The Calculus of Residues

Chapter 7

The Calculus of Residues

If f (z) has a pole of order m at z = z0, it can be written as Eq. (6.27), or

f (z)

=

(z)

=

a-1 (z - z0)

+

(z

a-2 - z0)2

+

...

+

(z

a-m - z0)m

,

(7.1)

where (z) is analytic in the neighborhood of z = z0. Now we have seen that if C encircles z0 once in a positive sense,

C

dz (z

1 - z0)n

=

2in,1,

(7.2)

where the Kronecker -symbol is defined by

m,n =

0, 1,

m m

= =

n, n.

.

(7.3)

Proof: By Cauchy's theorem we may take C to be a circle centered on z0. On the circle, write z = z0 + rei. Then the integral in Eq. (7.2) is

i rn-1

2

d ei(1-n),

0

(7.4)

which evidently integrates to zero if n = 1, but is 2i if n = 1. QED.

Thus if we integrate the function (7.1) on a contour C which encloses z0, while (z) is analytic on and within C, we find

f (z) dz = 2ia-1.

C

(7.5)

Because the coefficient of the (z - z0)-1 power in the Laurent expansion of f plays a special role, we give it a name, the residue of f (z) at the pole.

If C contains a number of poles of f , replace the contour C by contours , , , . . . encircling the poles singly, as shown in Fig. 7.1. The contour integral

63 Version of October 26, 2011

64 Version of October 26, 2011CHAPTER 7. THE CALCULUS OF RESIDUES

' ?i

?i ?i

$

&

C

%

Figure 7.1: Integration of a function f around the contour C which contains only poles of f may be reduced to the integrals around subcontours , , , etc., each of which contains but a single pole of f .

around C may be distorted to a sum of disjoint ones around , , . . . , so

f (z) dz = f (z) dz + f (z) dz + . . . ,

C

(7.6)

and since each small contour integral gives 2i times the reside of the single pole interior to that contour, we have established the residue theorem: If f be analytic on and within a contour C except for a number of poles within,

f (z) dz = 2i

residues,

C

poles within C

(7.7)

where the sum is carried out over all the poles contained within C. This result is very usefully employed in evaluating definite integrals, as the

following examples show.

7.1 Example 1

Consider the following integral over an angle:

I=

2 0

1

-

2p

d cos

+

p2

,

0 < p < 1.

Let us introduce a complex variable according to

z = ei, dz = iei d = iz d,

(7.8) (7.9)

so that

cos

=

1 2

z

+

1 z

.

(7.10)

Therefore, we can rewrite the angular integral as an integral around a closed contour C which is a unit circle about the origin:

I=

dz

1

C

iz 1 - p

z

+

1 z

+ p2

7.2. A FORMULA FOR THE RESIDUE

65 Version of October 26, 2011

=

dz

1

C i z - p(z2 + 1) + p2z

=

1 i

C

dz

(1

-

1 pz)(z

-

p)

.

(7.11)

The integrand exhibits two poles, one at z = 1/p > 1 and one at z = p < 1. Only the latter is inside the contour C, so since

1

1 - pz

z

1 -

p

=

z

1 -

p

+

1

p - pz

1

1 -

p2

,

(7.12)

we have from the residue theorem

I

=

2i

1 i

1

1 -

p2

=

1

2 - p2

.

(7.13)

Note that we could have obtained the residue without partial fractioning by evaluating the coefficient of 1/(z - p) at z = p:

1 1 - pz

z=p

=

1

1 -

p2

.

This observation is generalized in the following.

(7.14)

7.2 A Formula for the Residue

If f (z) has a pole of order m at z = z0, the residue of that pole is

a-1

=

1

dm-1

(m - 1)! dzm-1

[(z - z0)mf (z)]

.

z=z0

The proof follows immediately from Eq. (7.1).

(7.15)

7.3 Example 2

This time we consider an integral along the real line,

I=

dx

-

(x2

1 + 1)3

=

lim

R

R -R

dx

(x2

1 +

1)3

,

(7.16)

where we have made explicit the meaning of the upper and lower limits. We relate this to a contour integral as sketched in Fig. 7.2. Thus we have

C

dz (z2 + 1)3

=

R -R

dx (x2 + 1)3

+

dz (z2 + 1)3

,

(7.17)

66 Version of October 26, 2011CHAPTER 7. THE CALCULUS OF RESIDUES

d s

d R d

d

d ?i

-R

d

R

? -i

Figure 7.2: The closed contour C consists of the portion of the real axis between -R and R, and the semicircle of radius R in the upper half plane. Also shown in the figure are the location of the poles of the integrand in Eq. (7.17).

where we are to understand that the limit R is to be taken at the end

of the calculation. It is easy to see that the integral over the large semicircle vanishes in this limit:

dz (z2 + 1)3 =

0

R i eid (R2e2i + 1)3

0,

R .

(7.18)

Hence the integral desired is just the closed contour integral,

I=

C

dz (z2 + 1)3

=

2i(residue

at

i).

(7.19)

By the formula (7.15) the desired residue is

a-1

=

1 d2 2! dz2

(z

-

i)3

(z

-

1 i)3(z

+

i)3

z=i

=

1 d2 1 2! dz2 (z + i)3

z=i

1 (-3)(-4) = 2! (z + i)5 z=i

=

3 16i

,

(7.20)

so

I

=

3 8

.

(7.21)

7.4 Jordan's Lemma

The evaluation of a class of integrals depends upon this lemma. If f (z) 0 uniformly with respect to arg z as |z| for 0 arg z , and f (z) is analytic when |z| > c > 0 and 0 arg z , then for > 0,

lim eizf (z) dz = 0,

(7.22)

7.5. EXAMPLE 3

67 Version of October 26, 2011

where is a semicircle of radius above the real axis with center at the origin. (Cf. Fig. 7.2.)

Proof: Putting in polar coordinates,

eizf (z) dz =

ei( cos +i sin )f ei eii d.

0

(7.23)

If we take the absolute value of this equation, we obtain the inequality

eizf (z) dz

e- sin f ei

0

< e- sin d,

0

d

(7.24)

if f ei < for all when is sufficiently large. (This is what we mean by going to zero uniformly for large .) Now when

0

2

,

sin

2

,

(7.25)

which is easily verified geometrically. Therefore, the integral on the right-hand side of Eq. (7.24) is bounded as follows,

/2

e- sin d < 2

e-2/ d

0

0

=

1 - e-

.

(7.26)

Hence

eizf (z) dz

<

1 - e-

(7.27)

may be made as small as we like by merely choosing large enough (so 0). QED.

7.5 Example 3

Consider the integral

I=

0

cos x x2 + a2

dx.

The associated contour integral is

(7.28)

C

eiz z2 + a2

dz

=

R -R

eix x2 + a2

dx

+

eiz z2 + a2

dz,

(7.29)

where the contour is a large semicircle of radius R centered on the origin in the upper half plane, as in Fig. 7.2. (The only difference here is that the

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