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
(provided s > 0)
(b) f (t)= k. From equation (2),
L{k} = kL{1}
Hence L{k} = k
1
s
=
k
s
, from (a) above.
(c) f (t) = e at (where a is a real constant = 0).
From equation (1),
L{e
at
} =
∞
0
e
−st
(e
at
) dt =
∞
0
e
−(s−a)t dt,
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) = cos at (where a is a real constant).
From equation (1),
L{cos at } =
∞
0
e
−st cos at dt
=
e
−st
s 2 + a 2 (a sin at − s cos at )
∞
0
by integration by parts twice (see page 423),
=
e −s(∞)
s 2 + a 2 (a sin a(∞) − s cos a(∞))
−
e 0
s 2 + a 2 (a sin 0 − s cos 0)
=
s
s 2 + a 2 ( provided s > 0)
(e) f (t) = t. From equation (1),
L{t } =
∞
0
e
−st t dt =
t e −st
−s
−
e −st
−s
dt
∞
0
=
t e −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 (provided s > 0)
(f) f (t)= t n (where n =0, 1, 2, 3, …).
By a similar method to (e) it may be shown
that L{t 2 } =
2
s 3 and L{t 3 } =
(3)(2)
s 4 =
3!
s 4 . These
results can be extended to n being any positive
integer.
Thus L{t n } =
n!
s n+1 provided s > 0)
(g) f (t)= sinh at. From Chapter 5,
sinh at =
1
2
(e at − e −at ). Hence,
L{sinh at } = L
1
2
e
at
−
1
2
e
−at
=
1
2
L{e
at
} −
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 (provided s > 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 + 2t −
1
3
t
4
(b) L{5e
2t
− 3e
−t
}.