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Uniform Convergence

Thus the Laplace integral converges absolutely in this instance (and hence converges)

. . . . for Re(s) > α. 2
Let us define the class L as the set of those real- or complex- valued functions

defined on the open interval (0, ∞) for which the Laplace transform (defined in
terms of the Riemann integral) exists for some value of s. It is known that
whenever F(s) = Lf(t) exists for some value s , then F(s) exists for all s with
0
Re(s) > Re(s ), that is, the Laplace transform exists for all s in some right
0
half-plane. By Theorem 1, piecewise continuous functions on [0, ∞) having
exponential order belong to L. In what follows, we will denote the class of all
piecewise continuous functions with exponential order by PE. It can be proved
that any linear combination of functions in PE is also in PE. The same is true for

the product of two functions in PE.
However, there certainly are functions in L that do not satisfy one or both of
.
these conditions.

2
2
t
t
Example 1.9, Consider f(t) = 2te cos e . Prove that f ∈ L.
.
. . . . .
Solution. Then f(t) is continuous on [0, ∞) but not of exponential order.
However, the Laplace transform of f(t),

∫
∞ 2 2
t
t
L(f(t)) = e −st 2te cos e dt,
0
exists, since integration by parts yields (Re(s) > 0)

∫
∞ ∞ 2 2
2

t
t
L(f(t)) = e −st sin e t + s e −st sin e dt = − sin 1 + sL(sin e )

0 0
and the latter Laplace transform exists by Theorem 1. Thus we have
a continuous function that is not of exponential order yet nevertheless
. . . . possesses a Laplace transform.

1
Another example is the function f(t) = √ . Later we will compute its actual
t
Laplace transform in the context of the gamma function. While f(t) has
exponential order α = 0 (|f(t)| ≤ 1, t ≥ 1), it is not piecewise continuous on
[0, ∞) since f(t) → ∞ as t → 0+, that is, t = 0 is not a jump discontinuity.

Uniform Convergence

We have already seen by Theorem 1 that for functions f that are piecewise

continuous on [0, ∞) and of exponential order, the Laplace integral converges
∫ ∞
absolutely, that is, |e −st f(t)|dt converges. Moreover, for such functions the
0
Laplace integral converges uniformly.

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