Page 71 - Calculus with Complex Numbers

The lirst term on the right-hand side

As for the second term we have

1 ''<' (ïcln 1 ''<' rn c-r - 1
)-y n !
z s - )-y - = r ,
r
!

n
-
1 1
which gives
as r -+ 0. Therefore

as r --> 0.
This is a particular case of thehalf residuetheorem. lf y were a full circle centre
0, radius r then
d iz
- dz = ln'i
/4 Z

for all r > 0, since the residue of the integrand at z = 0 is 1. The half residue
theorem states that if y is a semicircle then the integral converges to half this value
as r --> 0 (see Appendix 2).
Piecing all this together we have

and hence
oo sin/ zr
dt = - .
o t l