Introduction to Laplace transforms 583
(a) f(t)= 1. From equation (1),
L{ 1 }=
∫∞
0
e−st( 1 )dt=
[
e−st
−s
]∞
0
=−
1
s
[e−s(∞)−e^0 ]=−
1
s
[0−1]
=
1
s
(provideds> 0 )
(b) f(t)=k. From equation (2),
L{k}=kL{ 1 }
HenceL{k}=k
(
1
s
)
=
k
s
,from (a) above.
(c) f(t)=eat(whereais a real constant=0).
From equation (1),
L{eat}=
∫∞
0
e−st(eat)dt=
∫∞
0
e−(s−a)tdt,
from the laws of indices,
=
[
e−(s−a)t
−(s−a)
]∞
0
=
1
−(s−a)
( 0 − 1 )
=
1
s−a
(provided(s−a)> 0 ,i.e.s>a)
(d) f(t)=cosat(where a is a real constant).
From equation (1),
L{cosat}=
∫∞
0
e−stcosatdt
=
[
e−st
s^2 +a^2
(asinat−scosat)
]∞
0
by integration by parts twice (see page 423),
=
[
e−s(∞)
s^2 +a^2
(asina(∞)−scosa(∞))
−
e^0
s^2 +a^2
(asin0−scos0)
]
=
s
s^2 +a^2
(provideds> 0 )
(e) f(t)=t. From equation (1),
L{t}=
∫∞
0
e−sttdt=
[
te−st
−s
−
∫
e−st
−s
dt
]∞
0
=
[
te−st
−s
−
e−st
s^2
]∞
0
by integration by parts,
=
[
∞e−s(∞)
−s
−
e−s(∞)
s^2
]
−
[
0 −
e^0
s^2
]
=( 0 − 0 )−
(
0 −
1
s^2
)
since(∞× 0 )= 0 ,
=
1
s^2
(provideds> 0 )
(f) f(t)=tn(wheren=0, 1, 2, 3, ...).
By a similar method to (e) it may be shown
thatL{t^2 }=
2
s^3
andL{t^3 }=
( 3 )( 2 )
s^4
=
3!
s^4
.These
results can be extended tonbeing any positive
integer.
ThusL{tn}=
n!
sn+^1
provideds> 0 )
(g) f(t)=sinhat. From Chapter 5,
sinhat=
1
2
(eat−e−at). Hence,
L{sinhat}=L
{
1
2
eat−
1
2
e−at
}
=
1
2
L{eat}−
1
2
L{e−at}
from equations (2) and (3),
=
1
2
[
1
s−a
]
−
1
2
[
1
s+a
]
from (c) above,
=
1
2
[
1
s−a
−
1
s+a
]
=
a
s^2 −a^2
(provideds>a)
A list of elementary standard Laplace transforms are
summarized in Table 61.1.
61.5 Worked problems on standard
Laplace transforms
Problem 1. Using a standard list of Laplace
transforms determine the following:
(a)L
{
1 + 2 t−
1
3
t^4
}
(b)L{5e^2 t−3e−t}.