2120587119909) ln1003816100381610038161003816100381610038161003816cos 120587119909

2

1003816100381610038161003816100381610038161003816119889119909

= minusint

1

12

cos (2120587119909) ln(cos 1205871199092) 119889119909

= minus1

4int

1

12

sin (2120587119909) tan 1205871199092119889119909

(26)

8 International Journal of Analysis

and further

minus1

4int

1

12

sin (2120587119909) tan 1205871199092119889119909

= minusint

1

12

sin21205871199092

cos (120587119909) 119889119909

=1

2int

1

12

(cos (120587119909) minus 1) cos (120587119909) 119889119909

=1

2int

1

12

(1 + cos (2120587119909)

2minus cos (120587119909)) 119889119909

=1

8minus1

2120587sin (120587119909)

1003816100381610038161003816100381610038161003816

1

12

=1

8+1

2120587

(27)

Similarly minusint32

1cos(2120587119909) ln | cos(1205871199092)|119889119909 =

minusint32

1cos(2120587119909) ln(minus cos(1205871199092))119889119909 = 18 + 12120587 while

integral over the period is equal to 14 (use the sameapproach as above or entry number 43843 of [14] withevident variable changes) so the total contribution of thesine factor in (24) to (20) is equal to 1198992 + 14 + 1120587

Collecting everything together we see the cancelling of119874(119899) and 119874(ln(119899)) terms Thus considering the limit 119899 rarrinfin we have established the following relation expressingcertain particular value of Fourier transform of the logarithmof the Riemann zeta-function

int

infin

12

cos (2120587119909) ln |120589 (119909)| 119889119909

=1

2Σ2120587 +

7

16+1

4120587+ln (4120587)81205872minus119888119894 (120587)

81205872+119904119894 (120587)

8120587

+1

2int

infin

0

(1

2+1

119890119905minus 1minus1

119905)119890minus1199052

1199052+ 41205872119889119905

(28)

Here we used standard notation 119904119894(119909) = minusintinfin119909(sin 119902119902)119889119902

119888119894(119909) = minus intinfin

119909(cos 119902119902)119889119902 [14] Of course similar equalities

can be established for many other such Fourier transformsRepeating what has been said above about the ldquomax-

imalityrdquo of the sum Σ2120587RH = sum120588119905119896gt0

119897119896119890minus2120587119905119896 we again

could formulate a statement of the type ldquoRH holds true ifand only if the integral equality (28) with Σ2120587 = Σ2120587RHholdsrdquo Numerical calculation taking into account first 100zeroes of the Riemann function gives the value Σ2120587 =26904854 sdot 10

minus39 Equality (28) has been tested numer-ically and shown to be true without the term Σ2120587 bothsides coincide up to 39 figures after decimal and are equalto 05620211964552684935093570395449423344 Then ofcourse they start to differ

5 Conclusions

In this paper we have established a number of new criteriainvolving the integrals of the logarithm of the Riemann 120589-function and equivalent to the Riemann hypothesis thistime with the exponential weight functions Exponential

weight functions lead to rather simple expressions for thecontribution of the Riemann function zeroes not lying onthe critical line to the contour integral value This enabledobtaining a rigorous estimation of the possible error whichwas shown to be extremely small

We also show how certain Fourier transforms of thelogarithm of the Riemann zeta-function taken along thereal (demi)axis are expressible via elementary functionsplus logarithm of the gamma-function and definite integralsthereof as well as certain sums over trivial and nontrivialRiemann function zeroes In our opinion further study ofsimilar Fourier transforms is interesting and might be usefulin the Riemann researches

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] S K Sekatskii S Beltraminelli and D Merlini ldquoOn equalitiesinvolving integrals of the logarithm of the Riemann-functionand equivalent to the Riemann hypothesisrdquo Ukrainian Mathe-matical Journal vol 64 no 2 pp 218ndash228 2012

[2] S Beltraminelli D Merlini and S Sekatskii ldquoA hidden symme-try related to the Riemann hypothesis with the primes into thecritical striprdquoAsian-European Journal ofMathematics vol 3 no4 pp 555ndash563 2010

[3] E C Titchmarsh and E R Heath-Brown The Theory of theRiemann Zeta-Function Clarendon Press Oxford UK 1988

[4] F T Wang ldquoA note on the Riemann zeta-functionrdquo Bulletin ofthe American Mathematical Society vol 52 pp 319ndash321 1946

[5] V V Volchkov ldquoOn an equality equivalent to the Riemannhypothesisrdquo Ukrainian Mathematical Journal vol 47 no 3 pp491ndash493 1995

[6] M Balazard E Saias and M Yor ldquoNotes sur la fonction 120577 deRiemann 2rdquo Advances in Mathematics vol 143 no 2 pp 284ndash287 1999

[7] D Merlini ldquoThe Riemann magneton of the primesrdquo Chaos andComplexity Letters vol 2 no 1 pp 93ndash98 2006

[8] E C Titchmarsh The Theory of Functions Oxford UniversityPress Oxford UK 1939

[9] C Ramare and Y Saouter ldquoShort effective intervals containingprimesrdquo Journal ofNumberTheory vol 98 no 1 pp 10ndash33 2003

[10] R J Backlund ldquoUber die Nullstellen der Riemannschen Zeta-funktionrdquo Acta Mathematica vol 41 no 1 pp 345ndash375 1916

[11] X Gourdon ldquoThe 1013 first zeroes of the Riemann ZetaFunction and zeroes computation at very large heightrdquo 2004httpnumberscomputationfreefrConstantsMiscellaneouszetazeros1e13-1e24pdf

[12] D J Platt ldquoComputing 120587(119909) analyticallyrdquo Mathematics ofComputation vol 84 no 293 pp 1521ndash1535 2015

[13] A Erdelyi EdHigher Transcendental Functions Based in PartOn Notes Left by Harry Bateman vol 1 McGraw-Hill NewYork NY USA 1953

[14] I S Gradshtein and I M Ryzhik Tables of Integrals Series andProducts Academic New York NY USA 1990

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Stochastic AnalysisInternational Journal of

4 International Journal of Analysis

possible contributions (2120587119886)sum120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 sin(119886(120590119896minus119887))

and minus(2120587119886)sum120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 cos(119886(120590119896minus119887)) from (4) and (5)

and also prevent them from being equal to zeroIn (6) it is interesting for 119887 = 12 to take 119886 = 2120587 thus

eliminating any problem with the integration over real axis(pole of the Riemann function coincides with zero of sine)

int

infin

0

119890minus2120587119905 ln

1003816100381610038161003816100381610038161003816120589 (1

2+ 119894119905)

1003816100381610038161003816100381610038161003816119889119905

minus int

infin

0

sin (2120587119909) ln1003816100381610038161003816100381610038161003816120589 (1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 = 0

(6a)

Similarly for (7) the choice 119886 = 120587 is useful

int

infin

0

119890minus120587119905 arg(120589 (1

2+ 119894119905)) 119889119905

+ int

infin

0

cos (120587119909) ln1003816100381610038161003816100381610038161003816120589 (1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 + 1 = 0

(7a)

Equalities (6a) (7a) are equivalent to the Riemannhypothesis They have been tested numerically using thestandard procedure Integrate in Mathematica Because thefirst integrand here decays very fast we can limit the inte-gration interval of the variable 119905 to [0 50] We need alsoto reduce the working precision (set to 50) and maximalrecursion (set to 70) For the second integral in (6a) and (7a)we set the integration interval for the variable 119909 to [0 200]and set the working precision and maximal recursion to50 After 15 hours of CPU time we obtained for (6a) theresult of 88044282 times 10minus32 and for (7a) after more than8 hours 114767406 times 10minus31 So in both cases we have acorrespondence for at least the first 30 figures

Next two similar theorems can be proved when we selectleft contour border line (119887 119887 + 119894infin) lying to the left to thecritical line

Theorem 2a Equality

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909 + 120587119886sin (119886 (1 minus 119887))

= 0

(8)

where 119886 119887 are real positive numbers such that 14 ge 119887 gt 0and 119886 = 120587119899(12 minus 119887) where 119899 is a positive integer such that119899 le 12119887 minus 1 holds true for some 119887 if and only if there are noRiemann function zeroes with 120590 le 119887

Proof In conditions of the theorem all contributions ofthe Riemann function zeroes lying on the critical line(2120587119886)sum

120588119905119896gt0120590119896gt119887119897119896119890minus119886119905119896 sin(119886(12 minus 119887)) to the contour inte-

gral value are equal to zero If there are a pair of Riemannfunction zeroes 120588 = 120590119896+119894119905119896 120588 = 1minus120590119896+119894119905119896 with 12 gt 120590119896 gt 119887both of them lie inside the contour and their contributionsexactly compensate each other thus contributing nothing tothe contour integral value If there is a zero with 120590119896 le 119887 it lies

outside the contour or on its border and hence contributesnothingwhile its partner zero 1minus120590119896+119894119905119896 lies inside the contourand contributes (2120587119886)119897119896 sin(119886(1 minus 120590119896 minus 119887)) In the conditionsof the theorem contributions of all such zeroes are nonzeroand have the same sign

Quite similarly we haveTheorem 2b whose proof one-to-one follows that of Theorem 2a

Theorem 2b Equality

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

+120587

119886(1 minus cos (119886 (1 minus 119887))) = 0

(9)

where 119886 119887 are real positive numbers such that 14 ge 119887 gt 0 and119886 = 120587(2119899+1)(1minus2119887) where 119899 is a positive integer or zero suchthat 119899 le 14119887 minus 1 holds true for some 119887 if and only if there areno Riemann function zeroes with 120590 le 119887

Unfortunately similar theorems cannot be formulated for12 gt 119887 gt 14 because no one value of 119886 can ensure thesame sign of all Riemann zeroes nonlying on the critical linecontributions for such a case

3 Rigorous Bound forthe Contribution of Riemann ZeroesNot Lying on the Critical Line

Exponential weight function appearing in the integralsconsidered in the previous section makes the problem ofestimation of themaximal possible contribution of remainingRiemann function zeroes nonlying on the critical line a rathersimple one For example for the real part we know ((4) for119887 = 12) that

int

infin

0

119890minus119886119905 ln

1003816100381610038161003816100381610038161003816120589 (1

2+ 119894119905)

1003816100381610038161003816100381610038161003816119889119905

minus int

infin

0

sin (119886119909) ln1003816100381610038161003816100381610038161003816120589 (1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 +

120587

119886sin(119886

2)

=2120587

119886sum

120588119905119896gt0

119897119896119890minus119886119905119896 sin 119886 (120590119896 minus

1

2)

(10)

where the sum is over all zeroes nonlying on the criticalline and having 120590 gt 12 How large the rhs here can beFirst we have that (2120587119886)| sum

120588119905119896gt0119897119896119890minus119886119905119896 sin 119886(120590119896 minus 12)| le

(2120587119886)sum120588119905119896gt0

119897119896119890minus119886119905119896 This last sum is given by Stieltjes inte-

gral 119868 = (2120587119886) intinfin0119890minus119886119905119889119873(119905) where119873(119905) is a discontinuous

counting function which counts the number of Riemannzeroes having the real part 120590119896 gt 12 and 119879 ge 119905119896 gt 0 takinginto account the order of zeroes The integration by partsreadily gives 119868 = (21205871198862) intinfin

0119890minus119886119905119873(119905)119889119905 About the function

119873(119905) we know that it is equal to zero when 119905 lt 119879 = 33 sdot 109[9] Further when 119905 gt 119879 it cannot exceed the function

International Journal of Analysis 5

(119905) minus (119879) where (119905) is a usual zero counting functionwhich counts zeroes in the strip 1 gt 120590119896 gt 0 119879 ge 119905119896 gt 0 (119873(119905)is equal to (119905) minus (119879) if all zeroes with 119905 gt 119879 do not lieon the critical line)Thus the contribution at question cannotexceed

119868 =120587

1198862int

infin

119879

119890minus119886119905( (119905) minus (119879)) 119889119905 (11)

Now for the function (119905) we use the known formula

(119905) =119905

2120587(ln 1199052120587minus 1) +

7

8+ 119876 (119905) (12)

with the estimation of the reminder given by Backlund in 1916[10]

|119876 (119905)| lt 0137 ln 119905 + 0443 ln ln 119905 + 4350 (13)

Then integration in (11) with (12) and (13) is trivial and weobtain that

|119868| lt1

21198863(2120587 sdot 0137 ln119879 + 1

119886ln119879 + 2120587 sdot 435) 119890minus119886119879

+2120587 sdot 0443

21198863

int

infin

119879

119890minus119886119905

ln2119905119889119905 + (

2120587 sdot 0137

21198863

+1

21198865)

sdot int

infin

119879

119890minus119886119905

119905119889119905

(14)

Of course intinfin119879(119890minus119886119905ln2119905)119889119905 lt (1119886)(119890

minus119886119879ln2119879)

intinfin

119879(119890minus119886119905119905)119889119905 lt (1119886)(119890

minus119886119879119879) and thus

|119868| lt1

21198863(0861 ln119879 + 1

119886ln119879 + 27332 + 2784

119886 ln2119879

+0861

119886119879+1

1198863119879) 119890minus119886119879

(15)

Thus for say 119886 = 1 we obtain |119868| lt 341119890minus33sdot10

9

=

341101433sdot10

9

and taking the maximal possible value of 119886119886 = 2120587 we get |119868| lt 0101119890minus2074sdot10

10

= 01011009sdot10

10

that isa precision with at least 143 milliard decimals in the first caseand of 9 milliards of decimals in the second case

To summarize we may say that combining a rigorousanalytical treatment with the known numerical results on theRiemann function zeros on the critical line we have found anequation of the form 119860 = 119861 + 120576 where 120576 = 0 is equivalent tothe RHWewere able to set a bound to 120576 of the order 1 over thenine milliardth power of 10 in this tiny factor lies the truth ofthe RH In a numerical experiment we verified the equationup to 30 decimals

Remark 3 There is a more recent calculation of Gourdonwhere it is reported that the first 119896 = 1013 Riemann zeroesare located on the critical line but we were unable to get anexact value of 119879 from the corresponding reference [11] See

also quite recent Platt paper [12] and references cited thereinfor a short review of the problem

It is also worthwhile to note that if we put the questionwhat is the attainable precision of equalities pertinent tocheck up whether there are no Riemann function zeroes with120590 lt 119887 lt 12 such a precision can be much larger because forsuch case in the conditions of Theorems 2 2a and 2b muchlarger values of 119886 can be taken for example for 119887 = 116 weare able to use the weight function 119890minus8120587119905

4 On Some Fourier Transforms ofLogarithm of the Riemann Zeta-FunctionTaken along the Real Axis and TheirRelation with the Riemann Hypothesis

We now discuss what is the situation with 119887 lt 0 that is ifwe move the left border of the contour further to the leftLet for a moment still 119887 gt minus2 First we select 119886 in such amanner that sin(119886(12 minus 119887)) = plusmn1 119886 le 120587 In doing so whenspeaking about integral equalities involving intinfin

0119890minus119886119905 ln |120589(119887 +

119894119905)|119889119905 we set all contributions of the Riemann function zeroeslying on the critical line to the ldquomaximal by modulusrdquo valueplusmn(2120587119886)sum

120588119905119896gt0120590119896=12119897119896119890minus119886119905119896 If we somehow know all ordi-

nates 119905119896 of the Riemann function zeroes 120588119896 = 120590119896+119894119905119896 we are ina position to calculate the value Σ119886RH = (2120587119886)sum120588119905119896gt0 119897119896119890

minus119886119905119896

and then we would be able to formulate a criterion equivalentto the Riemann hypothesis of the type ldquoRH holds true ifand only if the integral equality intinfin

0119890minus119886119905 ln |120589(119887 + 119894119905)|119889119905 minus

intinfin

0sin(119886119909) ln |120589(119887 + 119909)|119889119909 + (120587119886) sin(119886(1 minus 119887)) = Σ119886RH

holdsrdquo (evident changes taking into account an appearanceof trivial zeroes on the contour border are to be made if119887 lt minus2 see below) This is clear because from the formulaexpressing the contribution of zeroes to the contour integralvalue Σ119886 = (2120587119886)sum120588119905119896gt0120590119896gt119887 119897119896119890

minus119886119905119896 sin(119886(120590119896 minus 119887)) and ourchoice of 119886 119887 it immediately follows that for zeroes with120590119896 = 12 all contributions have the same sign and themoduleof contribution is smaller than it would be for zeroes with thesame ordinates but with 120590119896 = 12 | sin(119886(120590119896 minus 119887)) + sin(119886(1 minus120590119896 minus 119887))| lt |2 sin(119886(12 minus 119887))| = 2 Similar situation takesplace for an integral involving an argument of the Riemannfunction if we select cos(119886(12 minus 119887)) = plusmn1 119886 le 120587

Unfortunately we do not see how the constant Σ119886RH canbe efficiently calculated

The other way around we can select 119886 in such amanner that the condition cos(119886(12 minus 119887)) = 0 holdsThen for all such values of 119886 we obtain unconditionally(ie independent on the RH) true equalities all termsminus(2120587119886)sum

120588119905119896gt0120590119896gt119887119897119896119890minus119886119905119896 cos(119886(120590119896 minus 119887)) in (5) vanish they

are either equal to zero if 120590119896 = 12 or mutually compensateeach other for a pair of zeroes having 120590119896 = 12 plusmn 120572119896 = 12Similar unconditional integral equalities can be obtainedstarting from (4) for integrals involving logarithm of themodule of the Riemann function if we select sin(119886(12minus119887)) =0 For completeness we present these results as the followingsimple theorem

6 International Journal of Analysis

Theorem 4 For any real positive 119886 and any real 119887 such thatminus2 lt 119887 le 0 one has an integral equality

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

= minus120587

119886(1 minus cos (119886 (1 minus 119887)))

(16)

if cos(119886(12 minus 119887)) = 0 and integral equality

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909

= minus120587

119886sin (119886 (1 minus 119887))

(17)

if sin(119886(12 minus 119887)) = 0

If we move the left border of the contour further to theleft then contrary to the case 119887 gt minus2 on the lower border ofthe contour we have a number of trivial zeroes at the pointsminus2 minus4 minus6 and (4) is to be modified as

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909

=2120587

119886sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 sin (119886 (120590119896 minus 119887))

minus120587

119886sin (119886 (1 minus 119887)) + 120587

119886

119896lt[minus1198872]

sum

119896=1

sin (119886 (minus2119896 minus 119887))

(18)

Correspondingly (5) now reads

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+2120587

119886sum

120588119905119896gt0

119897119896119890minus119886119905119896 cos (119886 (120590119896 minus 119887))

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

+120587

119886(1 minus cos (119886 (1 minus 119887)))

minus120587

119886

119896lt[minus1198872]

sum

119896=1

(1 minus cos (119886 (minus2119896 minus 119887))) = 0

(19)

In this last equality we start from the following initial valueof an argument function arg(120589(119887 + 119894120576)) = minus120587 + [minus1198872] sdot120587 where real positive 120576 rarr 0 (As usual [minus1198872] means

thewhole part of positiveminus1198872number) Evident correspond-ing modifications should be done inTheorem 4

One of such unconditionally true equalities namely thatobtained from (5) putting 119886 = 2120587 and 119887 = minus72 thatis intinfin0119890minus2120587119905 arg(120589(minus72 + 119894119905))119889119905 + intinfin

0cos(2120587119909) ln |120589(minus72 +

119909)|119889119909 = 0 has also been checked up numerically (Note thathere 119887 lt minus2) In this case we calculated the first integralwith the same parameters as discussed above For the secondintegral we extended the integrationrsquos interval to [0 250] andthe maximal recursion to 100 The result 874533283 times 10minus30is similar to two previously mentioned cases Note that herearg(120589(minus72 + 119894120576)) = 0 and the contribution of the pole exactlycompensates that of the first trivial zero hence in fact the twointegrals when summing compensate each other up to 30digits as found in the computations

Interesting new possibilities appear if we take certainvalue of 119886 and introduce a sequence 119887119894 rarr minusinfin in such amanner that the condition cos(119886(12 minus 119887119894)) = 1 holds forany 119894 Below we illustrate such approach for a particular case119886 = 2120587 119887119899 = minus2119899 minus 12 (119899 is a positive integer) We have thefollowing particular case of (19)

int

infin

0

119890minus2120587119905 arg(120589 (minus2119899 minus 1

2+ 119894119905)) 119889119905 + Σ2120587

+ int

infin

0

cos (2120587119909) ln1003816100381610038161003816100381610038161003816120589 (minus2119899 minus

1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 +

1

2

minus119899

2= 0

(20)

where Σ2120587 = sum120588119905119896gt0 119897119896119890minus2120587119905119896 cos(2120587(120590119896+12)) is independent

on 119899Let us now analyze both integrals of (20) starting from

intinfin

0119890minus2120587119905 arg(120589(minus2119899 minus 12 + 119894119905))119889119905 All estimations below are

done with the 119874(1) precision because 119874(1119899) and smallerterms tend to zero for large 119899 To begin with we shouldinclude the contribution of an initial value of an argument ofthe Riemann function which is equal to arg(120589(119887 + 119894120576)) = (119899 minus1) sdot 120587 to (20) This contributes (119899 minus 1)2 to the integral valueNow let us change 120589(minus2119899minus12+119894119905) function to 120589(2119899+32minus119894119905)function using functional equation [3]

120589 (119904) =1

120587(2120587)119904 sin120587119904

2Γ (1 minus 119904) 120589 (1 minus 119904) (21)

Factor (2120587)minus2119899minus12+119894119905 in (21) contributes ln(2120587)41205872 tothe integral value For sine factor taking 119899 odd wehave sin((minus14 minus 119899 + 1198941199052)120587) = sin((14 minus 1198941199052)120587) =(radic22)(minus119894 sinh(1205871199052) + cosh(1205871199052)) Its argument isminusarctan(tanh(1205871199052)) hence we need to handle an integralminusintinfin

0119890minus2120587119905arctan(tanh(1205871199052))119889119905 Integrating by parts

we see that this contribution is equal to minus12120587 sdot (1205872) intinfin

0(eminus2120587t cosh(120587119905))119889119905 = minus(12) intinfin

0(119890minus2120587119905(119890120587119905+ 119890minus120587119905))119889119905

with the variable transform 119910 = 119890120587119905 this is easily calculated tobe equal to minus12120587 + 18

Integral involving an argument of Riemann 120589-functiontends to zero when 119899 rarr infin so next we evaluate the

International Journal of Analysis 7

contribution of the gamma-functionWe use Stirling formula[13]

ln Γ (32+ 2119899 minus 119894119905) = (1 + 2119899 minus 119894119905) ln(3

2+ 2119899 minus 119894119905)

minus (3

2+ 2119899 minus 119894119905) + 05 ln 2120587

+ 119892 (119909)

(22)

where 119892(119909) = intinfin0(12 minus 1119905 + 1(119890

119905minus 1))(119890

minus119905119909119905)119889119905 [13] and

have

Imintinfin

0

119890minus2120587119905(1 + 2119899 minus 119894119905) ln(3

2+ 2119899 minus 119894119905) 119889119905

= Im 1

2120587int

infin

0

119890minus2120587119905(minus119894 ln(3

2+ 2119899 minus 119894119905)

minus 1198941 + 2119899 minus 119894119905

32 + 2119899 minus 119894119905) 119889119905 = minus

1

41205872ln(32+ 2119899)

minus1

41205872+ 119874(

1

119899)

(23)

Further minus Imintinfin0119890minus2120587119905(32 + 2119899 minus 119894119905)119889119905 = 14120587

2 Factor05 ln 2120587 is real and contributes nothing It is not difficult tosee that integral involving 119892(119911) tends to zero with 119899 tendingto infinity

Now we need to analyze second integral in (20)that is the integral intinfin

0cos(2120587119909) ln |120589(minus12 minus 2119899 + 119909)|119889119909

With a variable change 119910 = minus12 minus 2119899 + 119909 it isequal to minusintinfin

minus12minus2119899cos(2120587119910) ln |120589(119910)|119889119910 which we divide into

minusint12

minus12minus2119899minusintinfin

12 For the first integral we use the variable

transform 119909 = 1 minus 119910 to write that it is equal tominusint32+2119899

12cos(2120587119909) ln |120589(1 minus 119909)|119889119909 and then use the same

functional equation (21) again so we need to estimate

1198682 = minusint

32+2119899

12

cos (2120587119909)

sdot ln10038161003816100381610038161003816100381610038161003816

1

120587(2120587)1minus119909 sin120587 (1 minus 119909)

2Γ (119909) 120589 (119909)

10038161003816100381610038161003816100381610038161003816

119889119909

(24)

First int32+211989912

cos(2120587119909) ln |(1120587)(2120587)1minus119909|119889119909 = 0 For anintegral involving logarithm of gamma-functionwe again use

the Stirling formula ln Γ(119909) = (119909minus12) ln(119909)minus119909 + 05 ln 2120587 +119892(119909) [13] and get

minus int

32+2119899

12

cos (2120587119909) (119909 minus 12) ln (119909) 119889119909

=1

2120587int

32+2119899

12

sin (2120587119909) (ln119909 + 1 minus 12119909) 119889119909

=1

2120587int

32+2119899

12

sin (2120587119909) (ln119909 minus 12119909) 119889119909

= minus1

2120587int

32+2119899

12

sin (2120587119909) 12119909119889119909

minus1

41205872cos (2120587119909) ln119909

1003816100381610038161003816100381610038161003816

32+2119899

12

+1

41205872int

32+2119899

12

cos (2120587119909) 1119909119889119909

=1

41205872(ln(3

2+ 2119899) + 2119899)

+1

41205872int

32+119899

12

1

119909cos (2120587119909) 119889119909

minus1

2120587int

32+2119899

12

sin (2120587119909) 12119909119889119909

(25)

To finish the consideration of the contribution of thelogarithm of gamma-function we need to add the contri-bution of 119892(119909) which is minusintinfin

12cos(2120587119909)119889119909 intinfin

0(12 minus 1119905 +

1(119890119905minus 1))(119890

minus119905119909119905)119889119905 Using cos(2120587119909) = (12)(1198902120587119909119894 + 119890minus2120587119909119894)

we can perform integration over 119909 minusintinfin12

cos(2120587119909)119890minus119905119909119889119909 =119890minus1199052(119905(1199052+ 41205872)) and thus the contribution at question is

intinfin

0(12 minus 1119905 + 1(119890

119905minus 1))(119890

minus1199052(1199052+ 41205872))119889119905

Finally we need to analyze the contribution of thesine factor in (24) which is minusint32+2119899

12cos(2120587119909) ln |sin(120587(1 minus

119909)2)|119889119909 = minusint32+211989912

cos(2120587119909) ln |cos(1205871199092)|119889119909 Integrationis to be applied with caution here due to the presence ofthe module To avoid errors it is reasonable to considerseparately integrals int1

12 int2119899+322119899+1

= int32

1 and int2

1 the latter is

an integral along the period Integrating by parts we have

minus int

1

12

cos (2120587119909) ln1003816100381610038161003816100381610038161003816cos 120587119909

2

1003816100381610038161003816100381610038161003816119889119909

= minusint

1

12

cos (2120587119909) ln(cos 1205871199092) 119889119909

= minus1

4int

1

12

sin (2120587119909) tan 1205871199092119889119909

(26)

8 International Journal of Analysis

and further

minus1

4int

1

12

sin (2120587119909) tan 1205871199092119889119909

= minusint

1

12

sin21205871199092

cos (120587119909) 119889119909

=1

2int

1

12

(cos (120587119909) minus 1) cos (120587119909) 119889119909

=1

2int

1

12

(1 + cos (2120587119909)

2minus cos (120587119909)) 119889119909

=1

8minus1

2120587sin (120587119909)

1003816100381610038161003816100381610038161003816

1

12

=1

8+1

2120587

(27)

Similarly minusint32

1cos(2120587119909) ln | cos(1205871199092)|119889119909 =

minusint32

1cos(2120587119909) ln(minus cos(1205871199092))119889119909 = 18 + 12120587 while

integral over the period is equal to 14 (use the sameapproach as above or entry number 43843 of [14] withevident variable changes) so the total contribution of thesine factor in (24) to (20) is equal to 1198992 + 14 + 1120587

Collecting everything together we see the cancelling of119874(119899) and 119874(ln(119899)) terms Thus considering the limit 119899 rarrinfin we have established the following relation expressingcertain particular value of Fourier transform of the logarithmof the Riemann zeta-function

int

infin

12

cos (2120587119909) ln |120589 (119909)| 119889119909

=1

2Σ2120587 +

7

16+1

4120587+ln (4120587)81205872minus119888119894 (120587)

81205872+119904119894 (120587)

8120587

+1

2int

infin

0

(1

2+1

119890119905minus 1minus1

119905)119890minus1199052

1199052+ 41205872119889119905

(28)

Here we used standard notation 119904119894(119909) = minusintinfin119909(sin 119902119902)119889119902

119888119894(119909) = minus intinfin

119909(cos 119902119902)119889119902 [14] Of course similar equalities

can be established for many other such Fourier transformsRepeating what has been said above about the ldquomax-

imalityrdquo of the sum Σ2120587RH = sum120588119905119896gt0

119897119896119890minus2120587119905119896 we again

could formulate a statement of the type ldquoRH holds true ifand only if the integral equality (28) with Σ2120587 = Σ2120587RHholdsrdquo Numerical calculation taking into account first 100zeroes of the Riemann function gives the value Σ2120587 =26904854 sdot 10

minus39 Equality (28) has been tested numer-ically and shown to be true without the term Σ2120587 bothsides coincide up to 39 figures after decimal and are equalto 05620211964552684935093570395449423344 Then ofcourse they start to differ

5 Conclusions

In this paper we have established a number of new criteriainvolving the integrals of the logarithm of the Riemann 120589-function and equivalent to the Riemann hypothesis thistime with the exponential weight functions Exponential

weight functions lead to rather simple expressions for thecontribution of the Riemann function zeroes not lying onthe critical line to the contour integral value This enabledobtaining a rigorous estimation of the possible error whichwas shown to be extremely small

We also show how certain Fourier transforms of thelogarithm of the Riemann zeta-function taken along thereal (demi)axis are expressible via elementary functionsplus logarithm of the gamma-function and definite integralsthereof as well as certain sums over trivial and nontrivialRiemann function zeroes In our opinion further study ofsimilar Fourier transforms is interesting and might be usefulin the Riemann researches

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] S K Sekatskii S Beltraminelli and D Merlini ldquoOn equalitiesinvolving integrals of the logarithm of the Riemann-functionand equivalent to the Riemann hypothesisrdquo Ukrainian Mathe-matical Journal vol 64 no 2 pp 218ndash228 2012

[2] S Beltraminelli D Merlini and S Sekatskii ldquoA hidden symme-try related to the Riemann hypothesis with the primes into thecritical striprdquoAsian-European Journal ofMathematics vol 3 no4 pp 555ndash563 2010

[3] E C Titchmarsh and E R Heath-Brown The Theory of theRiemann Zeta-Function Clarendon Press Oxford UK 1988

[4] F T Wang ldquoA note on the Riemann zeta-functionrdquo Bulletin ofthe American Mathematical Society vol 52 pp 319ndash321 1946

[5] V V Volchkov ldquoOn an equality equivalent to the Riemannhypothesisrdquo Ukrainian Mathematical Journal vol 47 no 3 pp491ndash493 1995

[6] M Balazard E Saias and M Yor ldquoNotes sur la fonction 120577 deRiemann 2rdquo Advances in Mathematics vol 143 no 2 pp 284ndash287 1999

[7] D Merlini ldquoThe Riemann magneton of the primesrdquo Chaos andComplexity Letters vol 2 no 1 pp 93ndash98 2006

[8] E C Titchmarsh The Theory of Functions Oxford UniversityPress Oxford UK 1939

[9] C Ramare and Y Saouter ldquoShort effective intervals containingprimesrdquo Journal ofNumberTheory vol 98 no 1 pp 10ndash33 2003

[10] R J Backlund ldquoUber die Nullstellen der Riemannschen Zeta-funktionrdquo Acta Mathematica vol 41 no 1 pp 345ndash375 1916

[11] X Gourdon ldquoThe 1013 first zeroes of the Riemann ZetaFunction and zeroes computation at very large heightrdquo 2004httpnumberscomputationfreefrConstantsMiscellaneouszetazeros1e13-1e24pdf

[12] D J Platt ldquoComputing 120587(119909) analyticallyrdquo Mathematics ofComputation vol 84 no 293 pp 1521ndash1535 2015

[13] A Erdelyi EdHigher Transcendental Functions Based in PartOn Notes Left by Harry Bateman vol 1 McGraw-Hill NewYork NY USA 1953

[14] I S Gradshtein and I M Ryzhik Tables of Integrals Series andProducts Academic New York NY USA 1990

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

International Journal of Analysis 5

(119905) minus (119879) where (119905) is a usual zero counting functionwhich counts zeroes in the strip 1 gt 120590119896 gt 0 119879 ge 119905119896 gt 0 (119873(119905)is equal to (119905) minus (119879) if all zeroes with 119905 gt 119879 do not lieon the critical line)Thus the contribution at question cannotexceed

119868 =120587

1198862int

infin

119879

119890minus119886119905( (119905) minus (119879)) 119889119905 (11)

Now for the function (119905) we use the known formula

(119905) =119905

2120587(ln 1199052120587minus 1) +

7

8+ 119876 (119905) (12)

with the estimation of the reminder given by Backlund in 1916[10]

|119876 (119905)| lt 0137 ln 119905 + 0443 ln ln 119905 + 4350 (13)

Then integration in (11) with (12) and (13) is trivial and weobtain that

|119868| lt1

21198863(2120587 sdot 0137 ln119879 + 1

119886ln119879 + 2120587 sdot 435) 119890minus119886119879

+2120587 sdot 0443

21198863

int

infin

119879

119890minus119886119905

ln2119905119889119905 + (

2120587 sdot 0137

21198863

+1

21198865)

sdot int

infin

119879

119890minus119886119905

119905119889119905

(14)

Of course intinfin119879(119890minus119886119905ln2119905)119889119905 lt (1119886)(119890

minus119886119879ln2119879)

intinfin

119879(119890minus119886119905119905)119889119905 lt (1119886)(119890

minus119886119879119879) and thus

|119868| lt1

21198863(0861 ln119879 + 1

119886ln119879 + 27332 + 2784

119886 ln2119879

+0861

119886119879+1

1198863119879) 119890minus119886119879

(15)

Thus for say 119886 = 1 we obtain |119868| lt 341119890minus33sdot10

9

=

341101433sdot10

9

and taking the maximal possible value of 119886119886 = 2120587 we get |119868| lt 0101119890minus2074sdot10

10

= 01011009sdot10

10

that isa precision with at least 143 milliard decimals in the first caseand of 9 milliards of decimals in the second case

To summarize we may say that combining a rigorousanalytical treatment with the known numerical results on theRiemann function zeros on the critical line we have found anequation of the form 119860 = 119861 + 120576 where 120576 = 0 is equivalent tothe RHWewere able to set a bound to 120576 of the order 1 over thenine milliardth power of 10 in this tiny factor lies the truth ofthe RH In a numerical experiment we verified the equationup to 30 decimals

Remark 3 There is a more recent calculation of Gourdonwhere it is reported that the first 119896 = 1013 Riemann zeroesare located on the critical line but we were unable to get anexact value of 119879 from the corresponding reference [11] See

also quite recent Platt paper [12] and references cited thereinfor a short review of the problem

It is also worthwhile to note that if we put the questionwhat is the attainable precision of equalities pertinent tocheck up whether there are no Riemann function zeroes with120590 lt 119887 lt 12 such a precision can be much larger because forsuch case in the conditions of Theorems 2 2a and 2b muchlarger values of 119886 can be taken for example for 119887 = 116 weare able to use the weight function 119890minus8120587119905

4 On Some Fourier Transforms ofLogarithm of the Riemann Zeta-FunctionTaken along the Real Axis and TheirRelation with the Riemann Hypothesis

We now discuss what is the situation with 119887 lt 0 that is ifwe move the left border of the contour further to the leftLet for a moment still 119887 gt minus2 First we select 119886 in such amanner that sin(119886(12 minus 119887)) = plusmn1 119886 le 120587 In doing so whenspeaking about integral equalities involving intinfin

0119890minus119886119905 ln |120589(119887 +

119894119905)|119889119905 we set all contributions of the Riemann function zeroeslying on the critical line to the ldquomaximal by modulusrdquo valueplusmn(2120587119886)sum

120588119905119896gt0120590119896=12119897119896119890minus119886119905119896 If we somehow know all ordi-

nates 119905119896 of the Riemann function zeroes 120588119896 = 120590119896+119894119905119896 we are ina position to calculate the value Σ119886RH = (2120587119886)sum120588119905119896gt0 119897119896119890

minus119886119905119896

and then we would be able to formulate a criterion equivalentto the Riemann hypothesis of the type ldquoRH holds true ifand only if the integral equality intinfin

0119890minus119886119905 ln |120589(119887 + 119894119905)|119889119905 minus

intinfin

0sin(119886119909) ln |120589(119887 + 119909)|119889119909 + (120587119886) sin(119886(1 minus 119887)) = Σ119886RH

holdsrdquo (evident changes taking into account an appearanceof trivial zeroes on the contour border are to be made if119887 lt minus2 see below) This is clear because from the formulaexpressing the contribution of zeroes to the contour integralvalue Σ119886 = (2120587119886)sum120588119905119896gt0120590119896gt119887 119897119896119890

minus119886119905119896 sin(119886(120590119896 minus 119887)) and ourchoice of 119886 119887 it immediately follows that for zeroes with120590119896 = 12 all contributions have the same sign and themoduleof contribution is smaller than it would be for zeroes with thesame ordinates but with 120590119896 = 12 | sin(119886(120590119896 minus 119887)) + sin(119886(1 minus120590119896 minus 119887))| lt |2 sin(119886(12 minus 119887))| = 2 Similar situation takesplace for an integral involving an argument of the Riemannfunction if we select cos(119886(12 minus 119887)) = plusmn1 119886 le 120587

Unfortunately we do not see how the constant Σ119886RH canbe efficiently calculated

The other way around we can select 119886 in such amanner that the condition cos(119886(12 minus 119887)) = 0 holdsThen for all such values of 119886 we obtain unconditionally(ie independent on the RH) true equalities all termsminus(2120587119886)sum

120588119905119896gt0120590119896gt119887119897119896119890minus119886119905119896 cos(119886(120590119896 minus 119887)) in (5) vanish they

are either equal to zero if 120590119896 = 12 or mutually compensateeach other for a pair of zeroes having 120590119896 = 12 plusmn 120572119896 = 12Similar unconditional integral equalities can be obtainedstarting from (4) for integrals involving logarithm of themodule of the Riemann function if we select sin(119886(12minus119887)) =0 For completeness we present these results as the followingsimple theorem

6 International Journal of Analysis

Theorem 4 For any real positive 119886 and any real 119887 such thatminus2 lt 119887 le 0 one has an integral equality

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

= minus120587

119886(1 minus cos (119886 (1 minus 119887)))

(16)

if cos(119886(12 minus 119887)) = 0 and integral equality

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909

= minus120587

119886sin (119886 (1 minus 119887))

(17)

if sin(119886(12 minus 119887)) = 0

If we move the left border of the contour further to theleft then contrary to the case 119887 gt minus2 on the lower border ofthe contour we have a number of trivial zeroes at the pointsminus2 minus4 minus6 and (4) is to be modified as

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909

=2120587

119886sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 sin (119886 (120590119896 minus 119887))

minus120587

119886sin (119886 (1 minus 119887)) + 120587

119886

119896lt[minus1198872]

sum

119896=1

sin (119886 (minus2119896 minus 119887))

(18)

Correspondingly (5) now reads

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+2120587

119886sum

120588119905119896gt0

119897119896119890minus119886119905119896 cos (119886 (120590119896 minus 119887))

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

+120587

119886(1 minus cos (119886 (1 minus 119887)))

minus120587

119886

119896lt[minus1198872]

sum

119896=1

(1 minus cos (119886 (minus2119896 minus 119887))) = 0

(19)

In this last equality we start from the following initial valueof an argument function arg(120589(119887 + 119894120576)) = minus120587 + [minus1198872] sdot120587 where real positive 120576 rarr 0 (As usual [minus1198872] means

thewhole part of positiveminus1198872number) Evident correspond-ing modifications should be done inTheorem 4

One of such unconditionally true equalities namely thatobtained from (5) putting 119886 = 2120587 and 119887 = minus72 thatis intinfin0119890minus2120587119905 arg(120589(minus72 + 119894119905))119889119905 + intinfin

0cos(2120587119909) ln |120589(minus72 +

119909)|119889119909 = 0 has also been checked up numerically (Note thathere 119887 lt minus2) In this case we calculated the first integralwith the same parameters as discussed above For the secondintegral we extended the integrationrsquos interval to [0 250] andthe maximal recursion to 100 The result 874533283 times 10minus30is similar to two previously mentioned cases Note that herearg(120589(minus72 + 119894120576)) = 0 and the contribution of the pole exactlycompensates that of the first trivial zero hence in fact the twointegrals when summing compensate each other up to 30digits as found in the computations

Interesting new possibilities appear if we take certainvalue of 119886 and introduce a sequence 119887119894 rarr minusinfin in such amanner that the condition cos(119886(12 minus 119887119894)) = 1 holds forany 119894 Below we illustrate such approach for a particular case119886 = 2120587 119887119899 = minus2119899 minus 12 (119899 is a positive integer) We have thefollowing particular case of (19)

int

infin

0

119890minus2120587119905 arg(120589 (minus2119899 minus 1

2+ 119894119905)) 119889119905 + Σ2120587

+ int

infin

0

cos (2120587119909) ln1003816100381610038161003816100381610038161003816120589 (minus2119899 minus

1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 +

1

2

minus119899

2= 0

(20)

where Σ2120587 = sum120588119905119896gt0 119897119896119890minus2120587119905119896 cos(2120587(120590119896+12)) is independent

on 119899Let us now analyze both integrals of (20) starting from

intinfin

0119890minus2120587119905 arg(120589(minus2119899 minus 12 + 119894119905))119889119905 All estimations below are

done with the 119874(1) precision because 119874(1119899) and smallerterms tend to zero for large 119899 To begin with we shouldinclude the contribution of an initial value of an argument ofthe Riemann function which is equal to arg(120589(119887 + 119894120576)) = (119899 minus1) sdot 120587 to (20) This contributes (119899 minus 1)2 to the integral valueNow let us change 120589(minus2119899minus12+119894119905) function to 120589(2119899+32minus119894119905)function using functional equation [3]

120589 (119904) =1

120587(2120587)119904 sin120587119904

2Γ (1 minus 119904) 120589 (1 minus 119904) (21)

Factor (2120587)minus2119899minus12+119894119905 in (21) contributes ln(2120587)41205872 tothe integral value For sine factor taking 119899 odd wehave sin((minus14 minus 119899 + 1198941199052)120587) = sin((14 minus 1198941199052)120587) =(radic22)(minus119894 sinh(1205871199052) + cosh(1205871199052)) Its argument isminusarctan(tanh(1205871199052)) hence we need to handle an integralminusintinfin

0119890minus2120587119905arctan(tanh(1205871199052))119889119905 Integrating by parts

we see that this contribution is equal to minus12120587 sdot (1205872) intinfin

0(eminus2120587t cosh(120587119905))119889119905 = minus(12) intinfin

0(119890minus2120587119905(119890120587119905+ 119890minus120587119905))119889119905

with the variable transform 119910 = 119890120587119905 this is easily calculated tobe equal to minus12120587 + 18

Integral involving an argument of Riemann 120589-functiontends to zero when 119899 rarr infin so next we evaluate the

International Journal of Analysis 7

contribution of the gamma-functionWe use Stirling formula[13]

ln Γ (32+ 2119899 minus 119894119905) = (1 + 2119899 minus 119894119905) ln(3

2+ 2119899 minus 119894119905)

minus (3

2+ 2119899 minus 119894119905) + 05 ln 2120587

+ 119892 (119909)

(22)

where 119892(119909) = intinfin0(12 minus 1119905 + 1(119890

119905minus 1))(119890

minus119905119909119905)119889119905 [13] and

have

Imintinfin

0

119890minus2120587119905(1 + 2119899 minus 119894119905) ln(3

2+ 2119899 minus 119894119905) 119889119905

= Im 1

2120587int

infin

0

119890minus2120587119905(minus119894 ln(3

2+ 2119899 minus 119894119905)

minus 1198941 + 2119899 minus 119894119905

32 + 2119899 minus 119894119905) 119889119905 = minus

1

41205872ln(32+ 2119899)

minus1

41205872+ 119874(

1

119899)

(23)

Further minus Imintinfin0119890minus2120587119905(32 + 2119899 minus 119894119905)119889119905 = 14120587

2 Factor05 ln 2120587 is real and contributes nothing It is not difficult tosee that integral involving 119892(119911) tends to zero with 119899 tendingto infinity

Now we need to analyze second integral in (20)that is the integral intinfin

0cos(2120587119909) ln |120589(minus12 minus 2119899 + 119909)|119889119909

With a variable change 119910 = minus12 minus 2119899 + 119909 it isequal to minusintinfin

minus12minus2119899cos(2120587119910) ln |120589(119910)|119889119910 which we divide into

minusint12

minus12minus2119899minusintinfin

12 For the first integral we use the variable

transform 119909 = 1 minus 119910 to write that it is equal tominusint32+2119899

12cos(2120587119909) ln |120589(1 minus 119909)|119889119909 and then use the same

functional equation (21) again so we need to estimate

1198682 = minusint

32+2119899

12

cos (2120587119909)

sdot ln10038161003816100381610038161003816100381610038161003816

1

120587(2120587)1minus119909 sin120587 (1 minus 119909)

2Γ (119909) 120589 (119909)

10038161003816100381610038161003816100381610038161003816

119889119909

(24)

First int32+211989912

cos(2120587119909) ln |(1120587)(2120587)1minus119909|119889119909 = 0 For anintegral involving logarithm of gamma-functionwe again use

the Stirling formula ln Γ(119909) = (119909minus12) ln(119909)minus119909 + 05 ln 2120587 +119892(119909) [13] and get

minus int

32+2119899

12

cos (2120587119909) (119909 minus 12) ln (119909) 119889119909

=1

2120587int

32+2119899

12

sin (2120587119909) (ln119909 + 1 minus 12119909) 119889119909

=1

2120587int

32+2119899

12

sin (2120587119909) (ln119909 minus 12119909) 119889119909

= minus1

2120587int

32+2119899

12

sin (2120587119909) 12119909119889119909

minus1

41205872cos (2120587119909) ln119909

1003816100381610038161003816100381610038161003816

32+2119899

12

+1

41205872int

32+2119899

12

cos (2120587119909) 1119909119889119909

=1

41205872(ln(3

2+ 2119899) + 2119899)

+1

41205872int

32+119899

12

1

119909cos (2120587119909) 119889119909

minus1

2120587int

32+2119899

12

sin (2120587119909) 12119909119889119909

(25)

To finish the consideration of the contribution of thelogarithm of gamma-function we need to add the contri-bution of 119892(119909) which is minusintinfin

12cos(2120587119909)119889119909 intinfin

0(12 minus 1119905 +

1(119890119905minus 1))(119890

minus119905119909119905)119889119905 Using cos(2120587119909) = (12)(1198902120587119909119894 + 119890minus2120587119909119894)

we can perform integration over 119909 minusintinfin12

cos(2120587119909)119890minus119905119909119889119909 =119890minus1199052(119905(1199052+ 41205872)) and thus the contribution at question is

intinfin

0(12 minus 1119905 + 1(119890

119905minus 1))(119890

minus1199052(1199052+ 41205872))119889119905

Finally we need to analyze the contribution of thesine factor in (24) which is minusint32+2119899

12cos(2120587119909) ln |sin(120587(1 minus

119909)2)|119889119909 = minusint32+211989912

cos(2120587119909) ln |cos(1205871199092)|119889119909 Integrationis to be applied with caution here due to the presence ofthe module To avoid errors it is reasonable to considerseparately integrals int1

12 int2119899+322119899+1

= int32

1 and int2

1 the latter is

an integral along the period Integrating by parts we have

minus int

1

12

cos (2120587119909) ln1003816100381610038161003816100381610038161003816cos 120587119909

2

1003816100381610038161003816100381610038161003816119889119909

= minusint

1

12

cos (2120587119909) ln(cos 1205871199092) 119889119909

= minus1

4int

1

12

sin (2120587119909) tan 1205871199092119889119909

(26)

8 International Journal of Analysis

and further

minus1

4int

1

12

sin (2120587119909) tan 1205871199092119889119909

= minusint

1

12

sin21205871199092

cos (120587119909) 119889119909

=1

2int

1

12

(cos (120587119909) minus 1) cos (120587119909) 119889119909

=1

2int

1

12

(1 + cos (2120587119909)

2minus cos (120587119909)) 119889119909

=1

8minus1

2120587sin (120587119909)

1003816100381610038161003816100381610038161003816

1

12

=1

8+1

2120587

(27)

Similarly minusint32

1cos(2120587119909) ln | cos(1205871199092)|119889119909 =

minusint32

1cos(2120587119909) ln(minus cos(1205871199092))119889119909 = 18 + 12120587 while

integral over the period is equal to 14 (use the sameapproach as above or entry number 43843 of [14] withevident variable changes) so the total contribution of thesine factor in (24) to (20) is equal to 1198992 + 14 + 1120587

Collecting everything together we see the cancelling of119874(119899) and 119874(ln(119899)) terms Thus considering the limit 119899 rarrinfin we have established the following relation expressingcertain particular value of Fourier transform of the logarithmof the Riemann zeta-function

int

infin

12

cos (2120587119909) ln |120589 (119909)| 119889119909

=1

2Σ2120587 +

7

16+1

4120587+ln (4120587)81205872minus119888119894 (120587)

81205872+119904119894 (120587)

8120587

+1

2int

infin

0

(1

2+1

119890119905minus 1minus1

119905)119890minus1199052

1199052+ 41205872119889119905

(28)

Here we used standard notation 119904119894(119909) = minusintinfin119909(sin 119902119902)119889119902

119888119894(119909) = minus intinfin

119909(cos 119902119902)119889119902 [14] Of course similar equalities

can be established for many other such Fourier transformsRepeating what has been said above about the ldquomax-

imalityrdquo of the sum Σ2120587RH = sum120588119905119896gt0

119897119896119890minus2120587119905119896 we again

could formulate a statement of the type ldquoRH holds true ifand only if the integral equality (28) with Σ2120587 = Σ2120587RHholdsrdquo Numerical calculation taking into account first 100zeroes of the Riemann function gives the value Σ2120587 =26904854 sdot 10

minus39 Equality (28) has been tested numer-ically and shown to be true without the term Σ2120587 bothsides coincide up to 39 figures after decimal and are equalto 05620211964552684935093570395449423344 Then ofcourse they start to differ

5 Conclusions

In this paper we have established a number of new criteriainvolving the integrals of the logarithm of the Riemann 120589-function and equivalent to the Riemann hypothesis thistime with the exponential weight functions Exponential

weight functions lead to rather simple expressions for thecontribution of the Riemann function zeroes not lying onthe critical line to the contour integral value This enabledobtaining a rigorous estimation of the possible error whichwas shown to be extremely small

We also show how certain Fourier transforms of thelogarithm of the Riemann zeta-function taken along thereal (demi)axis are expressible via elementary functionsplus logarithm of the gamma-function and definite integralsthereof as well as certain sums over trivial and nontrivialRiemann function zeroes In our opinion further study ofsimilar Fourier transforms is interesting and might be usefulin the Riemann researches

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] S K Sekatskii S Beltraminelli and D Merlini ldquoOn equalitiesinvolving integrals of the logarithm of the Riemann-functionand equivalent to the Riemann hypothesisrdquo Ukrainian Mathe-matical Journal vol 64 no 2 pp 218ndash228 2012

[2] S Beltraminelli D Merlini and S Sekatskii ldquoA hidden symme-try related to the Riemann hypothesis with the primes into thecritical striprdquoAsian-European Journal ofMathematics vol 3 no4 pp 555ndash563 2010

[3] E C Titchmarsh and E R Heath-Brown The Theory of theRiemann Zeta-Function Clarendon Press Oxford UK 1988

[4] F T Wang ldquoA note on the Riemann zeta-functionrdquo Bulletin ofthe American Mathematical Society vol 52 pp 319ndash321 1946

[5] V V Volchkov ldquoOn an equality equivalent to the Riemannhypothesisrdquo Ukrainian Mathematical Journal vol 47 no 3 pp491ndash493 1995

[6] M Balazard E Saias and M Yor ldquoNotes sur la fonction 120577 deRiemann 2rdquo Advances in Mathematics vol 143 no 2 pp 284ndash287 1999

[7] D Merlini ldquoThe Riemann magneton of the primesrdquo Chaos andComplexity Letters vol 2 no 1 pp 93ndash98 2006

[8] E C Titchmarsh The Theory of Functions Oxford UniversityPress Oxford UK 1939

[9] C Ramare and Y Saouter ldquoShort effective intervals containingprimesrdquo Journal ofNumberTheory vol 98 no 1 pp 10ndash33 2003

[10] R J Backlund ldquoUber die Nullstellen der Riemannschen Zeta-funktionrdquo Acta Mathematica vol 41 no 1 pp 345ndash375 1916

[11] X Gourdon ldquoThe 1013 first zeroes of the Riemann ZetaFunction and zeroes computation at very large heightrdquo 2004httpnumberscomputationfreefrConstantsMiscellaneouszetazeros1e13-1e24pdf

[12] D J Platt ldquoComputing 120587(119909) analyticallyrdquo Mathematics ofComputation vol 84 no 293 pp 1521ndash1535 2015

[13] A Erdelyi EdHigher Transcendental Functions Based in PartOn Notes Left by Harry Bateman vol 1 McGraw-Hill NewYork NY USA 1953

[14] I S Gradshtein and I M Ryzhik Tables of Integrals Series andProducts Academic New York NY USA 1990

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 International Journal of Analysis

Theorem 4 For any real positive 119886 and any real 119887 such thatminus2 lt 119887 le 0 one has an integral equality

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

= minus120587

119886(1 minus cos (119886 (1 minus 119887)))

(16)

if cos(119886(12 minus 119887)) = 0 and integral equality

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909

= minus120587

119886sin (119886 (1 minus 119887))

(17)

if sin(119886(12 minus 119887)) = 0

If we move the left border of the contour further to theleft then contrary to the case 119887 gt minus2 on the lower border ofthe contour we have a number of trivial zeroes at the pointsminus2 minus4 minus6 and (4) is to be modified as

int

infin

0

119890minus119886119905 ln |120589 (119887 + 119894119905)| 119889119905

minus int

infin

0

sin (119886119909) ln |120589 (119887 + 119909)| 119889119909

=2120587

119886sum

120588119905119896gt0120590119896gt119887

119897119896119890minus119886119905119896 sin (119886 (120590119896 minus 119887))

minus120587

119886sin (119886 (1 minus 119887)) + 120587

119886

119896lt[minus1198872]

sum

119896=1

sin (119886 (minus2119896 minus 119887))

(18)

Correspondingly (5) now reads

int

infin

0

119890minus119886119905 arg (120589 (119887 + 119894119905)) 119889119905

+2120587

119886sum

120588119905119896gt0

119897119896119890minus119886119905119896 cos (119886 (120590119896 minus 119887))

+ int

infin

0

cos (119886119909) ln |120589 (119887 + 119909)| 119889119909

+120587

119886(1 minus cos (119886 (1 minus 119887)))

minus120587

119886

119896lt[minus1198872]

sum

119896=1

(1 minus cos (119886 (minus2119896 minus 119887))) = 0

(19)

In this last equality we start from the following initial valueof an argument function arg(120589(119887 + 119894120576)) = minus120587 + [minus1198872] sdot120587 where real positive 120576 rarr 0 (As usual [minus1198872] means

thewhole part of positiveminus1198872number) Evident correspond-ing modifications should be done inTheorem 4

One of such unconditionally true equalities namely thatobtained from (5) putting 119886 = 2120587 and 119887 = minus72 thatis intinfin0119890minus2120587119905 arg(120589(minus72 + 119894119905))119889119905 + intinfin

0cos(2120587119909) ln |120589(minus72 +

119909)|119889119909 = 0 has also been checked up numerically (Note thathere 119887 lt minus2) In this case we calculated the first integralwith the same parameters as discussed above For the secondintegral we extended the integrationrsquos interval to [0 250] andthe maximal recursion to 100 The result 874533283 times 10minus30is similar to two previously mentioned cases Note that herearg(120589(minus72 + 119894120576)) = 0 and the contribution of the pole exactlycompensates that of the first trivial zero hence in fact the twointegrals when summing compensate each other up to 30digits as found in the computations

Interesting new possibilities appear if we take certainvalue of 119886 and introduce a sequence 119887119894 rarr minusinfin in such amanner that the condition cos(119886(12 minus 119887119894)) = 1 holds forany 119894 Below we illustrate such approach for a particular case119886 = 2120587 119887119899 = minus2119899 minus 12 (119899 is a positive integer) We have thefollowing particular case of (19)

int

infin

0

119890minus2120587119905 arg(120589 (minus2119899 minus 1

2+ 119894119905)) 119889119905 + Σ2120587

+ int

infin

0

cos (2120587119909) ln1003816100381610038161003816100381610038161003816120589 (minus2119899 minus

1

2+ 119909)

1003816100381610038161003816100381610038161003816119889119909 +

1

2

minus119899

2= 0

(20)

where Σ2120587 = sum120588119905119896gt0 119897119896119890minus2120587119905119896 cos(2120587(120590119896+12)) is independent

on 119899Let us now analyze both integrals of (20) starting from

intinfin

0119890minus2120587119905 arg(120589(minus2119899 minus 12 + 119894119905))119889119905 All estimations below are

done with the 119874(1) precision because 119874(1119899) and smallerterms tend to zero for large 119899 To begin with we shouldinclude the contribution of an initial value of an argument ofthe Riemann function which is equal to arg(120589(119887 + 119894120576)) = (119899 minus1) sdot 120587 to (20) This contributes (119899 minus 1)2 to the integral valueNow let us change 120589(minus2119899minus12+119894119905) function to 120589(2119899+32minus119894119905)function using functional equation [3]

120589 (119904) =1

120587(2120587)119904 sin120587119904

2Γ (1 minus 119904) 120589 (1 minus 119904) (21)

Factor (2120587)minus2119899minus12+119894119905 in (21) contributes ln(2120587)41205872 tothe integral value For sine factor taking 119899 odd wehave sin((minus14 minus 119899 + 1198941199052)120587) = sin((14 minus 1198941199052)120587) =(radic22)(minus119894 sinh(1205871199052) + cosh(1205871199052)) Its argument isminusarctan(tanh(1205871199052)) hence we need to handle an integralminusintinfin

0119890minus2120587119905arctan(tanh(1205871199052))119889119905 Integrating by parts

we see that this contribution is equal to minus12120587 sdot (1205872) intinfin

0(eminus2120587t cosh(120587119905))119889119905 = minus(12) intinfin

0(119890minus2120587119905(119890120587119905+ 119890minus120587119905))119889119905

with the variable transform 119910 = 119890120587119905 this is easily calculated tobe equal to minus12120587 + 18

Integral involving an argument of Riemann 120589-functiontends to zero when 119899 rarr infin so next we evaluate the

International Journal of Analysis 7

contribution of the gamma-functionWe use Stirling formula[13]

ln Γ (32+ 2119899 minus 119894119905) = (1 + 2119899 minus 119894119905) ln(3

2+ 2119899 minus 119894119905)

minus (3

2+ 2119899 minus 119894119905) + 05 ln 2120587

+ 119892 (119909)

(22)

where 119892(119909) = intinfin0(12 minus 1119905 + 1(119890

119905minus 1))(119890

minus119905119909119905)119889119905 [13] and

have

Imintinfin

0

119890minus2120587119905(1 + 2119899 minus 119894119905) ln(3

2+ 2119899 minus 119894119905) 119889119905

= Im 1

2120587int

infin

0

119890minus2120587119905(minus119894 ln(3

2+ 2119899 minus 119894119905)

minus 1198941 + 2119899 minus 119894119905

32 + 2119899 minus 119894119905) 119889119905 = minus

1

41205872ln(32+ 2119899)

minus1

41205872+ 119874(

1

119899)

(23)

Further minus Imintinfin0119890minus2120587119905(32 + 2119899 minus 119894119905)119889119905 = 14120587

2 Factor05 ln 2120587 is real and contributes nothing It is not difficult tosee that integral involving 119892(119911) tends to zero with 119899 tendingto infinity

Now we need to analyze second integral in (20)that is the integral intinfin

0cos(2120587119909) ln |120589(minus12 minus 2119899 + 119909)|119889119909

With a variable change 119910 = minus12 minus 2119899 + 119909 it isequal to minusintinfin

minus12minus2119899cos(2120587119910) ln |120589(119910)|119889119910 which we divide into

minusint12

minus12minus2119899minusintinfin

12 For the first integral we use the variable

transform 119909 = 1 minus 119910 to write that it is equal tominusint32+2119899

12cos(2120587119909) ln |120589(1 minus 119909)|119889119909 and then use the same

functional equation (21) again so we need to estimate

1198682 = minusint

32+2119899

12

cos (2120587119909)

sdot ln10038161003816100381610038161003816100381610038161003816

1

120587(2120587)1minus119909 sin120587 (1 minus 119909)

2Γ (119909) 120589 (119909)

10038161003816100381610038161003816100381610038161003816

119889119909

(24)

First int32+211989912

cos(2120587119909) ln |(1120587)(2120587)1minus119909|119889119909 = 0 For anintegral involving logarithm of gamma-functionwe again use

the Stirling formula ln Γ(119909) = (119909minus12) ln(119909)minus119909 + 05 ln 2120587 +119892(119909) [13] and get

minus int

32+2119899

12

cos (2120587119909) (119909 minus 12) ln (119909) 119889119909

=1

2120587int

32+2119899

12

sin (2120587119909) (ln119909 + 1 minus 12119909) 119889119909

=1

2120587int

32+2119899

12

sin (2120587119909) (ln119909 minus 12119909) 119889119909

= minus1

2120587int

32+2119899

12

sin (2120587119909) 12119909119889119909

minus1

41205872cos (2120587119909) ln119909

1003816100381610038161003816100381610038161003816

32+2119899

12

+1

41205872int

32+2119899

12

cos (2120587119909) 1119909119889119909

=1

41205872(ln(3

2+ 2119899) + 2119899)

+1

41205872int

32+119899

12

1

119909cos (2120587119909) 119889119909

minus1

2120587int

32+2119899

12

sin (2120587119909) 12119909119889119909

(25)

To finish the consideration of the contribution of thelogarithm of gamma-function we need to add the contri-bution of 119892(119909) which is minusintinfin

12cos(2120587119909)119889119909 intinfin

0(12 minus 1119905 +

1(119890119905minus 1))(119890

minus119905119909119905)119889119905 Using cos(2120587119909) = (12)(1198902120587119909119894 + 119890minus2120587119909119894)

we can perform integration over 119909 minusintinfin12

cos(2120587119909)119890minus119905119909119889119909 =119890minus1199052(119905(1199052+ 41205872)) and thus the contribution at question is

intinfin

0(12 minus 1119905 + 1(119890

119905minus 1))(119890

minus1199052(1199052+ 41205872))119889119905

Finally we need to analyze the contribution of thesine factor in (24) which is minusint32+2119899

12cos(2120587119909) ln |sin(120587(1 minus

119909)2)|119889119909 = minusint32+211989912

cos(2120587119909) ln |cos(1205871199092)|119889119909 Integrationis to be applied with caution here due to the presence ofthe module To avoid errors it is reasonable to considerseparately integrals int1

12 int2119899+322119899+1

= int32

1 and int2

1 the latter is

an integral along the period Integrating by parts we have

minus int

1

12

cos (2120587119909) ln1003816100381610038161003816100381610038161003816cos 120587119909

2

1003816100381610038161003816100381610038161003816119889119909

= minusint

1

12

cos (2120587119909) ln(cos 1205871199092) 119889119909

= minus1

4int

1

12

sin (2120587119909) tan 1205871199092119889119909

(26)

8 International Journal of Analysis

and further

minus1

4int

1

12

sin (2120587119909) tan 1205871199092119889119909

= minusint

1

12

sin21205871199092

cos (120587119909) 119889119909

=1

2int

1

12

(cos (120587119909) minus 1) cos (120587119909) 119889119909

=1

2int

1

12

(1 + cos (2120587119909)

2minus cos (120587119909)) 119889119909

=1

8minus1

2120587sin (120587119909)

1003816100381610038161003816100381610038161003816

1

12

=1

8+1

2120587

(27)

Similarly minusint32

1cos(2120587119909) ln | cos(1205871199092)|119889119909 =

minusint32

1cos(2120587119909) ln(minus cos(1205871199092))119889119909 = 18 + 12120587 while

integral over the period is equal to 14 (use the sameapproach as above or entry number 43843 of [14] withevident variable changes) so the total contribution of thesine factor in (24) to (20) is equal to 1198992 + 14 + 1120587

Collecting everything together we see the cancelling of119874(119899) and 119874(ln(119899)) terms Thus considering the limit 119899 rarrinfin we have established the following relation expressingcertain particular value of Fourier transform of the logarithmof the Riemann zeta-function

int

infin

12

cos (2120587119909) ln |120589 (119909)| 119889119909

=1

2Σ2120587 +

7

16+1

4120587+ln (4120587)81205872minus119888119894 (120587)

81205872+119904119894 (120587)

8120587

+1

2int

infin

0

(1

2+1

119890119905minus 1minus1

119905)119890minus1199052

1199052+ 41205872119889119905

(28)

Here we used standard notation 119904119894(119909) = minusintinfin119909(sin 119902119902)119889119902

119888119894(119909) = minus intinfin

119909(cos 119902119902)119889119902 [14] Of course similar equalities

can be established for many other such Fourier transformsRepeating what has been said above about the ldquomax-

imalityrdquo of the sum Σ2120587RH = sum120588119905119896gt0

119897119896119890minus2120587119905119896 we again

could formulate a statement of the type ldquoRH holds true ifand only if the integral equality (28) with Σ2120587 = Σ2120587RHholdsrdquo Numerical calculation taking into account first 100zeroes of the Riemann function gives the value Σ2120587 =26904854 sdot 10

minus39 Equality (28) has been tested numer-ically and shown to be true without the term Σ2120587 bothsides coincide up to 39 figures after decimal and are equalto 05620211964552684935093570395449423344 Then ofcourse they start to differ

5 Conclusions

In this paper we have established a number of new criteriainvolving the integrals of the logarithm of the Riemann 120589-function and equivalent to the Riemann hypothesis thistime with the exponential weight functions Exponential

weight functions lead to rather simple expressions for thecontribution of the Riemann function zeroes not lying onthe critical line to the contour integral value This enabledobtaining a rigorous estimation of the possible error whichwas shown to be extremely small

We also show how certain Fourier transforms of thelogarithm of the Riemann zeta-function taken along thereal (demi)axis are expressible via elementary functionsplus logarithm of the gamma-function and definite integralsthereof as well as certain sums over trivial and nontrivialRiemann function zeroes In our opinion further study ofsimilar Fourier transforms is interesting and might be usefulin the Riemann researches

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] S K Sekatskii S Beltraminelli and D Merlini ldquoOn equalitiesinvolving integrals of the logarithm of the Riemann-functionand equivalent to the Riemann hypothesisrdquo Ukrainian Mathe-matical Journal vol 64 no 2 pp 218ndash228 2012

[2] S Beltraminelli D Merlini and S Sekatskii ldquoA hidden symme-try related to the Riemann hypothesis with the primes into thecritical striprdquoAsian-European Journal ofMathematics vol 3 no4 pp 555ndash563 2010

[3] E C Titchmarsh and E R Heath-Brown The Theory of theRiemann Zeta-Function Clarendon Press Oxford UK 1988

[4] F T Wang ldquoA note on the Riemann zeta-functionrdquo Bulletin ofthe American Mathematical Society vol 52 pp 319ndash321 1946

[5] V V Volchkov ldquoOn an equality equivalent to the Riemannhypothesisrdquo Ukrainian Mathematical Journal vol 47 no 3 pp491ndash493 1995

[6] M Balazard E Saias and M Yor ldquoNotes sur la fonction 120577 deRiemann 2rdquo Advances in Mathematics vol 143 no 2 pp 284ndash287 1999

[7] D Merlini ldquoThe Riemann magneton of the primesrdquo Chaos andComplexity Letters vol 2 no 1 pp 93ndash98 2006

[8] E C Titchmarsh The Theory of Functions Oxford UniversityPress Oxford UK 1939

[9] C Ramare and Y Saouter ldquoShort effective intervals containingprimesrdquo Journal ofNumberTheory vol 98 no 1 pp 10ndash33 2003

[10] R J Backlund ldquoUber die Nullstellen der Riemannschen Zeta-funktionrdquo Acta Mathematica vol 41 no 1 pp 345ndash375 1916

[11] X Gourdon ldquoThe 1013 first zeroes of the Riemann ZetaFunction and zeroes computation at very large heightrdquo 2004httpnumberscomputationfreefrConstantsMiscellaneouszetazeros1e13-1e24pdf

[12] D J Platt ldquoComputing 120587(119909) analyticallyrdquo Mathematics ofComputation vol 84 no 293 pp 1521ndash1535 2015

[13] A Erdelyi EdHigher Transcendental Functions Based in PartOn Notes Left by Harry Bateman vol 1 McGraw-Hill NewYork NY USA 1953

[14] I S Gradshtein and I M Ryzhik Tables of Integrals Series andProducts Academic New York NY USA 1990

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

International Journal of Analysis 7

contribution of the gamma-functionWe use Stirling formula[13]

ln Γ (32+ 2119899 minus 119894119905) = (1 + 2119899 minus 119894119905) ln(3

2+ 2119899 minus 119894119905)

minus (3

2+ 2119899 minus 119894119905) + 05 ln 2120587

+ 119892 (119909)

(22)

where 119892(119909) = intinfin0(12 minus 1119905 + 1(119890

119905minus 1))(119890

minus119905119909119905)119889119905 [13] and

have

Imintinfin

0

119890minus2120587119905(1 + 2119899 minus 119894119905) ln(3

2+ 2119899 minus 119894119905) 119889119905

= Im 1

2120587int

infin

0

119890minus2120587119905(minus119894 ln(3

2+ 2119899 minus 119894119905)

minus 1198941 + 2119899 minus 119894119905

32 + 2119899 minus 119894119905) 119889119905 = minus

1

41205872ln(32+ 2119899)

minus1

41205872+ 119874(

1

119899)

(23)

Further minus Imintinfin0119890minus2120587119905(32 + 2119899 minus 119894119905)119889119905 = 14120587

2 Factor05 ln 2120587 is real and contributes nothing It is not difficult tosee that integral involving 119892(119911) tends to zero with 119899 tendingto infinity

Now we need to analyze second integral in (20)that is the integral intinfin

0cos(2120587119909) ln |120589(minus12 minus 2119899 + 119909)|119889119909

With a variable change 119910 = minus12 minus 2119899 + 119909 it isequal to minusintinfin

minus12minus2119899cos(2120587119910) ln |120589(119910)|119889119910 which we divide into

minusint12

minus12minus2119899minusintinfin

12 For the first integral we use the variable

transform 119909 = 1 minus 119910 to write that it is equal tominusint32+2119899

12cos(2120587119909) ln |120589(1 minus 119909)|119889119909 and then use the same

functional equation (21) again so we need to estimate

1198682 = minusint

32+2119899

12

cos (2120587119909)

sdot ln10038161003816100381610038161003816100381610038161003816

1

120587(2120587)1minus119909 sin120587 (1 minus 119909)

2Γ (119909) 120589 (119909)

10038161003816100381610038161003816100381610038161003816

119889119909

(24)

First int32+211989912

cos(2120587119909) ln |(1120587)(2120587)1minus119909|119889119909 = 0 For anintegral involving logarithm of gamma-functionwe again use

the Stirling formula ln Γ(119909) = (119909minus12) ln(119909)minus119909 + 05 ln 2120587 +119892(119909) [13] and get

minus int

32+2119899

12

cos (2120587119909) (119909 minus 12) ln (119909) 119889119909

=1

2120587int

32+2119899

12

sin (2120587119909) (ln119909 + 1 minus 12119909) 119889119909

=1

2120587int

32+2119899

12

sin (2120587119909) (ln119909 minus 12119909) 119889119909

= minus1

2120587int

32+2119899

12

sin (2120587119909) 12119909119889119909

minus1

41205872cos (2120587119909) ln119909

1003816100381610038161003816100381610038161003816

32+2119899

12

+1

41205872int

32+2119899

12

cos (2120587119909) 1119909119889119909

=1

41205872(ln(3

2+ 2119899) + 2119899)

+1

41205872int

32+119899

12

1

119909cos (2120587119909) 119889119909

minus1

2120587int

32+2119899

12

sin (2120587119909) 12119909119889119909

(25)

To finish the consideration of the contribution of thelogarithm of gamma-function we need to add the contri-bution of 119892(119909) which is minusintinfin

12cos(2120587119909)119889119909 intinfin

0(12 minus 1119905 +

1(119890119905minus 1))(119890

minus119905119909119905)119889119905 Using cos(2120587119909) = (12)(1198902120587119909119894 + 119890minus2120587119909119894)

we can perform integration over 119909 minusintinfin12

cos(2120587119909)119890minus119905119909119889119909 =119890minus1199052(119905(1199052+ 41205872)) and thus the contribution at question is

intinfin

0(12 minus 1119905 + 1(119890

119905minus 1))(119890

minus1199052(1199052+ 41205872))119889119905

Finally we need to analyze the contribution of thesine factor in (24) which is minusint32+2119899

12cos(2120587119909) ln |sin(120587(1 minus

119909)2)|119889119909 = minusint32+211989912

cos(2120587119909) ln |cos(1205871199092)|119889119909 Integrationis to be applied with caution here due to the presence ofthe module To avoid errors it is reasonable to considerseparately integrals int1

12 int2119899+322119899+1

= int32

1 and int2

1 the latter is

an integral along the period Integrating by parts we have

minus int

1

12

cos (2120587119909) ln1003816100381610038161003816100381610038161003816cos 120587119909

2

1003816100381610038161003816100381610038161003816119889119909

= minusint

1

12

cos (2120587119909) ln(cos 1205871199092) 119889119909

= minus1

4int

1

12

sin (2120587119909) tan 1205871199092119889119909

(26)

8 International Journal of Analysis

and further

minus1

4int

1

12

sin (2120587119909) tan 1205871199092119889119909

= minusint

1

12

sin21205871199092

cos (120587119909) 119889119909

=1

2int

1

12

(cos (120587119909) minus 1) cos (120587119909) 119889119909

=1

2int

1

12

(1 + cos (2120587119909)

2minus cos (120587119909)) 119889119909

=1

8minus1

2120587sin (120587119909)

1003816100381610038161003816100381610038161003816

1

12

=1

8+1

2120587

(27)

Similarly minusint32

1cos(2120587119909) ln | cos(1205871199092)|119889119909 =

minusint32

1cos(2120587119909) ln(minus cos(1205871199092))119889119909 = 18 + 12120587 while

integral over the period is equal to 14 (use the sameapproach as above or entry number 43843 of [14] withevident variable changes) so the total contribution of thesine factor in (24) to (20) is equal to 1198992 + 14 + 1120587

Collecting everything together we see the cancelling of119874(119899) and 119874(ln(119899)) terms Thus considering the limit 119899 rarrinfin we have established the following relation expressingcertain particular value of Fourier transform of the logarithmof the Riemann zeta-function

int

infin

12

cos (2120587119909) ln |120589 (119909)| 119889119909

=1

2Σ2120587 +

7

16+1

4120587+ln (4120587)81205872minus119888119894 (120587)

81205872+119904119894 (120587)

8120587

+1

2int

infin

0

(1

2+1

119890119905minus 1minus1

119905)119890minus1199052

1199052+ 41205872119889119905

(28)

Here we used standard notation 119904119894(119909) = minusintinfin119909(sin 119902119902)119889119902

119888119894(119909) = minus intinfin

119909(cos 119902119902)119889119902 [14] Of course similar equalities

can be established for many other such Fourier transformsRepeating what has been said above about the ldquomax-

imalityrdquo of the sum Σ2120587RH = sum120588119905119896gt0

119897119896119890minus2120587119905119896 we again

could formulate a statement of the type ldquoRH holds true ifand only if the integral equality (28) with Σ2120587 = Σ2120587RHholdsrdquo Numerical calculation taking into account first 100zeroes of the Riemann function gives the value Σ2120587 =26904854 sdot 10

minus39 Equality (28) has been tested numer-ically and shown to be true without the term Σ2120587 bothsides coincide up to 39 figures after decimal and are equalto 05620211964552684935093570395449423344 Then ofcourse they start to differ

5 Conclusions

In this paper we have established a number of new criteriainvolving the integrals of the logarithm of the Riemann 120589-function and equivalent to the Riemann hypothesis thistime with the exponential weight functions Exponential

weight functions lead to rather simple expressions for thecontribution of the Riemann function zeroes not lying onthe critical line to the contour integral value This enabledobtaining a rigorous estimation of the possible error whichwas shown to be extremely small

We also show how certain Fourier transforms of thelogarithm of the Riemann zeta-function taken along thereal (demi)axis are expressible via elementary functionsplus logarithm of the gamma-function and definite integralsthereof as well as certain sums over trivial and nontrivialRiemann function zeroes In our opinion further study ofsimilar Fourier transforms is interesting and might be usefulin the Riemann researches

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] S K Sekatskii S Beltraminelli and D Merlini ldquoOn equalitiesinvolving integrals of the logarithm of the Riemann-functionand equivalent to the Riemann hypothesisrdquo Ukrainian Mathe-matical Journal vol 64 no 2 pp 218ndash228 2012

[2] S Beltraminelli D Merlini and S Sekatskii ldquoA hidden symme-try related to the Riemann hypothesis with the primes into thecritical striprdquoAsian-European Journal ofMathematics vol 3 no4 pp 555ndash563 2010

[3] E C Titchmarsh and E R Heath-Brown The Theory of theRiemann Zeta-Function Clarendon Press Oxford UK 1988

[4] F T Wang ldquoA note on the Riemann zeta-functionrdquo Bulletin ofthe American Mathematical Society vol 52 pp 319ndash321 1946

[5] V V Volchkov ldquoOn an equality equivalent to the Riemannhypothesisrdquo Ukrainian Mathematical Journal vol 47 no 3 pp491ndash493 1995

[6] M Balazard E Saias and M Yor ldquoNotes sur la fonction 120577 deRiemann 2rdquo Advances in Mathematics vol 143 no 2 pp 284ndash287 1999

[7] D Merlini ldquoThe Riemann magneton of the primesrdquo Chaos andComplexity Letters vol 2 no 1 pp 93ndash98 2006

[8] E C Titchmarsh The Theory of Functions Oxford UniversityPress Oxford UK 1939

[9] C Ramare and Y Saouter ldquoShort effective intervals containingprimesrdquo Journal ofNumberTheory vol 98 no 1 pp 10ndash33 2003

[10] R J Backlund ldquoUber die Nullstellen der Riemannschen Zeta-funktionrdquo Acta Mathematica vol 41 no 1 pp 345ndash375 1916

[11] X Gourdon ldquoThe 1013 first zeroes of the Riemann ZetaFunction and zeroes computation at very large heightrdquo 2004httpnumberscomputationfreefrConstantsMiscellaneouszetazeros1e13-1e24pdf

[12] D J Platt ldquoComputing 120587(119909) analyticallyrdquo Mathematics ofComputation vol 84 no 293 pp 1521ndash1535 2015

[13] A Erdelyi EdHigher Transcendental Functions Based in PartOn Notes Left by Harry Bateman vol 1 McGraw-Hill NewYork NY USA 1953

[14] I S Gradshtein and I M Ryzhik Tables of Integrals Series andProducts Academic New York NY USA 1990

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 International Journal of Analysis

and further

minus1

4int

1

12

sin (2120587119909) tan 1205871199092119889119909

= minusint

1

12

sin21205871199092

cos (120587119909) 119889119909

=1

2int

1

12

(cos (120587119909) minus 1) cos (120587119909) 119889119909

=1

2int

1

12

(1 + cos (2120587119909)

2minus cos (120587119909)) 119889119909

=1

8minus1

2120587sin (120587119909)

1003816100381610038161003816100381610038161003816

1

12

=1

8+1

2120587

(27)

Similarly minusint32

1cos(2120587119909) ln | cos(1205871199092)|119889119909 =

minusint32

1cos(2120587119909) ln(minus cos(1205871199092))119889119909 = 18 + 12120587 while

integral over the period is equal to 14 (use the sameapproach as above or entry number 43843 of [14] withevident variable changes) so the total contribution of thesine factor in (24) to (20) is equal to 1198992 + 14 + 1120587

Collecting everything together we see the cancelling of119874(119899) and 119874(ln(119899)) terms Thus considering the limit 119899 rarrinfin we have established the following relation expressingcertain particular value of Fourier transform of the logarithmof the Riemann zeta-function

int

infin

12

cos (2120587119909) ln |120589 (119909)| 119889119909

=1

2Σ2120587 +

7

16+1

4120587+ln (4120587)81205872minus119888119894 (120587)

81205872+119904119894 (120587)

8120587

+1

2int

infin

0

(1

2+1

119890119905minus 1minus1

119905)119890minus1199052

1199052+ 41205872119889119905

(28)

Here we used standard notation 119904119894(119909) = minusintinfin119909(sin 119902119902)119889119902

119888119894(119909) = minus intinfin

119909(cos 119902119902)119889119902 [14] Of course similar equalities

can be established for many other such Fourier transformsRepeating what has been said above about the ldquomax-

imalityrdquo of the sum Σ2120587RH = sum120588119905119896gt0

119897119896119890minus2120587119905119896 we again

could formulate a statement of the type ldquoRH holds true ifand only if the integral equality (28) with Σ2120587 = Σ2120587RHholdsrdquo Numerical calculation taking into account first 100zeroes of the Riemann function gives the value Σ2120587 =26904854 sdot 10

minus39 Equality (28) has been tested numer-ically and shown to be true without the term Σ2120587 bothsides coincide up to 39 figures after decimal and are equalto 05620211964552684935093570395449423344 Then ofcourse they start to differ

5 Conclusions

In this paper we have established a number of new criteriainvolving the integrals of the logarithm of the Riemann 120589-function and equivalent to the Riemann hypothesis thistime with the exponential weight functions Exponential

weight functions lead to rather simple expressions for thecontribution of the Riemann function zeroes not lying onthe critical line to the contour integral value This enabledobtaining a rigorous estimation of the possible error whichwas shown to be extremely small

We also show how certain Fourier transforms of thelogarithm of the Riemann zeta-function taken along thereal (demi)axis are expressible via elementary functionsplus logarithm of the gamma-function and definite integralsthereof as well as certain sums over trivial and nontrivialRiemann function zeroes In our opinion further study ofsimilar Fourier transforms is interesting and might be usefulin the Riemann researches

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] S K Sekatskii S Beltraminelli and D Merlini ldquoOn equalitiesinvolving integrals of the logarithm of the Riemann-functionand equivalent to the Riemann hypothesisrdquo Ukrainian Mathe-matical Journal vol 64 no 2 pp 218ndash228 2012

[2] S Beltraminelli D Merlini and S Sekatskii ldquoA hidden symme-try related to the Riemann hypothesis with the primes into thecritical striprdquoAsian-European Journal ofMathematics vol 3 no4 pp 555ndash563 2010

[3] E C Titchmarsh and E R Heath-Brown The Theory of theRiemann Zeta-Function Clarendon Press Oxford UK 1988

[4] F T Wang ldquoA note on the Riemann zeta-functionrdquo Bulletin ofthe American Mathematical Society vol 52 pp 319ndash321 1946

[5] V V Volchkov ldquoOn an equality equivalent to the Riemannhypothesisrdquo Ukrainian Mathematical Journal vol 47 no 3 pp491ndash493 1995

[6] M Balazard E Saias and M Yor ldquoNotes sur la fonction 120577 deRiemann 2rdquo Advances in Mathematics vol 143 no 2 pp 284ndash287 1999

[7] D Merlini ldquoThe Riemann magneton of the primesrdquo Chaos andComplexity Letters vol 2 no 1 pp 93ndash98 2006

[8] E C Titchmarsh The Theory of Functions Oxford UniversityPress Oxford UK 1939

[9] C Ramare and Y Saouter ldquoShort effective intervals containingprimesrdquo Journal ofNumberTheory vol 98 no 1 pp 10ndash33 2003

[10] R J Backlund ldquoUber die Nullstellen der Riemannschen Zeta-funktionrdquo Acta Mathematica vol 41 no 1 pp 345ndash375 1916

[11] X Gourdon ldquoThe 1013 first zeroes of the Riemann ZetaFunction and zeroes computation at very large heightrdquo 2004httpnumberscomputationfreefrConstantsMiscellaneouszetazeros1e13-1e24pdf

[12] D J Platt ldquoComputing 120587(119909) analyticallyrdquo Mathematics ofComputation vol 84 no 293 pp 1521ndash1535 2015

[13] A Erdelyi EdHigher Transcendental Functions Based in PartOn Notes Left by Harry Bateman vol 1 McGraw-Hill NewYork NY USA 1953

[14] I S Gradshtein and I M Ryzhik Tables of Integrals Series andProducts Academic New York NY USA 1990

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of