11.5. THE METHOD OF RESIDUES 263

In taking the limit, you can use L’Hospital’s rule provided you have an indeterminate formbecause you know the limit exists. Thus you could take a limit along a one dimensional linea+ t as t → 0 and you can apply L’Hospital’s rule to the real and imaginary parts. Thus,you can get the answer by pretending z is a real variable and using the usual techniquesfor functions of a real variable.

I want to consider contours like the above and functions which are of the form

f (z) = g(z)+m

∑j=1

S (a j,z) (11.8)

where g(z) is analytic and the S (a j,z) are the terms which yield a pole at a j. Consider γ

centered at a with radius r ∫γ

S (a j,z)dz =p

∑k=1

∫γ

bk

(z−a j)k dz.

One of these terms is∫γ

bk

(z−a j)k dz = bk

∫ 2π

0

1rkeikt ireitdt =

{0 if k > 1b1i∫ 2π

0 dt = b12πi if k = 1

Thus, if the (z−a j)−1 coefficient in 11.8 is b j

1,∫γ0

f (z)dz =m

∑j=1

∫γ j

g(z)dz+m

∑j=1

∫γ j

S (a j,z)dz

=m

∑j=1

∫γ j

b j1

z−a jdz = 2πi res( f ,a j)

This is the residue method. You can use it to compute very obnoxious improper integralsand this is illustrated next.

Letting p(x) ,q(x) be polynomials, you can use the above method of residues to evalu-ate obnoxious integrals of the form∫

∞

−∞

p(x)q(x)

dx≡ limR→∞

∫ R

−R

p(x)q(x)

dx

provided the degree of p(x) is two less than the degree of q(x) and the zeros of q(z) involveIm(z) > 0. Of course if the degree of p(x) is larger than that of q(x) , you would do longdivision. The contour to use for such problems is γR which goes from (−R,0) to (R,0)along the real line and then on the semicircle of radius R from (R,0) to (−R,0).

x

y

Letting CR be the circular part of this contour, for large R,∣∣∣∣∫CR

p(z)q(z)

dz∣∣∣∣≤ πR

CRk

Rk+2