Complex Analysis - Maths KU

378 Chapter 6 Series and Residues
C R
s n
s 2
s 4
s 3
y
s1 O
γ + iR
γ
L R
γ – iR
Figure 6.23 Possible contour that
could be used to evaluate (7)
x
The limit in (7), which defines a principal value of the integral, is usually
written as
f(t) =� −1 {F (s)} = 1
2πi
� γ+i∞
γ−i∞
e st F (s) ds, (8)
where the limits of integration indicate that the integration is along the infinitely
long vertical-line contour Re(s) =x = γ.Here γ is a positive real
constant greater than c and greater than all the real parts of the singularities
in the left half-plane.The integral in (8) is called a Bromwich contour integral.Relating
(8) back to (3), we see that the kernel of the inverse transform
is H(s, t) =est /2πi.
The fact that F (s) has singularities s1, s2, ...,sn to the left of the line
x = γ makes it possible for us to evaluate (7) by using an appropriate closed
contour encircling the singularities.A closed contour C that is commonly
used consists of a semicircle CR of radius R centered at (γ, 0) and a vertical
line segment LR parallel to the y-axis passing through the point (γ, 0) and
extending from y = γ −iR to y = γ +iR .See Figure 6.23.We take the radius
R of the semicircle to be larger than the largest number in set of moduli of
the singularities {|s1|, |s2|, ...,|sn|}, that is, large enough so that all the
singularities lie within the semicircular region.With the contour C chosen in
this manner, (7) can often be evaluated using Cauchy’s residue theorem.If
we allow the radius R of the semicircle to approach ∞, the vertical part of
the contour approaches the infinite vertical line that is the contour in (8).
We use the contour just described in the proof of the following theorem.
Theorem 6.25 Inverse Laplace Transform
Suppose F (s) is a Laplace transform that has a finite number of poles
s1, s2, ... , sn to the left of the vertical line Re(s) =γ and that C is the
contour illustrated in Figure 6.23. If sF (s) is bounded as R →∞, then
� −1 {F (s)} =
n�
k=1
Res � e st �
F (s),sk . (9)
Proof From Figure 6.23 and Cauchy’s residue theorem, we have
or
�
e st �
F (s) ds +
CR
� γ+iR
1
2πi
γ−iR
LR
e st F (s) ds =
e st F (s) ds =2πi
n�
k=1
n�
k=1
Res � e st �
F (s), sk
Res � e st �
� 1
F (s), sk −
2πi
CR
e st F (s) ds. (10)
The �theorem
is justified by letting R → ∞ in (10) and showing that
lim CR R→∞
estF (s) ds = 0.Now if the semicircle CR is parametrized by