Singularities/Residues/ Applicationsprovided thatlira J estf(s)ds = 0
R->co
r+721(9.11-60)A sufficient condition for (9.11-60) to hold is that f(s) = 0(1/lsi°'), a:> 0,
as isl = R -t oo, 81 ~ 8 ~ 27r - 81.
If f( s) has infinitely many isolated singularities bk, the preceding method
still can be applied if r+ is replaced by a sequence of arcs r;t with increasing
radii Rn chosen so that the corresponding C;t ~ill not pass through any
singularity. In this case, letting Rn -t oo in the equation corresponding
to (9.11-58), we obtainF(t) = 2 :i 1~:~co estf(s) ds = ~ ~i~ estf(s) (9.11-61)
which is valid as long as (9.11-60) holds and the series of residues on the
right of (9.11-61) converges.
Example Suppose that f(s) = sinhas/(s^2 coshs), 0 < a < 1. Here
f(s) = 0(1/lsi^2 ) as Rn -t oo, and f(s) has a simple pole at s = 0. The
residue of est f ( s) at this pole is
est sinhaslira =a
s->O s coshsAlso, f(s) has simple poles at the zeros of coshs, namely, at the points
bk=^1 /i2k + 1)7ri (k = 0,±1,±2, ... ), and
ebkt sinh abk e<^1!^2 H^2 k+1)11"it sinh %a(2k + l)7ri
Res estf(s) = = --------'--=---'----'--
s=bk b~ sinhbk -^1 /i2k + 1)^2 7r^2 sinh %(2k + l)7ri
= 4(-1 )k+l e(l/2)(2k+l)'ll"it sin ~ a(2k + 1 )7r
(2k + 1 )^2 7r^2 2Hence
4 +co (-l)k+^1 · 1
F(t) =a+
2
L e<^1 /^2 )(^2 k+l)'ll"it sin -a(2k + l)7r
7r k=-co (2k + 1)^2 2s co c-1r 1 1
=a+ 2 L ( ) 2 cos -(2m - l)7rtsin -(2m - l)a7r
7r m=l 2m - 1 2 2
Clearly, the series above converges absolutely and uniformly with respect
tot since it is dominated by I::= 1 l/(2m -1)^2.
Type X. J: f(x) dx, where f(x) has an algebraic multiple-valued exten-
sion to the complex plane and the points a and b are algebraic branch points