6.4 Poles on the Real Axis

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A problem which can easily arise is when there is a pole actually on the

x-axis. We have a di_erent sort of improper integral in this case, and if f(x)

goes o_ to in_nity at b in the interval [a; c], we say that the integral R c

a is

de_ned provided that

lim

y!b􀀀 Z y

a

exists,

lim

y!b+ Z c

y

exists, and the improper integral over [a; c] is de_ned to be

lim

y!b􀀀 Z y

a

+ lim

y!b+ Z c

y

In the same way as we symmetrised the de_nition of the other improper

integrals, we can take a Cauchy Principal Value for these also, and we have

that for a pole at b 2 [a; c]. the Cauchy Principal Value of

Z c

a

f(x) dx

exists and is

lim

_!0+ _Z b􀀀_

a

f(x) dx + Z c

b+_

f(x) dx_

162 CHAPTER 6. RESIDUES

providing the limit exists. Again, the integral may not actually exist, but still

have a Cauchy Principal Value. One could wish for more carefully thought

out terminology. If it does exist, then the Cauchy Principal Value is the

value of the integral. This is rather a drag to keep writing out, so I shall

just go on writing an ordinary integral sign. So mentally, you should adapt

the de_nition of the integral from the Riemann integral to the symmetrised

integral and all will be well.

If the (symmetrised) integral exists, then it can be evaluated as for the case

where the poles are o_ the axis, except in one respect: we count the residue

from a pole on the axis as only 'half a residue'. We have the more general

case:

Theorem 6.6 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, and if u1; u2; u` are isolated zeros of order one of P,

then

Z 1

􀀀1

f(z) dz = 2_i

k

Xj=1

Res[f(z); wj] + _i

`

Xj=1

Res[f(z); uj]

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

poles u1; _ _ _; u` of f(z) on the real axis. 2

Similarly for the trigonometric functions:

Theorem 6.7 When P and Q are real polynomials with degree of Q at least

one greater than the degree of P, and when Q has ` isolated zeros of order

one, the integral of the function

f(x) =

P(x)

Q(x)

cos(ax)

for a > 0 is given by extending f to the complex plane by putting

f(z) =

P(z)eiaz

Q(z)

6.4. POLES ON THE REAL AXIS 163

whereupon

Z 1

􀀀1

P(x)

Q(x)

cos(ax) dx = 􀀀2_

k

Xj=1

=(Res[f(z); wj])􀀀_

`

Xj=1

=(Res[f(z); uj])

and

Z 1

􀀀1

P(x)

Q(x)

sin(ax) dx = 2_

k

Xj=1

<(Res[f(z); wj]) + _

`

Xj=1

<(Res[f(z); uj])

where w1; _ _ _; wk are the poles in the top half of the complex plane, and

u1; _ _ _ ; u` are the poles on the real axis. 2

Example 6.4.1 Evaluate

Z 1

􀀀1

sin x

x

dx

Solution

The above theorem tells us that the integral is

_<_Res[

eiz

z

; 0]_

Now at z = 0 the residue is just the coe_cient ei0 = 1 so the result is _.

Example 6.4.2 The same calculation shows that

Z 1

􀀀1

cos x

x

dx = 0

It is easy to see that this is true for the Cauchy symmetrised integral, but not

for the unreconstructed Riemann integral, which does not exist.

Exercise 6.4.1 Sketch the graph of

cos x

x

and verify that the integral

Z 1

0

cos x

x

dx

does not exist.

164 CHAPTER 6. RESIDUES

Figure 6.2: The bites provide half the residues

The last two theorems depend for the proof on taking little bites out of the

path around the poles in the top half plane in the neighbourhood of each of

the singularities on the real axis. The diagram of _gure 6.2 gives the game

away. Those of you with the persistence should try to prove the results. For

those without, there are proofs in all the standard texts.

We have to take the limits as the radii of the bites get smaller and the big

semi-circle gets bigger. The reason for the half of the 2_i Res[f,w] should be

obvious.

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