Singularities/Residues/ Applications 729
replace the circle 'Yt of that figure (see Fig. 9.29). Then we have
J
fo(z) dz= 1-S fo(x) dx + J fo(z) dz+ rb-£ fo(x) dx
. a+£ ls
r+ -y~+
+ J fo(z)dz+ l~/o(x)dx+ J fo(z)dz
'Yt 'Yd
l
a+e J
+ _
8
f 0 (x)dx+ f 0 (z)dz (9.11-65)
'Yt
As in case (i), we obtain
J
27rB
fo(z) dz= y'=A
r+
lim e->0 J fo(z) dz= e->0 lim J fo(z) dz= 0
'Yt 'Yt
and
y
r+
R x
Fig. 9.29