6.2 In_nite Integrals of rational functions

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There is a very nice application of the above ideas to integrating real functions

from􀀀1 to1. These are some of the so called 'improper' integrals, so called

because respectable integrals have real numbers at the limits of the integrals,

and functions which are bounded on those bounded intervals.

In _rst year, you did the Riemann Integral, and it was all about chopping

the domain interval up into little bits and taking limits of sums of heights

of functions over the little bits. I remind you that we de_ne, for any real

number b,

Z 1

b

f(x) dx = lim

y!1Z y

b

f(x) dx

when the limit exists, and

Z a

􀀀1

f(x) dx = lim

y!􀀀1Z a

y

f(x) dx

for any real number a. Then the doubly in_nite integral exists if, for some a,

lim

y!1Z y

a

f(x) dx

and

lim

y!􀀀1Z a

y

f(x) dx

both exist, in which case

Z 1

􀀀1

f(x) dx = lim

y!1Z y

a

f(x) dx + lim

y!􀀀1Z a

y

f(x) dx

This is the Riemann Integral for the case when the domain is unbounded.

There are plenty of cases where it doesn't exist. For example,

Z 1

􀀀1

x dx

6.2. INFINITE INTEGRALS OF RATIONAL FUNCTIONS 155

clearly does not exist.

On the other hand, there is a case for saying that for the function f(x) = x,

the area above the X-axis on the positive side always cancels out the area

below it on the negative side, so we ought to have

Z 1

􀀀1

x dx = 0

There are two approaches to this sort of problem; one is the rather repressive

one favoured by most schoolteachers and all bureaucrats, which is to tell you

what the rules are and to insist that you follow them. The other is favoured

by engineers, mathematicians and all those with a bit of go in them, and it

is to make up a new kind of integral which behaves the way your intuitions

think is reasonable2.

We therefore de_ne a new improper integral for the case where the function

is bounded and the domain is the whole real line:

C Z 1

􀀀1

f(x) dx = lim

y!1Z y

􀀀y

f(x) dx

This means that C RR x = 0, although RR x does not exist.

The C stands for Cauchy, but I shan't call this the Cauchy Integral because

that term could cause confusion. It is often called the Cauchy Principal

Value, but this leads one to think it is something possessed by an integral

which does not exist.

Note that if the integral does exist, then so does the C R and they have the

same value. The converse is obviously false.

In what follows, I shall just use the integral sign, without sticking a C in

front of it, to denote this Cauchy Principal Value. We are using the new,

symmetrised integral instead of the Riemann integral: they are the same

when the Riemann integral is de_ned, but the new symmetrised integral

exists for functions where there is no Riemann Integral3.

2This may turn out to be impossible, in which case your intuitions need a bit of straight-

ening out. You should not assume that absolutely anything goes; only the things that make

sense work.

3This sort of thing happens a good deal more than you might have been led to believe

156 CHAPTER 6. RESIDUES

a b

Figure 6.1: The integral around the outer semicircle tends to zero if f dies

away fast enough

The idea of the application of Complex Analysis to evaluation of such integrals

is indicated by the diagram _gure 6.1.

The idea is that the integral along the line segment from a to b plus the

integral around the semicircle is a loop integral which can be evaluated by

using residues. But as a gets more negative and b more positive, the line

integral gets closer to the integral

Z 1

􀀀1

f(x) dx

and the arc gets further and further away from the origin. Now if f(z) ! 0

fast enough to overcome the arc length getting longer, the integral around

the arc tends to zero. So for some functions at least, we can integrate f over

the real line by making f the real part of a complex function and counting

residues in the top half of the plane.

I hope you will agree that this is a very cool idea and deserves to work.

Example 6.2.1 Evaluate the area under the 'poor man's gaussian':

Z 1

􀀀1

dx

1 + x2

in _rst year. There are, still to come, Riemann-Stieltjes integrals and Lebesgue integrals,

both of which also extend Riemann integrals in useful ways. But not in this course.

6.2. INFINITE INTEGRALS OF RATIONAL FUNCTIONS 157

Solution 1

The old fashioned way is to substitute x = tan _ when we get the inde_nite

integral arctan _, and evaluating from x = 0 to x = 1 is to go from _ = 0 to

_ = _=2. So the answer is just _.

Solution 2

We argue that we the result will be the same as

Ic

dz

1 + z2

where c is the in_nite semi-circle in the positive plane, because the path length

of the semi-circle will go up linearly with the radius of the semi-circle, but

the value of the integral will go down as the square at each point, so the limit

of the integral around the semi-circle will be zero, and the whole contribution

must come from the part along the real axis.

Now there is a pole at _i, and the pole at 􀀀i is outside the region. So we

factorise

1

1 + z2

= (

1

z +i

) (

1

z 􀀀i

)

whereupon the Laurent expansion about z = i has coe_cient

1

2i

and the integral is

2_i

1

2i

= _

This gives us the right answer, increasing con_dence in the reasoning.

It is about the same amount of work whichever way you do this particular

case, but the general situation is that you probably won't know what substitution

to make. The contour integral approach means you don't have to

know.

Example 6.2.2 Evaluate

Z 1

􀀀1

dx

(x2 􀀀 4x + 5)2

158 CHAPTER 6. RESIDUES

Solution

First we recognise that this is

IH

dz

(z2 􀀀 4z + 5)2

where H is a semicircle in the upper half plane big enough to contain all poles

of the function with positive imaginary part.

We factorise

1

(z2 􀀀 4z + 5)2

= _ 1

(z 􀀀(2 + i))2_ _ 1

(z 􀀀 (2 􀀀 i))2_

and note that there is one pole at 2 + i of order 2 in the upper half plane.

We evaluate the residue by taking

lim

z!2+i

d

dz _ 1

(z 􀀀 (2 􀀀 i))2_ = 􀀀2

((2 + i) 􀀀 (2 􀀀 i))3 = 􀀀i=4

Then it follows that the integral is 2_i times this, i.e. _=2.

Exercise 6.2.1 Evaluate

Z 1

􀀀1

dx

(x2 + 4)3

Exercise 6.2.2 Evaluate

Z 1

􀀀1

x dx

(x2 + 1)2

Verify your answer by drawing the graph of the function.

Exercise 6.2.3 Make up a few integrals of this type and solve them. If this

is beyond you, try the problems in Mathews and Howell, p 260.

The ideas should be now su_ciently clear to allow you to see the nature of

the arguments required to prove:

6.3. TRIGONOMETRIC AND POLYNOMIAL FUNCTIONS 159

Theorem 6.4 If

f(x) =

P(x)

Q(x)

for real non-zero polynomials P;Q and if the degree of Q is at least two more

than the degree of P, then

Z 1

􀀀1

f(z) dz = 2_i

k

Xj=1

Res[f(z); wj]

where there are k poles w1; _ _ _; wk of f(z) in the top half plane of C . 2

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