1550251515-Classical_Complex_Analysis__Gonzalez_

692 Chapter9

More generally, for any integral of the form

1

(^00) xmdx
o xn +an

where m and n are positive integers with m :::; n - 2 and a > 0, we may

proceed in a similar manner by using now as a contour c+ the boundary

of a circular sector with the initial ray along the real axis (as in Fig. 9.14)

and the terminal side with argument 27r /n. If R > a, this sector contains

only one pole of f(z) = zm /(zn +an), namely, b 0 = aein:/n, and the residue

of f at bo is

Res J = ___!!L = .! b:J'-n+l = .!_ am-n+l ei(m-n+l)n:/n

z=bo nb~-l n n

Hence with the same notation as above, we have

1

R _xm dx + J zm dz - ei(m+1)2n:/n 1R xm dx

o xn + an zn + an o xn + an

r+

= 27ri am-n+lei(m-n+I)n:/n (9.11-7)

n

Letting R --t oo, the second integral on the left tends to zero (by

Corollary 9.5), and (9.11-7) yields

{oo xm dx

Jo xn +an

27ri ei(m+I)n:/ne-in:

= -am-n+l __ ~~-~

n · 1 _ ei(m+l)2n:/n

7r am-n+l

=

;:; sin[(m + l)7r/n]

(9.11-8)

Note The preceding method does not require that the integrand be even.

Formula (9.11-8) is due to 0. J. Farrell and B. Ross [4]. For

generalizations of this result, see [17] and [22].

Type II. J!°: f( x) dx, where f ( x) satisfies the same conditions as in

type I, except that [ is not even.

Both integrals f_ 00 f ( x) dx and f 0

00

f ( x) dx exist because of condi-

tions (2) and ( 4). Hence the value of the given integral is the same as

limR-+oo J!:R f( x) dx. Thus we get from (9.11-2),

1

+00 m

-oo f(x)dx=27ri~!l~~f(z)

k=l

(9.11-9)

As before, this formula applies, in particular, to rational functions f(z) =

P(z)/Q(z) with the degree of Q at least two units greater than that of P.

Example To evaluate J~:[(x + 1)/(x^4 + 1)] dx. The poles of the inte-

grand lying on the upper half-plane are bk = ei(n:/4+kn:/Z) (k = 0, 1), and