Lecture Notes in Differential Equations - Bruce E. Shapiro

332 LESSON 32. THE LAPLACE TRANSFORM
Solution. From the definition of the Laplace Transform
L [ e 2t] =
∫ ∞
0
e 2t e −st dt =
= 1
2 − s e(2−s)t ∣ ∣∣∣
∞
= 1
s − 2
0
∫ ∞
0
e (2−s)t dt (32.5)
(32.6)
so long as s > 2 (32.7)
Remark. As a generalization of the last example we can observe that
L [ e at] = 1
s − a , s > a (32.8)
Definition 32.2 (Inverse Transform). If F (s) is the Laplace Transform of
f(t) then we say that f(t) is the Inverse Laplace Transform of F (s) and
write
f(t) = L −1 [F (s)] (32.9)
Example 32.3. From the example 32.1 we can make the following observations:
1. The Laplace Transform of the function f(t) = t is the function
L[f(t)] = 1/s 2 .
2. The Inverse Laplace Transform of the function F (s) = 1/s 2 is the
function f(t) = t.
In order to prove a condition that will guarantee the existence of a Laplace
transform we will need some results from Calculus.
Definition 32.3 (Exponentially Bounded). Suppose that there exist some
constants K > 0, a, and M > 0, such that
|f(t)| ≤ Ke at (32.10)
for all t ≥ M. Then f(t) is said to be Exponentially Bounded.
Definition 32.4 (Piecewise Continuous). A function is said to be Piecewise
Continuous on an interval (a, b) if the interval can be partitioned
into a finite number of subintervals
a = t 0 < t 1 < t 2 < · · · < t n = b (32.11)
such that f(t) is continuous on each sub-interval (t i , t i+1 ) (figure 32).