0
1/t
Figure 17.4An infinite discontinuity.
Note that the value of the integrals involved is unaffected by the absence of the isolated
points such ast=1, 2. The Laplace transform off(t)is
L[f(t)]=
∫∞
0
f(t)e−stdt=
∫ 2
1
e−stdt=
[
e−st
−s
] 2
1
=
e−s−e−^2 s
s
This result again illustrates the useful property of the Laplace (and other integral) trans-
forms, that the transform of a discontinuous function may be continuous. Taking the
transformimprovesthe behaviour of the function.
Exercises on 17.3
- Derive each of the Laplace transforms in Table 17.1. The results for sin and cos are
obtained by integration by parts and from these differentiations with respect toωwill
give the results fortsinωtandtcosωt. - Find the Laplace transform of the piecewise continuous function
f(t)=− 10 <t< 2
=t 2 <t< 3
= 03 <t
Answers
(
1
s^2
+
3
s
)
(e−^2 s−e−^3 s)−
1
s