1550251515-Classical_Complex_Analysis__Gonzalez_

710 Chapter^9

lim jzaf(z)dz = 0

r-•0

(9.11-38)

'i't

lim J za f(z) dz= -i?T Reszaf(z)

6-•0 .z=p

"Y2

and

lim J za f(z)dz = 0

R-+oo

(9.11-39)

r+

by Lemmas 9.5, 9.4 and 9.3, respectively. Thus, by using (9.11-37)
in (9.11-36) (with the primes dropped), and letting S--+ 0, we obtain

ei11"a1R xaf(-x)dx+(PV) 1R xaf(x)dx+ J zaf(z)dz

'i't

-i?r~;zaf(z)-1-J zaf(z)dz=21rit~1,~zaf(z) (9.11-40)

r+ k=l

Next, we wish to let r --+ 0 and then R--+ oo. However, the first two
integrals in (9.11-4,0) involver as well as R, and we need to show that one
of them converges as r --+ O, and also as R ..:.+ oo [the convergence of the
other being assured by (9.11-40) together with (9.11-38) and (9.11-39)]. To
this effect, consider

(9.11-41)

Choose 'ff such that 0 < 'T/ < 1 and a + 'T/ > 0. Then for x > 0 we have

0 ::; xa+'I If (-x) I ::; xa+l IJ(-x )I --+ 0

as x --+ o+. Then, for any given B > 0 there is X1 > 0 such that 0 < x ::; X1
implies that

or

But J;^1 (B/x'1)dx converges as r--+ o+. Hence so does (9.11-41).