Chapter 4 Integral transforms In mathematics, an integral transform is any transform T of a given function f of the following form: Tf (s)= x 2 x 1 K (x, s)f (x)dx. (4.1) The input is a function f (x) and the output is another function Tf (s). There are different integral transforms, depending on the kernel function K (x, s). The transforms we consider in this chapter are the Laplace transform and the Fourier transform. 4.1 Laplace transform 4.1.1 Basic definition and properties To obtain the Laplace transform of a given function f (x) we use the kernel K (x, s)= e −sx , namely: L{f } = F (s)= ∞ 0 f (x)e −sx dx. (4.2) Here s can also be a complex variable, namely the Laplace transform maps a real function to a complex one. For our purposes it is enough to consider for the moment s real. We can easily verify that L is a linear operator. In fact: L{af + bg } = ∞ 0 [af (x)+ bg (x)]e −sx dx = a ∞ 0 f (x)e −sx dx + b ∞ 0 g (x)e −sx dx ⇒ L{af + bg } = aL{f } + bL{g }. (4.3) 123
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Chapter 4
Integral transforms
In mathematics, an integral transform is any transform T of a given function f of
the following form:
Tf(s) =
∫ x2
x1
K(x, s)f(x)dx. (4.1)
The input is a function f(x) and the output is another function Tf(s). There
are different integral transforms, depending on the kernel function K(x, s). The
transforms we consider in this chapter are the Laplace transform and the Fourier
transform.
4.1 Laplace transform
4.1.1 Basic definition and properties
To obtain the Laplace transform of a given function f(x) we use the kernel K(x, s) =
e−sx, namely:
L{f} = F (s) =
∫ ∞
0
f(x)e−sxdx. (4.2)
Here s can also be a complex variable, namely the Laplace transform maps a real
function to a complex one. For our purposes it is enough to consider for the moment
s real. We can easily verify that L is a linear operator. In fact:
L{af + bg} =
∫ ∞
0
[af(x) + bg(x)]e−sxdx = a
∫ ∞
0
f(x)e−sxdx + b
∫ ∞
0
g(x)e−sxdx
⇒ L{af + bg} = aL{f} + bL{g}. (4.3)
123
124 CHAPTER 4. INTEGRAL TRANSFORMS
Example 4.1.1 Find the Laplace transform of the function f(x) = 1.
It is
L{1} =
∫ ∞
0
e−sxdx = −1
s
[
e−sx]∞
0=
1
s,
provided that s > 0 (this ensures us that limx→∞
e−sx = 0, namely that the integral∫ ∞0
e−sxdx converges).
Example 4.1.2 Find the Laplace transform of f(x) = xn, with n positive integer.
We integrate by parts and obtain:
L{xn} =
∫ ∞
0
xne−sxdx = −1
s
[
xne−sx]∞
0+
n
s
∫ ∞
0
xn−1e−sxdx =n
sL{xn−1}.
We have assumed also in this case that s > 0 (otherwise the integral∫ ∞0
xne−sxdx
does not converge). To obtain L{xn−1} we proceed the same way and obtain L{xn−1} =n−1
sL{xn−2}. We iterate n times and obtain:
L{xn} =n(n − 1)(n − 2) . . .
snL{1} =
n!
sn+1.
Example 4.1.3 Find the Laplace transform of f(x) = sin(mx).
It is L{f(x)} =∫ ∞0
e−sx sin(mx)dx. By using the relation sin(mx) = eimx−e−imx
2iwe
obtain:
L{f(x)} =1
2i
(∫ ∞
0
e(im−s)xdt −∫ ∞
0
e−(im+s)xdx
)
=1
2i
([
e(im−s)x
im − s
]∞
0
−[
e−(im+s)x
−im − s
]∞
0
)
=1
2i
(
1
s − im− 1
s + im
)
=m
s2 + m2,
for s > 0. In fact, the terms e(im−s)x and e−(im+s)x can be written as e−sx [cos(mx) ± sin(mx)].
In the limit for x → ∞, only the term e−sx matters (the term [cos(mx) ± sin(mx)]
oscillates) and it tends to zero for any s > 0.
In these three simple cases we have seen that the integral 4.2 was convergent
for any possible value of s > 0. This is not always the case, as the two following
examples show.
4.1. LAPLACE TRANSFORM 125
Example 4.1.4 Find the Laplace transform of f(x) = eax.
L{eax} =
∫ ∞
0
eaxe−sxdx = limA→∞
∫ A
0
e(a−s)xdx = limA→∞
[
e(a−s)x
a − s
]A
0
.
It is clear that this limit exists and is finite only if a < s (a < Re (s) if s ∈ C),
namely we can define the Laplace transform of the function f(x) = eax only if Re
(s) > a. In this case it is:
L{eax} =1
s − a.
Example 4.1.5 Find the Laplace transform of the function f(x) = cosh(mx).
It is L{f(x)} =∫ ∞0
e−sx cosh(mx)dx. By using the relation cosh(mx) = emx+e−mx
2
we obtain:
L{f(x)} =1
2
(∫ ∞
0
e(m−s)xdt +
∫ ∞
0
e−(m+s)xdx
)
=1
2
([
e(m−s)x
m − s
]∞
0
−[
e−(m+s)x
m + s
]∞
0
)
=1
2
(
1
s − m+
1
s + m
)
=s
s2 − m2.
This result holds as long as e(m−s)x and e−(m+s)x tend to zero for x → ∞, namely it
must be s > |m|.
There are a few properties of the Laplace transform that help us finding the
transform of more complex functions. If we know that F (s) is the Laplace transform
of f(x), namely that L{f(x)} = F (s), then:
•L
{
ecxf(x)}
= F (s − c) (4.4)
This property comes directly from the definition of Laplace transform, in fact:
L{
ecxf(x)}
=
∫ ∞
0
f(x)ecxe−sxdx =
∫ ∞
0
f(x)e−(s−c)xdx = F (s − c).
•L{f(cx)} =
1
cF
(s
c
)
, (c > 0) (4.5)
To show that it is enough to substitute cx with t. In this way is x = tc, dx = dt
c
and therefore:
126 CHAPTER 4. INTEGRAL TRANSFORMS
L{f(cx)} =
∫ ∞
0
e−sxf(cx)dx =1
c
∫ ∞
0
e−s
ctf(t)dt =
1
cF
(s
c
)
.
•L{uc(x)f(x − c)} = e−scF (s) (4.6)
Here is uc(x) the Heaviside or step function, namely:
uc(x) =
0 x < c
1 x ≥ c(4.7)
The function uc(x)f(x − c) is thus given by:
uc(x)f(x − c) =
0 x < c
f(x − c) x ≥ c
We have thus:
L{uc(x)f(x − c)} =
∫ ∞
c
e−sxf(x − c)dx.
With the substitution t = x − c we obtain:
L{uc(x)f(x − c)} =
∫ ∞
0
e−s(c+t)f(t)dt = e−scF (s).
•L{xnf(x)} = (−1)nF (n)(s) (4.8)
It is enough to derive F (s) with respect to s, to obtain:
F ′(s) =d
ds
∫ ∞
0
e−sxf(x)dx = −∫ ∞
0
xe−sxf(x)dx = −L{xf(x)}.
If we now differentiate n times F (s) with respect to s we obtain:
F (n)(s) = (−1)nL{xnf(x)}.From it, Eq. 4.8 is readily obtained.
•L{f ′(x)} = −f(0) + sF (s) (4.9)
This property can be obtained integrating e−sxf ′(x) by parts, namely:
L{f ′(x)} =
∫ ∞
0
e−sxf ′(x)dx =[
f(x)e−sx]∞
0+s
∫ ∞
0
e−sxf(x)dx = −f(0)+sF (s),
provided that limx→∞
f(x)e−sx = 0.
4.1. LAPLACE TRANSFORM 127
Example 4.1.6 Find the Laplace transform of cos(mx).
We could calculate this transform directly but it is easier to use the Laplace transform
of sin(mx) that we have calculated in Example 4.1.3 (L{sin(mx)} = ms2+m2 ). From
Eq. 4.9 (and reminding that L is a linear operator) we have:
L{
d
dxsin(mx)
}
= mL{cos(mx)} = − sin(0) + s · m
s2 + m2.
⇒ L{cos(mx)} =s
s2 + m2.
Example 4.1.7 Find the Laplace transform of x cosh(mx).
We remind from Example 4.1.5 that L{cosh(mx)} = F (s) = ss2−m2 (s > |m|). Eq.
4.8 tells us that F ′(s) is the Laplace transform of −x cosh(mx). We have therefore:
L{x cosh(mx)} = −F ′(s) = −s2 − m2 − 2s2
(s2 − m2)2=
s2 + m2
(s2 − m2)2.
Example 4.1.8 Find the Laplace transform of the function f(x) defined in this way:
f(x) =
x x < π
x − cos(x − π) x ≥ π
By means of the step function (Eq. 4.7) we can rewrite f(x) as f(x) = x −uπ(x) cos(x−π). The Laplace transform of this function can be found by means of Eq.
4.6 and of the known results L{xn} = n!sn+1 (Example 4.1.2) and L{cos(mx)} = s
s2+m2
(Example 4.1.6).
L{f(x)} = L{x} − L{uπ(x) cos(x − π)} =1
s2− e−πsL{cos x} =
1
s2− se−πs
s2 + 1.
128 CHAPTER 4. INTEGRAL TRANSFORMS
4.1.2 Solution of initial value problems by means of Laplace
transforms
We have seen (Eq. 4.9) that the Laplace transform of the derivative of a function
is given by L{f ′(x)} = −f(0) + sF (s), where F (s) = L{f(x)}. If we consider the
Laplace transform of higher order derivatives we obtain (always integrating by parts):
L{f ′′(x)} =
∫ ∞
0
e−sxf ′′(x)dx =[
e−sxf ′(x)]∞
0+ s
∫ ∞
0
e−sxf ′(x)dx
= −f ′(0) − sf(0) + s2F (s)
L{f ′′′(x)} =
∫ ∞
0
e−sxf ′′′(x)dx =[
e−sxf ′′(x)]∞
0+ s
∫ ∞
0
e−sxf ′′(x)dx
= −f ′′(0) − sf ′(0) − s2f(0) + s3F (s)
...
L{f (n)(x)} = snF (s) − sn−1f(0) − sn−2f ′(0) − · · · − sf (n−2)(0) − f (n−1)(0), (4.10)
provided that limx→∞
f (m)(x)e−sx = 0, with m = 0 . . . n − 1. This result allows us to
simplify considerably linear ODEs. Let us take for instance an initial value problem
consisting of a second-order inhomogeneous ODE with constant coefficients (but the
method can be applied also to more complex ODEs):
a2y′′(x) + a1y
′(x) + a0y(x) = f(x)
y(0) = y0
y′(0) = y0′
If we now make the Laplace transform of both members of this equation (calling Y (s)
the Laplace transform of y(x) and F (s) the Laplace transform of f(x)), we obtain:
a2
[
s2Y (s) − sy0 − y0′]
+ a1 [sY (s) − y0] + a0Y (s) = F (s)
⇒ Y (s)(a2s2 + a1s + a0) = F (s) + a1y0 + a2(sy0 + y0
′)
⇒ Y (s) =F (s) + a1y0 + a2(sy0 + y0
′)
a2s2 + a1s + a0. (4.11)
Namely, we have transformed an ODE into an algebraic one, which is of course easier
to solve. Moreover, the particular solution (satisfying the given initial conditions)
is automatically found, without need to search first the general solution and the
look for the coefficients that satisfy the initial conditions. Further, homogeneous
and inhomogeneous ODEs are handled in exactly the same way; it is not necessary
to solve the corresponding homogeneous ODE first. The price to pay for these
4.1. LAPLACE TRANSFORM 129
advantages is that Eq. 4.11 is not yet the solution of the given ODE; we should
invert this relation and find the function y(x) whose Laplace transform is given by
Y (s). This function is called the inverse Laplace transform of Y (s) and it is indicated
with L−1{Y (s)}.Since the operator L is linear, it is easy to show that also the inverse operator
L−1 is linear. In fact, given two functions f1(x) and f2(x) whose Laplace transforms
are F1(s) and F2(s), respectively, the linearity of the operator L ensures us that:
L{c1f1(x) + c2f2(x)} = c1F1(s) + c2F2(s).
If we apply now the operator L−1 to both members of this equation we obtain: