Complex Analysis - Maths KU

380 Chapter 6 Series and Residues
s n
s 2
s 4
s 3
y
s1 O
γ + iR
γ
L R
γ – iR
C R
Figure 6.24 Contour for inversion
integral (7) for t 0.
Solution Considered as a function of a complex variable s, the function
F (s) =1 � s 3 has a pole of order 3 at s = 0.Thus by (9) and (2) of Section
6.5:
f(t) =� −1
�
1
s3 � �
= Res e
= 1
2 lim
d
s→0
2
est
ds2 = 1
2 lim
s→0 t2 e st
= 1
2 t2 .
�
st 1
, 0 =
s3 1
2 lim
d
s→0
2
ds
s 3
2 (s − 0)3 est
Those readers familiar with the Laplace transform recognize that the answer
in Example 1 is consistent (for n = 2) with the result � {tn } = n!/sn+1 found in all tables of Laplace transforms.
The Laplace transform (1) utilizes only the values of a function f(t) for
t>0, and so f is often taken to be 0 for t0.Although we shall not delve into details, the inversion integral
(7) can be derived from a result known as the Fourier integral formula.In
that analysis it is shown that
� γ+i∞
1
e
2πi γ−i∞
st ⎧
⎨ f(t), t > 0
F (s)ds =
(16)
⎩ 0, t < 0.
This result is hinted�at in the proof of Theorem 6.25. Notice from (15) that
the conclusion lim CR R→∞
estF (s) ds =0isnot valid for t