1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/Applicatio11s

y

y=0

y=20hr

(^0) e
Fig. 9.9

Then if a > 0, we have

'

lim j eiaz f(z) dz= 0

R--+oo

'Y

(Fig. 9.10).

Proof For every e > 0 there exist· Rf such that

lf(z)I < e for R >Rf

. Hence if R > max( Ro, Rf), we have

J eiaz J(z) dz S eR lp e-aRsin e d() S eR 1-n: e-aRsin 8 d()

'Y

681

(9.10-2)

{1/2-n: 11/2-n:

= 2eR Jo e-aRsin8 d() S 2eR o e-aR(29/-n:) d()

_ - f.7r - (1 -e -aR) < -f.7r

a a

x

Fig. 9.10