Page 132 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

122 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
2 3
4 5
1 s ς s ς
Since the function sinh ς s = ς + + + ... the function
s 3! 5!
1
sinh ς s
s
is analytic and Y (s ) can be written as the ratio of two analytic functions
1 sinh ς s
s
Y (s ) =
s cosh s
Y (s ) therefore has a simple pole at s = 0 and the residue there is
2 3
s ς
lim lim ς + 3! + ...
R(s = 0) = sY (s ) = = gς
s → 0 s → 0 cosh s
The remaining poles are the singularities of cosh s . But cosh s = cosh x cos y + i sinh x sin y,
so the zeros of this function are at x = 0and cosy = 0.
Hence, s n = i(2n − 1)π/2. The residues at these points are

lim g sinh ς s s τ g sinh ς s n s n τ
R(s = s n ) = d e = 2 e (n =±1, ±2, ±3 ...)
s → s n s (s cosh s ) s n sinh s n
ds
Since
2n − 1 2n − 1

sinh i (πς) = i sin (πς)
2 2
we have
2n−1
gi sin (πς) 2n − 1
R(s = s n ) = 2 exp i πτ
2n−1 2n−1

2
2
− π i sin π
2 2
and
2n − 1

sin π = (−1) n+1
s
The exponential function can be written as
2n − 1 2n − 1 2n − 1

exp i πτ = cos πτ + i sin πτ
2 2 2
Note that for the poles on the negative imaginary axis (n < 0) this expression can be
written as

2m − 1 2m − 1 2m − 1
exp i πτ = cos πτ − i sin πτ
2 2 2
where m =−n > 0. This corresponds to the conjugate poles.