the riemann zetafunction chandrasekharan tata
TRANSCRIPT
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Lectures on
The Riemann ZetaFunction
By
K. Chandrasekharan
Tata Institute of Fundamental Research, Bombay
1953
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Lectures on the Riemann Zeta-Function
By
K. Chandrasekharan
Tata Institute of Fundamental Research
1953
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The aim of these lectures is to provide an intorduc-
tion to the theory of the Riemann Zeta-function for stu-dents who might later want to do research on the subject.
The Prime Number Theorem, Hardys theorem on the
Zeros of (s), and Hamburgers theorem are the princi-
pal results proved here. The exposition is self-contained,
and required a preliminary knowledge of only the ele-
ments of function theory.
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Contents
1 The Maximum Principle 1
2 The Phragmen-Lindelof principle 9
3 Schwarzs Lemma 17
4 Entire Functions 25
5 Entire Functions (Contd.) 35
6 The Gamma Function 45
1 Elementary Properties . . . . . . . . . . . . . . . . . . . 45
2 Analytic continuation of(z) . . . . . . . . . . . . . . . 49
3 The Product Formula . . . . . . . . . . . . . . . . . . . 51
7 The Gamma Function: Contd 55
4 The Bohr-Mollerup-Artin Theorem . . . . . . . . . . . . 55
5 Gausss Multiplication Formula . . . . . . . . . . . . . 58
6 Stirlings Formula . . . . . . . . . . . . . . . . . . . . . 59
8 The Zeta Function of Riemann 63
1 Elementary Properties of(s) . . . . . . . . . . . . . . . 63
9 The Zeta Function of Riemann (Contd.) 692 Elementary theory of Dirichlet Series . . . . . . . . . . 69
v
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vi Contents
10 The Zeta Function of Riemann (Contd) 75
2 (Contd). Elementary theory of Dirichlet series . . . . . . 75
11 The Zeta Function of Riemann (Contd) 87
3 Analytic continuation of(s). First method . . . . . . . 87
4 Functional Equation (First method) . . . . . . . . . . . . 90
5 Functional Equation (Second Method) . . . . . . . . . . 92
12 The Zeta Function of Riemann (Contd) 97
6 Some estimates for (s) . . . . . . . . . . . . . . . . . . 97
7 Functional Equation (Third Method) . . . . . . . . . . . 99
13 The Zeta Function of Riemann (Contd) 105
8 The zeros of (s) . . . . . . . . . . . . . . . . . . . . . 105
14 The Zeta Function of Riemann (Contd) 113
9 Riemann-Von Magoldt Formula . . . . . . . . . . . . . 113
15 The Zeta Function of Riemann (Contd) 121
10 Hardys Theorem . . . . . . . . . . . . . . . . . . . . . 121
16 The Zeta Function of Riemann (Contd) 127
17 The Zeta Function of Riemann (Contd) 135
12 The Prime Number Theorem . . . . . . . . . . . . . . . 13513 Prime Number Theorem and the zeros of(s) . . . . . . 145
14 Prime Number Theorem and the magnitude ofpn . . . . 146
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Lecture 1
The Maximum Principle
Theorem 1. If D is a domain bounded by a contour C for which Cauchys 1
theorem is valid, and f is continuous on C regular in D, then |f| Mon C implies |f| M in D, and if|f| = M in D, then f is a constant.Proof. (a) Let zo D, n a positive integer. Then
|f(zo)|n = 12i
C
{f(z)}ndzz zo
lc M
n
2,
where lc = length ofC, = distance ofzo from C. As n
|f(z)| M.
(b) If|f(zo)| = M, then f is a constant. For, applying Cauchys inte-gral formula to
d
dz
{f(z)}n, we get|n{f(zo)}n1 f(zo)| =
1
2i
C
fndz
(z zo)2 lCMn
22,
1
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2 1. The Maximum Principle
so that
|f(zo)| l
cM
22
1
n 0, as n
Hence
|f(zo)| = 0.
(c) If|f(zo)| = M, and |f(zo)| = 0, then f(zo) = 0, for2
d2
dz2{f(z)}n = n(n 1){f(z)}n2{f(z)}2 + n{f(z)}n1f(z).
At zo we have
d2
dz2{f(z)}nz=zo = n fn1(Z0)f(z0),
so that
|nMn1f(z0)| = 2!2i
C
{f(z)}ndz(z zo)3
2!lc
23Mn,
and letting n
, we see that f(zo) = 0. By a similar reasoning
we prove that all derivatives of f vanish at z0 (an arbitrary pointofD). Thus f is a constant.
Remark. The above proof is due to Landau [12, p.105]. We shall now
show that the restrictions on the nature of the boundary C postulated in
the above theorem cab be dispensed with.
Theorem 2. If f is regular in a domain D and is not a constant, and
M = maxzD
|f|, then
|f(zo)
|< M, zo
D.
For the proof of this theorem we need a
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1. The Maximum Principle 3
Lemma. If f is regular in |z z0| r, r > 0, then
|f(z0)| Mr,
where Mr = max|zzo|=r|f(z)|, and |f(zo)| = Mr only if f(z) is constant for3
|z zo| = r.Proof. On using Cauchys integral formula, we get
f(zo) =1
2i
|zzo|=n
f(z)
z z0dz
=
1
2
20
f(zo + rei
)d. (1)
Hence
|f(zo)| Mr.Further, if there is a point (such that) | zo| = r, and |f()| 0 and we should have|f(zo)| < Mr; so that |f(zo)| = Mr implies |f(z)| = Mr everywhere on|z zo| = r. That is |f(zo + rei)| = |f(zo)| for 0 2, or
f(zo + rei
) = f(zo)ei
, 0 2On substituting this in (1), we get
1 =1
2
20
cos d
Since is a continuous function ofand cos 1, we get cos 1 forall , i.e. = 0, hence f(s) is a constant.
Remarks. The Lemma proves the maximum principle in the case of a 4
circular domain. An alternative proof of the lemma is given below [7,Bd 1, p.117].
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4 1. The Maximum Principle
Aliter. If f(z0 + rei) = () (complex valued) then
f(z0) =1
2
20
()d
Now, if a and b are two complex numbers, |a| M, |b| M anda b, then |a + b| < 2M. Hence if() is not constant for 0 < 2,then there exist two points 1, 2 such that
|(1) + (2)| = 2Mr 2, say, where > 0
On the other hand, by regularity,
|(1 + t) (1)| < /2, for 0 < t < |(2 + t) (2)| < /2, for 0 < t <
Hence
|(1 + t) + (2 + t)| < 2Mr , for 0 < t < .
Therefore
2
0
()d =
1
0
+
1+
1
+
2
1+
+
2+
2
+
2
2+
=
1
0
+
21+
+
22+
()d+
0
[(1 + t) + (2 + t)] dt
Hence
|2
0
() d| Mr(2 2) + (2Mr ) = 2Mr
|1
2
20
()d| < Mr
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1. The Maximum Principle 5
Proof of Theorem 2. Let zo D. Consider the set G1 of points z such5that f(z) f(zo). This set is not empty, since f is non-constant. Itis a proper subset of D, since zo D, zo G1. It is open, because f iscontinuous in D. Now D contains at least one boundary point ofG1. For
if it did not, G1 D would be open too, and D = (G1 G1)D would be
disconnected. Let z1 be the boundary point ofG1 such that z1 D. Thenz1 G1, since G1 is open. Therefore f(z1) = f(zo). Since z1 BdG1and z1 D, we can choose a point z2 G1 such that the neighbourhoodofz1 defined by
|z z1| |z2 z1|lies entirely in D. However, f(z2) f(z1), since f(z2) f(zo), and
f(zo) = f(z1). Therefore f(z) is not constant on
|z
z1
|= r, (see
(1) on page 3) for, if it were, then f(z2) = f(z1). Hence if M =max
|zz1|=|z2z1||f(z)|, then
|f(z1)| < M M = maxzD
|f(z)|i.e. |f(zo)| < M
Remarks. The above proof of Theorem 2 [7, Bd I, p.134] does not make
use of the principle of analytic continuation which will of course provide
an immediate alternative proof once the Lemma is established.
Theorem 3. If f is regular in a bounded domain D, and continuous
in D, then f attains its maximum at some point in Bd D, unless f is a
constant.
Since D is bounded, D is also bounded. And a continuous function
on a compact set attains its maximum, and by Theorem 2 this maximum 6
cannot be attained at an interior point of D. Note that the continuity of
|f| is used.Theorem 4. If f is regular in a bounded domain D, is non-constant, and
if for every sequence {zn}, zn D, which converges to a point Bd D,we have
lim supn
|f(zn)| M,
then |f(z)| < M for all z D. [17, p.111]
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6 1. The Maximum Principle
Proof. D is an F, for define sets Cn by the property: Cn consists of
all z such that |z| n and such that there exists an open circular neigh-bourhood of radius 1/n with centre z, which is properly contained in D.
Then Cn Cn+1, n = 1, 2, . . .; and Cn is compact.
Define
maxZCn
|f(z)| mn.
By Theorem 2, there exists a zn Bd Cn such that |f(zn)| = mn. Thesequence {mn} is monotone increasing, by the previous results; and thesequence {zn} is bounded, so that a subsequence {znp} converges to alimit
Bd D. Hence
lim|f(znp )| Mi.e. lim mnp M
or mn < M for all n
i.e. |f(z)| < M. z D.
N.B. That Bd D can be seen as follows. If D, then thereexists an No such that CNo D, and |f()| < mNo lim mn, whereas7we have f(znp ) f(), so that |f(znp )| |f()|, or lim mn = |f()|.
Corollaries. (1) If f is regular in a bounded domain D and contin-uous in D, then by considering ef(z) and ei f(z) instead of f(z),it can be seen that the real and imaginary parts of f attain their
maxima on Bd D.
(2) If f is an entire function different from a constant, then
M(r) l.u.b|z|=r
|f(z)|, and
A(r) l.u.b|z|=r
Re{f(z)}
are strictly increasing functions of r. Since f(z) is bounded if M(r)is, for a non-constant function f , M(r) with r.
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1. The Maximum Principle 7
(3) If f is regular in D, and f 0 in D, then f cannot attain a
minimum in D.Further if f is regular in D and continuous in D, and f 0 in D,
and m = minzBd D
|f(z)|, then |f(z)| > m for all z D unless f is aconstant. [We see that m > 0 since f 0, and apply the maximum
principle to 1/f ].
(4) If f is regular and 0 in D and continuous in D, and |f| is aconstant on Bd D, then f = in D. For, obviously 0, and,
by Corollary 3, |f(z)| for z D, se that f attains its maximumin D, and hence f = .
(5) If f is regular in D and continuous in
D, and |f| is a constant on 8Bd D, then f is either a constant or has a zero in D.
(6) Definition: If f is regular in D, then is said to be a boundary
value of f at Bd D, if for every sequence zn Bd D,zn D, we have
limn f(zn) =
(a) It follows from Theorem 4 that if f is regular and non - con-
stant in D, and has boundary-values {}, and || H foreach , then |f(z)| < H, z D. Further, if M = max
b.v|| then
M=
M maxzD |f(z)|For, || is a boundary value of |f|, so that || M, henceM M. On the other hand, there exists a sequence {zn},zn D, such that |f(zn)| M. We may suppose thatzn zo (for, in any case, there exists a subsequence {znp}with the limit zo, say, and one can then operate with the
subsequence). Now zo D, for otherwise by continuity
f(zo) = limn f(zn) = M, which contradicts the maximum
principle. Hence zo Bd D, and M is the boundary-valueof |f| at zo. Therefore M M, hence M = M.
(b) Let f be regular and non-constant in D, and have boundary-
values {} on Bd D. Let m = minzD
|f|; m = minb.v.
|| > 0.
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8 1. The Maximum Principle
Then, if f 0 in D, by Corollary (6)a, we get
maxzD
1
|f(z)| =1
m=
1
m;
so that|f| > m = m in D. Thus if there exists a point zo D9such that |f(zo)| m, then f must have a zero in D, so thatm > 0 = m. We therefore conclude that, if m > 0, thenm = m if and only if f 0 in D [6, Bd. I, p.135].
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Lecture 2
The Phragmen-Lindelof
principle
We shall first prove a crude version of the Phragmen-Lindelof theorem 10
and then obtain a refined variant of it. The results may be viewed as
extensions of the maximum-modulus principle to infinite strips.
Theorem 1. [6, p.168] We suppose that
(i) f is regular in the strip < x < ; f is continuous in x
(ii) |f| M on x = and x = (iii) f is bounded in x
Then, |f| M in < x < ; and |f| = M in < x < only if f isa constant.
Proof. (i) If f(x + iy) O as y uniformly in x, x ,then the proof is simple. Choose a rectangle x , |y| with sufficiently large to imply |f|(x i) M for x
. Then, by the maximum-modulus principle, for any zo in the
interior of the rectangle,
|f(zo)| M.
9
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10 2. The Phragmen-Lindelof principle
(ii) If |f(x + iy)| 0, as y , consider the modified functionfn(z) = f(z)e
z2/n
. Now
|fn(z)| 0 as y uniformly in x, and|fn(z)| Mec
2/n
on the boundary of the strip, for a suitable constant c.
Hence11
|fn(zo)| Mec2/n
for an interior point zo of the rectangle, and letting n , we get
|f(zo)
| M.
If |f(zo)| = M, then f is a constant. For, if f were not a constant, in theneighbourhood ofzo we would have points z, by the maximum-modulus
principle, such that |f(z)| > M, which is impossible. Theorem 1 can be restated as:
Theorem 1: Suppose that
(i) f is continuous and bounded in x (ii) |f( + iy)| M1, |f( + iy)| M2 for all y
Then|f(xo + iyo)| ML(xo)1 M
1L(xo)2
for < xo < , |yo| < , where L(t) is a linear function oftwhich takesthe value 1 at and 0 at . If equality occurs, then
f(z) = cML(z)
1M
1L(z)2
; |c| = 1.
Proof. Consider
f1(z) =f(z)
ML(z)
1M
1L(z)2
and apply Theorem 1 to f1
(z).12
More generally we have [12, p.107]
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2. The Phragmen-Lindelof principle 11
Theorem 2. Suppose that
(i) f is regular in an open half-strip D defined by:
z = x + iy, < x < , y >
(ii) limzin D
|f| M, for Bd D
(iii) f = Oexp
e
|z|
, < D
uniformly in D.
Then, |f| M in D; and |f| < M in D unless f is a constant.Proof. Without loss of generality we can choose
= 2
, = +/2, = 0
Set
g(z) = f(z). exp(eikz) f(z).{(z)}, say,
where > 0, < k < 1. Then, as y +,
g = O expe|z| eky cos k2 uniformly in x, and 13
e|z| eky cos k2
e(y+/2) eky cos k2
, as y ,
uniformly in x, since < k. Hence
|g| 0 uniformly as y ,
so that |g| M, for y > y, < x < .
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12 2. The Phragmen-Lindelof principle
Let zo D. We can then choose H so that
|g(z)| M
for y = H, < x < , and we may also suppose that H > y0 = im(zo).
Consider now the rectangle defined by < x < , y = 0, y = H.
Since || 1 we havelim
zin D
|g| M
for every point on the boundary of this rectangle. Hence, by the
maximum-modulus principle,
|g(zo)| < M, unless g is a constant;or |f(zo)| < M|ee
ikzo |
Letting 0, we get14
|f(zo)| < M.
Remarks. The choice of in the above proof is suggested by the critical
case = 1 of the theorem when the result is no longer true. For take
= 1, = /2, = /2, = 0, f = eeiz
Then
|eeiz | = eRe(eiz) = eRe{eyix} = eey cos x
So
|f| = 1 on x = /2, and f = eey on x = 0.
Corollary 1. If f is regular in D, continuous on Bd D, |f| M on BdD, and
f = O ee
|z| , < 1,
then |f| M in D.
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2. The Phragmen-Lindelof principle 13
Corollary 2. Suppose that hypotheses (i) and (iii) of Theorem 2 hold;
suppose further that f is continuous on Bd D,
f = O(ya) on x = , f = O(yb) on x = . Then
f = O(yc) on x = ,
uniformly in , where c = p+ q, and px + q is the linear function which
equals a at x = and b at x = .
Proof. Let > 0. Define (z) = f(z)(z), where is the single-valued 15
branch of (zi)(pz+q) defined in D, and apply Theorem 2 to . We firstobserve that is regular in D and continuous in D. Next
|(z)| = |(y ix)(px+q)ipy |
= |y ix|(px+q) exp(py im logy ixy
)
= y(px+q)|1 + O
1
y
|O(1) exp
py
x
y+ O
1
y2
= y(px+q)epx{1 + O
1
y
},
the Os being uniform in the xs.Thus = O(1) on x = and x = . Since = O(yk) = O(|z|k), we
see that condition (iii) of Theorem 2 holds for if is suitably chosen.Hence = O(1) uniformly in the strip, which proves the corollary.
Theorem 3. [12, p.108] Suppose that conditions (i) and (iii) of Theorem
2 hold. Suppose further that f is continuous in Bd D, and
limy|f| M on x = , x =
Then
limy
|f| M uniformly in x .
Proof. f is bounded in
D, by the previous theorem. Let 0, > 0. 16Let H = H() be the ordinate beyond which |f| < M+ on x = , x = .
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14 2. The Phragmen-Lindelof principle
Let h > 0 be a constant. Then zz + hi
< 1in the strip. Choose h = h(H) so large thatf(z) z
z + hi
< M+ on y = H(possible because
f zz + hi
|f| < M). Then the functiong(z) = f z
z + hi
satisfies the conditions of Theorem 2 (with M + in place of M) in the
strip above y = H. Thus
|g| M+ in this stripand so lim
y|g| M+ uniformly.That is,
lim|f| M+ ,since
z
z + hi 1 uniformly in x. Hence
lim|f| M.
Corollary 1. If conditions (i) and (iii) of Theorem 2 hold, if f is contin-17
uous on Bd D, and if
lim|f| a on x = ,b on x = ,
where a 0, b 0, then
limy|f| e
px+q,
uniformly, p and q being so chosen that epx+q
=a for x
= and
=b for
x = .
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2. The Phragmen-Lindelof principle 15
Proof. Apply Theorem 3 to g = f e(pz+q)
Corollary 2. If f = O(1) on x = , f = o(1) on x = , then f = o(1) on
x = , < .[if conditions (i) and (iii) of Theorem 2 are satisfied].
Proof. Take b = in Corollary 1, and note that ep+q 0 as 0 forfixed , provided p and q are chosen as specified in Corollary 1.
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Lecture 3
Schwarzs Lemma
Theorem 1. Let f be regular in |z| < 1, f(o) = 0. 18If, for |z| < 1, we have |f(z)| 1, then
|f(z)| |z|, for |z| < 1.
Here, equality holds only if f(z) c z and |c| = 1. We further have|f(o)| 1.
Proof.f(z)
zis regular for |z| < 1. Given o < r < 1 choose such that
r < < 1; then since
|f(z)
| 1 for
|z
|= , it follows by the maximum
modulus principle that f(z)z
1
,
also for |z| = r. Since the L.H.S is independent of 1, we let andobtain |f(z)| |z|, for |z| < 1.
If for z0, (|z0| < 1), we have |f(z0)| = |z0|, then f(z0)
z0
= 1, (by themaximum principle applied to
f(z)
z) hence f = cz, |c| = 1.
Since f(z) = f(o)z + f(o)z2 + in a neighbourhood of the origin,and since f(z)
z 1 in z 1, we get |f(o)| 1. More generally, we
have
17
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18 3. Schwarzs Lemma
Theorem 1: Let f be regular and |f| M in |z| < R, and f(o) = 0.Then
|f(z)| M|z|R
, |z| < R.In particular, this holds if f is regular in |z| < R, and continuous in19|z| R, and |f| M on |z| = R.Theorem 2 (Caratheodorys Inequality). [12, p.112] Suppose that
(i) f is regular in |z| < R f is non-constant(ii) f(o) = 0
(iii) Re f
in
|z
|< R. (Thus
> 0, since f is not a constant).
Then,
|f| 2 R , for|z| = < R.
Proof. Consider the function
(z) =f(z)
f(z) 2 .
We have Re{f 2} < 0, so that is regular in |z| < R.If f = u + iv, we get
|| =
u2 + v2
(2 u)2 + v2
so that
|| 1, since 2 u |u|.But (o) = o, since f(o) = o. Hence by Theorem 1,
|(z)| |z|R
, |z| < R.
But f =
2
1 . Now take
|z|
= < R.
Then, we have,20
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3. Schwarzs Lemma 19
|f
|
2 ||
1 ||
2
R
N.B. If |f(zo)| =2
R , for |zo| < R, then
f(z) =2U cz1 cz , |c| = 1.
More generally, we have
Theorem 2: Let f be regular in |z| < R; f non-constant
f(o) = ao = + i
Re f (|z| < R), so that Then
|f(z) ao| 2( )
R for |z| = < RRemark . If f is a constant, then Theorem 2 is trivial. We shall now
prove Borels inequality which is sharper than Theorem 2.
Theorem 3 (Borels Inequality). [12, p.114] Let f(z) =
n=0anz
n, be
regular in |z| < 1. LetRe f . f(o) = ao = + i.Then
|an| 2( ) 2 1, say, n > 0.Proof. We shall first prove that |a1| < 1, and then for general an.
(i) Let f1
n=1anz
n. Then Re f1 1. 21
For |z| = < 1, we have, by Theorem 2,
|f1| 21
1 i.e.
f1(z)z 211
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20 3. Schwarzs Lemma
Now letting z 0, we get
|a1| 21
(ii) Define = e2i/k, ka +ve integer.
Then
kv=1
m = o
ifm 0& ifm a multiple ofk= k
ifm = 0
or if m = a multiple ofk.
We have
1
k
k1r=0
f1(rz) =
n=1
ankznk
= g1(zk) g1(), say.
The series for g1 is convergent for |z| < 1, and so for || < 1. Henceg1 is regular for || < 1. Since Re g1 1, we have
| coefficient of | 21 by the first part of the proof i.e. |ak| 2221. Corollary 1. Let f be regular in |z| < R, andRe{f(z)f(o)} 1. Then
|f(z)| 2 1 R(R )2 , |z| = < R
Proof. Suppose R = 1. Then
|f(z)|
n|an|n1 2 1
nn1
This argument can be extended to the nth derivative.
Corollary 2. If f is regular for every finite z, and
ef(z) =
O(e|z|
k
), as
|z
| ,
then f(z) is a polynomial of degree k.
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3. Schwarzs Lemma 21
Proof. Let f(z) = anzn. For large R, and a fixed , || < 1, we have
Re{f(R)} < cRk.By Theorem 3, we have
|anRn| 2(2Rk Re ao), n > 0.Letting R , we get
an = 0 ifn > k.
Apropos Schwarzs Lemma we give here a formula and an inequal-
ity which are useful.
Theorem 4. Let f be regular in |z zo| r 23f = P(r, ) + Q(r, ).
Then
f(zo) =1
r
2o
P(r, )eid.
Proof. By Cauchys formula,
(i) f(zo) =1
2i |zzo|=rf(z)dz
(z
zo
)2=
1
2r
2
0
(P + iQ)eid
By Cauchys theorem,
(ii) 0 =1
r2 1
2i
|zzo|=r
f(z)dz =1
2r
20
(P + iQ)eid
We may change i to i in this relation, and add (i) and (ii).Then
f(zo) =1
r
2
0
P(r, )eid
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22 3. Schwarzs Lemma
Corollary. If f is regular, and | Re f M in |z z0| r, then
|f(zo)| 2Mr
.
Aliter: We have obtained a series of results each of which depended
on the preceding. We can reverse this procedure, and state one general
result from which the rest follow as consequences.
Theorem 5. [11, p.50] Let f(z)
n=0
cn(zz0)n be regular in |zz0| < R,24andRe f < . Then
|en| 2( Re co)
Rn, n = 1, 2, 3, . . . (5.1)
and in |z zo| < R, we have
|f(z) f(z0)| 2
R { Re f(zo)} (5.2) f(n)(z)
n!
2R(R )n+1 { Re f(zo)} n = 1, 2, 3, . . . (5.3)Proof. We may suppose zo = 0.
Set (z) = f(z) = c0 1
cnzn
0
bnzn, |z| < R.
Let denote the circle |z| = < R. Then
bn =1
2i
(z)dz
zn+1=
1
2n
(P + iQ)eind, n 0, (5.4)
where (r, ) = P(r, ) + iQ(r, ). Now, ifn 1, then (z)zn1 is regularis , so that
0 =
n
2r
(P + iQ)ein
d, n 1 (5.5)
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3. Schwarzs Lemma 23
Changing i to i in (5.5) and adding this to (5.4) we get
bnn
=1
Penid, n 1.
But P = Re f 0 in |z| < R and so in . Hence, ifn 1, 25
|bn|n 1
|Peni|d = 1
Pd = 2 Re bo, (5.6)
on using (5.4) with n = 0. Now letting R, we get
|bn|Rn 2o 2 Re bo.
Since bo = co, and bn = cn, n 1, we at once obtain (5.1). Wethen deduce, for |z| < R
|(z) (o)| = |1
bnzn|
1
20
R
n=
2o
R (5.7)
and ifn 1,
|(n)(z)
|
r=n
r(r
1) . . . (r
n + 1)
20rn
Rr
=
d
d
n n=0
20
R
r
=
d
d
n2eR
R = 2
oR n!(R )n+1 (5.8)
(5.7) and (5.8) yield the required results on substituting = f and0 = Re f(0).
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Lecture 4
Entire Functions
An entire function is an analytic function (in the complex plane) which 26
has no singularities except at . A polynomial is a simple example. Apolynomial f(z) which has zeros at z1, . . . ,zn can be factorized as
f(z) = f(0)
1 z
z1
. . .
1 z
zn
The analogy holds for entire functions in general. Before we prove this,
we wish to observe that ifG(z) is an entire function with no zeros it can
be written in the form eg(z) where g is entire. For considerG(z)G(z)
; every
(finite) point is an ordinary point for this function, and so it is entireand equals g1(z), say. Then we get
log
G(z)
G(z0)
=
z1z0
g1()d = g(z) g(z0), say,
so that
G(z) = G(z0)eg(z)g(z0) = eg(z)g(z0)+log G(zo)
As a corollary we see that if G(z) is an entire function with n zeros,
distinct or not, then
G(z) = (z z1) . . . (z zn)eg(z)
25
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26 4. Entire Functions
where g is entire. We wish to uphold this in the case when G has an
infinity of zeros.
Theorem 1 (Weierstrass). [15, p.246] Given a sequence of complex27
numbers an such that
0 < |a1| |a2| . . . |an| . . .
whose sole limit point is , there exists an entire function with zeros atthese points and these points only.
Proof. Consider the function
1
z
an
eQ(z)
,
where Q(z) is a polynomial of degree q. This is an entire function whichdoes not vanish except for z = an. Rewrite it as
1 zan
eQ(z) = eQ(z)+log
1 zan
= e z
an z2
2a2...+Q(z),
and choose
Q(z) =
z
an+ . . . +
z
an
so that 1 z
an
eQ(z) = e
zan
+11
+1
...
1 + n(z), say.
We wish to determine in such a way that
1
1 zan
eQ(z)
is absolutely and uniformly convergent for |z| < R, however large R may28
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4. Entire Functions 27
be.
Choose an R > 1, and an such that 0 < < 1; then there exists aq (+ ve integer) such that
|aq| R
, |aq+1| >
R
.
Then the partial product
q1
1 z
an
eQ(Z)
is trivially an entire function of z. Consider the remainder
q+1
1 z
an
eQ(z)
for |z| R.We have, for n > q, |an| >
R
or
zan < < 1. Using this fact we
shall estimate each of the factors {1 + n(z)} in the product, for n > q.
| n (z)| =e 1
+1
z
an
+1... 1 e
1+1 zan
+1... 1
Since |em 1| e|m| 1, for em 1 = m + m2
2!+ . . . or |em 1| 29
|m| + |m2|
2. . . = e|m| 1.
|Un(z)| e zan
+11+ zan+... 1,
e
zan
+1
(1++2+ ) 1
= e1
1 zan +1 1
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28 4. Entire Functions
1
1 z
an+1
e
11
zan
+1
, since if x is
real, ex 1 xex, for ex 1 = x
1 +x
2!+
x2
3!+
x(1 + x + x
2
2!+
) = xexHence
|Un(z)| e1
1
1 zan
+1Now arise two cases: (i) either there exists an integer p > 0 such30
that
n=11
|an|p < or (ii) there does not. In case (i) take = p 1, so
that
|Un(z)| Rp
|an|p e
11a
1 , since |z| R;
and hence |Un(z)| < for |z| R, i.e.
q+1
(1 + Un(z)) is absolutely and
uniformly convergent for |z| R.In case (ii) take = n 1, so that
|Un(z)| q
zan < 1|z| R|an|
Then by the root-test |Un(z)| < , and the same result follows
as before. Hence the product
1
1 z
an
eQ(z)
is analytic in |z| R; since R is arbitrary and does not depend on R we31see that the above product represents an entire function.
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4. Entire Functions 29
Remark. If in addition z = 0 is also a zero ofG(z) thenG(z)
zm
G(z)
has no
zeros and equals eg(z), say, so that
G(z) = eg(z) zm
n=1
1 z
an
eQ(z)
In the above expression for G(z), the function g(z) is an arbitrary
entire function. IfG is subject to further restrictions it should be possible
to say more about g(z); this we shall now proceed to do. The class of
entire functions which we shall consider will be called entire functions
of finite order.
Definition. An entire function f(z) is of finite order if there exists a con-stant such that |f(z)| < er for |z| = r > r0. For a non-constant f , of
finite order, we have > 0. If the above inequality is true for a certain
, it is also true for > . Thus there are an infinity ofs o satisfyingthis. The lower bound of such s is called the order of f . Let usdenote it by . Then, given > 0, there exists an ro such that
|f(z)| < er+
for |z| = r > r0.This would imply that
M(r) = max|z|=r |f(z)| < er+
, r > r0,
while 32
M(r) > er
for an infinity of values ofrtending
to +Taking logs. twice, we get
log log M(r)
log r< +
and
log log M(r)log r
> for an infinity of values ofr
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30 4. Entire Functions
Hence
= limr log log M(r)logR
Theorem 2. If f(z) is an entire function of order < , and has aninfinity of zeros, f(0) 0, then given > 0, there exists an R such that
for R Ro, we have
n
R
3
1
log2 log e
R+
|f(o)|
Here n(R) denotes the number of zeros of f(z) in |z| R.
Proof. We first observe that if f(z) is analytic in |z| R, and a1 . . . anare the zeros of f inside |z| < R/3, then for
g(z) =f(z)
1 za1
. . .
1 z
an
we get the inequality33
|g(z)| M2n
where M is defined by: |f(z)| M for |z| = R. For, if |z| = R, then since
|ap| R
3 , p = 1 . . . n, we get zap 3 or 1
z
ap
2By the maximum modules theorem,
|g(o)| M2n
, i.e. |f(o)| M2n
or
n nR
3
1
log2 log
M
f(o)
If, further, f is of order , then for r > r, we have M < er+
which
gives the required results.N.B. The result is trivial if the number of zeros is finite.
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4. Entire Functions 31
Corollary.
n(R) = O(R+
).
For
n
R
3
1
log2 R+ log |f(o)| ,
and log |f(0)| < R+, say.Then for R R
n
R
3
, then
1|an|
<
Proof. Arrange the zeros in a sequence:
|a1| |a2| . . .Let
p = |ap|Then in the circle |z| r = n, there are exactly n zeros. Hence
n < c
+
n , for n Nor
1
n>
1
c 1
+n
Let > + (i.e. choose 0 < < ). Then1
n/+>
1
c/+ 1
n,
or1
n< c/+ 1
n/+for n N
Hence
1n
< .
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32 4. Entire Functions
Remark. (i) There cannot be too dense a distribution of zeros, since
n < c < +
n . Nor can their moduli increase too slowly, since, forinstance,
1(log n)p
does not converge.
(ii) The result is of course trivial if there are only a finite number of35
zeros.
Definition. The lower bound of the numbers for which 1
|an|< ,
is called the exponent of convergence of{an}. We shall denote it by 1.Then
1|an|1+ < , 1
|an|1
=
, > 0
By Theorem 1, we have 1
N.B. If the an s are finite in number or nil, then 1 = 0. Thus 1 > 0implies that f has an infinity of zeros. Let f(z) be entire, of order < ;f(o) 0, and f(zn) = 0, n = 1, 2, 3, . . .. Then there exists an integer
(p + 1) such that 1
|zn|+1<
(By Theorem 1, any integer > will serve for + 1). Thus, by
Weierstrasss theorem, we have
f(z) = eQ(z)
1
1 z
zn
e
z
zn +
z2
2zn2 ++z
zn ,
where = p (cf. proof of Weierstrasss theorem).
Definition. The smallestinteger p for which 1
|zn|p+1 < is the genusof the canonical product
1 z
zn
e
zzn
++ zppz
pn , with zeros zn. If z
n s36
are finite in number, we (including nil) define p = 0, and we define the
product as
1 z
zn
Examples. (i) zn=
n, p=
1 (ii) zn=
en
, p=
0(iii) z1 =
12
log 2, zn = log n, n 2, no finite p.
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4. Entire Functions 33
Remark. If > 1, then 1|zn| < Also, 1|zn|p+1
<
But 1|zn|p
=
Thus, if1 is not an integer, p = [1], and if1 is an integer, then either 1|zn|1 0, let N be defined by the property:
|zN| R, |zN+1| > R.We need the following
Lemma. Let0 < < 1. Thenf(z)f(z)
N
n+1
1
z zn
< cR1+
for |z| = R2
> R0, and |z| = R2
is free from any zn.
35
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36 5. Entire Functions (Contd.)
For we can choose an R0 such that for |z| = R > R0,(i) |f(z)| < eR+ ,38
(ii) no zn lies on |z| = R2
, and
(iii) log1
|f(0)| < (2R)+
For such an R > R0, define
gR(z) =f(z)
f(0)
N
n=1
1 zzn
1
Then for |z| = 2R, we have
|gR(z)| 1.43
Hence, for all u, we have
|E(u)| e|2u|+log(1+|u|),
if p p + 1.It is possible to choose in such a way that
(i) p p + 1,
(ii) > 0,
(iii) 1
|zn|< , (if there are an infinity of zn s) and
(iv) 1 1 +
For, if 1
|zn|1< (if the series is infinite this would imply that
1 > 0), we have only to take = > 0. And in all other cases we must
have p 1 < p + 1, [for, if 1 = p + 1, then 1
|zn|1< ] and so we
choose such that1 < < p + 1; this implies again that1
|zn|
0. For such a we can write the above inequality as
|E(u)| ec1 |u| , c1 is a constant.
Hence44
|f(z)| e|Q(z)|+c1
n=1
zzn
i.e. M(r) < ec2rq
+c3r
, |z| = r > 1, and c1
|1/zn| = c3.
Hence
log M(r) = O(rq) + O(r), as r ,
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5. Entire Functions (Contd.) 41
= O(rmax(q,)), as r (2.1)
so that
max(q, )Since is arbitrarily near1, we get
max(q, 1),
and this proves the theorem.
Theorem 3. If the order of the canonical product G(z) is, then = 1
Proof. As in the proof of Theorem 2, we have
|E(u)| ec1|u|
p p + 1; > 0 1
|zn| 0
Hence
c1
n=1
z
zn
|G(z)| e ec3r , |z| = r.
If M(r) = max|z|=r
|G(z)| , then 45
M(r) < ec3r
Hence
which implies
1; but 1
. Hence = 1
Theorem 4. If, either
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42 5. Entire Functions (Contd.)
(i) 1 < q or
(ii) 1
|zn|1 < (the series being infinite),
then
log M(r) = O(r), for = max(1, q).
Proof. (i) If1 < q, then we take < q and from (2.1) we get
log M(r) = O(rq) = O(rmax(1,q))
(ii) If
n=11
|zn
|1
< , then we may take = 1 > 0 and from (2.1) we
get
log M(r) = O(rq) + O(r1 )
= O(r)
Theorem 5. (i) We have
= max(1, q)
(The existence of either side implies the other)
(ii) If > 0, and if log M(r) = O(r) does not hold, then 1 = , f(z)46
has an infinity of zeros zn, and |zn| is divergent.
Proof. The first part is merely an affirmation of Hadamards theorem,
1 and Theorem 2.For the second part, we must have 1 = ; for if1 < , then = q
by the first part, so that 1 < q, which implies, by Theorem 4, that
logM(r) = O(rq) = O(r) in contradiction with our hypothesis. Since
> 0, and 1 = we have therefore 1 > 0, which implies that f(z) has
an infinity of zeros. Finally
1|zn|1 is divergent, for if
|zn|1 < ,
then by Theorem 4, we would have log M(r)=
O(r
) which contradictsthe hypothesis.
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5. Entire Functions (Contd.) 43
Remarks. (1) 1 is called the real order of f(z), and the apparent
order Max (q, p) is called the genus of f(z).
(2) If is not an integer, then = r, for q is an integer and =
max(1, q). In particular, if is non-integral, then f must have an
infinity of zeros.
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Lecture 6
The Gamma Function
1 Elementary Properties [15, p.148]47
Define the function by the relation.
(x) =
0
tx1etdt, x real.
(i) This integral converges at the upper limit , for all x, sincetx1et = t2tx+1et = O(t2) as t ; it converges at the lowerlimit o for x > 0.
(ii) The integral converges uniformly for 0 < a x b. For
0
tx1etdt =
10
+
1
= O
1
0
ta1dt
+ O
1
tb1etdt
= O(1), independently of x.
Hence the integral represents a continuous function for x > 0.
(iii) Ifz is complex,
0
tz1etdtis again uniformly convergent over any
finite region in which Rez
a > 0. For ifz = x + iy, then
|tz1| = tx1,
45
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46 6. The Gamma Function
and we use (ii). Hence (z) is an analytic function forRez > 0
(iv) Ifx > 1, we integrate by parts and get
(x) =tx1et
0
+ (x 1)
0
tx2etdt
= (x 1)(x 1).
However,48
(1) =
0
etdt = 1; hence
(n) = (n 1)!,
ifn is a +ve integer. Thus (x) is a generalization ofn!
(v) We have
log (n) = (n 12
) log n n + c + O(1),
where c is a constant.
log (n) = log(n 1)! =n1r=1
log r.
We estimate log rby an integral: we have
r+ 12
r 12
log t dt =
12
12
log(s + r)ds =
12
0
+
0 1
2
=
1/2
0
log(t+ r) + log(r t) dt
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1. Elementary Properties 47
=
1/2
0
log r
2
+ log
1 t2
r2
dt
= log r+ cr, where cr = O
1
r2
.
Hence
log (n) =
n1r=1
r+ 1
2r 1
2
log t dt cr
=
n 12
12
log t dtn
1
r=1
cr
=
n 1
2
log
n 1
2
1
2log
1
2
(n 1
2) +
1
2
r=1
cr +
n
Cr
= (n 12
) log n n + c + o(1), where c is a constant.49
Hence, also(n!) = a.ennn+
12 eo(1),
where a is a constant such that c = log a
(vi) We have
(x)(y)
(x +y)=
O
ty1
(1 + t)x+ydt, x > 0, y > 0.
For
(x)(y) =
0
tx
1
et
dt
0
sy
1
es
ds, x > o, y > 0
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48 6. The Gamma Function
Put s = tv; then
(x)(y) =
0
tx1et dt
0
tyvy1etv dv
=
0
vy1dv
0
tx+y1et(1+v)dt
=
0
vy1dv
0
ux+y1eu
(1 + v)x+ydu
= (x +y)
0
vy1
(1 + v)x+y dv.
The inversion of the repeated integral is justified by the use of the50
fact that the individual integrals converge uniformly for x > 0,y > 0.
(vii) Putting x = y =1
2in (vii), we get [using the substitution t =
tan2 ]
{(
1
2
)
}2
= 2(1)
/2
0
d = .
Since ( 12
) > 0, we get ( 12
) =
Putting y = 1 x, on the other hand, we get the important relation
(x)(1 x) =
0
ux
1 + udu =
sin(1 x) , 0 < x < 1,
since
0
xa1
1 + x dx =
sin a for 0 < a < 1, by contour
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2. Analytic continuation of (z) 49
integration. Hence
(x)(1 x) = sin x
, 0 < x < 1.
2 Analytic continuation of(z) [15, p. 148]
We have seen that the function
(z) =
o
ettz1 dt
is a regular function for Rez > 0. We now seek to extend it analytically
into the rest of the complex z-plane. Consider
I(z) C
e()z1d,
where C consists of the real axis from to > 0, the circle || = in 51the positive direction and again the real axis from to .
We define
()z1 = e(z1) log()by choosing log() to be real when =
The integral I(z) is uniformly convergent in any finite region of the
Z-plane, for the only possible difficulty is at =
, but this was covered
by case (iii) in 1. Thus I(z) is regular for all finite values of z.We shall evaluate I(z). For this, set
= ei
so that
log() = log + i( ) on the contour[so as to conform with the requirement that log() is real when = ]
Now the integrals on the portion of C corresponding to (, ) and(,) give [since = 0 in the first case, and = 2 in the second]:
e+(z
1)
(log
i)
d +
e+(z
1)(log+i)
d
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50 6. The Gamma Function
=
e+(z1)(logi) + e+(z1)(log+i) d=
e+(z1)loge(z1)i e(z1)i
d
= 2i sinz
e z1d
On the circle || = , we have
|()z
1
|=
|e(z
1)
log()|=
|e(z
1)[log +i(
)]
|= e(x1)log y()
= O(x1).
52
The integral round the circle || = therefore gives
O(x) = O(1), as 0, if x > 0.
Hence, letting 0, we get
I(z) = 2i sin z
0
ez1d, Rez > 0
= 2i sin z(z), Rez > 0.
We have already noted that I(z) is regular for all finite z; so
1
2iI(z) cosec z
is regular for all finite z, except (possibly) for the poles of cosec z
namely, z = 0, 1, 2, . . .; and it equals (z) for Rez > 0.Hence -
1
2 iI(z) cosecz is the analytic continuation of
(z) all overthe z-plane.
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3. The Product Formula 51
We know, however, by (iii) of 1, that (z) is regular for z =1, 2, 3, . . .. Hence the only possible poles of 1
2iI(z) cosec x are z =
0, 1, 2, . . .. These are actually poles of (z), for if z is one of thesenumbers, say n, then ()z1 is single-valued in 0 and I(z) can be eval-uated directly by Cauchys theorem. Actually
(1)n+1C
e
n+1d =
2i
n!(1)n+n+1 = 2i
n!
i.e. I(n) = 2in!
.
So the poles of cosec z, at z = 0,
n, are actually poles of(z). The 53
residue at z = n is therefore
limzn
2in!
z + n
2i sinz
=(1)n
n!
The formula in (viii) of 1 can therefore be upheld for complex values.Thus
(z) (1 z) = cosec zfor all non-integral values of z. Hence, also,
1
(z)is an entire function.
(Since the poles of(1 z) are cancelled by the zeros of sin z)
3 The Product Formula
We shall prove that1
(z)is an entire function of order 1.
Since I(z) = 2i sinz (z), and
(z) (1 z) = sin z
,
we get
I(1 z) = 2i sin( z)(1 z)
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52 6. The Gamma Function
=
2i sinz
sin z
1
(z)
=
2i
(z)
or1
(z)=
I(1 z)2i .
Now54
I(z) = O
e|z|
10
ettx1dt +
1
ettx1dt
= O
e|z|
1 +
1
ett|z|dt
= O
e|z|1 +
1+|z|1
+
1+|z|
= Oe|z|(|z| + 1)|z|+1
= O
e|z|
1+
, > 0.
Hence1
(z)is of order 1. However, the exponent of convergence
of its zeros equals 1. Hence the order is = 1, and the genus of the
canonical product is also equal to 1.
Thus1
(z)= z, eaz+b
n=1
(1 +z
n)e
zn
by Hadamards theorem. However, we know that limz0+
1
z(z)= 1 and
(1) = 1. These substitutions give b = 0, and
1 = ea
1 +
1
e1/n
or,
0 = a +
1
log
1 + 1
n 1n
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3. The Product Formula 53
= a
limm
m
1log 1 + 1
n
1
n
= a + limm
m1
log(n + 1) log n m
1
1
n
= a + lim
m
log(m + 1) m1
1
n
= a , where is Eulers constant.
55
Hence
1(z)
= ezz n=1
1 + z
n
ez/n
As a consequence of this we can derive Gausss expression for (z).
For
(z) = limn
ez(log n1
12... 1
n )1
z e
z/1
1 + z1
. . .ez/n
1 + zn
= limn
nz 1
z 1
z + 1 2
z + 2. . .
n
z + n
= lim
n nz n!
z(z + 1) . . . (z + n)
We shall denote
nz n!z(z + 1) . . . (z + n)
n(z),
so that 56
(z) = limn n(z).
Further, we get by logarithmic-differentiation,
(z)
(z)=
1
z
n=1
1z + n
1
n
.
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54 6. The Gamma Function
Thus
d2
dz2[log (z)] =
n=0
1(n +z)2
;
Henced2
dz2[log (z)] > 0 for real, positive z.
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Lecture 7
The Gamma Function:
Contd
4 The Bohr-Mollerup-Artin Theorem
[7, Bd.I, p.276]
We shall prove that the functional equation of (x), namely (x + 1) = 57
x(x), together with the fact thatd2
dx2[log (x)] > 0 for x > 0, deter-
mines (x) essentially uniquely-essentially, that is, except for a con-
stant of proportionality. If we add the normalization condition (1) = 1,
then these three properties determine uniquely. N.B. The functional
equation alone does not define uniquely since g(x) = p(x)(x), where
p is any analytic function of period 1 also satisfies the same functional
equation.
We shall briefly recall the definition of Convex functions. A real-
valued function f(x), defined for x > 0, is convex, if the corresponding
function
(y) =f(x +y) f(x)
y,
defined for all y > x, y 0, is monotone increasing throughout therange of its definition.
If 0 < x1 < x < x2 are given, then by choosing y1 = x1 x and
55
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56 7. The Gamma Function: Contd
y2 = x2 x, we can express the condition of convexity as:
f(x) x2 xx2 x1
f(x1) +x x1
x2 x1f(x2)
Letting x x1, we get f(x1 + 0) f(x1); and letting x2 x, we58get f(x) f(x + 0) and hence f(x1 + 0) = f(x1) for all x1. Similarlyf(x1 0) = f(x1) for all x1.
Thus a convex function is continuous.
Next, if f(x) is twice (continuously) differentiable, we have
(y) =y f(x +y) f(x +y) + f(x)
y2
and writing x +y = u, we get
f(x) = f(u y) = f(u) y f(u) + y2
2f[u (1 )y], 0 1
which we substitute in the formula for (y) so as to obtain
(y) =1
2f[x + y].
Thus if f(x)
0 for all x > 0, then (y) is monotone increasing and fis convex. [The converse is also true].
Thus log (x) is a convex function; by the last formula of the previ-
ous section we may say that (x) is logarithmically convex for x > 0.
Further (x) > 0 for x > 0.
Theorem. (x) is the positive function uniquely defined for x > 0 by the
conditions:
(1.) (x + 1) = x(x)
(2.) (x) is logarithmically convex
(3.) (1) = 1.
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4. The Bohr-Mollerup-Artin Theorem 57
Proof. Let f(x) > 0 be any function satisfying the above three condi-
tions satisfied by (x).Choose an integer n > 2, and 0 < x 1.59Let
n 1 < n < n + x n + 1.By logarithmic convexity, we get
log f(n 1) log f(n)(n 1) n 0, we then have
dx
dx2[log f(x)] > 0.
Further
f(x + 1) = x f(x).
Hence by the Bohr-Artin theorem, we get
f(x) = ap(x).
Put x = 1. Then we get
ap = p
1
p
2
p
. . .
p
p
(5.1)
[a1 = 1, by definition]. Put k = 1, 2, . . . p, in the relation
n k
p =nk/pn!pn+1
k(k+ p) . . . (k+ np),
and multiply out and take the limit as n . We then get from (5.1),
ap = p. limn
np+1
2 (n!)ppnp+p
(np + p)!
However
(np + p)! = (np)!(np)p
1 +1
np
1 +
2
np
. . .
1 +
p
np
Thus, for fixed p, as n , we get
ap = p limn
(n!)p
pnp
(np)!n(p1)/2
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6. Stirlings Formula 59
Now by the asymptotic formula obtained in (v) of 1, [see p.49]61
(n!)p = apnnp+p/2enpeo(1)
(np)! = annp+12pnp+
12 enp eo(1)
Hence
ap =
p ap1.Putting p = 2 and using (5.1) we get
a =12
, a2 =12
2
1
2
(1) =
2 (5.2)
Hence
ap = p(2)(p1)/2.Thus
px
x
p
. . .
x + p 1
p
=
p(2)(p1)/2(x)
For p = 2 and p = 3 we get:
x
2
x + 1
2
=
2x1(x);
x
3
x + 1
3
x + 2
3
=
2
3x12
(x).
6 Stirlings Formula [15, p.150]
From (v) of 1, p.49 and (5.2) we get
(n!) =
2 en nn+ 12 eo(1),
which is the same thing as
log(n 1)! =
n 12
log n n + log
2 + o(1) (6.1)
Further 62
1 + 12
+ + 1N 1 logN = + 0(1), as N (6.2)
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60 7. The Gamma Function: Contd
and
log(N +z) = logN + log
1 +
z
N
= logN +
z
N+ O
1
N2
, as N . (6.3)
We need to use (6.1)-(6.3) for obtaining the asymptotic formula for
(z) where z is complex.
From the product-formula we get
log (z) =
n=1 z
n log 1 +
z
n z logz, (6.4)each logarithm having its principal value.
Now
N0
[u] u + 12
u +zdu =
N1n=0
n+1n
n + 12 +zu +z 1 du
=
N1n=0
(n +1
2+z)[log(n + 1 +z) log(n z)] N
=
N1n=0
n[log(n+
1+
z) log(n +z)] + z + 12 N1n=0
log(n + 1 +z) log(n +z) N= (N 1) log(N +z)
N1n=1
log(n +z) +
z +
1
2
log(N +z)
z +
1
2
logz N
=
N 1
2+z
log(N +z)
z +
1
2
logz
NN1n=1
log n + log
1 +
z
n
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6. Stirlings Formula 61
= N1
2
+z log(N +z) z +1
2 logz N log(N 1)!
+
N1n=1
z
n log
1 +
z
n
z
N11
1
n
63
Now substituting for log(N +z) etc., from (6.1)-(6.3), we get
N0
[u] u + 12
u +zdu = (N 1
2+z)
logN +
z
N+ O
1
N2
z +
1
2
logz N N1
2 logN + N c + O(1)+N1n=1
z
n log
1 +
z
n
z
N11
1
n
= z
logNN1
1
1
n
+z + O(1) log 2z 1
2
logz logz +
N1n=1
z
n log
1 +
z
n
Now letting N , and using (6.4), and (6.2), we get 64
0
[u] u + 12u +z
du + 12
log 2 z +z 1
2
logz = log (z) (6.5)
If(u) =u
0
([v] v + 12
)dv, then
(u) = O(1), since (n + 1) = (n),
ifn is an integer.
Hence
0
[u]
u + 1
2
u +z du =
0
d(u)
u +z =
0
(u)
(u +z)2
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62 7. The Gamma Function: Contd
=O
0
dv
|u +z|2 If we write z = rei, and u = ru1, then
0
[u] u + 12
u +zdu = O
1r
0
du1
|u1 + i|2
The last integral is finite if . Hence
0
[u] u + 12
u+
z
du = O 1
r , uniformly for |
arg z
|
<
Thus we get from (6.5):
log (z) =
z 1
2
logz z + 1
2log2 + O
1
|z|
for |arg z| < Corollaries. (1) log (z + ) = (z + 1
2)logz z + 1
2log2 + O
1|z|
as |z| uniformly for|arg z| < . bounded.65(2) For any fixed x, as y
|(x + iy)| e 12 |y||y|x 12 2
(3)
(z)(z)
= logz 12z
+ O
1
|z|2, |arg z| < [11, p.57]
This may be deduced from (1) by using
f(z) =1
2i
C
f()
(z)2 d,
where f(z) = log (z)
z
12 logz +z
12
log2 and C is a circle
with centre at = z and radius |z| sin /2.
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Lecture 8
The Zeta Function of
Riemann
1 Elementary Properties of(s) [16, pp.1-9]
We define (s), for s complex, by the relation 66
(s) =
n=1
1
ns, s = + it, > 1. (1.1)
We define xs
, for x > 0, as es log x
, where log x has its real determination.Then |ns| = n, and the series converges for > 1, and uniformlyin any finite region in which 1 + > 1. Hence (s) is regularfor 1 + > 1. Its derivatives may be calculated by termwisedifferentiation of the series.
We could express (s) also as an infinite product called the Euler
Product:
(s) =
p
1 1
ps
1, > 1, (1.2)
where p runs through all the primes (known to be infinite in number!).
It is the Euler product which connects the properties of with the prop-erties of primes. The infinite product is absolutely convergent for > 1,
63
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64 8. The Zeta Function of Riemann
since the corresponding series
p
1ps =
p
1
p< , > 1
Expanding each factor of the product, we can write it as
p
1 +
1
ps+
1
p2s+
Multiplying out formally we get the expression (1.1) by the unique67
factorization theorem. This expansion can be justified as follows:
pp
1 +
1
ps+
1
p2s+
= 1 +
1
ns1
+1
ns2
+ ,
where n1, n2, . . . are those integers none of whose prime factors exceed
P. Since all integers P are of this form, we get
n=1
1
nspP
1 1
p2
1 =
n=1
1
ns 1 1
ns1
1ns
2
1
(P + 1)+
1
(P + 2)+
0, as P , if > 1.Hence (1.2) has been established for > 1, on the basis of (1.1).
As an immediate consequence of (1.2) we observe that (s) has no zeros
for > 1, since a convergent infinite product of non-zero factors is
non-zero.
We shall now obtain a few formulae involving (s) which will be of
use in the sequel. We first have
log (s) = p
log
1 1
p2
, > 1. (1.3)
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1. Elementary Properties of(s) 65
If we write (x) = px1, then
log (s) =
n=2
{(n) (n 1)} log
1 1ns
=
n=2
(n)
log
1 1
ns
log
1 1
(n + 1)s
=
n=2
(n)
n+1n
s
x(xs 1) dx
68
Hence
log (s) = s
2
(x)
x(xs 1) dx, > 1. (1.4)
It should be noted that the rearrangement of the series preceding the
above formula is permitted because
(n) n and log
1 1ns
= O(n).
A slightly different version of (1.3) would be
log (s) =
p
m
1
mpms, > 1 (1.3)
where p runs through all primes, and m through all positive integers.
Differentiating (1.3), we get
(s)
(s)=
p
ps log p1 ps =
p,m
logpms
,
or
(s)(s) =
n=1
(n)
ns , > 1 , (1.5)
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66 8. The Zeta Function of Riemann
where
(n) = log p, ifn is a + ve power of a prime p0, otherwise.
69
Again
1
(s)=
p
1 1
ps
=
n=1
(n)
ns, > 1 (1.6)
(1.6) where (1) = 1, (n) = (1)k if n is the product of k differentprimes, (n) = 0 ifn contains a factor to a power higher than the first.
We also have
2(s) =
n=1
d(n)
ns, > 1
where d(n) denotes the number of divisors ofn, including 1 and n. For
2(s) =
m=1
1
ms
n=1
1
ns=
=1
1
s
mn=
1
More generally
k(s) =
n=1dk(n)
ns, > 1 (1.7)
where k = 2, 3, 4, . . ., and dk(n) is the number of ways of expressing n70
as a product ofkfactors.
Further
(s) (s a) =
m=1
1
ms
n=1
na
ns,
=
=1
1
s
mn=
na,
so that
(s) (s a) =
=1
2()
(1.8)
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1. Elementary Properties of(s) 67
where a() denotes the sum of the ath powers of the divisors of .
If a 0, we get, from expanding the respective terms of the Eulerproducts,
(a)(s a) =
p
1 +
1
ps+
1
p2s+
1 +
pa
ps+
p2a
p2a+
=
p
1 +
1 + pa
ps+
1 + pa + p2a
p2s+
=
p
1 +
1 p2a1 pa
1
ps+
Using (1.8) we thus get
2(n) =1 p(m1+1)a
1
1 pa1
. . .1 p(mr+1)ar
1 par(1.9)
ifn = pm11
, pm22
. . . pmrr by comparison of the coefficients of
1
ns. 71
More generally, we have
(s) (s a) (s b) (s a b)(2s a b) =
p
1 p2s+a+b(1 ps)(1 ps+a)(1 ps+b)
(1 ps+a+b)
for > max {1, Re a + 1, Re b + 1, Re(a + b) + 1}. Putting ps = z, weget the general term in the right hand side equal to
1 pa+bz2(1 z) (1 paz) (1 pbz)(1 pa+bz)
=1
(1 pa)(1 pb)
1
1 z pa
1 paz pb
1 pbz +pa+b
1 pa+bz
=1
(1 pa)(1 pb)
m=0
1 p(m+1)a p(m+1)b + p(m+1)(a+b)
zm
= 1(1 pa)(1 pb)
m=o
1 p(m+1)a 1 p(m+1)bzm
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68 8. The Zeta Function of Riemann
Hence 72
(s) (s a)(s b) (s a h)(2s a b)
=
p
m=0
1 p(m+1)a1 pa
1 p(m+1)b1 pb
1
pms
Now using (1.9) we get
(s) (s a) (s b)(s a b)(2s a b) =
n=1
a(n)b(n)
ns(1.10)
> max{1, Re a+
1, Re b+
1, Re(a+
b)+
1}Ifa = b = 0, then
{(s)}4(2s)
=
n=1
{d(n)2}ns
, > 1. (1.11)
If is real, and 0, we write i for a and i for b in (1.10), andget
2(s)(s i)(s + i)(2s)
=
n=1
|i(n)|2ns
, > 1, (1.12)
where i(n)=
d/n di
.
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Lecture 9
The Zeta Function of
Riemann (Contd.)
2 Elementary theory of Dirichlet Series [10]73
The Zeta function of Riemann is the sum-function associated with the
Dirichlet series 1
ns. We shall now study, very briefly, some of the
elementary properties of general Dirichlet series of the form
n=1
anen s, o 1 < 2 < . . .
Special cases are: n = log n; n = n, es
= x. We shall first prove a
lemma applicable to the summation of Dirichlet series.
Lemma. Let A(x) =
nxan, where an may be real or complex. Then, if
x 1, and (x) has a continuous derivative in (0, ), we have
n a
n(n) =
1
A(x)(x)dx + A()().
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70 9. The Zeta Function of Riemann (Contd.)
If A()() 0 as , then
n=1
an(n) =
1
A(x)(x)dx,
provided that either side is convergent.
Proof.
A()()
nan(n)
= n
an{()
(n)
}
=
n
n
an(x)dx
=
1
A(x)(x)dx
74
Theorem 1. If
n=1
anen s converges for s = s0
0 + ito, then it
converges uniformly in the angular region defined by
|am(s s0)| < /2, for 0 < < /2
Proof. We may suppose that so = 0. Forane
n s =
anen s0 en(ss0)
bne
n s , say,
and the new series converges at s = 0. By the above lemma, we get
+1
anen s =
1
1
anens
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2. Elementary theory of Dirichlet Series 71
=
A(x) exs
sdx + A()e s
A()e s
= s
{A(x) A()}exs dx + [A() A()]e s
We have assumed that
an converges, therefore, given > 0, there 75
exists an n0 such that for x > n0, we have
|A(x) A()| <
Hence, for such , we have
+1
anen s
|s|
exdx + e
2 |s|
+
< 2 (sec + 1),
if 0, since|s|
< sec , and this proves the theorem. As a conse-
quence of Theorem 1, we deduce
Theorem 2. If
anens converges for s = s0, then it converges for
Re s > o, and uniformly in any bounded closed domain contained in
the half-plane > o. We also have
Theorem 3. If
anen s converges for s = s0 to the sum f(s0), then
f(s) f(s0) as s s0 along any path in the region |am(s s0)| 1, and diverges for Re s 1.
If a Dirichlet series converges for some, but not all, values of s, and76
if s1 is a point of convergence while s2 is a point of divergence, then,
on account of the above theorems, we have 1 > 2. All points on the
real axis can then be divided into two classes L and U; U if theseries converges at , otherwise L. Then any point ofU lies to theright of any point ofL, and this classification defines a number o such
that the series converges for every > o and diverges for < o,
the cases = 0 begin undecided. Thus the region of convergence is a
half-plane. The line = 0 is called the line of convergence, and the
number o is called the abscissa of convergence. We have seen that 0may be . On the basis of Theorem 2 we can establish
Theorem 4. A Dirichlet series represents in its half-plane of conver-
gence a regular function of s whose successive derivatives are obtained
by termwise differentiation.
We shall now prove a theorem which implies the uniqueness of
Dirichlet series.
Theorem 5. If
ane
sn converges for s = 0, and its sum f(s) = 0 foran infinity of values of s lying in the region:
> 0, |am s| < /2,
then an = 0 for all n.
Proof. f(s) cannot have an infinite number of zeros in the neighbour-
hood of any finite point of the giver region, since it is regular there.
Hence there exists an infinity of valves sn = n + itn, say, with n+1 >
, lim n = such that f(sn) = 0.However,77
g(s) e1 sf(s) = a1 +2
ane(n1)s,
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2. Elementary theory of Dirichlet Series 73
(here we are assuming 1 > 0) converges for s = 0, and is therefore
uniformly convergent in the region given; each term of the series on theright 0, as s and hence the right hand side, as a whole, tendsto a1 as s . Thus g(s) a1 as s along any path in the givenregion. This would contradict the fact that g(sn) = 0 for an infinity of
sn , unless a1 = 0. Similarly a2 = 0, and so on. Remark. In the hypothesis it is essential that > 0, for if = 0, the
origin itself may be a limit point of the zeros of f, and a contradiction
does not result in the manner in which it is derived in the above proof.
The above arguments can be applied to
ane
n s so as to yield theexistence of an abscissa of absolute convergence , etc. In general,
0. The strip that separates from o may comprise the wholeplane, or may be vacuous or may be a half-plane.
Ex.1. 0 = 0, = 1
(1 21s)(s) =
1
1s+
1
2s+
1
3s+
2
1
2s+
1
4s+
=
1
1s 1
2s+
1
3s 1
4s+
Ex.2. (1)n
n(log n)s converges for all s but never absolutely. In the 78
simple case n = log n, we have
Theorem 6. 0 1.For if
ann
s < , then |an| n = O(1), so that |an|
ns+1+<
for > 0
While we have observed that the sum-function of a Dirichlet series
is regular in the half-plane of convergence, there is no reason to assume
that the line of convergence contains at least one singularity of the func-
tion (see Ex 1. above!). In the special case where the coefficients are
positive, however we can assert the following
Theorem 7. If an 0, then the point s = 0 is a singularity of thefunction f(s).
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74 9. The Zeta Function of Riemann (Contd.)
Proof. Since an 0, we have 0 = , and we may assume, withoutloss of generality, that 0 = 0. Then, if s = 0 is a regular point, theTaylor series for f(s) at s = 1 will have a radius of convergence > 1.
(since the circle of convergence of a power series must have at least one
singularity). Hence we can find a negative value of s for which
f(s) =
=0
(s 1)!
f()(1) =
=0
(1 s)1
n=1
anne
n
Here every term is positive, so the summations can be interchanged and
we get79
f(s) =
n=1
anen
=0
(1 s)n!
=
n=1
anesn
Hence the series converges for a negative value of s which is a contra-
diction.
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Lecture 10
The Zeta Function of
Riemann (Contd)
2 (Contd). Elementary theory of Dirichlet series80
We wish to prove a formula for the partial sum of a Dirichlet series. We
shall give two proofs, one based on an estimate of the order of the sum-
function [10], and another [14, Lec. 4] which is independent of such an
estimate. We first need the following
Lemma 1. If > 0, then
1
2i
+ii
eus
sds =
1, if u > 0
0, if u < 012
, if u = 0
where+i
1= lim
T
+iTiT
.
Proof. The case u = 0 is obvious. We shall therefore study only the
cases u 0. Now
+iTiT
eusds
s =
eus
us
+iTiT
+
+iTiT
eus
us2 ds
75
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76 10. The Zeta Function of Riemann (Contd)
However
|eu(+iT)
| = eu
and |s| > T; hence, letting T , we get+i
i
eusds
s=
+ii
eus
us2ds I, say.
We shall evaluate I on the contour suggested by the diagram. If81
u < 0 , we use the contour L + CR; and ifu > 0, we use L + CL.
Ifu < 0, we have, on CR,
eus
s2 eu
r2, since Re s > .
Ifu > 0, we have, on CL,euss2 eur2 , since Re s < .
Hence
CR,CL
eusds
us2
2reu
|u|2 0, as r .
By Cauchys theorem, however, we have
L
= CR
;
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2. (Contd). Elementary theory of Dirichlet series 77
hence
(1)1
2i
+ii
eus
us2ds = 0, ifu < 0.
If, however, u > 0, the contour L + CL encloses a pole of the second
order at the origin, and we getL
= CL
+1,
so that
(2)1
2i
+ii
eus
us2ds = 1, ifu > 0.
Remarks. We can rewrite the Lemma as: if a > 0, 82
1
2i
a+i
ai
e(n)s
sds =
0, if < n
1, if > n1
2 , if = n
Formally, therefore, we should have
1
2i
a+iai
f(s)es
sds =
1
2i
1
an
a+iai
e(n)s
sds
=
n
an
the dash denoting that if = n the last term should be halved. We wish
now to establish this on a rigorous basis.
We proceed to prove another lemma first.
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78 10. The Zeta Function of Riemann (Contd)
Lemma 2. Let anen s converge for s = real. Then
n1
ae s = o(|t|),
the estimate holding uniformly both in + > and in n. Inparticular, for n = , f(s) = o(|t|) uniformly for + > Proof. Assume that = 0, without loss of generality. Then
a = O(1) and A() A() = O(1) for all and If 1 < N < n, then
n1
ae s =N1
+
nN+1
ae s
=
N1
ae s + s
nN
{A(x) A(N)}exsdx
{A(n) A(N)}esn
= O(N) + O(1) + O
|s|
nN
exdx
= O(N) + O |s| eN
= O(N) + O|t|eN
since |s|2 = t2 + 2 and .83
Hence
n1
ae s = O(N) + O(|t|eN), if 1 < N < n
IfN n, then, trivially,n1
ae s = O(N).
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2. (Contd). Elementary theory of Dirichlet series 79
If we choose N as a function of|t|, tending to more slowly than |t|, weget
n1
ae s = O(|t|)
in either case. 84
Theorem (Perrons Formula). Let0 be the abscissa of convergence ofane
n s whose sum-function is f(s). Then for > 0 and > 0, wehave
an =1
2i
+i
i
f(s)
sesds,
where the dash denotes the fact that if = n the last term is to be
halved.
Proof. Let m < m+1, and
g(s) = es
f(s) m1
anen s
(This has a meaning since Re s > 0). Now
1
2i
+ii
g(s)
sds
=1
2i
+ii
f(s)es
sds
n
an
by the foregoing lemma. We have to show that the last expression is
zero.
Applying Cauchys theorem to the rectangle
iT1, + iT2, + iT2, iT1,
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80 10. The Zeta Function of Riemann (Contd)
where T1, T2 > 0, and > , we get
1
2i
+ii
g(s)
sds = 0,
for85
1
2i
+iT2iT1
=1
2i
iT1
iT1
+
+iT2iT1
+
+iT2+iT2
= I1 + I2 + I3, say.
Now, if we fix T1 and T2 and let
, then we see that I2
0,
for g(s) is the sum-function of 1
bn sn , where bn = am+n, n = m+n ,
with 1 > 0, and hence g(s) = o(1) as s in the angle |ams| 2
.Thus
1
2i
+iT2iT1
g(s)
sds =
1
2i
iT1
iT1
+iT2
+iT2
if the two integrals on the right converge.
Now if
h(s)
g(s)e(m+1)s = am+1 + am+2e(m+2m+1)s +
then by lemma 2, we have|h(s)| < T2
for s = + iT2, 0 + , T2 T0. Therefore for the second integralon the right we have86
+iT2+iT2
g(s)
sds
0. This proves not only that the integral converges for a finite T2,
but also that as T2 , the second integral tends to zero. Similarly forthe first integral on the right.
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2. (Contd). Elementary theory of Dirichlet series 81
N.B. An estimate for g(s) alone (instead ofh(s)) will not work!
Remarks. More generally we have for > 0 and >
n
ane
n s =1
2i
+ii
f(s)
s s e(ss)ds,
for
anenS en(ss) =
ane
n s = f(s),
and writing s = s s and bn = anen s , we have f(s) f(s) =bne
n s .We shall now give an alternative proof of Persons formula without
using the order of f(s) as s [14, Lec. 4]
Aliter. If f(s) =1
anen s has 0 as its abscissa of convergence,
and ifa > 0, a > 0, then for n, we have
n 0.By Lemma 1, as applied to the first integral on the right hand side, we
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82 10. The Zeta Function of Riemann (Contd)
get
(B) limT
1
2i
a+iTaiT
f(s)es
sds
n . For all such m the last relation holds.
Now write bn = anen , (x) = ex(s
), so that
anen s =
bn(). Write B(x) =
nxbn.
Then applying the Lemma of the 9th lecture, we get
Nm+1
anen s =
N1
m1
= (s )N
m
B(x)ex(s) dx
+ B(N)eN(s) B(m)em(s)
= {B(N) B(m)}eN(s)
+(s )
N
m
{B(x) B(m)}ex(s
)
dx.
88
Now choose > 0 and a > > 0. [If o 0 the secondcondition implies the first; if0 < 0, then since a > 0, choose 0 <
. Hence, lettingN , we get
rm(s) = (s )
m
{B(x) B(m)}ex(s
)
dx,
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2. (Contd). Elementary theory of Dirichlet series 83
or
|rm(s)| c |s
| em()
Now consider1
2i
L+CR
rm(s)es
sds
where L + CR is the contour indicated in the diagram.
rm(s) is regular in this contour, and the integral is therefore zero. 89
Hence L
=
CR
On CR, however, s =
+ rei and
a, so that
|rm(s)| c ra e
mrcos
|es| = e(+rcos ), |s| r.
Hence1
2i
CR
rm(s)es
sds
cr
2(a ) e
/2/2
e(m)rcos d
=
cr
(a ) e
/2
0
e(
m)rsin
d
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84 10. The Zeta Function of Riemann (Contd)
However, sin
2
for 0
/2; hence
1
2i
a+iTaiT
rm(s)es
sds
cre
(a )
/20
e2
(m)rd
cre
2(a )(m )r,
for all T. Hence
(C) limT
a+iT
aiT
rm(s)es
sds
c1
(m
)
for all m mo.90Now reverting to relation (A), we get
(D) limT
1
2i
a+iTaiT
f(s)es
sds 1
2i
a+iTaiT
fm(s)es
sds
= lim
T
1
2i
a+iTaiT
rm(s)es
sds
for every m. However,
limT
1
2i
a+iTaiT
fm(s)es
sds =
n
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2. (Contd). Elementary theory of Dirichlet series 85
limT
1
2i
a+iTaiT
f(s)es
sds
n mo.
Letting m we get the result. 91Remarks. (i) If = n the last term in the sum
n 0) then in the contour-integration the residue
at the origin has to be taken, and this will contribute a term - f(0).
In the proof, the relation between |s| and rhas to be modified.
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Lecture 11
The Zeta Function of
Riemann (Contd)
3 Analytic continuation of (s). First method [16,
p.18]92
We have, for > 0,
(s) =
0
xs1exdx.
Writing nx for x, and summing over n, we get
n=1
(s)
ns=
n=1
n=0
xs1enxdx
=
0
n=1
enxxs1dx, if > 1.
since
n=1
xs
1
enx
dx
n=1
0
x
1
enx
dx =
n=1
()
n < ,
87
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88 11. The Zeta Function of Riemann (Contd)
if > 1. Hence
n=1
(s)
ns=
0
ex
1 ex xs1dx =
0
xs1dxex 1
or
(s)(s) =
0
xs1
ex 1 dx, > 1
In order to continue (s) analytically all over the s-plane, consider
the complex integral
I(s) =C
zs1
e 1 dz
where C is a contour consisting of the real axis from + to , 0 < A|z|;Hence, for fixed s,
|
|z|=| 2
1
A e2|t| 0 as 0, if > 1.
Thus, on letting 0, we get, if > 1,
I(s) =
0
xs1
ex 1 dx +
0
(xe2i)s1
ex 1 dx
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3. Analytic continuation of (s). First method 89
= (s)(s) + e2is (s)(s)= (s)(s)(e2s 1).
Using the result
(s)(1 s) = sin s
we get 94
I(s) =(s)
(1 s) 2i.e2is 1
eis eis
=(s)
(1 s) 2i eis,
or
(s) = eis
(1 s)2i
C
zs
1
dzez 1 , > 1.
The integral on the right is uniformly convergent in any finite region of
the s-plane (by obvious majorization of the integrand), and so defines
an entire function. Hence the above formula, proved first for > 1,
defines (s), as a meromorphic function, all over the s-plane. This only
possible poles are the poles of(1 s), namely s = 1, 2, 3, . . .. We knowthat (s) is regular for s = 2, 3, . . . (As a matter of fact, I(s) vanishes at
these points). Hence the only possible pole is at s = 1.
Hence
I(1) =C
dz
ez 1 = 2i,
while
(1 s) = 1s 1 +
Hence the residue of(s) at s = 1 is 1.
We see in passing, since
1
ez 1 =1
z 1
2+ B1
z
2! B2
z3
4!+
that 95
(0) = 12
, (2m) = 0, (1 2m) = (1)m
Bm2m
, m = 1, 2, 3, . . .
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90 11. The Zeta Function of Riemann (Contd)
4 Functional Equation (First method) [16, p.18]
Consider the integral zs1dzez 1
taken along Cn as in the diagram.
Between C and Cn, the integrand has poles at
2i , . . . , 2ni.
The residue at
2mi is (2me2
i)s1
while the residue at 2mi is (2me3/2i)s1; taken together they amountto
(2m)ei(s1)e
i2 (s 1) + e i2 (s1)
= (2m)s1ei(s1)2cos
2(s 1)
= 2(2m)s1eis sin 2
s
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4. Functional Equation (First method) 91
Hence96
I(s) =Cn
zs
1
dzez 1 + 4i sin s2 e
is
nm=1
(2m)s1
by the theorem of residues.
Now let < 0, and n . Then, on Cn,
|zs1| = O(|z|1) = O(n1),
and1
ez 1 = O(1),
for
|ez 1|2 = |ex+iy 1|2 = |ex(cosy + i siny) 1|2= e2x 2ex cosy + 1,
which, on the vertical lines, is (ex 1)2 and, on the horizontal lines,= (ex + 1)2. (since cosy = 1 there).
Also the length of the square-path is O(n). Hence the integral round
the square 0 as n .Hence
I(s) = 4ie
is
sin s
2
m=1
(2m)s1
= 4ieis sins
2 (2)s1(1 s), if < 0.
or 97
(s)(s)(e2is 1) = (4i)(2)s1eis sin s2
(1 s)
= 2ieis(s)
(1 s)Thus
(s) = (1 s)(1 s)2ss1 sin s2
,
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92 11. The Zeta Function of Riemann (Contd)
for < 0, and hence, by analytic continuation, for all values of s (each
side is regular except for poles!).This is the functional equation.
Since
x
2
x + 1
2
=
2x1(x),
we get, on writing x = 1 s,
2s
1 s2
1 s
2
=
(1 s);
also
s
2 1 s
2 =
sin s
2Hence
(1 s) = 2s
1 s2
s
2
1 {sin s
2}1
Thus
s/2(s/2)(s) = (1s)
2
1 s
2
(1 s)
If98
(s) 12
s(s 1)s/2(s/2)(s)
1
2 s(s 1)(s),then (s) = (1 s) and (s) = (1 s). If (z) = ( 1
2+ iz), then
(z) (z).
5 Functional Equation (Second Method) [16, p.13]
Consider the lemma given in Lecture 9, and write n = n, an = 1,
(x) = xs in it. We then get
nx n
s
= s
X1
[x]
xs+1 dx +
[X]
Xs , if X 1.
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5. Functional Equation (Second Method) 93
=
s
s 1 s
(s 1)Xs1 s
X
1
x
[x]
xs+1 dx +
1
Xs1 X
[X]
Xs
Since 1Xs1 = 1/X1, and
X [X]Xs 1/X,
we deduce, on making X ,
(s) =s
s 1 s
1
x [x]xs+1
dx, if > 1
or
(s) = s
1
[x] x + 12
xs+1dx +
1
s 1 +1
2, if > 1
5.1 Since [x] x + 12
is bounded, the integral on the right hand side 99
converges for > 0, and uniformly in any finite region to the right of
= 0. Hence it represents an analytic function of s regular for > 0,
and so provides the continuation of(s) up to = 0, and s = 1 is clearly
a simple pole with residue 1.
For 0 < < 1, we have, however,
10
[x] xxs+1
dx = 1
0
xsdx =1
s 1 ,
and
s
2=
1
dx
xs+1=
1
2
Hence
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94 11. The Zeta Function of Riemann (Contd)
5.2
(s) = s
0
[x] xxs+1
dx, 0 < < 1
We have seen that (5.1) gives the analytic continuation up to = 0.
By refining the argument dealing with the integral on the right-hand side
of (5.1) we can get the continuation all over the s-plane. For, if
f(x) [x] x + 12
, f1(x) x
1
f(y)dy,
then f1(x) is also bounded, since
n+1n
f(y)dy = 0 for any integer n.
Hence100
x2x1
f(x)
xs+1dx =
f1(x)
xs+1
x2
x1
+ (s + 1)
x2x1
f1(x)
xs+2dx
0, as x1 , x2 ,if > 1.
Hence the integral in (5.1) converges for > 1.
Further
s
10
[x] x + 12
xs+1dx =
1
s 1 +1