Complex Analysis - Maths KU

376 Chapter 6 Series and Residues
y
a t1 t2 t3 b
Figure 6.20 Piecewise continuity on
[0, ∞)
y
T
Me ct (c > 0)
f(t)
Figure 6.21 Exponential order
y
e t
t
cos t
Figure 6.22 f(t) = cos t is of
exponential order c =0.
t
t
EXAMPLE 1 Existence of a Laplace Transform
The Laplace transform of f(t) =1,t ≥ 0is
� ∞
� {1} =
0
e −st � b
(1) dt = lim
b→∞ 0
−e
= lim
b→∞
−st
s
If s is a complex variable, s = x + iy, then recall
0
e −st dt
�b
�
�
1 − e
� = lim
b→∞
−sb
. (5)
s
e −sb = e −bx (cos by + i sin by). (6)
From (6) we see in (5) that e−sb → 0asb→∞if x>0.In other words, (5)
gives � {1} = 1
, provided Re(s) > 0.
s
Existence of � {f(t)} Conditions that are sufficient to guarantee
the existence of � {f(t)} are that f be piecewise continuous on [0, ∞) and
that f be of exponential order.Recall from elementary calculus, piecewise
continuity on [0, ∞) means that on any interval there are at most a finite
number of points tk, k = 1, 2, ... , n, tk−1 < tk, at which f has finite
discontinuities and is continuous on each open interval tk−1 0 so that | f(t) |≤Me ct , for t>T.The
condition | f(t) |≤Me ct for t>T states that the graph of f on the interval
(T, ∞) does not grow faster than the graph of the exponential function Me ct .
See Figure 6.21. Alternatively, e −ct | f(t) | is bounded; that is, e −ct | f(t) |≤M
for t>T.As can be seen in Figure 6.22, the function f(t) = cos t, t ≥ 0
is of exponential order c = 0 for t>0.Indeed, it follows that all bounded
functions are necessarily of exponential order c =0.
Theorem 6.22 Sufficient Conditions for Existence
Suppose f is piecewise continuous on [0, ∞) and of exponential order c
for t>T.Then � {f(t)} exists for Re(s) >c.
Proof By the additive interval property of definite integrals,
� {f(t)} =
� T
0
e −st � ∞
f(t) dt + e −st f(t) dt = I1 + I2.
The integral I1 exists since it can be written as a sum of integrals over intervals
on which e −st f(t) is continuous.To prove the existence of I2, we let s be a
T