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Laplace Transform Basics

exists. If Lf(t) does converge absolutely, then

∫ τ ′ ∫ τ ′

e −st |e −st f(t)|dt → 0
f(t)dt ≤

τ tau
as τ → ∞, for all τ > τ. This then implies that Lf(t) also converges in the
′
ordinary sense of (1.1).
There is another form of convergence that is of the utmost importance from a
mathematical perspective. The integral (1.1) is said to converge uniformly for s

in some domain Ω in the complex plane if for any ε > 0, there exists some number
τ such that if τ ≥ τ , then
0
0
∫
∞
−st
e f(t)dt < ε

τ
for all s in Ω. The point here is that τ can be chosen sufficiently large in order to
0
make the “tail” of the integral arbitrarily small, independent of s.

Continuity Requirements

Since we can compute the Laplace transform for some functions and not others,
2
t
such as e , we would like to know that there is a large class of functions that do
have a Laplace transform. There is such a class once we make a few restrictions
on the functions we wish to consider.

Definition 1.2✓ A function f has a jump discontinuity at a point

t if both the limits lim f(t) = f(t − 0) and lim f(t) = f(t + 0)
0
0
0
t→t 0 −0 t→t 0 +0
exist (as finite numbers) and f(t − 0) ̸= f(t + 0).
0
0
. . . . .
Here, t → t − 0 and t → t + 0 mean that t → t from the left and right,
0
0
0
.
respectively.
Example 1.6, The function f(t) = 1 has a discontinuity at t = 3,
t−3
but it is not a jump discontinuity since neither lim f(t) nor
t→3−0
lim f(t) exists.
. . . . . . t→3+0

Example 1.7, The function

{ 2
t
e − 2 , t > 0,
f(t) =
0, t < 0

has a jump discontinuity at t = 0 and is continuous elsewhere.
. . . . .

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