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An Engineer’s Guide to Complex Integration

David Sirajuddin

Itcanbeshown.com

April 15, 2008



CONTENTS CONTENTS

Contents

1 Introduction 3

2 Complex Variables and Functions 3

2.1 Number Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.2 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.2.1 Complex Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.2.2 Analyticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2.3 Branch Cuts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2.4 Power Series Representation . . . . . . . . . . . . . . . . . . . . . . . 5

3 Introduction to Complex Integration 6

4 Cauchy’s Integral Theorem 7

5 Cauchy’s Integral Formula 7

6 Residue Theorem 8

6.1 Definition of a Residue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

6.2 Finding Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

6.2.1 Functional replacement . . . . . . . . . . . . . . . . . . . . . . . . . . 10

6.2.2 Quotient rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

6.2.3 Multiplication of Series . . . . . . . . . . . . . . . . . . . . . . . . . . 106.2.4 Division of Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

6.2.5 Computing Coefficients Directly . . . . . . . . . . . . . . . . . . . . . 12

7 General Approach to Solving Complex Integrals 12

8 General Approach to Solving for Real Integrals 14

8.1 Indented Paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

8.2 Paths About and Involving Branch Cuts . . . . . . . . . . . . . . . . . . . . 18

8.3 Proving Complex Integrals Tend to Zero . . . . . . . . . . . . . . . . . . . . 18

8.3.1 ML Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

8.3.2 Jordan’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

8.4 Integrals Containing Trigonometric Functions . . . . . . . . . . . . . . . . . 21

9 References 21

1



CONTENTS CONTENTS

10 Acknowledgements 22

11 Appendix 22

11.1 Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

11.2 Useful Series Representations . . . . . . . . . . . . . . . . . . . . . . . . . . 22

11.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

11.3.1 Real Integration: Functional Replacement and ML Bound . . . . . . 24

11.3.2 Real Integration: Functional Replacement and Jordan’s Lemma . . . 26

11.3.3 Real Integration: Integrals Containing Trigonometric Functions . . . 29

11.3.4 Real Integration: Strange Contours (Fresnel Integral) . . . . . . . . . 31

11.3.5 Complex Integration: Quotient Rule . . . . . . . . . . . . . . . . . . 35

11.3.6 Finding Residues: Multiplication of Series . . . . . . . . . . . . . . . 36

11.3.7 Finding Residues: Division of Series . . . . . . . . . . . . . . . . . . . 37

2



2 COMPLEX VARIABLES AND . . .

1 Introduction

Complex integration is an intuitive extension of real integration. Since a complex numberrepresents a point on a plane while a real number is a number on the real line, the analog of asingle real integral in the complex domain is always a path integral . For some special functions

and domains, the integration is path independent, but this should not be taken to be thecase in general. Given the sensitivity of the path taken for a given integral and its result,parametrization is often the most convenient way to evaluate such integrals. Of particularimportance in many physics and engineering problems is that of the closed path integral,where both the beginning and ending points of the path are identical. As with real calculus,such integrals are called contour integrals , and are denoted by a circle in the integration sign(

). This guide will discuss solution strategies for complex closed contour integrals, withparticular emphasis on their use in solving difficult real-valued integrals. The material will bepresented to the reader assuming minimal knowledge of complex mathematics. While formaldefinitions and theorems will be provided so as to make this document self-contained, proofswill not be included. The reader is recommended to see Brown and Churchill’s book (see

References ) for discussion and proofs of theorems used throughout this document. Generalstrategies of complex integration are discussed, while fully worked examples are provided inthe Appendix .

2 Complex Variables and Functions

2.1 Number Sets

At this point, it is worth explicitly delineating the fundamental differences between realand complex numbers. A number x is said to be an element of the real number set, denoted

by xR, if it represents any number on the real number line. Alternatively, a number z issaid to be a complex number, written as zC, if it is of the form z = x + iy, where {x, y}Rand the imaginary number i =

√ −1.This notation is perhaps more readable when it is informally written down in vector

form, or as an ordered pair: z = xx + iyy ≡ (x, y), where x and y are unit vectors inthe x, and y directions respectively. This representation is not conventional for complexnumbers, but it may be helpful at first to keep these notations in mind. Furthermore,these representations convey an important distinction between real and complex numbers:while real numbers denote scalar numbers on a number line, complex numbers representpoints on a complex plane . As in the case of ordered pairs, it is also possible to equivalentlyexpress complex numbers in polar form. Thus, a complex number z = x + iy = reiθ canbe fully represented in either cartesian or polar coordinates, where r is the radius measuredfrom the origin, and θ is the angle of z measured counterclockwise from the positive x-axis.Numbers on the complex plane that lie on the abscissa, are the conventional real numbers,while numbers on the ordinate axis are called imaginary numbers (owing to these numbersbeing multiples of the imaginary unit i). That is to say, real numbers are a subset of thecomplex number set (R ⊂ C). The observation that complex numbers extend a scalar to apoint on a plane immediately admits the result that one extra dimension is inherited by the

3



2 COMPLEX VARIABLES AND . . . 2.2 Functions

adoption of this number set. This extra dimension can sometimes change things considerablyin regards to complex mathematics and its real counterpart. In general, most results fromreal mathematics carry over into complex mathematics; however, the few aspects that aredifferent – are grossly so. Thus, the reader is cautioned when assuming any property fromreal numbers is translatable into complex numbers.

2.2 Functions

A complex function f of an independent complex variable z = x + iy = reiθ, where{r ≥ 0, (r,θ,x,y)R}, is of the general form:

f (z ) = u(x, y) + iv(x, y) = u(r, θ) + iv(r, θ)

the latter representation of z involving radius r and angle θ of a complex number followsfrom the transformation from cartesian to polar coordinates, and the use of Euler’s identityto replace the resulting trigonometric functions with a complex exponential. By convention,

the variable z is used to denote a complex variable. It is often convenient to group a functionin both real and imaginary parts. These parts are represented in the contained functionsu, and v, which are both real-valued, but themselves correspond to the real and imaginary

parts of the function f respectively, and both of which are functions of either x and y or rand θ.

2.2.1 Complex Differentiability

A function f (z ) has a derivative f (z 0) at a point z 0, given by the following limit providedit exists:

f (z 0) = limz→z0f (z ) − f (z 0)z − z 0

(1)

Since the complex number z 0 = x0 + iy0 can be approached from any number z = x + iy byway of an infinite number of paths, the limit is said to exist if and only if the same, uniquelimit is obtained when approached from any path that originates from any complex numberz . The criteria that a function f (x, y) = u(x, y) + iv(x, y) = u(r, θ) + iv(r, θ) must meet inorder for this to be true are known as the Cauchy-Riemann Equations . These equations canbe expressed in either cartesian or polar form.

∂u

∂x =

∂v

∂y and

∂u

∂y = −∂v

∂x Cartesian Form (2)

∂u

∂r =

1

r

∂v

∂θ and

1

r

∂u

∂θ = −∂v

∂r Polar Form (3)

Thus, while real differentiability only demands the continuity of a function at a point x0

in order for it to be said that its derivative exists at that point, a complex-valued functionis complex differentiable at a point z 0 if and only if it is both analytic and satisfies theCauchy-Riemann equations at that point.

4



2 COMPLEX VARIABLES AND . . . 2.2 Functions

2.2.2 Analyticity

Complex differentiability is a much more robust property than real differentiability. If a function is complex differentiable at a point z 0, it is also said to be analytic at z 0. Afunction satisfying these conditions inherits another consequence in that the function isinfinitely differentiable. The terms regular and holomorphic are often used interchangeablywith analytic.

2.2.3 Branch Cuts

By virtue of the polar representation of complex numbers, certain functions are inherentlymultiple-valued. This is to say, that in the sense, that there exists an infinite number of anglesθi that correspond to an angle θ0 of a complex number z 0 = r0eiθ0. This is to say that, anangle θi is equivalent to an angle θ0 if θi = θ0 + 2niπ, where n is an integer. Thus, functionscan take on the same value for different coordinates. Functions that meet this qualificationlose their analytic properties, since they are no longer one-to-one. However, what is often

done so as to restore a function’s analyticity is to define a curve of discontinuity, or branch cut . Branch cuts are often made in regards to the angle, and are usually taken to be eitherthe positive/negative real or imaginary axis. A common example is the complex logarithmfunction log z = ln r + iθ, where the value of the angle θ is restricted so that only values overan equivalent interval of 2π are permitted. Under the adoption of a branch cut, a functionwill hold distinct values for distinct radii and angles representing numbers in the complexplane and is hence one-to-one, and thereby analytic.

2.2.4 Power Series Representation

An analytic function can equivalently be expressed as a power series. It is discussed here

to provide a straightforward definition of a residue in subsequent sections, and because itcan sometimes be quicker to find residues by computing coefficients of a power series ratherthan through other methods (See Section 6.2).

The most general case is if a function f is analytic throughout an annular domain R1 <|z − z 0| < R2, centered at z 0, the function is said to be meromorphic and has the convergentpower series representation

f (z ) =∞n=0

an(z − z 0)n +∞n=1

bn(z − z 0)n

(4)

where

an = 1

2πi

C

f (z )

(z − z 0)n+1 (n = 0, 1, 2,...) (5)

bn = 1

2πi

C

f (z )

(z − z 0)1−n (n = 0, 1, 2,...) (6)

for some positively oriented simple closed contour C containing z 0 inside of the domain. Eqn.(4) is sometimes equivalently written as the doubly infinite sum

5



3 INTRODUCTION TO COMPLEX . . .

f (z ) =∞

n=−∞

cn(z − z 0)n (R1 < |z − z 0| < R2) (7)

where the coefficient cn is given by

cn = 12πi

C

f (z )(z − z 0)n+1

(n = 0, ±1, ±2,...) (8)

Eqns. (4) and (7) are Laurent series representations, and can be used to express any functionthat is analytic inside an annular region.

Alternatively, it is important to note that if the function f is analytic throughout thedisk , |z −z 0| < R2, then the bn coefficients given in Eqn. (4) are all zero by Cauchy’s integraltheorem since the function is analytic interior to and on the contour C . Thus, if a functionis analytic throughout a disk, the power series representation only holds nonzero coefficientsfor the an terms, which are then seen to be equivalent to the Taylor series coefficients:

an = 12πi

C

f (z )(z − z 0)n+1

= f (n)(z 0)n!

(n = 0, 1, 2,...)

It is noted that only negative powers of z − z 0 are found in functions that hold singularitiesinterior to an otherwise analytic disk. In this case, a punctured disk , or annular domain,must be defined that does not contain the singularities, and where the function is complexdifferentiable. This will be important in the discussion of residues .

For completeness, different series representations will now be discussed and summarized.A function that is analytic inside a disk with the exception of a finite number of singularitiescan be made meromorphic if these points are omitted in the construction of an appropriateannular domain. A function that is analytic in such a domain is given by a power series

representation involving both positive and negative powers of z − z 0, called a Laurent series.A function that is analytic inside a disk |z − z 0| < R2, centered at z 0 is given by a Taylorseries, which includes only positive powers of z − z 0. Finally, a function that is analyticinside a disk |z | < R2, centered at z 0 = 0, is given by its Maclaurin series .

* By the way, power series can be integrated or differentiated term by term, or in generalizedsummation notation.

3 Introduction to Complex Integration

A function is not integrable, in a complex sense, unless it is analytic at all points on thecontour along which it is to be integrated. This is similar to a real function only being inte-grable if it is continuous on the interval of integration. However, since in order for a functionto be analytic it must be both continuous and satisfy the Cauchy-Riemann equations, it isevident that complex-valued functions require much more to be integrable than real-valuedfunctions. It will further be seen in subsequent sections that while the integrability of acomplex function is only dependent on its analyticity at all points along the path of inte-gration, the value of the integral is dependent on the function’s analyticity for all points in

6



5 CAUCHY’S INTEGRAL FORMULA

the domain interior to the closed path of integration. As previously stated, the integral of a complex function f (z ) is always a path integral, since f (z ) → f (x, y) is a function of twovariables. While integrals of a real function can have physical interpretations, such as thearea under a curve, or the volume prescribed an area, no such direct explanations exists forcomplex integrals. The only integrals discussed in this guide will be closed-contour integrals.

The integral of a function f (z ) over a simple closed contour C is denoted as C

f (z )dz

where C is often taken to be circular, centered about a point z 0, with the convenientparametrization z (t) = z 0 + Reit(a ≤ t ≤ b) for some nonnegative radius R. The con-tour C has a direction, or sense , of integration with the convention of counterclockwisebeing positive and clockwise paths as negative. If the contour is depicted on axes, its senseis indicated with a directed line segment, i.e. an arrow. Sometimes the sense is indicatedin the integral symbol (

) with the arrow being drawn in the circle. These integrals will

be solved via residue theorem (Section 6 ) or parametrization (Section 8 ); however, beforea discussion of residue theorem is presented, two important theorems must first be cited inorder to build its theoretical foundation. The two theorems are presented in the subsequentsections, followed by their generalization into the so-called residue theorem.

4 Cauchy’s Integral Theorem

Theorem. If a function is analytic at all points interior to and on a simple closed contourC , then

C f (z )dz = 0. (9)

This same result is present in real calculus. Whenever an integrable function f (x) is inte-grated over a closed path, the value of the integral is zero. This is also true in complexmathematics provided that, in addition to a complex-valued function being analytic at allpoints on its contour, it must also be analytic for all points interior to it. That is to say,this single result should be expected in some sense; however, the theorem presented in thefollowing section does not necessarily follow from intuition, and exhibits the first sign of astrong separation between real and complex calculus.

5 Cauchy’s Integral FormulaTheorem. Let a function f be analytic for all points interior to and on a simple,

positively oriented, closed contour C . If z 0 is any point interior to C , then C

f (z )

z − z 0dz = 2πif (z 0). (10)

This result is curious, and the reader is redirected to the proof published in the Derivations

7



6 RESIDUE THEOREM

section of Itcanbeshown.com for further explanation if desired. The statement in the theoremimplies that the value of a complex integral is dependent not only on the path of integration,but also all points interior to its closed contour.

6 Residue TheoremResidue theorem is a generalization of both Cauchy’s integral theorem and formula to the

scenario when there exists a single, or multiple, number of singularities within a prescribeddomain of interest. From Cauchy’s integral theorem formula, and a function’s Laurent series,you can probably postulate this theorem on your own, to verify your own intuition kindlycheck with the theorem, stated in full, below:

Theorem. Let C be a simple closed contour, describe in the positive sense. If a functionf is analytic inside and on C except for a finite number of singular points z k(k = 1, 2, . . . , n)inside C , then

C

f (z )dz = 2πin

k=1

Resz=zk

f (z ) (11)

In words, the integral of a function f over any contour containing the singular points z k isequal to the product of 2πi and the sum of the residues of all the singular points interior toC . Note the simplicity of the result, by invoking this theorem there is no need for any kindof tricky, or resourceful, integration. All one need do is compute all the residues of a functionf (z ) at all of the contained singularities within the prescribed contour of integration, andmultiple this sum by 2πi! Thus, the real work is only in finding these residues, whateverthose are. A precise definition of a residue follows.

6.1 Definition of a Residue

Recall from Section 2.2.3 that a function that is analytic at every point inside a disk|z − z 0| < R2 with the exception of a finite number of singular points can be made holo-morphic inside a redefined domain that omits these singularities. This new domain can becircumscribed to be an annulus with inner and outer radii R1 and R2 respectively. That isto say, a function f that is analytic interior to an annular domain R1 < |z − z 0| < R2 it canbe equivalently expressed as a Laurent series

f (z ) =

∞

n=0

an(z − z 0)n

+

b1

z − z 0 +

b2

(z − z 0)2 + . . . +

bn

(z − z 0)n + . . .

where the coefficients an and bn are defined as in Eqns. (5) and (6). In order to integratef (z ), both sides of the above equation are integrated. It follows from Cauchy’s integralformula that all of the an and bn coefficients will be zero, except the coefficient b1. That isto say,

an = 0 (n = 1, 2, . . .)

8



6 RESIDUE THEOREM 6.2 Finding Residues

bn = 0 (n = 1)

Letting n = 1 in the definition of bn (Eqn. (6)), it is found to be identical to the form of Cauchy’s integral formula which enables the following to be written down

C

f (z )dz = 2πib1

Thus, the only contribution to the integral of the function f (z ) along the simple closedcontour C that encloses a finite number of singular points comes from the b1 coefficient of thepower series expansion of the function for each contour centered at each of the singularities.The coefficient b1 is called a residue . A residue is the b1 coefficient of a function centered aboutthat singular point. In general, in order to extract the residue of a function directly from apower series representation, one must first express the expansion centered about that point.For example, in the form of Cauchy’s integral formula in Eqn. (10), the residue correspondsto that a point z 0. There exist residues for each contained singular point z k(k = 1, 2, . . .)

inside of a prescribed contour C of a function f . The notation for a residue at z = z k for afunction f (z ) is given by,

Resz=zk

f (z )

It follows from intuition that the expansion must be written down in an expansion of theform z − z k, as the integral containing a singular point z k is independent of the radius size of the contour. The derivation of Cauchy’s integral formula, and the subsequent generalizationfor residue theorem, follows from centering a contour at each singularity, and letting theseradii tend to zero so as to approach the constant b1 as expressed in Cauchy’s integral formula.

6.2 Finding Residues

There are several methods to finding residues, and I have – of course – not described themall below. I have listed some of the methods I have found to be the most useful. It should benoted that the following names are not used widely throughout complex mathematics, butI have given them names for identification purposes rather than leave them all as numberedtheorems. As a general strategy, I suggest using methods described in Sections 6.2.1 and6.2.2 if possible, as they will allow for the quickest extraction of residues with the leastamount of work. If these methods are not permitted, please proceed to directly manipulatingpower series expansions of the function (Sections 6.2.3 and 6.2.4 ). And, only as a lastresort, use the definition of the coefficients (Section 6.2.5 ), as this method will be most timeconsuming (and confusing).

As previously mentioned, if the method of functional replacement is not permitted, andthe function cannot be expressed as a quotient, the next method demanding the least amountof work is direct manipulations of the power series, this can be done by using basic arithmeticof known series expansions of the constituent terms of a function, in order to properly combinethem so as to obtain the power series expansion for the function as a whole. Since addition of subtraction is straightforward, it will not be covered. However, multiplication and division

9




is worth presenting to eliminate any potential curiosities about their workings. Techniquesfor finding residues are now presented.

6.2.1 Functional replacement

Theorem. An isolated singular point z 0 of a function f is a pole of order m if and onlyif f (z ) can be written in the form

f (z ) = φ(z )

(z − z 0)m,

where φ(z ) is analytic and nonzero at z 0. Moreover,

Resz=z0

f (z ) = φ(z 0) if m = 1 (12)

and,

Resz=z0

f (z ) = φ(m−1)

(z 0)(m − 1)!

if m ≥ 2 (13)

where φ(z )(m−1) denotes the (m − 1)th derivative of φ(z ).

6.2.2 Quotient rule

Theorem. Suppose the functions p and q be analytic at z 0. If

p(z 0) = 0, q (z 0) = 0, and q (z 0) = 0,

then z 0 is a simple pole of the quotient p(z )/q (z ) and

Resz=z0

p(z )

q (z ) =

p(z 0)

q (z 0) (14)

A simple pole is defined in the above section, where m = 1 (i.e. a singular point of orderone).

6.2.3 Multiplication of Series

Sometimes one may not know the power series of a particular function, but often thefunction can be broken up so that one can locate in literature, or easily derive, series ex-pansions of each part. If a function, or part of a function, is able to be separated into a

product then the combined series involving both terms can be found by simply multiplyingboth series term-by-term.

Let a function f (z ) be given by its series expansion about point z = z k,

f (z ) =∞n=0

an(z − z k)n

10




and assume that it can be separated such that f (z ) = G(z )H (z ). Both functions G and H have the series representations centered about z k:

G(z ) = g1 + g2 + g3 + · · · + gn + . . .

H (z ) = h1 + h2 + h3 + . . . + hn + . . .

where the terms gn and hn contain both their corresponding coefficients and powers of z − z k. Note that the coefficient of the product gnhn is equal to a2n. Since the series formsare centered about z = z k, if there exists a residue in the combined series it will correspondto that at z = z k (if the residue about a different point is desired, try to obtain an expansioncentered about this point). The function f can be found by multiplying the known series,

f (z ) = G(z )H (z )

= (g1 + g2 + g3 + . . . )(h1 + h2 + h3 + . . .)

f (z ) = g1h1 + g1h2 + g1h3 + g2h1 + . . .Furthermore, suppose that some term gih j in the combined series contains the dependence(z − z k)−1, for the appropriate i and j. Then, the residue at z = z k is the coefficient of this term. These series may be long, or even infinite, and in order to find any residues itis important to notice that it is not necessary to multiply even most of the terms out, butonly those that matter. The reader is strongly cautioned to make sure they know whichterms matter, so as to not inadvertently, and detrimentally, neglect relevant terms. Lastly,a specific coefficient an can be found by Leibniz’s rule

an =n

r=0 n

rG(r)(z )H (n−r)(z ), (15)

where G(r) and H (n−r) denote the rth and (n − r)th derivatives of G and H respectively.This result follows from the combined series G(z )H (z ) = f (z ) actually being a Taylor seriesexpansion.

6.2.4 Division of Series

If a function, or part of a function, f is able to expressed in terms of f (z ) = A(z )/B(z ),and these constituent functions have the series representations centered about some numberz k:

A(z ) = a1 + a2 + a3 + · · · + an + . . .

B(z ) = b1 + b2 + b3 + . . . + bn + . . .

where the terms an and bn contain both their corresponding coefficients and powers of z −z k.The power series for the function f can be found by dividing the series representations of Aand B by way of conventional polynomial division. For the sake of example, let the functionshave the following series expansions, centered about the point z k = 0:

11



7 GENERAL APPROACH TO . . .

A(z ) = α−1

z + α1z + α3z 3 + . . .

B(z ) = β 0 + β 1z + β 2z 2 + . . .

where the terms αn and β n represent the numerical constants of the series. Then, the series

of f (z ) = A(z )/B(z ) can be found by division

α−1

β 0z −1 − α−1

β 1β 20

+ . . .

β 0 + β 1z + β 2z 2 + . . .

α−1

z + α1z + α3z 3 + . . .

α−1

z + α−1β −1

β 0+ α−1

β 2β 0

z + . . .

− α−1β −1

β 0+

α1 − α−1β 2β 0

z + . . .

− α−1β −1

β 0− α−1

β 21

β 20

z + . . .

α1 + α−1

β 0 β 21

β 0

−β 2 z + . . .

Notice how if the goal is to find a residue that the computation of only a few terms arerequired, and how the actual work could have stopped after finding only the first term of the quotient. This is because the first term contains the dependence z −1, and its coefficient(α−1/β 0) is exactly that of the function’s residue at z = 0.

6.2.5 Computing Coefficients Directly

I have never done this in my life, but on the other hand – by the date of this documentbeing written I have also yet to enter graduate school so I may be speaking too soon. Themethod is to solve directly for the coefficients, if possible, by way of Eqn. (6). I do not

recommend anyone ever do this, and in most cases – you will not be able to. But, if everything else fails: go ahead.

7 General Approach to Solving Complex Integrals

With all of the necessary information described, solutions to complex integrals are nowable to solved. While the solution strategy for using complex integrals to solve for real in-tegrals is a bit more involved (See Section 8 ), the methodology of solving complex integralsis straightforward. Little resourcefulness is demanded, and depending on where one encoun-ters a particular complex integral, the functions to be integrated as well as their paths of

integration usually have little flexibility and are – in this sense – clearly defined. For clarity,a representative function f (z ) is taken to be integrated over the simple closed contour C ,with singular points located at z 1 and z 2. It is enforced that the f is analytic at all pointsalong the path of integration. The solution to the following integral will be described so asto make the following discussion more tractable.

C

f (z )dz

12




Where the contour C and the singular points of f are shown in Figure 1.

Figure 1 - A semicircular contour C is chosen in order to integrate the function f (z). The contourencloses the singularity at the point z1 only, while the singularity at z2 is exterior to the contour. The

residue at z2, therefore, does not contribute to the value of the integral in the system drawn above.

A systematic approach to solving complex-valued integrals is enumerated below:

1. Locate any singular points : Begin by examining the function f and locate the relevantsingular points. The only singularities that contribute to the result will be those thatlie interior to the closed contour. Furthermore, in order to determine whether thefunction is integrable, these singularities must not lie on the contour of integration.If any singular points are present on the contour, then the function is not integrablealong this contour and no solution can be obtained. In the case of the example, onlythe singularity z 1 need be taken into account since z 2 is outside of the closed contour.

Note that if both points z 1 and z 2 were outside of the contour, the contour wouldenclose no singularities, and the integral would then be identical to zero by Cauchy’sintegral theorem (Section 4 ).

2. Integrate : If the function contains singularities interior to its closed contour of integra-tion, then the integration can be accomplished in one of two ways, by the application of residue theorem or by parameterizing the contour. It is recommended that whichevermethod demands the least amount of work, and whichever is more practical, be used.If the contour is too strange, then parametrization of the contour can become toocomplicated to make this method practical. Both strategies are outlined below.

(a) Residue Theorem – Begin by writing down the statement of residue theorem. Inthe case of the example, this becomes

C

f (z )dz = 2πi Resz=z0

f (z )

Next, find the residue(s) by the methods described in Section 6.2 , insert the resultsinto the above equation, and the integral is solved.

13




(b) Parametrization – If the simple closed contour is circular, then apply the parametriza-tion z (t) = z 0 + Reit(0 ≤ t ≤ 2π), where R is the radius of the contour, and z 0 isits center. Differentiation of z (t) with respect to t then admits the substitutiondz = iReitdt. Insert these terms into the integral and proceed with conventionalintegration to obtain a solution. In the example, this process yields the following

C

f (z )dz = iR

2π0

f (z 0 + Reit)eitdt

.

After integration of the above equation, the solution has been arrived at.

These strategies can be applied to any complex integral, and usually necessitate little inte-gration, in the conventional sense. The most difficulty will arise in finding the residues of the function contained in the integrand (as parametrization is usually not the easiest way tosolve most integrals of this kind).

8 General Approach to Solving for Real Integrals

The method to be outlined will provide a robust strategy in solving difficult, semi-infiniteand infinite, real integrals. It begins with a real-valued function of a single variable that isto be integrated. To be explicit, a real function f (x) is taken as an example to be integratedover the bounds x(−∞, ∞). The transposed function f (z ) in the complex domain is takento be analytic at all points in the complex plane except at both z 0 and z 1 located in the half plane y ≥ 0. A walkthrough using this function as an example will be given so as to makethe following discussion more understandable.

In thinking about this method, it is important to realize that the aim is to integrate thecomplex version of a function in the complex domain (which is obtained by substituting thesingle real variable with a complex variable z , i.e. f (x) → f (z )), then by way of residuetheorem and parametrization, the solution to the integral of the real function is extracted.Furthermore, close attention should be payed to these steps, especially in the proper diagnosisof the specific scenario an integral may present by way of identifying the singular points of a function. In this respect, while there are no special cases for the integration of complexfunctions, there are many circumstances that need to be taken into account when attemptingto extract the values of real integrals from complex integrals. The steps outlining the methodwill forewarn the reader of any such pitfalls, and solutions to these problems are presentedin the subsequent subsections. The method proceeds as the following:

1. Go complex : transpose all single variables of the same kind to a complex variable (e.g.f (x) → f (z ))

2. Locate any singular points : For the complex-valued function f (z ), find its poles and/orsingularities. These points are located where the function becomes boundless, that isto say, where the denominator equals zero or if the function contains a logarithm thispoint will be for z = 0. This is an important preliminary step, as the location of thesingular points will diagnose which situation is present and indicate the strategy that

14




should be used to go about evaluating the integral. The singular points play a role inthe choice of contour for the complex integral. The contour chosen (step 3) will alwaysinvolve the real line since it is desired to obtain the result of a real integral, rather thana complex integral. There are two distinct categories of paths. An indented path is onein which at least one singular point is located on the real line, and it is evident that in

this scenario a contour involving the real line will intersect this point, but is not ableto be integrated directly through this point due to the function failing to be analytic.In this case, a different form of residue theorem must be incorporated in addition tothe standard theorem (Section 8.1 ). The steps in this section should still be followed,but it changes step x, which needs to be accounted for in order to obtain the correctresult. Alternatively, if the singular points are all either above or below the real line,the path is said to be unindented . Proceed with the following steps in this section, anduse the conventional form of residue theorem. The example function used throughoutthis section falls under this category. Finally, be mindful of branch cuts that must bemade for reasons of preserving analyticity (see Glossary ). It is not possible to integrateacross a branch cut, but it is possible to integrate around the branch cut.

3. Choose an appropriate contour with symbolic bounds : In choosing a proper contour, itis important to think before you draw. Since a real integral is aimed to be extractedfrom a complex closed contour integral, at the very least choose a contour that containsa segment on the real line. At this point, it is only necessary to draw a contourthat will allow for a parametrization of the complex function f (z ) on the real lineportion of the contour to be f (x) (I recommend you draw the singular points and yourproposed contour on axes for clarity). For now, do not worry about the actual boundsof integration of the real function, let them be symbolic. That is, denote the limitson the real line as in the above example as −R and R. An illustration of a choice of contour along with the functions singular points z

0 and z

1 is shown in Figure 2.

Figure 2 - A semicircular contour C is chosen in order to integrate the function f (z). Note howthe contour encloses two singularities at z0 and z1. These points will contribute to the value of the

contour integral C f (z)dz by way of residue theorem.

The contour was chosen to contain the singularities located at z 0 and z 1, and a simplesemicircle (y ≥ 0) does the trick since: (1) it encloses the singular points, and (2)

15




involves the real axis. Often it is in your best interest to choose a contour that enclosesthe least amount of singularities (as this will necessitate the least amount of residuecomputations). Bear in mind, however, that limits will be taken (step 5), and a contourshould be chosen such that it encloses all singularities before such limits are taken. Asthese singular points will contribute to the value of the real integral before, or after,

limits are taken. That is to say, whether you decide count them or not. Contourscontaining no singularities should generally be avoided, as this will admit the trivialresult of the complex integral being equal to zero by consequence of Cauchy’s integraltheorem (e.g. a semicircle in the region y ≤ 0). It is important to realize that in thisinstance by breaking up the complex integral into a sum of path integrals (where thesum of the paths are equal to the original contour), the problem can still be solved;however, doing this often makes the real integral result more difficult to retrieve.

4. Apply residue theorem : Look at the contour you have drawn, and write down theresidue theorem statement for the closed contour integral being considered. For theexample,

C

f (z )dz = 2πi

Resz=z0

f (z ) + Resz=z1

f (z )

where C is the entire semicircular contour. Next, break up the integral into a summa-tion of a parameterized real integral and complex integrals that makeup the remainderof the contour. The parametrization of the complex function f (z ) into the real functionf (x) follows from the idea that z = x + iy → x on the real line. This is the integralto which a solution is desired. As in the example, the integral is broken up in thefollowing way

C

f (z )dz = R−R

f (x)dx + C R

f (z )dz = 2πi

Resz=z0

f (z ) + Resz=z1

f (z )

where the contour C R is the upper half of the semicircle contour (as shown in Figure2). For convenient bookkeeping, move any complex integrals to the side of the equationthat contains the residues. The result from above is then a direct statement of thevalue of the real integral,

R−R

f (x)dx = 2πi

Resz=z0

f (z ) + Resz=z1

f (z )

− C R

f (z )dz

It is then only necessary to adjust the limits (step 5), and to solve the RHS of theabove equation to find the value of the integral (steps 6 and 7).

5. Take appropriate limits : Let the bounds tend to those which the original real integralcontained. In the example, this corresponds to letting R → ∞, such that f (x) isintegrated from −∞ < x < ∞.

16



8 GENERAL APPROACH TO . . . 8.1 Indented Paths

∞−∞

f (x)dx = 2πi

Resz=z0

f (z ) + Resz=z1

f (z )

− lim

R→∞

C R

f (z )dz

6. Find the value of the complex integrals or prove they tend to zero in this limit : Deter-mine the value of these complex integrals by either parametrization, or other means(Section 8.1 ). If any integral cannot be evaluated, prove they tend to zero by usingeither an ML bound or Jordan’s lemma given the limits of the previous step ( Section

8.2 ). If you cannot prove they all tend to zero, repeat this procedure beginning withstep 3 in order to choose a different contour that may work.

7. Find the residues : Use the methods described in Section 6.2 to find the residues of thefunction to be integrated. Reserve this step as the last in order to save time. If it isnot possible to find values for all the complex integrals, a different contour must bechosen, and any residues that were computed prior to checking this would have beenfor naught.

8. Express the final answer : Depending on whether the path of integration was indentedor unindented, the result obtained will be the sum of the residues multiplied by either2πi and/or −πi. The solution to the example is

∞−∞

f (x)dx = 2πi

Resz=z0

f (z ) + Resz=z1

f (z )

Furthermore, suppose that the function f (x) is even, such that f (−x) = f (x), it isthen possible to determine the value of the semi-infinite integral by dividing the resultby two.

∞0

f (x)dx = πi

Resz=z0

f (z ) + Resz=z1

f (z )

This is allowed only for even functions, and is permitted due to symmetry. This isnot the only method to solve for semi-infinite real integrals, but it is often the mostconvenient. For an example of a semi-infinite integral solution that is achieved directlyfrom complex integration, see Section 11.3.5 , the Fresnel integral solution.

The solution to the original real integral has been found! This practice is fairly easy, with themost difficult parts being a proper choice of a contour (subsequent examples in the Appendix

will demonstrate this), and showing that the complex integrals tend to zero. Make sureto mind the scenario carefully, only a few exceptions exist (which have been outlined), butneglecting to account for any exception will cause the result to be incorrect.

8.1 Indented Paths

Theorem. Suppose that(i) a function f (z ) has a simple pole at a point z = x0 on the real axis, with a Laurent

series representation in a punctured disk 0 < |z − x0| < R;17



8 GENERAL APPROACH TO . . . 8.2 Paths About and Involving Branch Cuts

(ii) C ρ denotes the upper half of a circle |z − x0| = ρ, C R is the upper half of the circle|z − x0| = R, where ρ < R and the clockwise direction is taken,

Then

limρ→0 C ρ f (z )dz =

−πi Res

z=x0

f (z ) (16)

Notice how the clockwise direction is taken, as opposed to the counterclockwise direction asper usual. The theorem is written in this form because contours are usually convenientlychosen containing this segment in this direction, so as to enable the path C R, and the entirepath as a whole, to be counterclockwise.

Although a figure is not included in this section, the scenario should be clear. A positivelyoriented semicircular contour of radius R is desired to be used; however, its path along thereal axis contains a point z = x0 at which the function is not analytic, and therefore notintegrable. Thus, the desired contour is obtained by way of choosing paths that initiallyavoid the singular point, but which – upon taking the proper limits – collapse to the desired

contour. A common way of accomplishing this is by choosing a counterclockwise semicircularpath C R. Continuing along its counterclockwise direction, the path traverses the real line,at which point the path evades the singularity at z = x0 by indenting the path along thereal line. That is to say, at some some distance ρ from the singularity z = x0, the path alongthe real axis is rerouted through a semicircular clockwise path C ρ. At the intersection of C ρand the real axis, the path continues along the real line in the + x direction so as to close thecontour. The new contour can be described as tracing out a counterclockwise half-annulus .During the subsequent analysis, the radius ρ is allowed to tend to zero, such that the valueof the path integral at that point is known by way of this theorem. In this manner, it ispossible to integrate the function through its singular point, and hence along the entirety of the originally desired contour. Alternatively, if an improper real integral result is desired,

then the steps in Section 8 can be followed henceforth.

8.2 Paths About and Involving Branch Cuts

The idea is straightforward. Since it is not possible to integrate directly through a branchcut, it is possible to integrate around it. Furthermore, the presence of a point on the branchcut in the path of integration provides the same effect as a singular point, it cannot beintegrated directly. However, by using an indented path (Section 8.1 ) a path can be madeto incorporate such a point in a roundabout manner.

8.3 Proving Complex Integrals Tend to ZeroThis section should go without saying; however, it has been my experience that people

tend to wrecklessly skip this step, particularly when dealing with physics and engineeringproblems. This step only corresponds to the scenario when a real integral result is attemptedto be extracted by way of complex integration. As previously discussed, the complex contourintegral must be broken up into a sum of path integrals that are equivalent to the contourintegral, and in which one path is along the real line such that it corresponds to the desiredreal integral. It is necessary to obtain values for each of the complex contributions. If a

18



8 GENERAL APPROACH TO . . . 8.3 Proving Complex Integrals Tend to Zero

result for any of these integrals is unable to be determined, the integral must be shown totend to zero when taking proper limits so as to validly neglect the term. This is to say, if it cannot directly be shown that a complex integral is identical to zero, the actual answerto the real integral could involve other contributions that have not been accounted for. Thefailure to prove this may hint that a different contour should be chosen. For most purposes,

a conventional ML limit will show that the integrands will tend to zero. If this fails, and theintegral is able to factored in a form equivalent to Jordan’s lemma, it then can be employedto demonstrate the result is zero. For most purposes, these two lemmas should be all thatis needed to adequately prove this.

8.3.1 ML Limit

Let C denote a contour represented by the parametrization z = z (t)(a ≤ t ≤ b), it thenfollows that the integration of a function f can be expressed as

C f (z )dz = b

a

f [z (t)]z (t)dt

In order to obtain a maximum value, a modulus is taken

C

f (z )dz

=

ba

f [z (t)]z (t)dt

=

ba

|f [z (t)]||z (t)|dt (17)

the function f is taken to be bounded on the contour C , such that |f (z )| ≤ M , for anynonnegative constant M . By taking the maximum value of f on the contour, and factoringit out of the integrand, Eqn. (17) becomes

C f (z )dz = M b

a |z (t)

|dt

The integral on the RHS is identified as the arc length L of the contour, thus it can bewritten as

C

f (z )dz

≤ ML (18)

And, an upper bound to a function along any contour has been reached! Since a complexvalue function is dependent on z = x + iy, it is evident that such functions are mappedinto R3, making plotting difficult. It is easier to understand the concept of an ML boundby downsizing by one dimension, and considering a real function f (x). The ML bound is

depicted in Figure 3.

19



8 GENERAL APPROACH TO . . . 8.3 Proving Complex Integrals Tend to Zero

Figure 3 - A real-valued function of one variable f (x) is plotted. In the figure, the maximum value M

which occurs at a point x = x0 is labelled along with length L of the “contour.” The product ML (i.e. thearea), has been shaded. As evident from the figure, the integration of f over the interval must be less than

the maximum value M of the function multiplied by the length of the contour C .

In this scenario, the maximum value M , occurring at x = x0, is labelled alongside the length

of the “contour” C . The contour in this single real variable case is the interval 0 ≤ x ≤ x1.The value of the integral x10

f (x)dx is the area under the curve, and it is evident that thevalue of this integral must always be less than or equal to the product of maximum value M of the function and the length L of the contour (shown as the shaded region above). Whilesome of the details in the derivation are not the same in the complex case, and it is not

justifiable to transpose this result from real to complex calculus, the result and idea remainsthe same in both real and complex mathematics.

This limit will prove handy when certain integrals must be shown to tend to zero whenapplying residue theorem. This can be shown if the upper bound ML = 0 for a complexintegral. Since a modulus is always a positive quantity, the inequality must then actually bean equality, such that the integral is identical to zero.

8.3.2 Jordan’s Lemma

Theorem. Suppose that(i) a function f (z ) is analytic at all points z in the upper half plane y ≥ 0 that are

exterior to a circle |z | = R0;(ii) C R denotes a semicircle z = Reiθ(0 ≤ θ ≤ π), where R > R0;(iii) for all points z on C R there is a positive constant M R such that |f (z )| ≤ M R, where

limR→∞

M R = 0

Then, for every positive constant a,

limR→∞

C R

f (z )eiazdz = 0 (19)

This to say, if an appropriate function g(z ) can be factored into a form of Eqn. (19), wherethe constituent function f (z ) meets the above criterion, then the integral is equal to zero.

20



9 REFERENCES 8.4 Integrals Containing Trigonometric . . .

8.4 Integrals Containing Trigonometric Functions

It is demonstrated in the examples contained in Section 11.3 that a convenient transposi-tion of a real-valued cosine or sine function to the complex domain is the complex exponentialfunction eiaz, for some constant a. As Euler would say, the reason for this is “immediatelyobvious,” and can be seen by examining his famous identity,

eix = cos x + i sin x

thus,

Re

eix

= cos x and Im

eix

= sin x

where the above operators denote taking the real (Re[ ]) and imaginary (Im[ ]) parts of thefunction. However, the extraction of either a cosine or a sine is only convenient so long as thecomplex exponential is not complicatedly embedded within the function. One of the morecommon scenarios wherein this arises is the case when trigonometric function(s) are involved

in sums and/or differences in the denominator. In these instances, it is not possible to easilytake the real or imaginary part of the complex version of the function so as to isolate eitherthe cosine or sine function. Thus, these integrals may prove difficult in moving to complexspace in order to integrate, and ultimately obtain the desired real result. However, theseintegrals can be solved under special circumstances provided that the integral is taken to thecomplex domain by slightly different means.

If the limits of integration encompass all 2π radians of a circle, then the real-valuedlimits can be taken into the complex domain by way of introducing a circular contour. Thisis accomplished by introducing the following parametrization:

z = eiθ (0

≤θ

≤2π)

and,

dz = ieiθdθ = izdθ → dθ = dz

iz

Then, by using the identities relating the sine and cosine functions to z :

cos θ = z + z −1

2 and sin θ =

z − z −1

2i

the integral can be taken into the complex domain and solved with residue theorem. Thisresult will be equivalent to the original value of the real integral.

9 References

1. Barrett, David E. Math 555: Complex Variables and Applications Class Notes . Septem-ber 2007 - December 2007.

2. Brown, James Ward. Churchill, Ruel V. Complex Variables and Applications, Seventh

Edition . McGraw-Hill Companies, Inc. New York, NY. 2004.21



11 APPENDIX

10 Acknowledgements

I would like to thank Benjamin Tong Yee, who at the time of this writing is attendingthe University of Michigan for his graduate degree in Nuclear Engineering & RadiologicalSciences, for his suggestion on how to typeset the Residue notation in LATEX. Furthermore, I

found it silly to scour the internet and/or paraphrase theorems. Thus, many of the theorems,and definitions cited in this guide are taken verbatim from Chuchill’s Complex Variables and

Applications , and are the sole property of the authors. This was the book from which I stud-ied complex mathematics of a single variable, and I wish to thank the authors for a splendidtext. Finally, a special thanks to my Math 555 professor, Dr. David E. Barrett, at theUniversity of Michigan for making the course so accessible to an engineering, and not math-ematics, student like myself as well as for his excellent instruction. I have attempted to dofor all of you what he did for me. This guide, under no circumstances, is to be used for profit,especially since many of the theorems are attributed entirely to the authors. Please do nothesitate to send any enquiries, corrections, and/or concerns to me at [email protected].

11 Appendix

11.1 Glossary

1. Analytic: a function is analytic at a point z 0 if it is both continuous, and satisfies theCauchy-Riemann equations at z 0.

2. Branch cut: Typically this is a restriction that is applied on the angular value θ of a multiple-valued function so as to enforce that it is single-valued (e.g. the complexlogarithm function), and inherits analytic properties. The cut can be either an infinite

ray (constant angle), or a curve. The only requirement is that the function along thecut is discontinuous such that it preserves its single-valuedness.

3. Complex differentiable: a function is complex differentiable if it is analytic. A complexdifferentiable function is also infinitely differentiable.

4. Holomorphic: See analytic , strictly speaking these two terms are not entirely synony-mous, although the small difference between them is of no consequence for applicationsof the type described in this guide.

5. Meromorphic: A function that is analytic inside an annular domain.

11.2 Useful Series Representations

The following series representations are useful in invoking the methods described in Sec-

tions 6.2.3 and 6.2.4.

1. 11−z

= ∞

n=0 z n (|z | < 1)

2. ez = ∞

n=0zn

n! (|z | < ∞)

22



11 APPENDIX 11.2 Useful Series Representations

3. sin z = ∞

n=0(−1)n z2n+1

(2n+1)! (|z | < ∞)

4. cos z = ∞

n=0(−1)n z2n

(2n)! (|z | < ∞)

5. sinh z = ∞n=0

z2n+1

(2n+1)! (|z | < ∞)

6. cosh z = ∞

n=0z2n

(2n)! (|z | < ∞)

These series can be obtained from each other. For instance, the cos z series is yielded fromdifferentiation of the sin z series, and that the cosh z expansion can be found via the relationcosh z = cos(iz ). One can find series for tan z by polynomial division of the series expansionsfor sin z/ cos z , etc. Of the above, one of the more useful expressions is series (1), which canbe used to find expressions for any form of the reciprocal of any difference or sum involvinga constant and any power of z . This is accomplished by substitution of z for an appropriateform to find the desired expansion.

For example, in order to obtain the series representation for 1/(4 + z 2), begin with series

(1)

1

1 − w =

∞n=0

wn (|w| < 1)

Enforcing the substitution w = −z 2/4, the above equation becomes

1

1 + (z 2/4) =

∞n=0

z 2n

4n (|z | < 2)

Multiplying and dividing both sides of the equation by a form of unity (i.e. 4 /4)

4 1

4 + z 2 =

∞n=0

4

4

z 2n

4n

1

4 + z 2 =

∞n=0

z 2n

4n+1 (|z | < 1)

And, the desired expression has been obtained.

*N.B. Something that was not mentioned in this guide was the idea of the radius of conver-

gence for power series expansions. This is to say that a function f (z ) has a valid convergentpower series representation only within a certain radius |z |. The radius of convergence foreach of the above series has been noted parenthetically, and must be payed with close at-tention. It is important when finding series expansions to mind the radius of convergence,elsewise the series expression arrived at will not be correct. Luckily, this is of little setback,as all that is needed to do is to express the function in a fashion such that for all points

23



11 APPENDIX 11.3 Examples

z of a considered domain, the expanded term holds values that are within the radius of convergence. This is easily accomplished by factorization.

11.3 Examples

11.3.1 Real Integration: Functional Replacement and ML Bound

The real integral ∞0

dx

2x2 + 8 =

1

2

∞0

dx

x2 + 4

can be evaluated by first transposing the function f (x) contained in the integrand into acomplex function f (z ),

f (z ) = 1

z 2 + 4 =

1

(z + 2i)(z − 2i)

The singularities at z = ±2i are evident from the factorization on the RHS. A contour C is chosen to be a semicircle in the upper-half plane centered at the origin with a radiusR chosen to be large enough to contain the singularity z = 2i. The semicircular arc thatexcepts the real line is denoted by C R. The factor of (1/2) in the problem statement will beneglected for now, and incorporated at the end. Residue theorem implies

C

dz

z 2 + 4 =

R−R

dx

x2 + 4 +

C R

dz

z 2 + 4 = 2πi Res

z=2if (z )

Thus, the real integral can be expressed as

R−R

dx

x2 + 4 = 2πiResz=2i f (z ) − C R

dz

z 2 + 4

Letting R tend to infinity ∞−∞

dx

x2 + 4 = 2πiRes

z=2if (z ) − lim

R→∞

C R

dz

z 2 + 4 (20)

An expression similar to the original integral is arrived at. Next, prove the the complexintegral on the RHS is equal to zero by way of an ML bound. That is to say, find theupperbound of the integral by way of the following inequality

C R

dz

z 2 + 4 ≤ ML

where M is the maximum value of the function on the contour, and L is the length of thecontour. Beginning with the representation of R before the limits are applied, by inspection,the length of the contour L = πR. Furthermore, the maximum value of the function isobtained in the following manner:

24




1

z 2 + 4

= 1

|z 2 + 4|=

1

|z |2

+ 4=

1

R2 + 4= M

Then the product M L, C R

dz

z 2 + 4

≤ ML = πR

R2 + 4

As R → ∞, the M L bound becomes

ML = limR→∞

πR

R2 + 4

= limR→∞

π/R

1 + 4/R2

= 0

Since the upper bound of the integral is equal to zero, then the integral itself must be zero, C R

dz

z 2 + 4 = 0

The only term left to compute is residue in Eqn. (20). This is done by functional replacementof the complex function f (z )

f (z ) = 1

(z + 2i)(z − 2i) =

φ(z )

z − 2i

where

φ(z ) = 1

z + 2i

The residue at z = 2i is then equivalent to φ(2i)

Resz=2i

f (z ) = φ(2i)

= 1

(2i) + 2i

= 1

4i

25




Inserting these results into Eqn. (20)

∞−∞

dx

x2 + 4 =

2π i

4 i

= π

2

Notice that the function in the integrand of the real integral is even (f (−x) = f (x)), thisallows one to obtain the result for the semi-infinite integral by dividing the result by a factorof 2, that is to say ∞

0

dx

x2 + 4 =

π

4

Finally, incorporating the factor of (1/2) that was contained in the original problem state-ment on the RHS, the above equation is multiplied by (1/2)

1

2

∞0

dx

x2 + 4 =

1

2

π

4

Multiplying through on the left and right hand sides, yields the final result in its originalform ∞

0

dx

2x2 + 8 =

π

8

*N.B. The integral could have been solved via real integration by introducing the trigono-metric substitutions

x = 2 tan θ and dx = 2 sec2 θdθ

followed by invoking the identity tan2 θ + 1 = sec2 θ. This would have yielded the result ∞0

dx

2x2 + 8 =

1

4 tan−1

x

2

∞0

= π

8

Since the inverse tangent function tends to (π/2) as x → ∞.

11.3.2 Real Integration: Functional Replacement and Jordan’s Lemma

In order to integrate ∞−∞

x sin xdx

x2 + 9 .

begin by taking the function f in the integrand into the complex domain (i.e. f (x) → f (z )).The complex-valued function f (z ) is then

26




f (z ) = zeiz

z 2 + 9

It is more convenient to transpose the sine function into a complex exponential rather thanto a complex-valued sine function, as the sine term can be easily extracted by taking the

imaginary part of this integral, while a complex sine function would provide no such conve-nience. The singular points are located at the zeros of the polynomial in the denominator,that is to say, when z 2 + 9 = 0. Solutions to this equation are found easily by factorization

zeiz

z 2 + 9 =

zeiz

(z + 3i)(z − 3i)

Singularities are located at z = ±3i. A semicircular contour C is chosen, centered at z = 0,with a sufficiently large radius R chosen such that the point z = 3i is contained within thesemicircular domain. The semicircular arc is denoted as C R. Note, that when choosing thiscontour, the point z = −3i will not contribute to the value of the integral and is therefore

neglected in the subsequent analysis.Applying residue theorem, and breaking up the integral into complex and real-valuedparts admits the following

C

zeizdz

z 2 + 9 = 2πi Res

z=3if (z )

R−R

xeixdx

x2 + 9 +

C R

zeizdz

z 2 + 9 = 2πi Res

z=3if (z )

R−R

xeixdx

x2 + 9 = 2πi Res

z=3if (z ) −

C R

zeizdz

z 2 + 9

Adjustments are then made to the limits of integration so that they fit those given in theoriginal problem, i.e. let R → ∞.

limR→∞

R−R

xeixdx

x2 + 9 = 2πi Res

z=3if (z ) − lim

R→∞

C R

zeizdz

z 2 + 9 ∞−∞

xeixdx

x2 + 9 = 2πi Res

z=3if (z ) − lim

R→∞

C R

zeizdz

z 2 + 9 (21)

Where the bounds have been informally represented as explicit limits of integration for thereal integral, and left in front of the complex integral on the RHS. The limit is not writtenas being applied to the residue term, as the residue contains no further dependence on theradius of the contour, provided that it was all ready contained in the domain interior toit before the limit was taken. In order to solve the RHS of the above equation, begin byproving the complex integral is equal to zero.

Since the complex integral contains a complex exponential, and given the limit of theradius R of the contour tending to infinity, it is evident that the integral cannot be easily

27




shown to be zero by way of an ML bound. Jordan’s lemma is invoked to prove this, butbefore it can be used the function contained in the integrand must first be factored in thesame form as the statement of the theorem. That is to say, the function f (z ) must be showncapable of being expressed as f (z ) = g(z )eiz, for some complex function g. This is easilyaccomplished by examining the function in the integrand.

zeiz

z 2 + 9 = g(z )eiz

where

g(z ) = z

z 2 + 9

The function g immediately meets all the requirements of the theorem, there exists a radiusR0 < R for a positive semicircle in the upper half plane where g is analytic to all pointsexterior. Then, by way of Jordan’s lemma, it can be said that

limR→∞

C R

zeiz

dz z 2 + 9

= limR→∞

C R

g(z )eizdz = 0

The residue at z = 3i is determined by the so-called functional replacement. Rewriting f (z )as

f (z ) = zeiz

z 2 + 9

= zeiz

(z + 3i)(z − 3i)

= φ(z )z − 3i

where

φ(z ) = zeiz

z + 3i

The resulting expression shows a simple pole, the residue of the function f is equivalent tothe value of φ(z ) at z = 3i.

Resz=3i f (z ) = φ(3i)

= (3i)e3i

2

(3i) + 3i

= i3e−3

6 i

→ = e−3

2

28




Inputting this result into Eqn. (21)

∞−∞

xeixdx

x2 + 9 = 2πi

e−3

2

= iπe−3

Taking the imaginary part of the above equation,

Im

∞−∞

xeixdx

x2 + 9

= Im

iπe−3

∞−∞

x sin xdx

x2 + 9 =

π

e3

The desired result is obtained.

11.3.3 Real Integration: Integrals Containing Trigonometric Functions

Prove the integration formula π−π

dθ

5 + 4 sin θ =

π

3 (22)

The function is difficult to integrate due to the sine function’s location in the denominator.The standard method of replacing the real sine function with a complex exponential providesno easy method to isolate the sine function in order to retrieve the real-valued result. Thus,since the limits of integration encompass all 2π radians of a circle, the methods of Section

8.4 can be used. Letting z = eiθ, the following substitutions are identified:

sin θ = z − z −1

2i

and

dθ = dz

iz

inserting these expressions into the original integral allows the integral to be taken into thecomplex domain.

|z|=1

dz 5 + (4/2i)(z − z −1)

1iz

The function contained in the integrand is manipulated until the residues can be recognized.

29




1

5 + (4/2i)(z − z −1)

1

iz =

1

5 + (2/i)(z − z −1)

1

iz

= 1

5 − 2i(z − z −1

)

1

iz =

1

5 − 2iz + 2iz −1

1

iz

= − i

5z − 2iz 2 + 2i

= i

2iz 2 − 5z − 2i

= 1/2

z 2 + (5i/2)z − 1

the quadratic formula is used to find the roots, z 1 and z 2 of the polynomial in the denomi-

nator.

z 1 = 1

4i and z 2 = −11

4 i

Since the contour is designated as |z | = 1, it is evident that the only root, and hencesingularity, that is interior to it is the root z 1. Given these roots, the integral can beexpressed as

|z|=1

(1/2)dz

(z − z 1)(z − z 2)

which can be written in the form of functional replacement,φ(z )

z − z 1where φ(z ) =

(1/2)

z − z 2

Thus, the integral in the problem statement is equal to the product of 2πi and the residueat z = z 1. The residue is equivalent to φ(z 1):

Resz=z1

(1/2)

(z − z 1)(z − z 2) = φ(z 1)

= (1/2)

z 1 − z 2

= −1

6i

Thus, |z|=1

(1/2)dz

(z − z 1)(z − z 2) = 2πi Res

z=z1f (z ) = 2πi(−1

6i) =

π

3

30




Since this integral is equivalent to the original integral, it can then be said that π−π

dθ

5 + 4 sin θ =

π

3

11.3.4 Real Integration: Strange Contours (Fresnel Integral)The following integral ∞

0

cos x2dx (23)

can prove difficult only due to the odd choice of contour one must use in order to obtaina solution. This integral is known as a Fresnel integral and arises in the field of opticsin the description of near field Fresnel diffraction. While the function is transcendentalwhen evaluating the integral over definite limits, a solution can be found when the boundsare treated as semi-infinite. In fact, the convergence of the real integral over semi-infinite

bounds is suggested when looking at a trace of the function (Figure 4).

Figure 4 - A plot is shown of the Fresnel cosine function. The frequency increases with x.

As evident from above, the frequency of the function increases with x, until the wavelengthof the function tends to zero as x

→ ∞. The function oscillates above and below the x-axis,

suggesting that it is possible in the limit for large x, that sufficient contributions from thearea sweeped out by the function will be cancelled out by its negative and positive portions,leading to a finite result. This is precisely the case, and this finite value of the integral isfound via complex integration.

The real-valued function f (x) = cos x2 is transposed into the complex domain as acomplex exponential f (z ) = exp(iz 2). The complex function f is identified to hold nosingularities; however, this only suggests that the complex closed contour integral of thisfunction about any domain is zero, not that all path integrals making up the closed contour

31




are themselves zero. Thus, a solution could still be obtained in this manner. In orderto choose a proper contour, begin by examining the behavior of the complex exponentialfunction in the complex plane, this is qualitatively shown in the figure below:

Figure 5 - The behavior of the function eiz2

is examined. In (a) the function is described along a positivelyoriented closed circular contour. The regions where the function decays and increases are labelled. In (b), aproposed quarter circle contour is shown that evades the increasing regions of the function. And, in (c) an

eighth circle is given, which is a contour that allows for the integration to yield a proper, finite value.

It is seen by inspection that, when examined along a positively oriented closed circular con-

tour, the complex exponential function exp(iz 2) increases and decays in different quadrants.The integral of the function in a region of growth is not capable of admitting a finite resultsince the limits of integration (and hence the radius of any circular path) must eventually beextended to infinity in accordance with the original problem. However, it would seem thatit is still possible to integrate, so long as the increasing regions of the function are avoided.Thus, a first choice of contour could be the quarter-circle illustrated in Figure 5(b). However,when choosing this contour, applying residue theorem, and taking limits, the solution to theproblem does not yield a value, but rather an identity: ∞

0

cos x2dx =

∞0

sin x2dx (24)

This is certainly nice to know, but a value of both these functions integrated over thestipulated limits is still yet to be found. In order to integrate the function over the desiredbounds so as to obtain a finite result, the 1/8th circle, shown in Figure 5(c), is used.

Applying residue theorem to the system provides the following statement C

eiz2

dz = 0

32




where C is used to denote the closed contour shown in Figure 5(c). Notice how since thereare no residues contained within the contour, the contour integral of the function is equalto zero. That is to say, residue theorem is reduced to Cauchy’s integral theorem in thisinstance. The integral can further be broken up into a summation of path integrals, wherethe sum of the paths is equivalent to the contour C .

C 1

eiz2

dz I

+

C 2

eiz2

dz II

− C 3

eiz2

dz III

= 0 (25)

The integrals have been labelled as I, II, and III for convenient referencing, and each pathC 1, C 2, and C 3 are as shown in Figure 5(c). Each integral is evaluated below.

Integral I

The only work that needs to be done on this integral is to parameterize it along the real

line, and take limits. Thus, C 1

eiz2

dz =

R0

eix2

dx

since f (z ) = f (x, y), and y = 0 for all points on the real line. Letting R → ∞, Integral Ibecomes ∞

0

eix2

dx. (26)

where it is noted that the real part of this integral can be taken to make this equation of thesame form as the original integral in the problem statement (Eqn. (22)).

Integral II

Since integral II has a complex-valued, nonconstant path, it would be convenient to provethis integral tends to zero. This integral does – indeed – turn out to be identical to zero, byway of Jordan’s lemma. This is shown by initially factoring the function in the integrand inthe following way

C 2

eiz2

dz =

C 2

ei(z2−z)eizdz =

C 2

g(z )eizdz.

The above equation is now in the form of the statement in Jordan’s lemma, where g(z ) =eiz(z−1). Since the function g is entire , that is, it is analytic for all points in the complexplane, it meets the requirements for Jordan’s lemma. Thus, it can be said that

C 2

eiz2

dz = 0 (27)

Integral III

33




For the third integral, the following parametrization is introduced

z (t) = ei(π/4)t = 1 + i√

2t (0 ≤ t ≤ t0)

it then follows that

dz = 1 + i√

2dt

for an appropriate value of t0, such that dist|0, z (t0)| = R. Inputting these substitutions intointegral III in Eqn. (25)

C 3

eiz2

dz = 1 + i√

2

t00

exp

i

eiπ4t2

dt

= 1 + i

√ 2 t00

expieiπ

4 t22 dt

= 1 + i√

2

t00

exp(i2t2)dt

= 1 + i√

2

t00

e−t2

dt

Furthermore, if t0 is allowed to extend to infinity then C 3

eiz2

dz = 1 + i√

2

∞0

e−t2

dt

The parametrization has rendered the original integral into a Gaussian integral . The aboveintegral can be evaluated in a number of ways. Such methods include Feynmann’s so-calledparametric integration, or by integrating the square of the integral in polar coordinates, orby using the gamma function Γ(t). It is identified that the integral above is equivalent toΓ(3/2) =

√ π/2. Thus, the solution to the above equation can be written as

1 + i√ 2

∞0

e−t2

dt =

1 + i√

2

√ π

2 =

1

2

π

2 +

i

2

π

2 (28)

Inserting Eqns. (25), (26), and (27) into (24), and solving for the real integral reveals

∞0

eix2dx = 12 π

2 + i

2 π

2

Taking the real part of the above equation, and combining this result with the identity shownin Eqn. (24), the solution is found to be

∞0

cos x2dx =

∞0

sin x2dx = 1

2

π

2 (29)

34




Thus, demonstrating the importance of choosing an appropriate contour.

11.3.5 Complex Integration: Quotient Rule

To integrate

C

coth zdz = C

cosh z

sinh z dz

with the simple closed contour C taken to be the circle |z | = 1 is straightforward in therespect that the contour is defined, as well as the function to be integrated. Singular pointsof the function can be located by expressing the complex hyperbolic sine function in termsof x and y .

sinh z = sinh x cos y + i cosh x sin y

It is clear that the zeroes of this functions are periodic since it involves trigonometric func-

tions, with zeroes located at

sinh(inπ) = 0

where n is an integer. From this statement, all zeroes of the function lie on the y-axiswhereas only one singularity (corresponding to n = 0) is contained within the contour|z | = 1. Application of residue theorem implies

C

coth zdz = 2πi Resz=0

coth z

where the residue at z = 0 can be found by invoking the quotient rule of Section 6.2.2 with p(z ) = cosh z and q (z ) = sinh z .

coth z = cosh z

sinh z

= p(z )

q (z )

Since p(0) = 0, q (0) = 0, and q (0) = p(0) = 0, using the theorem is valid and the residue isgiven by

Resz=0 coth z =

cosh(0)

cosh(0) = 1

Thus, |z|=1

coth zdz = 2πi.

35




11.3.6 Finding Residues: Multiplication of Series

Suppose the residues of the function

f (z ) = ez

z (1 + z 2)

are to be found. It is evident that these residues are located at z = 0 and z = ±i. While all of these residues can be found via the method of functional replacement, replete examples haveall ready been shown illustrating this strategy. Thus, rather than focusing on this method,this example will instead show how direct extraction of the residue from the function’s seriesexpansion, corresponding to z = 0, is accomplished. The need for only this residue couldarise when considering a complex circular closed contour integral, centered at the origin,with a contour of radius R < 1.

The methodology is straightforward: find a series representation centered about z = 0,and locate the coefficient that corresponds to the series’ z −1 dependence, this coefficient isthe residue at that point. In order to find the series expansion of the whole function, the

function is first written as the following for clarity:

f (z ) = 1

z · ez · 1

1 + z 2

A series representation centered about z = 0 for the entire function above can be obtained bymultiplication of the known Maclaurin series among the constituent terms of the function.That is to say, known expansions for each term in the above equation can be multipliedtogether in order to obtain a combined series expansion for the whole function. From thelist given in Section 11.2 , the following series are noted,

ez

=

∞

n=0

z n

n!

and

1

1 + z 2 =

n=0

(−1)nz 2n

where the latter series follows from the replacement of z with −z 2 in series (1). The term z −1

is all ready in the form of a series expansion centered about z = 0. Beginning by multiplyingthe first few terms of following two known Maclaurin series

ez 11 + z 2

=

1 + z + 12

z 2 + 16

z 3 + . . .

1 − z 2 + z 4 + . . .

= 1 + z + 1

2z 2 +

1

6z 3 + . . .

− z 2 − z 3 + . . .

ez 1

1 + z 2 = 1 + z − 1

2z 2 − 5

6z 3 + . . .

36




Finally, the z −1 term is multiplied through yielding the full expansion of the function f (z )

ez

z (1 + z 2) =

1

z + 1 − 1

2z − 5

6z 2 + . . .

Identifying the coefficient in front of the z −1 term as the residue provides the desired result.

Resz=0

ez

z (1 + z 2) = 1

Thus, while in this particular case it was not the quickest way to find the function’s residue (asfunctional replacement would have required much less work), direct extraction of the residuefrom the power series expansion was shown to still be a possible method. Furthermore, forsome functions (e.g. exp(cos z )), this method can actually prove to be less tasking, so it isimportant to not forget where residues come from.

11.3.7 Finding Residues: Division of Series

The Laurent series representation, centered about z = 0, of

f (z ) = 1

ez − 1

can be found by division of series. Recall from the list in Section 11.2 , that

ez =∞n=0

z n

n! = 1 + z +

1

2z 2 +

1

6z 3 + . . .

then the denominator of f becomes

ez − 1 = z + 12

z 2 + 16

z 3 + . . .

the method proceeds by dividing this series into unity.

1z − 1

2 + . . .

z + 12

z 2 + 16

z 3 + . . .

1

1 + 12

z + 16

z 2 + . . .

− 12z − 1

6z 2 + . . .− 1

2z − 1

4z 2 + . . .

1

12

z 2 + 1

24

z 3 + . . .

The first term of the Laurent series that appears in the quotient of this division has thenecessary z −1 dependence, its coefficient then must be the residue.

Resz=0

1

ez − 1 = 1

As discussed earlier, usually only a small number of terms of the series need be computed in

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