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Infinitely Many Poles
.

writing (a , b ̸= 0)
n
m
n
a s + a s n−1 + · · · + a a + a n−1 + · · · + a 0
n
n
F(s) = n n−1 0 = ( s s ),
m
b s + b m−1 s m−1 + · · · + b 0 s m−n b + b m−1 + · · · + s b 0
m
m
m
s
it is enough to observe that for |s| suitably large,
a
0
n−1 a
a + + · · · + ≤ |a | + |a n−1 | + · · · + |a | = c ,
n
n
0
1
s s n

b b |b
0
m
m−1 0 m−1 | |b | |b |
b + + · · · + ≥ |b | − − · · · − ≥ = c ,
m
m
2
s s m |s| |s| m 2
and thus
c /c 2
1
|F(s)| ≤ .
|s| m−n
Hence,
z k t
e P(z )
Res(z ) = k , k = 1, 2, . . . , m,
k
Q (z )
′
k
and
m
∑ P(z )
k
z k t
f(t) = e .
Q (z )
′
k=1 k
. . . . .
Infinitely Many Poles
∞
Suppose that F(s) has infinitely many poles at {z } all to the left of the line
k k=1
Re(s) = x > 0, and that
0
|z | ≤ |z | ≤ · · · ,
2
1
where |z | → ∞ as k → ∞. Choose a sequence of contours Γ = C ∪ [x −
n
n
k
0
iy , x + iy ] enclosing the first n poles z , z , . . . , z as in Figure 4.3. Then by the
2
0
n
n
1
n
Cauchy residue theorem,
∫ n
1 ∑
ts
e F(s)ds = Res(z ),
k
2πi
Γ n
k=1
ts
where as before, Res(z ) is the residue of e F(s) at the pole s = z .
k
k
Hence
n ∫ ∫
∑ 1 x 0 +iy n 1
ts
ts
Res(z ) = e F(s)ds + e F(s)ds.
k
2πi 2πi
k=1 x 0 −iy n C n
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