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Laplace Transform Basics

To see this, suppose that

αt
|f(t)| ≤ Me , t ≥ t .
0
Then

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

= = ,
−(x − α) x − α
t 0
provided Re(s) > α. Taking x ≥ x > α gives an upper bound for the last
0
expression:
Me −(x−α)t 0 M
≤ e (x 0 −α)t 0 . (1.5)
x − α x − α
0
By choosing t sufficiently large, we can make the term on the right- hand side
0
of (1.5) arbitrarily small; that is, given any ε > 0, there exists a value T > 0 such

that
∫
∞

e −st f(t)dt < ε,

t 0
whenever t ≥ T for all values of s with Re(s) ≥ x > α. This is precisely the
0
0
condition required for the uniform convergence of the Laplace integral in the
region Re(s) ≥ x 0 > α. The importance of the uniform convergence of the
Laplace transform cannot be overemphasized, as it is instrumental in the proofs
of many results.
A general property of the Laplace transform that becomes apparent from an

inspection of the table of Laplace transforms is the following.
Theorem 1.2È

If f is piecewise continuous on [0, ∞) and has exponential order α, then

F(s) = L(f(t)) → 0

as Re(s) → ∞.
. . . . . . .
In fact, by (1.4)

∫
∞
−st M
e f(t)dt ≤ , (Re(s) = x > α),

x − α
0
and letting x → ∞ gives the result.

Remark 1.1E As it turns out, F(s) → 0 as Re(s) → ∞ whenever the Laplace
transform exists, that is, for all f ∈ L. As a consequence, any function F(s) without
s
2
this behaviour, say (s − 1)/(s + 1), e /s, or s , cannot be the Laplace transform of
. . . . .

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