1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/Applications 701

line calls for an indentation '"ts· Proceeding as in types III and IV, we

find that

(PV) /

00

eiax f(x) dx = 27ri [f ~s eiaz f(z) + % t ~es eiaz f(z)]

-oo k=l z-b1c s=l Z-Ps

(9.11-23)

where, as before, the bk (k = 1, ... , m) are the poles of eiaz f(z) lying in

the upper half-plane. Because eiaz is entire and never vanishes, the poles

of eiazf(z) are precisely those of f(z). Since

(PV) 1: [F(x) + G(x )] dx = (PV) l: F(x) dx + (PV) l: G(x) dx

it suffices to equate real and imaginary parts in (9.11-23) to evaluate the

given integrals.

The reader will note that if in either

(PV) l: f(x) cos ax dx or (PV) l: f(x)sinaxdx

all the poles of f ( x) on the real axis are simple and coincide with some of

the zeros of either cos ax or sin ax, then those singularities are removable
and the PV indication before the corresponding integral can be omitted,
the integrals as improper integrals of the first kind being convergent in the
ordinary sense, as shown in type III. In particular, (9.11-23) applies to
rational functions f(z) = P(z)/Q(z) where P and Q are prime to each
other and the 'degree of Q exceeds that of P at least by one.

Examples 1. To evaluate

and (PV) l: [sinax/(x^2 - p^2 )] dx,

where a > 0 and p > 0.

Consider the integral Ia+ [eiax /(z^2 - p^2 )] dz, where the contour c+ is

as shown in Fig. 9.18 with R > p and indentations at the simple poles p

and -p lying on the real axis.

In this case the integrand has no poles enclosed by c+' and the residues

at the real poles are

eiaz eiap

Res =

z=p z2 - p2 2p '

eiaz e-iap

Res =

z=-p z^2 - p2 -2p

Hence by (9.11-23) we have

(PV) Joo :iax 2 dx =

-oo x - p

7ri.. 7r

• ( e•ap - e-iap) = - - sin ap
  2p p