Stone M., Goldbart P. Mathematics for Physics: A Guided Tour for Graduate Students

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17.7 Further exercises and problems 665

with the ± being selected depending on whether n is odd or even. Where would you put

cuts to ensure that w is a single-valued function?

Problem 17.22: Cutting open a genus-2 surface. The Riemann surface for the function

y =



(z − a

)(z − a

)

has genus g = 2. Such a surface M is sketched in Figure 17.22, where the four inde-

pendent 1-cycles α

1,2

and β

1,2

that generate H

(M ) have been drawn so that they share

a common vertex.

(a) Realize the genus-2 surface as two copies of C ∪ {∞} cross-connected by three

square-root branch cuts. Sketch how the 1-cycles α

and β

, i = 1, 2 of Figure 17.22

appear when drawn on your thrice-cut plane.

(b) Cut the surface open along the four 1-cycles, and convince yourself that the resulting

surface is homeomorphic to the octagonal region appearing in Figure 17.23.

Hence show that for closed 1-forms a, b, on the surface we have



a ∧ b =



i=1





b −





Applications of complex variables

In this chapter we will find uses for what we have learned of complex variables. The

applications will range from the elementary to the sophisticated.

18.1 Contour integration technology

The goal of contour integration technology is to evaluate ordinary, real-variable, definite

integrals. We have already met the basic tool, the residue theorem:

Theorem: Let f (z) be analytic within and on the boundary  = ∂D of a simply connected

domain D, with the exception of a finite number of points at which the function has

poles. Then



f (z ) dz =



poles ∈ D

2πi (residue at pole).

18.1.1 Tricks of the trade

The effective application of the residue theorem is something of an art, but there are

useful classes of integrals which we can learn to recognize.

Rational trigonometric expressions

Integrals of the form



2π

F(cos θ , sin θ)dθ (18.1)

are dealt with by writing cos θ =

(z +z), sin θ =

(z −z) and integrating around the

unit circle. For example, let a, b be real and b < a. Then

I =



2π

dθ

a + b cos θ

|z|=1

+ 2az + b

(z − α)(z − β)

. (18.2)

Since αβ = 1, only one pole is within the contour. This is at

α = (−a +



− b

)/b. (18.3)

666

18.1 Contour integration technology 667

The residue is

α −β

√

− b

. (18.4)

Therefore, the integral is given by

I =

2π

√

− b

. (18.5)

These integrals are, of course, also do-able by the “t” substitution t = tan(θ/2), whence

sin θ =

1 + t

, cos θ =

1 − t

1 + t

, dθ =

2dt

1 + t

, (18.6)

followed by a partial fraction decomposition. The labour is perhaps slightly less using

the contour method.

Rational functions

Integrals of the form



∞

−∞

R(x) dx, (18.7)

where R(x) is a rational function of x with the degree of the denominator exceeding

the degree of the numerator by two or more, may be evaluated by integrating around

a rectangle from −A to +A, A to A + iB, A + iB to −A + iB, and back down to −A.

Because the integrand decreases at least as fast as 1/|z|

as z becomes large, we see that

if we let A, B →∞, the contributions from the unwanted parts of the contour become

negligible. Thus

I = 2π i





Residues of poles in upper half-plane

. (18.8)

We could also use a rectangle in the lower half-plane with the result

I =−2πi





Residues of poles in lower half-plane

. (18.9)

This must give the same answer.

For example, let n be a positive integer and consider

I =



∞

−∞

(1 + x

)

. (18.10)

668 18 Applications of complex variables

The integrand has an n-th order pole at z =±i. Suppose we close the contour in the

upper half-plane. The new contour encloses the pole at z =+i and we therefore need to

compute its residue. We set z −i = ζ and expand

(1 + z

)

[(i + ζ)

+ 1]

(2iζ)



1 −

iζ



−n

(2iζ)



1 + n



iζ



n(n + 1)



iζ



+···



. (18.11)

The coefficient of ζ

−1

(2i)

n(n + 1) ···(2n − 2)

(n − 1)!





n−1

2n−1

(2n − 2)!

((n − 1)!)

. (18.12)

The integral is therefore

I =

2n−2

(2n − 2)!

((n − 1)!)

. (18.13)

These integrals can also be done by partial fractions.

18.1.2 Branch-cut integrals

Integrals of the form

I =



∞

α−1

R(x)dx, (18.14)

where R(x) is rational, can be evaluated by integration round a slotted circle (or “key-

hole”) contour (Figure 18.1). A little more work is required to extract the answer, though.

For example, consider

I =



∞

α−1

1 + x

dx,0< Re α<1. (18.15)

The restrictions on the range of α are necessary for the integral to converge at its upper

and lower limits.

We take  to be a circle of radius  centred at z = 0, with a slot indentation designed

to exclude the positive real axis, which we take as the branch cut of z

α−1

, and a small

circle of radius  about the origin. The branch of the fractional power is defined by

setting

α−1

= exp[(α − 1)(ln |z|+iθ)], (18.16)

18.1 Contour integration technology 669

–1

Figure 18.1 A slotted circle contour  of outer radius  and inner radius .

where we will take θ to be zero immediately above the real axis, and 2π immediately

below it. With this definition the residue at the pole at z =−1ise

iπ(α−1)

. The residue

theorem therefore tells us that



α−1

1 + z

dz = 2π ie

πi(α−1)

. (18.17)

The integral decomposes as



α−1

1 + z

dz =

|z|=

α−1

1 + z

dz + (1 − e

2πi(α−1)

)







α−1

1 + x

dx −

|z|=

α−1

1 + z

dz.

(18.18)

As we send  off to infinity we can ignore the “1” in the denominator compared to the

z, and so estimate

|z|=

α−1

1 + z

→

|z|=

α−2

≤ 2π × 

Re (α)−2

. (18.19)

This tends to zero provided that Re α<1. Similarly, provided 0 < Re α, the integral

around the small circle about the origin tends to zero with . Thus

−e

πiα

2πi =



1 − e

2πi(α−1)

I. (18.20)

We conclude that

I =

2πi

πiα

− e

−πiα

)

sin πα

. (18.21)

670 18 Applications of complex variables

Figure 18.2 The contour 

Exercise 18.1: Using the slotted circle contour, show that

I =



∞

p−1

1 + x

dx =

2 sin(πp/2)

cosec (πp/2),0< p < 2.

Exercise 18.2: Integrate z

a−1

/(z − 1) around a contour 

consisting of a semicircle

in the upper half-plane together with the real axis indented at z = 0 and z = 1 (see

Figure 18.2)toget

0 =



a−1

z − 1

dz = P



∞

a−1

x − 1

dx − iπ + (cos π a + i sin π a)



∞

a−1

x + 1

dx.

As usual, the symbol P in front of the integral sign denotes a principal-part integral,

meaning that we must omit an infinitesimal segment of the contour symmetrically dis-

posed about the pole at z = 1. The term −iπ comes from integrating around the small

semicircle about this point. We get −1/2 of the residue because we have only a half-

circle, and that traversed in the “wrong” direction. Warning: this fractional residue

result is only true when we indent to avoid a simple pole – i.e. one that is of order one.

Now take real and imaginary parts and deduce that



∞

a−1

1 + x

dx =

sin πα

,0< Re a < 1,

and



∞

a−1

1 − x

dx = π cot πa,0< Re a < 1.

18.1 Contour integration technology 671

18.1.3 Jordan’s lemma

We often need to evaluate Fourier integrals

I(k) =



∞

−∞

ikx

R(x) dx (18.22)

with R(x) a rational function. For example, the Green function for the operator −∂

is given by

G(x) =



∞

−∞

2π

ikx

+ m

. (18.23)

Suppose x ∈ R and x > 0. Then, in contrast to the analogous integral without the

exponential function, we have no flexibility in closing the contour in the upper or lower

half-plane. The function e

ikx

grows without limit as we head south in the lower half-

plane, but decays rapidly in the upper half-plane. This means that we may close the

contour without changing the value of the integral by adding a large upper-half-plane

semicircle (Figure 18.3).

The modified contour encloses a pole at k = im, and this has residue i/(2m)e

−mx

. Thus

G(x) =

−mx

, x > 0. (18.24)

For x < 0, the situation is reversed, and we must close in the lower half-plane. The

residue of the pole at k =−im is −i/(2m)e

, but the minus sign is cancelled because

the contour goes the “wrong way” (clockwise). Thus

G(x) =

+mx

, x < 0. (18.25)

–im

Figure 18.3 Closing the contour in the upper half-plane.

672 18 Applications of complex variables

We can combine the two results as

G(x) =

−m|x|

. (18.26)

The formal proof that the added semicircles make no contribution to the integral when

their radius becomes large is known as Jordan’s lemma:

Lemma: Let  be a semicircle, centred at the origin, and of radius R. Suppose that

(i) f (z) is meromorphic in the upper half-plane;

(ii) f (z) tends uniformly to zero as |z|→∞for 0 < arg z <π;

(iii) the number λ is real and positive.

Then





iλz

f (z ) dz → 0, as R →∞. (18.27)

To establish this, we assume that R is large enough that |f | <on the contour, and

make a simple estimate





iλz

f (z ) dz

< 2R



π/2

−λR sin θ

dθ

< 2R



π/2

−2λRθ/π

dθ

π

(1 − e

−λR

π

. (18.28)

In the second inequality we have used the fact that (sin θ )/θ ≥ 2/π for angles in the

range 0 <θ <π/2. Since  can be made as small as we like, the lemma follows.

Example: Evaluate

I(α) =



∞

−∞

sin(αx)

dx. (18.29)

We have

I(α) = Im





∞

−∞

exp iαz



. (18.30)

If we take α>0, we can close in the upper half-plane, but our contour must exclude the

pole at z = 0. Therefore

0 =



|z|=R

exp iαz

dz −



|z|=

exp iαz

dz +



−

−R

exp iαx

dx +





exp iαx

dx.

(18.31)

18.1 Contour integration technology 673

As R →∞, we can ignore the big semicircle, the rest, after letting  → 0, gives

0 =−iπ + P



∞

−∞

iαx

dx. (18.32)

Again, the symbol P denotes a principal-part integral. The −iπ comes from the small

semicircle. We get −1/2 of the residue because we have only a half circle, and that

traversed in the “wrong” direction. (Remember that this fractional residue result is only

true when we indent to avoid a simple pole – i.e one that is of order one.)

Reading off the real and imaginary parts, we conclude that



∞

−∞

sin αx

dx = π, P



∞

−∞

cos αx

dx = 0, α>0. (18.33)

No “P” is needed in the sine integral, as the integrand is finite at x = 0.

If we relax the condition that α>0 and take into account that sine is an odd function

of its argument, we have



∞

−∞

sin αx

dx = π sgn α. (18.34)

This identity is called Dirichlet’s discontinuous integral.

We can interpret Dirichlet’s integral as giving the Fourier transform of the principal-

part distribution P(1/x) as



∞

−∞

iωx

dx = iπ sgn ω. (18.35)

This will be of use later in the chapter.

Example: We will evaluate the integral

a−1

dz (18.36)

about the first-quadrant contour shown in Figure 18.4. Observe that when 0 < a < 1

neither the large nor the small arc makes a contribution, and that there are no poles.

Hence, we deduce that

0 =



∞

a−1

dx − i



∞

−y

a−1

(a−1)

dy ,0< a < 1. (18.37)

Taking real and imaginary parts, we find



∞

a−1

cos xdx = (a) cos



,0< a < 1,



∞

a−1

sin xdx = (a) sin



,0< a < 1, (18.38)

674 18 Applications of complex variables

Figure 18.4 Quadrant contour.

where

(a) =



∞

a−1

−y

dy (18.39)

is the Euler Gamma function.

Example: Fresnel integrals. Integrals of the form

C(t) =



cos(πx

/2) dx, (18.40)

S(t) =



sin(πx

/2) dx, (18.41)

occur in the theory of diffraction and are called Fresnel integrals after Augustin Fresnel.

They are naturally combined as

C(t) + iS(t) =



iπ x

dx. (18.42)

The limit as t →∞exists and is finite. Even though the integrand does not tend to zero

at infinity, its rapid oscillation for large x is just sufficient to ensure convergence.

We can exhibit this convergence by setting x

= s and then integrating by parts to get



iπ x

dx =



iπ s/2

1/2

iπ s/2

πis

1/2

2πi



iπ s/2

3/2

The right-hand side is now manifestly convergent as t →∞.