The Laplace transform is an integral transform that can be used to solve linear differential equations. It transforms a function f(t) into another function F(s) called its Laplace transform. To take the Laplace transform, the function f(t) must be piecewise continuous and of exponential order. If f(t) satisfies these conditions, its Laplace transform F(s) will exist for all values of s greater than the constant a in the exponential order condition. Some common elementary functions and their Laplace transforms are provided as examples.

1. The Laplace Transform
Among the tools are useful for solving differential linear equations are integral
transforms.
Definition: An integral transform is a relation of the form
F(s) = K(s,t)f(t)dt
a
b
!
where K(s,t) is a given function of the variables s and t, called the kernel of the
transformation and the limits of integration a and b are given and it is possible that a = -∞
and b = ∞.
The relation transforms the function f(t) into another function F(s), which is called the
transform of f.
The general idea of using an integral transform to solve a linear differential equation is to
transform a problem for an unknown function f into a simpler problem for the function F.
Then, solve the simpler problem to find F and then recover the function f from its
transform F. This last step is known as inverting the transform.
There are several integral transform that are useful to solve differential equation, but in
this chapter we will use the Laplace transform.
Definition: Let f(t) be given for t ≥ 0 and assume the function satisfy certain conditions
to be stated later on. The Laplace transform of f(t), that it is denoted by L{f(t)} or F(s),
is defined by the equation
L{f(t)} = F(s) = e!st
f(t)dt
0
"
#
whenever the improper integral converges.
Remark: The kernel of the Laplace transform is K(s,t) = e-st
.
Let’s compute the of Laplace transform of some important elementary functions, before
discussing the restriction that have to be imposed on f(t) so it has a Lapalce transform.
Examples:
1) Let f(t) = 1, t ≥ 0
L{1} = F(s) = e!st
dt = lim
A"#0
#
$ e!st
dt =
0
A
$ lim
A"#
!e!st
s
0
A
= lim
A"#
1
s
!
e!As
s
%
&
'
(
)
* =
1
s
, for all s > 0
2) Let f(t) = t, t ≥ 0
L{t} = F(s) = e!st
t!dt = lim
A"#0
#
$ e!st
t!dt =
0
A
$ lim
A"#
!e!st
s2
(st +1) 0
A
=
=! lim
A!"
1
s2
#
e#As
s2
sA +1( )
$
%
&
'
(
) =
1
s2
, for all s > 0

2. 3) Let f(t) = eat
, t ≥ 0
L{eat
} = F(s) = e!st
eat
dt = lim
A"#0
#
$ e!(s!a)t
dt =
0
A
$
lim
A!"
#e#(s#a)t
s # a
0
A
= lim
A!"
1
s # a
#
e#(s#a)A
s # a
$
%
&
'
(
) =
1
s # a
, for all s > a.
4) Let f(t) = sin at, t ≥ 0
L{sin at} = F(s) = e!st
sinat!dt = lim
A"#0
#
$ e!st
sin!at!dt =
0
A
$
lim
A!"
#
e#st
cosat
a
0
A
#
s
a
e#st
cosat!dt
0
A
$
%
&
'
(
)
* =
1
a
#
s
a
e#st
cosat!dt =
0
"
$
1
a
# lim
A!"
#
e#st
sinat
a
0
A
+
s2
a2
e#st
sinat!dt
0
A
$
%
&
'
(
)
* =
1
a
#
s2
a2
F(s)
then,
F(s) =
1
a
#
s2
a2
F(s),!!F(s)
a2
+ s2
a2
+
,
-
.
/
0 =
1
a
,!F(s) =
a
s2
+ a2
,!s 1 0
5) Let f(t) = cos at, t ≥ 0,
From example 4, we have
L{sin at} =
a
s2
+ a2
=
1
a
!
s
a
L{cos at},
then,
L{cos at} =
1
a
!
a
s2
+ a2
"
#$
%
&'
a
s
=
s2
+ a2
! a2
s2
+ a2
( )a
a
s
=
s
s2
+ a2
,!s > 0
Definition: A function f(t) is bounded if there exists a positive constant M such that
|f(t)| ≤ M
for all t in the domain of f.
Example:
1) f(t) = sin t is bounded since |sin t | ≤ 1 for all t in its domain.
2) f(t) = e-at
, a > 0, t ≥ 0 is bounded, since |e-at
| ≤ 1, for all t ≥ 0.
Definition: A function f is said of exponential order if there exists a constant a and
positive constants t0 and M such that
|f(t)| ≤ M eat
for all t > t0 at which f(t) is defined.

3. Remark: If f(t) is of exponential order, there is a constant a such that e-at
|f(t)| < M for all
t > t0, meaning that the product e-at
|f(t)| is bounded for all sufficient large values of t.
If f(t) is of exponential order with the constant a, then for any constant b such that b > a,
then f(t) is of exponential order with the constant b.
Remark: Every bounded function is of exponential order.
Example:
1) Given f(t) = ebt
sin ct,
since |f(t)| = |ebt
sin ct| ≤ 1ebt
, so M = 1 and a = b
2) Given f(t) = tn
where n > 0.
For a > 0, lim
t!"
e#at
tn
= 0 (use L”Hopital rule n times), then there exists constants M > 0
and t0 such that
e-at
tn
≤ M, for all t > t0.
Therefore, |f(t)| < M eat for t > t0, hence f(t) = tn
is of exponential order with the constant
a any positive number.
3) Given f(t) = et2
is NOT of exponential order, since
e-at
|f(t)| = e!at
et2
= et2
!at
" #!when!t " #
no matter what is the value of a.
Theorem: Existence of the Laplace Transform
Let f(t), t > 0 be a real-valued function that has the following properties:
1) f is piecewise continuous in every finite closed interval 0 ≤ t ≤ b, b > 0.
2) f is of exponential order; there exists a, M > 0 and t0 > 0 such that |f(t) | ≤ M eat
for t >
t0.
Conclusion:
The Lapalce transform
L{f(t)} = F(s) = e!st
f(t)dt
0
"
#
exists for s > a.
To prove the theorem we must show that the improper integral converges for s > a.
Splitting the improper integral into two parts, we have
e!st
f(t)!dt =
0
"
# e!st
f(t)!dt +
0
t0
# e!st
f(t)!dt
t0
"
#
The first integral on the right side exists by hypothesis 1, hence the existence of the
transform F(s) depends on the convergence of the second integral.
By hypothesis 2, we have
e!st
f(t) " Me!st
eat
= Me(a!s)t
Now, Me!(s!a)t
dt = M
t0
"
# e!(s!a)t
dt
t0
"
# and it is convergent if s > a.
Then, by the comparison test for improper integrals theorem, F(s) exists for s > a