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Laplace Transform Basics

Definition 1.4✓ A function f has exponential order α if there
exist constants M > 0 and α such that for some t ≥ 0,
0

αt
|f(t)| ≤ Me , t ≥ t .
0
. . . . .
at
Clearly, the exponential function e has exponential order α = a, whereas t n
has exponential order α for any α > 0 and any n ∈ N, and bounded functions like
2
t
sin t, cos t have exponential order 0, whereas e −t has order −1. However, e does
not have exponential order. Note that if β > α, then exponential order α implies
βt
exponential order β, since e αt ≤ e , t ≥ 0. We customarily state the order as the
smallest value of α that works, and if the value itself is not significant it may be
suppressed altogether.

The Class L

We now show that a large class of functions possesses a Laplace transform.
Theorem 1.1È

If f is piecewise continuous on [0, ∞) and of exponential order α, then the
Laplace transform L(f) exists for Re(s) > α and converges absolutely.
. . . . . . .

PROOF. First,
αt
|f(t)| ≤ M e , t ≥ t ,
0
1
for some real α. Also, f is piecewise continuous on [0, t ] and hence bounded there (the
0
bound being just the largest bound over all the subintervals), say

|f(t)| ≤ M , 0 < t < t .
0
2
αt
Since e has a positive minimum on [0, t ], a constant M can be chosen sufficiently
0
large so that
αt
|f(t)| ≤ Me , t > 0.
Therefore,

τ τ Me −(x−α)t
∫ ∫ τ
|e −st f(t)|dt ≤ M e −(x−α)t dt = =
0 0 −(x − α) 0
M Me −(x−α)τ
= − .
x − α x − α

Letting τ → ∞ and noting that Re(s) = x > α yield

∫
∞ M
|e −st f(t)|dt ≤ . (1.4)
. . . . 0 x − α

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