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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)= 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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Page 1: Chapter 4 Integral transforms - Universität Wien · Chapter 4 Integral transforms ... transforms we consider in this chapter are the Laplace transform and the Fourier ... the Heaviside

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

Page 2: Chapter 4 Integral transforms - Universität Wien · Chapter 4 Integral transforms ... transforms we consider in this chapter are the Laplace transform and the Fourier ... the Heaviside

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.

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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:

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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.

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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.

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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

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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:

L−1L{c1f1(x) + c2f2(x)} = L−1{c1F1(s) + c2F2(s)} = c1L−1{F1(s)}+ c2L−1{F2(s)}.

To invert the function F (s) it is therefore enough to split it into many (possibly

simple) addends and find for each of them the inverse Laplace transform. Based

on the examples in Sect. 4.1.1 (and others that we do not have time to calculate,

but that can be found in the mathematical literature) it is possible to construct a

“dictionary” of basic functions/expressions and corresponding Laplace transforms,

as in Table 4.1. Any time we face a particular F (s), we can look at the dictionary and

check whether it is possible to recover the function f(x) whose Laplace transform is

F (s).

Since the Laplace transform of the solution y(x) is always in the form of a

fraction (see Eq. 4.11), the method we will always use to split a function F (s) into

simple factors is the method of the partial fractions. It is worth reminding it briefly.

We assume that F (s) = Pn(s)Qm(s)

is the quotient between two polynomials Pn(s) and

Qm(s), with degrees n and m respectively. We will also assume m > n. It is always

possible to factorize the polynomial Qm(s) at the denominator into factors of the

type as + b or factors of the type cs2 + ds + e. Sometimes, when we factorize Qm(s),

we obtain factors of the type (as + b)k (that means that s = − ba

is a root with

multiplicity k of the polynomial Qm(s)) or of the type (cs2 +ds+e)k. The methos of

the partial fractions consists in writing the fraction Pn(s)/Qm(s) as sum of simpler

fractions of the type A(as+b)k or As+B

(cs2+ds+e)k . The partial fractions we seek depend on

the factor at the denominator, namely:

• If a factor (as+b) is present at the denominator, then we seek a partial fraction

of the type:A

as + b

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130 CHAPTER 4. INTEGRAL TRANSFORMS

Table 4.1: Summary of elementary Laplace transforms

f(x) = L−1{F (s)} F (s) = L{f(x)} Convergence

1 1s

s > 0

emx 1s−m

s > m

xn n!sn+1 s > 0

sin(mx) ms2+m2 s > 0

cos(mx) ss2+m2 s > 0

sinh(mx) ms2−m2 s > m

cosh(mx) ss2−m2 s > m

emx sin(px) p(s−m)2+p2 s > m

emx cos(px) s−m(s−m)2+p2 s > m

xnemx n!(s−m)n+1 s > m

x−1/2√

πs

s > 0√x 1

2

πs3 s > 0

δ(x − c) e−cs c > 0

uc(x) e−cs

ss > 0

uc(x)f(x − c) e−csF (s)

ecxf(x) F (s − c)

f(cx) 1cF

(

sc

)

c > 0∫ x

0f(x̃)dx̃ F (s)

s∫ x

0f(x − ξ)g(ξ)dξ F (s)G(s)

(−1)nxnf(x) F (n)(s)

f (n)(x) snF (s) − sn−1f(0) − · · · − f (n−1)(0)

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4.1. LAPLACE TRANSFORM 131

• If a factor (as + b)k is present at the denominator, then we seek k partial

fractions of the type:Ai

(as + b)i, i = 1 . . . k

• If a factor (cs2 + ds + e) is present at the denominator, then we seek a partial

fraction of the type:As + B

cs2 + ds + e

• If a factor (cs2 + ds+ e)k is present at the denominator, then we seek k partial

fractions of the type:

Ais + Bi

(cs2 + ds + e)i, i = 1 . . . k

Example 4.1.9 Use the method of the partial fraction to split the function

F (s) =s3 + s2 + 1

s2(s2 + s + 1).

We must determine the coefficients A, B, C, D so that:

A

s+

B

s2+

Cs + D

s2 + s + 1=

s3 + s2 + 1

s2(s2 + s + 1).

If we multiply this equation by s2(s2 + s + 1), we obtain:

As(s2 + s + 1) + B(s2 + s + 1) + Cs3 + Ds2 = s3 + s2 + 1.

We must now compare the coefficients with like power of s, obtaining the system of

equations:

A + C = 1

A + B + D = 1

A + B = 0

B = 1

, ⇒

A = −1

B = 1

C = 2

D = 1

.

The given fraction can be thus decomposed in this way:

s3 + s2 + 1

s2(s2 + s + 1)= −1

s+

1

s2+

2s + 1

s2 + s + 1.

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132 CHAPTER 4. INTEGRAL TRANSFORMS

Example 4.1.10 Find the inverse Laplace transform of the function

F (s) =s2 + 5

s3 − 9s

We can write the given function as:

F (s) =s2 + 5

s(s2 − 9)=

s2 + 5

s(s − 3)(s + 3).

To invert this function we have to apply the method of the partial fractions, namely:

s2 + 5

s(s − 3)(s + 3)=

A

s+

B

s − 3+

C

s + 3=

As2 − 9A + Bs2 + 3Bs + Cs2 − 3Cs

s(s − 3)(s + 3).

Now we can compare terms with like power of s, obtaining the following system of

equations:

A + B + C = 1

3B − 3C = 0

−9A = 5

From the second we obtain B = C, from the last A = −59. From the first equation:

2B =14

9⇒ B = C =

7

9.

Now we can invert all the terms of the given function and obtain:

f(x) = L−1

{

s2 + 5

s3 − 9s

}

= L−1

{

−5

9

1

s+

7

9

[

1

s − 3+

1

s + 3

]}

= −5

9+

14

9L−1

{

s

s2 − 9

}

= −5

9+

14

9cosh(3x).

Example 4.1.11 Solve the initial value problem

y′′(x) + 4y(x) = ex

y(0) = 0

y′(0) = −1

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4.1. LAPLACE TRANSFORM 133

We have to apply the operator L to both members of the given ODE. Since this is

a second-order ODE with constant coefficients, we can apply directly Eq. 4.11 to

obtain:

Y (s) =F (s) − 1

s2 + 4=

1s−1

− 1

s2 + 4=

2 − s

(s − 1)(s2 + 4).

We apply now the method of the partial fractions to decompose this function:

A

s − 1+

Bs + C

s2 + 4=

2 − s

(s − 1)(s2 + 4)

⇒ As2 + 4A + Bs2 + Cs − Bs − C = 2 − s.

By equating the terms with like power of s we obtain the system of equations:

A + B = 0

C − B = −1

4A − C = 2

A = −B

C = B − 1

−4B − B = 1

B = −15

A = 15

C = −65

The decomposed Y (s) is thus given by:

Y (s) =1

5

1

s − 1− 1

5

s

s2 + 4− 6

5

1

s2 + 4.

With the help of Table 4.1 we can easily identify the inverse Laplace transforms of

these addends, obtaining therefore:

y(x) =ex

5− cos(2x)

5− 3 sin(2x)

5.

The method of the Laplace transform is sometimes more convenient, sometimes less

convenient compared to traditional methods of ODE resolution. It proves however

to be always more convenient in the case in which the inhomogeneous function is

a step function. In fact, in this case the only available traditional method is the

laborious variation of constants, whereas the Laplace transform of the step function

can be readily found.

Example 4.1.12 Find the solution of the initial value problem:

y′′(x) + y(x) = g(x)

y(0) = 0

y′(0) = 0

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134 CHAPTER 4. INTEGRAL TRANSFORMS

where g(x) is given by:

g(x) =

0 0 ≤ x < 1

x − 1 1 ≤ x < 2

1 x ≥ 2

(also known as ramp loading).

The function g(x) can be written as:

g(x) = u1(x)(x − 1) − u2(x)(x − 2),

where uc(x) is the Heaviside function (Eq. 4.7). In fact, for x < 1 both u1 and u2 are

zero. For x between 1 and 2 is u1 = 1 but u2 is still zero. For x ≥ 2 both functions

are 1 and therefore u1(x)(x− 1)− u2(x)(x− 2) = x− 1− x + 2 = 1. If we make the

Laplace transform of both members of the given ODE we obtain:

s2Y (s) + Y (s) =e−s

s2− e−2s

s2

⇒ Y (s) =(

e−s − e−2s) 1

s2(s2 + 1)=

(

e−s − e−2s) 1 + s2 − s2

s2(s2 + 1)=

e−s − e−2s

s2− e−s − e−2s

s2 + 1.

To invert this function Y (s) we use again the relation L{uc(x)f(x− c)} = e−csF (s)

(and therefore L−1{e−csF (s)} = uc(x)f(x − c)) to obtain:

f(x) = u1(x)(x − 1) − u2(x)(x − 2) − u1(x) sin(x − 1) + u2(x) sin(x − 2).

Among the results presented in Table 4.1 very significant is the one concerning the

Dirac delta function δ(x−c). We remind here briefly what is the Dirac delta function

and what are its properties. Given a function g(x) defined in the following way:

g(x) = dξ(x) =

12ξ

− ξ < x < ξ

0 x ≤ −ξ or x ≥ ξ(4.12)

it is clear that the integral of this function is 1 for any possible choice of ξ, in fact:

∫ ∞

−∞g(x)dx =

∫ ξ

−ξ

1

2ξdx = 1.

It is also clear that if ξ tends to zero, the interval of values of x in which g(x) is

different from zero becomes narrower and narrower until it disappears. Analogously,

the function g(x − c) = dξ(x − c) is non-null only in a narrow interval of x centered

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4.1. LAPLACE TRANSFORM 135

on c that disappears for ξ tending to zero. The limit of the function g(x) = dξ(x)

for ξ → 0 is called Dirac delta function and is indicated with δ(x). It is therefore

characterized by the properties:

δ(x − c) = 0 ∀x 6= c (4.13)∫ ∞

−∞δ(x)dx = 1. (4.14)

Given a generic function f(x), if we integrate f(x)δ(x − c) between −∞ and ∞ we

obtain:

∫ ∞

−∞f(x)δ(x − c)dx = lim

ξ→0

1

∫ c+ξ

c−ξ

f(x)dx = limξ→0

1

2ξ[2ξf(x̃)] , x̃ ∈ [c − ξ, c + ξ].

The last step is justified by the mean value theorem for integrals. But the interval

of values in which x̃ must be taken collapses to the point c for ξ → 0, therefore we

obtain the important property of the Dirac delta function:

∫ ∞

−∞f(x)δ(x − c)dx = f(c). (4.15)

To calculate the Laplace transform of δ(x − c) (with c ≥ 0) it is conveninet to

calculate first the Laplace transform of the function dξ(x−c) and then take the limit

ξ → 0, namely:

L{δ(x − c)} = limξ→0

∫ ∞

0

e−sxdξ(x − c)dx = limξ→0

∫ c+ξ

c−ξ

e−sx

2ξdx

= limξ→0

e−s(c+ξ) − e−s(c+ξ)

−2sξ= e−sc lim

ξ→0

esξ − e−sξ

2sξ

= e−sc limξ→0

ξ(esξ + e−sξ)

2ξ= e−sc.

The last step is justified by the de l’Hopital’s rule for limits. In this way we have

found the result reported in Table 4.1 about the Laplace transform of δ(x − c). In

the case that c = 0 we have L{δ(x)} = 1.

4.1.3 The Bromwich integral

Although for most of the practical purposes the inverse Laplace transform of a given

function F (s) can be found by means of the “dictionary” provided by Tab. 4.1 (or

of more extended tables that can be found in the literature), a general formula for

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136 CHAPTER 4. INTEGRAL TRANSFORMS

Figure 4.1: The infinite line L along which the Bromwich integral must be performed.

the inversion of F (s) can be found treating F (s) as a complex function and is given

by the so-called Bromwich integral:

f(x) = L−1{F (s)} =1

2πi

∫ λ+i∞

λ−i∞esxF (s)ds, (4.16)

where λ is a real positive number and is larger that the real parts of all the singu-

larities of esxF (s). Since F (s) has been defined as the integral of e−sxf(x) between

x = 0 and x = ∞, we will consider in this formula only positive values of x, as well.

In practice, the integral must be performed along the infinite line L, parallel to the

imaginary axis, indicated in Fig. 4.1. At this point, a curve must be chosen in order

to close the contour C. Possible completion paths are for instance the curves Γ1 or

Γ2 indicated in Fig. 4.2, namely the half-circles with radius R on the left and on

the right of L, respectively. For R → ∞ these curves make with L a closed contour.

The Bromwich integral can be evaluated by means of the residue theorem provided

that the integral of the function esxF (s) tends to zero for R (radius of the chosen

half-circle) tending to infinity. If we choose the completion path Γ1, then the residue

theorem ensures us that:

f(x) =1

2πi· 2πi

C

Rj =∑

C

Rj , (4.17)

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4.1. LAPLACE TRANSFORM 137

Figure 4.2: Possible contour completions for the integration path L to use in the

Bromwich integral.

where the sum is extended to all the residues of the function esxF (s) in the complex

plane. In fact, by construction L lies on the right of each singularity of esxF (s) and

on the limit R → ∞ the closed curve C = L+Γ1 will enclose them all (including for

instance the singularity z1 that in Fig. 4.2 is not yet enclosed in C). If we instead

have to choose the completion path Γ2, then the closed curve L + Γ2 will enclose no

singularities and therefore f(x) will be zero.

Example 4.1.13 Find the inverse Laplace transform of the function

F (s) =2e−2s

s2 + 4.

From the relation L{uc(x)f(x − c)} = F (s)e−cs we can already derive the inverse

Laplace transform of the given function, namely u2(x) sin[2(x − 2)]. We check if we

can obtain the same result be means of the Bromwich integral. We have to evaluate

the integral

1

2πi

∫ λ+i∞

λ−i∞

2es(x−2)

s2 + 4ds.

We notice first that the given function has two simple poles at s = 2i and s = −2i

(in fact it is s2 + 4 = (s + 2i)(s − 2i)), both of which have Re (z) = 0. We can

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138 CHAPTER 4. INTEGRAL TRANSFORMS

therefore take an arbitrarily (but positive) small value of λ. We can distinguish two

cases: i) x < 2 and ii) x > 2. For x < 2 the exponent s(x − 2) has negative real

part if Re (s) > 0. We notice here that es(x−2) = e(x−2)Re(s)ei(x−2)Im(s), therefore

what determines the behavior of this function at infinity is e(x−2)Re(s) (ei(x−2)Im(s) has

modulus 1 and does not create problems). That means that, for Re (s) → +∞ the

function es(x−2) tends to zero. At the same time the denominator s2 + 4 diverges as

Re (s) → +∞ and this means that also the term 1s2+4

that multiplies es(x−2) tends to

zero along the curve Γ2 for R → ∞. Therefore the integral of the function F (s)esx

tends to zero along the curve Γ2 of Fig. 4.2 (for R → ∞) and we can calculate the

Bromwich integral by means of the contour C = L + Γ2. For what we have learned,

since the given closed contour does not enclose the poles, the function f(x) is zero.

For x > 2, the function es(x−2) tends to zero for Re (s) → −∞. That means

that the integral of the function es(x−2)

s2+4tends to zero (for R → ∞) along the curve Γ1

of Fig. 4.2 and we take therefore Γ1 as a completion of L to calculate the Bromwich

integral. For the residue theorem, this integral is given by the sum of the residues of

the function esxF (s) at all the poles, namely:

f(x) = Res(2i) + Res(−2i).

We have:

Res(2i) = lims→2i

(s − 2i)2es(x−2)

s2 + 4= lim

s→2i

2es(x−2)

s + 2i=

e2i(x−2)

2i

Res(−2i) = lims→−2i

(s + 2i)2es(x−2)

s2 + 4= lim

s→−2i

2es(x−2)

s − 2i=

e−2i(x−2)

−2i

By summing up these two residues we obtain:

f(x) =1

2i

[

e2i(x−2) − e−2i(x−2)]

= sin[2(x − 2)].

This is what we obtain if x > 2 whereas, as we have seen, if x is smaller than 2

the function is zero. Recalling the definition of the Heaviside function uc(x) we can

conclude that the inverse Laplace transform of the given function is:

f(x) = L−1

{

2e−2s

s2 + 4

}

= u2(x) sin[2(x − 2)].

Example 4.1.14 Find the inverse Laplace transform of the function:

F (s) =√

s − a,

with a ∈ R.

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4.1. LAPLACE TRANSFORM 139

The function esx√

s − a has no poles, but the function√

z is multiple-valued in the

complex plane, therefore, as we have seen, a branch point is present at the point

z = 0, namely at s = a. This is the only singularity of our F (s)esx and therefore,

in order to evaluate the Bromwich integral, we have to take λ larger than a. The

integral to calculate will be:

L−1{√

s − a} =1

2πi

∫ λ+i∞

λ−i∞

√s − aesxds.

By means of the substitution z = s − a we obtain:

L−1{√

s − a} =1

2πi

∫ λ+i∞

λ−i∞

√ze(z+a)xdz =

eax

2πi

∫ λ+i∞

λ−i∞

√zezxdz.

In this case, the branch point is at zero, therefore λ can be arbitrarily small (but

always larger than zero). Since z = 0 is a branch point of the function to integrate,

we have to introduce a branch cut to evaluate the integral. Although we have taken

so far the positive real axis as a branch cut, we have also said that this choice is

arbitrary and to make the function√

z singe value it is enough that closed curves are

not allowed to enclose the origin. We can therefore take as branch cut the negative

real axis. In Fig. 4.3 we indicate the contour we must use to integrate the given

function. Since the closed contour C = L + Γ1 + r1 + γ + r2 + Γ2 does not enclose

singularities, its integral is zero. To evaluate the Bromwich integral (namely the

integral along L) we have to calculate the integral along the arcs Γ1 and Γ2, along

the straight lines r1 and r2 and along the circumference γ.

Since the function√

zezx tends to zero for Re (z) → −∞ (the term√

z cannot

contrast the exponential decay of ezx; remind that x must be positive), the integral

along the arcs Γ1 and Γ2 disappears.

To evaluate the integral along γ we take as usual z = εeiθ and we take the limit

for ε → 0. The interval of values of θ is [π,−π], in fact, as we arrive at γ the first

argument will be π. Then, we rotate clockwise around the origin and after a whole

circuit the argument will be −π. Since dz = iεeiθdθ we have:

γ

√zezxdz =

∫ −π

π

√εei θ

2 exεeiθ · iεeiθdθ.

The integrating function clearly tends to zero for ε → 0, therefore there is no contri-

bution from the integral over γ.

Along the straight lines r1 and r2 we can assume that the arguments of the

complex numbers lying on them are π (along r1) and −π (along r2) and that their

imaginary parts tend to zero, therefore we have z = reiπ (r1) and z = re−iπ (r2).

Consequently, dz = eiπdr (r1) and dz = e−iπdr (r2). Notice here that, although we

are on the negative real axis, r is positive. In fact, eiπ = e−iπ = −1. The parameter

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140 CHAPTER 4. INTEGRAL TRANSFORMS

Figure 4.3: Contour to use in Example 4.1.14.

r runs between +∞ and 0 (r1) and between 0 and +∞ (r2). The integral of the given

function along r1 turns out to be:

r1

√zezxdz =

∫ 0

√rei π

2 exreiπ · eiπdr =

∫ 0

√r · i · e−xr · (−1)dr = i

∫ ∞

0

√re−xrdr.

Along r2 we have:∫

r2

√zezxdz =

∫ ∞

0

√re−i π

2 exre−iπ ·e−iπdr =

∫ ∞

0

√r·(−i)·e−xr·(−1)dr = i

∫ ∞

0

√re−xrdr.

In the end we have:

f(x) = L−1{√

s − a} = − eax

2πi

r1+r2

√zezxdz = −eax

π

∫ ∞

0

√re−xrdr.

The sign minus is due to the fact that, as we have said, the integral along the whole

closed curve C is zero, therefore∫

LF (s)esxds = −

r1+r2F (s)esxds. To evaluate

the integral∫ ∞0

√re−xrdr we make the substitution xr = t2, therefore r = t2

xand

dr = 2tdtx

. We obtain:

∫ ∞

0

√re−xrdr =

1

x3/2

∫ ∞

0

te−t2 · 2tdt.

Since −2te−t2 is the differential of e−t2 we can integrate the given function by parts

and obtain:∫ ∞

0

√re−xrdr = − 1

x3/2

{

[

te−t2]∞

0−

∫ ∞

0

e−t2dt

}

.

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4.2. FOURIER TRANSFORMS 141

The term under square brackets is zero. By using the known result∫ ∞0

e−t2dt =√

π2

we obtain:

∫ ∞

0

√re−xrdr =

√π

2x3/2.

This result completes our inversion of the function F (s) =√

s − a, namely we have:

f(x) = L−1{√

s − a} = − eax

2√

πx3.

4.2 Fourier transforms

Fourier transforms are widely used in physics and astronomy because they allow to

express a function (not necessarily periodic) as a superposition of sinusoidal func-

tions, therefore we devote this section to them. Since the Fourier transforms are used

mostly to represent time-varying functions, we shall use t as independent variable

instead of x. On the other hand, the transformed variable represents for most of the

application a frequency and will be indicated with ω instead of s.

4.2.1 Fourier series

For some physical applications, we might need to expand in series some functions

that are not continuous or not differentiable and that therefore do not admit a

Taylor series. Fourier series allow to represent periodic functions, for which a Taylor

expansion does not exist, as superposition of sine and cosine functions. Given a

periodic function f(t) with period T such that the integral of |f(t)| over one period

converges, f(t) can be expressed in this way:

f(t) =a0

2+

∞∑

n=1

[

an cos

(

2πnt

T

)

+ bn sin

(

2πnt

T

)]

,

where the constant coefficients an, bn are called Fourier coefficients. Defining the

angular frequency ω = 2πT

we simplify this expression into:

f(t) =a0

2+

∞∑

n=1

[an cos(ωnt) + bn sin(ωnt)] , (4.18)

namely the function f(t) can be expressed as a superposition of an infinite number

of sinusoidal functions having periods Tn = 2πωn

.

It can be shown that these coefficients are given by:

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142 CHAPTER 4. INTEGRAL TRANSFORMS

an =2

T

∫ T

2

−T

2

f(t) cos(ωnt)dt (4.19)

bn =2

T

∫ T

2

−T

2

f(t) sin(ωnt)dt (4.20)

Example 4.2.1 Find the Fourier series expansion of the function

f(t) =

−1 − T2

+ kT ≤ t < kT

1 kT ≤ t < T2

+ kT

This is a square wave: a series of positive impulses followed periodically by negative

impulses of the same intensity. We can notice immediately that the function f(t)

is odd (f(t) = −f(−t)). Since the function cos(ωnt) is even, the whole function

f(t) cos(ωnt) is odd and its integral between −T/2 and T/2 is zero. That means that

the coefficients an are zero.

To find the coefficients bn we apply Eq. 4.20 obtaining:

bn =2

T

∫ T

2

−T

2

f(t) sin(ωnt)dt =2

T

[

−∫ 0

−T

2

sin(ωnt)dt +

∫ T

2

0

sin(ωnt)dt

]

=4

T

∫ T

2

0

sin(ωnt)dt = − 2

nπ[cos(ωnt)]

T

20

=2

nπ[1 − cos(nπ)] .

Here we have used the relation ωT = 2π. We can notice here that cos(nπ) is 1 if

n is even and -1 if n is odd, namely cos(nπ) = (−1)n. We could find the same

result by means of the de Moivre’s theorem applied to the complex number z = eiπ.

The coefficients bn are equal to zero if n is even and to 4nπ

if n is odd. The Fourier

expansion we looked at is therefore:

f(t) =4

π

(

sin(ωt) +sin(3ωt)

3+

sin(5ωt)

5+ . . .

)

.

By using the identities cos z = (eiz + e−iz)/2 and sin z = (eiz − e−iz)/2i the Fourier

expansion of a function f(t) can also be written as:

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4.2. FOURIER TRANSFORMS 143

f(t) =a0

2+

∞∑

n=1

[

aneiωnt + e−iωnt

2+ bn

eiωnt − e−iωnt

2i

]

=a0e

iω0t

2+

1

2

∞∑

n=1

[

(an − ibn)eiωnt + (an + ibn)e−iωnt]

.

In this way we can see that the function f(t) can be expressed as sum, extending

from −∞ to +∞, of terms of the form eiωnt, where ωn = ω · n, namely we have:

f(t) =∞

−∞

cneiωnt; cn =

12(an − ibn) n ≥ 0

12(an + ibn) n < 0

. (4.21)

This compact representation of the periodic function f(t) is called complex Fourier

series. If we combine the coefficients an and bn as indicated in Eq. 4.21 we find that,

irrespective of the sign of n, we have:

cn =1

T

∫ T

2

−T

2

f(t)e−iωntdt. (4.22)

4.2.2 From Fourier series to Fourier transform

We have seen that the Fourier series allow us to describe periodic functions as su-

perpositions of sinusoidal functions characterized by angular frequencies ωn. To

represent non-periodic functions, what we can do is to extend the period T to infin-

ity (every function can be considered periodic if the period is large enough). That

corresponds to consider a vanishingly small “frequency quantum” ∆ω = ωn

n= 2π

T

and therefore a continuous spectrum of angular frequencies ωn. Given a function

f(t) =∑∞

n=−∞ cneiωnt, with cn = 1T

∫T

2

−T

2

f(u)e−iωnudu, we want to see what happens

in the limit T → ∞ (or, analogously, ∆ω = 2πT

→ 0). We have:

f(t) =∞

n=∞

1

T

∫ T

2

−T

2

f(u)e−iωnudu · eiωnt =∞

n=∞

∆ω

∫ T

2

−T

2

f(u)e−iωnudu · eiωnt.

In the limit for T → ∞ and ∆ω → 0 the limits of the integration extend to infinity,

the sum becomes an integral and the discrete values ωn become a continuous variable

ω (with ∆ω → dω). We have thus:

f(t) =1

∫ ∞

−∞dωeiωt

∫ ∞

−∞duf(u)e−iωu. (4.23)

From this relation we can define the Fourier transform of a function f(t) as:

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144 CHAPTER 4. INTEGRAL TRANSFORMS

f̃(ω) = F{f(t)} =1√2π

∫ ∞

−∞f(t)e−iωtdt. (4.24)

Here we require, in order this integration to be possible, that∫ ∞−∞ |f(t)|dt is finite.

Unlike the Laplace transform, the Fourier transform is very easy to invert. In fact,

we can directly see from Eq. 4.23 that:

f(t) =1√2π

∫ ∞

−∞f̃(ω)eiωtdω. (4.25)

Example 4.2.2 Find the Fourier transform of the normalized Gaussian distribution

f(t) =1

τ√

2πe−

t2

2τ2 .

By definition of Fourier transform we have:

f̃(ω) =1√2π

∫ ∞

−∞f(t)e−iωtdt =

1

2πτ

∫ ∞

−∞e−iωt− t

2

2τ2 dt.

We can modify the exponent of e in the integral as follows:

−iωt − t2

2τ 2= − 1

2τ 2

[

t2 + 2iωtτ 2 + (iωτ 2)2 − (iωτ 2)2]

.

The first 3 addends inside the square brackets are the square of t + iωτ 2, namely we

obtain:

−iωt − t2

2τ 2= −(t + iωτ 2)2

2τ 2+

(iωτ 2)2

2τ 2= −

(

t + iωτ 2

√2τ

)2

− 1

2ω2τ 2.

Since the term e−12ω2τ2

does not depend on t we obtain:

f̃(ω) =1

2πτe−

12ω2τ2

∫ ∞

−∞e−

t+iωτ2

”2

dt.

This is the integral of a complex function, therefore we should use the methods of

complex integration we have learned so far. However, we can see that the integration

simplifies significantly by means of the substitution:

t + iωτ 2

√2τ

= s, dt =√

2τds.

In this way we obtain:

f̃(ω) =1√2π

e−12ω2τ2

∫ ∞

−∞e−s2

ds =1√2π

e−12ω2τ2

,

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4.2. FOURIER TRANSFORMS 145

where we have made use of the known result∫ ∞−∞ e−s2

ds =√

π. It is important to

note that the Fourier transform of a Gaussian function is another Gaussian function.

The Fourier transform allows us to express the Dirac delta function in an elegant

and useful way. We recall Eq. 4.23

f(t) =1

∫ ∞

−∞dωeiωt

∫ ∞

−∞duf(u)e−iωu.

By exchanging the variable of integration we obtain:

f(t) =1

∫ ∞

−∞dω

∫ ∞

−∞duf(u)eiω(t−u)

=1

∫ ∞

−∞du

∫ ∞

−∞dωf(u)eiω(t−u)

=

∫ ∞

−∞duf(u)

[

1

∫ ∞

−∞eiω(t−u)dω

]

,

where the exchange of the order of integration has been made possible by the Fubini’s

theorem. Recalling Eq. 4.15 we can immediately recognize that:

δ(t − u) =1

∫ ∞

−∞eiω(t−u)dω. (4.26)

Analogously to the Laplace transform, it is easy to calculate the Fourier trans-

form of the derivative of a function. It is:

F{f ′(t)} =1√2π

∫ ∞

−∞f ′(t)e−iωtdt

=1√2π

[

f(t)e−iωt]∞

−∞− (−iω)√

∫ ∞

−∞f(t)e−iωtdt

= iωF{f(t)}. (4.27)

Here we have assumed that the function f(t) tends to zero for t → ±∞ (as it should

be since∫ ∞−∞ |f(t)|dt is finite). It is easy to iterate this procedure and show that:

F{f (n)(t)} = (iω)nF{f(t)}. (4.28)

This relation can be used in some cases to solve ODEs analogously to what done by

means of Laplace transforms, namely we transform both members of an ODE, solve

the obtained algebraic equation as a function of F{y(x)} (the Fourier transform of

the solution y(x) we seek) and then invert the function we have obtained. However,

for most of the practical cases, it is more convenient to use Laplace transformation

methods to solve ODEs. Fourier transformation methods can be extremely useful

instead to solve partial differential equations (see Sect. 6.3.1).