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Properties of Laplace Transform

where F(s) = L(f(t)). Thus, for example

( ) ( ) ′
s sinh at
−1
L = = cosh at.
2
s − a 2 a
It may be the case that f has a jump discontinuity other than at the origin. This
can be treated in the following way.

Theorem 2.6È

Suppose that f is continuous on [0, ∞) except for a jump discontinuity

at t = t > 0, and f has exponential order α with f piecewise continuous
′
1
on [0, ∞). Then
+
L(f (t)) = sL(f(t)) − f(0) − e −t 1 s (f(t ) − f(t )) (Re(s) > α).
′
−
1
1
. . . . . . .
PROOF.
∫ ∫
∞ τ
f (t)dt = lim
e −st ′ e −st ′
f (t)dt =
0 τ→∞ 0
[ ]
t 1 τ τ
− ∫

= lim e −st f(t) + e −st f(t) + s e −st f(t)dt =

τ→∞ +
0 t 0
1
[ ∫ ]
τ
+
−
= lim e −st 1 f(t ) + f(0) + e −sτ f(τ) − e −st 1 f(t ) + s e −st f(t)dt . 2
1
1
τ→∞
0
Hence
+
′
−
L(f (t)) = sL(f(t)) − f(0) − e −st 1 (f(t ) − f(t )).
1
1
. . . .
If 0 = t < t < · · · < t are a finite number of jump discontinuities, the
n
0
1
formula becomes
n
∑
′ + −st k + −
L(f (t)) = sL(f(t)) − f(0 ) − e (f(t ) − f(t )). (2.4)
k k
k=1
Remark 2.2E If we assume that f is continuous [0, ∞) and also of exponential
order, then it follows that the same is true of f itself.
. . . . .
To see this, suppose that
αt
′
|f (t)| ≤ Me , t ≥ t , α ̸= 0.
0
Then
∫
t
′
f(t) = f (τ)dτ + f(t )
0
t 0
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