9 De nite integrals using the residue theorem

Topic 9 Notes

Jeremy Orloff

9 Definite integrals using the residue theorem

9.1 Introduction

In this topic we'll use the residue theorem to compute some real definite integrals.

b

f (x) dx

a

The general approach is always the same

1. Find a complex analytic function g(z) which either equals f on the real axis or which is closely connected to f , e.g. f (x) = cos(x), g(z) = eiz.

2. Pick a closed contour C that includes the part of the real axis in the integral. 3. The contour will be made up of pieces. It should be such that we can compute

g(z) dz over each of the pieces except the part on the real axis.

4. Use the residue theorem to compute g(z) dz.

C

5. Combine the previous steps to deduce the value of the integral we want.

9.2 Integrals of functions that decay

The theorems in this section will guide us in choosing the closed contour C described in the introduction.

The first theorem is for functions that decay faster than 1/z.

Theorem 9.1. (a) Suppose f (z) is defined in the upper half-plane. If there is an a > 1

and M > 0 such that

|f (z)|

<

M |z|a

for |z| large then

lim f (z) dz = 0,

R CR

where CR is the semicircle shown below on the left.

Im(z) CR

Im(z)

-R

R Re(z)

CR

Re(z)

-R

R

1

9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM

2

Semicircles: left: Rei, 0 < < right: Rei, < < 2. (b) If f (z) is defined in the lower half-plane and

|f (z)|

<

M |z|a

,

where a > 1 then lim f (z) dz = 0,

R CR

where CR is the semicircle shown above on the right.

Proof. We prove (a), (b) is essentially the same. We use the triangle inequality for integrals and the estimate given in the hypothesis. For R large

f (z) dz

CR

|f (z)| |dz|

CR

CR

M |z|a

|dz|

=

0

M Ra

R

d

=

M Ra-1

.

Since a > 1 this clearly goes to 0 as R . QED

The next theorem is for functions that decay like 1/z. It requires some more care to state and prove.

Theorem 9.2. (a) Suppose f (z) is defined in the upper half-plane. If there is an M > 0

such that

|f (z)|

<

M |z|

for |z| large then for a > 0

lim

f (z)eiaz dz = 0,

x1, x2 C1+C2+C3

where C1 + C2 + C3 is the rectangular path shown below on the left.

C3 -x2

Im(z) C2

i(x1 + x2)

C1

x1 Re(z)

-x2

Im(z)

C3

-i(x1 + x2)

C2

Rectangular paths of height and width x1 + x2. (b) Similarly, if a < 0 then

x1 Re(z) C1

lim

f (z)eiaz dz = 0,

x1, x2 C1+C2+C3

where C1 + C2 + C3 is the rectangular path shown above on the right. Note. In contrast to Theorem 9.1 this theorem needs to include the factor eiaz.

Proof. (a) We start by parametrizing C1, C2, C3.

C1: 1(t) = x1 + it, t from 0 to x1 + x2

9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM

3

C2: 2(t) = t + i(x1 + x2), t from x1 to -x2 C3: 3(t) = -x2 + it, t from x1 + x2 to 0.

Next we look at each integral in turn. We assume x1 and x2 are large enough that

|f (z)|

<

M |z|

on each of the curves Cj.

f (z)eiaz dz

C1

|f (z)eiaz| |dz|

C1

C1

M |z|

|eiaz

|

|dz|

x1+x2

=

0

M |eiax1-at| dt x21 + t2

M x1

x1+x2

e-at dt

0

=

M x1

(1

-

e-a(x1+x2))/a.

Since a > 0, it is clear that this last expression goes to 0 as x1 and x2 go to .

f (z)eiaz dz

C2

|f (z)eiaz| |dz|

C2

C2

M |z|

|eiaz

|

|dz

|

x1

=

M

|eiat-a(x1+x2)| dt

-x2 t2 + (x1 + x2)2

M e-a(x1+x2) x1 + x2

x1+x2

dt

0

M e-a(x1+x2)

Again, clearly this last expression goes to 0 as x1 and x2 go to . The argument for C3 is essentially the same as for C1, so we leave it to the reader. The proof for part (b) is the same. You need to keep track of the sign in the exponentials and make sure it is negative. Example. See Example 9.16 below for an example using Theorem 9.2.

9.3 Integrals and

-

0

Example 9.3. Compute

I=

-

(1

1 + x2)2

dx.

Solution: Let

f (z) = 1/(1 + z2)2.

9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM

4

It is clear that for z large

f (z) 1/z4.

In particular, the hypothesis of Theorem 9.1 is satisfied. Using the contour shown below we have, by the residue theorem,

f (z) dz = 2i residues of f inside the contour.

(1)

C1+CR

Im(z)

CR i

-R C1

Re(z) R

We examine each of the pieces in the above equation.

f (z) dz: By Theorem 9.1(a),

CR

lim f (z) dz = 0.

R CR

f (z) dz: Directly, we see that

C1

R

lim f (z) dz = lim f (x) dx = f (x) dx = I.

R C1

R -R

-

So letting R , Equation 1 becomes

I = f (x) dx = 2i

-

residues of f inside the contour.

Finally, we compute the needed residues: f (z) has poles of order 2 at ?i. Only z = i is inside the contour, so we compute the residue there. Let

g(z)

=

(z

-

i)2f (z)

=

(z

1 +

i)2 .

Then So,

Res(f, i)

=

g

(i)

=

-

2 (2i)3

=

1 4i

I = 2i Res(f, i) =

2

.

Example 9.4. Compute

I=

-

1 x4 +

1

dx.

9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM

5

Solution: Let f (z) = 1/(1 + z4). We use the same contour as in the previous example Im(z)

CR

ei3/4 -R C1

ei/4

Re(z) R

As in the previous example,

lim f (z) dz = 0

R CR

and

lim f (z) dz = f (x) dx = I.

R C1

-

So, by the residue theorem

I = lim

f (z) dz = 2i

R C1+CR

The poles of f are all simple and at

residues of f inside the contour.

ei/4, ei3/4, ei5/4, ei7/4.

Only ei/4 and ei3/4 are inside the contour. We compute their residues as limits using L'Hospital's rule. For z1 = ei/4 :

Res(f, z1)

=

lim (z

zz1

-

z1)f

(z)

=

lim

zz1

z - z1 1 + z4

=

lim

zz1

1 4z3

=

1 4ei3/4

=

e-i3/4 4

and for z2 = ei3/4 :

Res(f, z2)

=

lim (z

zz2

- z2)f (z)

=

lim

zz2

z - z2 1 + z4

=

lim

zz2

1 4z3

=

1 4ei9/4

=

e-i/4 4

So, I = 2i(Res(f, z1) + Res(f, z2)) = 2i

-1- i + 1 - i 42 42

= 2i - 2i 42

=

2 2

Example 9.5. Suppose b > 0. Show

0

cos(x) x2 + b2

dx

=

e-b 2b

.

Solution: The first thing to note is that the integrand is even, so

I

=

1 2

-

cos(x) x2 + b2

.

9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM

6

Also note that the square in the denominator tells us the integral is absolutely convergent.

We have to be careful because cos(z) goes to infinity in either half-plane, so the hypotheses of Theorem 9.1 are not satisfied. The trick is to replace cos(x) by eix, so

I~ =

-

eix x2 + b2

dx,

with

I

=

1 2

Re(I~).

Now let For z = x + iy with y > 0 we have

f (z)

=

eiz z2 +

b2

.

|f (z)|

=

|ei(x+iy)| |z2 + b2|

=

e-y |z2 + b2|

.

Since e-y < 1, f (z) satisfies the hypotheses of Theorem 9.1 in the upper half-plane. Now we can use the same contour as in the previous examples

Im(z)

CR ib

-R C1

Re(z) R

We have

lim f (z) dz = 0

R CR

and

lim f (z) dz =

f (x) dx = I~.

R C1

-

So, by the residue theorem

I~ = lim

f (z) dz = 2i

R C1+CR

residues of f inside the contour.

The poles of f are at ?bi and both are simple. Only bi is inside the contour. We compute

the residue as a limit using L'Hospital's rule

Res(f, bi)

=

lim (z

zbi

-

bi)

z2

eiz +

b2

=

e-b 2bi

.

So, Finally, as claimed.

I~ =

2i Res(f, bi)

=

e-b b

.

I

=

1 2

Re(I~)

=

e-b 2b

,

Warning: Be careful when replacing cos(z) by eiz that it is appropriate. A key point in

the

above

example

was

that

I

=

1 2

Re(I~).

This

is

needed

to

make

the

replacement

useful.

9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM

7

9.4 Trigonometric integrals

The trick here is to put together some elementary properties of z = ei on the unit circle.

1. e-i = 1/z.

2.

cos() =

ei + e-i 2

=

z

+ 1/z 2

.

3.

sin() =

ei - e-i 2i

=

z

- 1/z 2i

.

We start with an example. After that we'll state a more general theorem.

Example 9.6. Compute Assume that |a| = 1.

2 0

1

+

a2

d - 2a

cos() .

Solution: Notice that [0, 2] is the interval used to parametrize the unit circle as z = ei. We need to make two substitutions:

cos()

=

z

+ 1/z 2

dz = iei d

d

=

dz iz

Making these substitutions we get

I=

2

d

0 1 + a2 - 2a cos()

=

|z|=1

1

+

a2

-

1 2a(z

+

1/z)/2

?

dz iz

=

|z|=1

i((1

+

a2)z

1 -

a(z2

+

1))

dz.

So, let

f (z)

=

i((1

+

a2)z

1 -

a(z2

+

1)) .

The residue theorem implies

I = 2i residues of f inside the unit circle.

We can factor the denominator:

f (z)

=

ia(z

-

-1 a)(z

-

1/a)

.

The poles are at a, 1/a. One is inside the unit circle and one is outside.

If

|a| > 1

then

1/a

is

inside

the

unit

circle

and

Res(f, 1/a) =

1 i(a2 - 1)

If |a| < 1 then a is inside the unit circle and

Res(f,

a)

=

i(1

1 -

a2)

9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM

8

We have

I=

2 a2-1

2 1-a2

if |a| > 1 if |a| < 1

The example illustrates a general technique which we state now.

Theorem 9.7. Suppose R(x, y) is a rational function with no poles on the circle

x2 + y2 = 1

then for we have

f (z)

=

1 iz

R

z

+ 1/z 2

,

z

- 1/z 2i

2

R(cos(), sin()) d = 2i

0

residues of f inside |z| = 1.

Proof. We make the same substitutions as in Example 9.6. So,

2

R(cos(), sin()) d =

0

R

|z|=1

z

+ 1/z 2

,

z

- 1/z 2i

dz iz

The assumption about poles means that f has no poles on the contour |z| = 1. The residue theorem now implies the theorem.

9.5 Integrands with branch cuts

Example 9.8. Compute

I=

0

x1/3 1 + x2

dx.

Solution: Let

f (x)

=

x1/3 1 + x2

.

Since this is asymptotically comparable to x-5/3, the integral is absolutely convergent. As

a complex function

f (z)

=

z1/3 1 + z2

needs a branch cut to be analytic (or even continuous), so we will need to take that into account with our choice of contour.

First, choose the following branch cut along the positive real axis. That is, for z = rei not on the axis, we have 0 < < 2.

Next, we use the contour C1 + CR - C2 - Cr shown below.

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