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Exponential Order
.

Example 1.8, The function

{
0, t < 0,
f(t) =
1
cos , t > 0
t
is discontinuous at t = 0, but lim f(t) fails to exist, so f does
t→0+
not have a jump discontinuity at t = 0.
. . . . .

The class of functions for which we consider the Laplace transform defined will
have the following property.

Definition 1.3✓ A function f is piecewise continuous on the
interval [0, ∞) if (i) lim f(t) = f(0+) exists and (ii) f is continuous
t→0+
on every finite interval (0, b) except possibly at a finite number
of points τ , τ , . . . , τ in (0, b) at which f has a jump discontinuity.
1
2
n
. . . . .
The function in Example 1 is not piecewise continuous on [0, ∞). Nor is the
function in Example 1. However, the function in Example 1 is piecewise
continuous on [0, ∞).
An important consequence of piecewise continuity is that on each subinterval
the function f is also bounded. That is to say,

|f(t)| ≤ M , τ < t < τ i+1 , i = 1, 2, . . . , n − 1,
i
i
for finite constants M . Exercises 1.3 11 In order to integrate piecewise continuous
i
functions from 0 to b, one simply integrates f over each of the subintervals and

takes the sum of these integrals. This can be done since the function f is both
continuous and bounded on each subinterval and thus on each has a well-defined
(Riemann) integral.

Exponential Order

The second consideration of our class of functions possessing a well-defined
Laplace transform has to do with the growth rate of the functions. In the

definition
∫
∞
L(f(t)) = e −st f(t)dt,
0
when we take s > 0 or Re(s) > 0, the integral will converge as long as f does not
2
t
grow too rapidly. It can be proved that f(t) = e does grow too rapidly for our
purposes. A suitable rate of growth can be made explicit.

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