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z

z = 0 and use this expansion to determine the residue.

Unfortunately, this method does not always work. For some special functions other tricks

can be used. Here we will consider functions with a simple pole, these are functions where

in the expansion (13.3) the terms for n < do not contribute:

;1

1

X

an(z z0 )n :

h(z) = (13.13)

;

n=;1

h(z) = cos z .

The residue at the point z0 follows by

An example of such a function is z

\extracting" the coe cient a;1 from the series (13.13).

Problem h: Multiply (13.13) with (z z0 ), take the limit z z0 to show that:

; !

Res h(z0 ) = z!z0(z z0 )h(z0 ) (simple pole) :

lim (13.14)

;

However, remember that this only works for functions with a simple pole, this recipe gives

the wrong answer (in nity) when applied to a function that has nonzero coe cients an

for n < in the expansion (13.3).

;1

In the treatment of this section we considered an integration in the counter-clockwise

direction around the pole z0 .

Problem i: Redo the derivation of this section for a contour integration in the clockwise

direction around the pole z0 and show that in that case

I

h(z)dz = iRes h(z0 ) (clockwise direction) : (13.15)

;2

C

Find out in which step of the derivation the minus sign is picked up!

13.3. APPLICATION TO SOME INTEGRALS 165

Problem j: It may happen that a contour encloses not a single pole but a number of

poles at points zj . Find for this case a contour similar to the contour C in the

right panel of gure (13.1) to show that the contour integral is equal to the sum of

the contour integrals around the individual poles zj . Use this to show that for this

situation:

I X

h(z)dz = 2 i Res h(zj ) (counter clockwise direction) : (13.16)

;

C j

13.3 Application to some integrals

The residue theorem has some applications to integrals that do not contain a complex

variable at all! As an example consider the integral

Z1 1

I= dx : (13.17)

;1 1 + x2

If you know that 1=(1 + x2 ) is the derivative of arctan x it is not di cult to solve this

integral:

I = arctan x]1 = 2 2=: (13.18)

;1 ; ;

Now suppose you did not know that arctan x is the primitive function of 1=(1 + x2 ). In

that case you would be unable to see that the integral (13.17) is equal to . Complex

integration o ers a way to obtain the value of this integral in a systematic fashion.

First note that the path of integration in the integral can be viewed as the real axis

in the complex plane. Nothing prevents us from viewing the real function 1=(1 + x2 ) as a

complex function 1=(1 + z 2 ) because on the real axis z is equal to x. This means that

Z 1 dz

I= (13.19)

Creal 1 + z 2

where Creal denotes the real axis as integration path. At this point we cannot apply the

residue theorem yet because the integration is not over a closed contour in the complex

plane, see gure (13.2). Let us close the integration path using the circular path CR in

the upper half plane with radius R, see gure (13.2). In the end of the calculation we will

let R go to in nity so that the integration over the semicircle moves to in nity.

Problem a: Show that

Z 1 dz = I 1 dz Z 1 dz :

I= (13.20)

Creal 1 + z 2 C 1 + z2 CR 1 + z 2

;

The circular integral is over the closed contour in gure (13.2). What we have done is that

we have closed the contour and that we subtract the integral over the semicircle that we

added to obtain a closed integration path. This is the general approach in the application

of the residue theorem one adds segments to the integration path to obtain an integral

over a closed contour in the complex plane, and corrects for the segments that one has

added to the path. Obviously, this is only useful when the integral over the segment that

CHAPTER 13. COMPLEX INTEGRATION

166

R

z=i x

z = -i x

Figure 13.2: De nition of the integration paths in the complex plane.

one has added can be computed easily or when it vanishes. In this example the integral

over the semicircle vanishes as R This can be seen with the following estimations:

! 1.

Z 1 dz Z Z

= R2 R 1 0 as R

1 1 :

CR 2 1

CR 1 + z 2 CR 1 + z 2

jdzj jdzj ! !1

;

jzj ;

(13.21)

Problem b: Justify each of these steps.

The estimate (13.21) implies that the last integral in (13.20) vanishes in the limit R ! 1.

I1

This means that

I = 1 + z 2 dz : (13.22)

C

Now we are in the position to apply the residue theorem because we have reduced the

problem to the evaluation of an integral along a closed contour in the complex plane. We

know from section (13.2) that this integral is determined by the poles of the function that

is integrated within the contour.

Problem c: Show that the function 1=(1 + z2 ) has poles for z = +i and z = ;i.

Only the pole at z = +i is within the contour C, see gure (13.2). Since 1=(1 + z 2 ) =

1=f(z i)(z + i)g this pole is a simple (why?).

;

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