Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 8 potx


x
2α+1
y



+ λx
2α−1
y = 0, y(a) = y(b) = 0
Now we verify that the Sturm-Liouville properties are satisfied.
• The eigenvalues
λ
n
= α
2
+

nπ
ln(b/a)

2
, n ∈ Z
are real.
• There are an infinite number of eigenvalues
λ
1
< λ
2
< λ

3
< ··· ,
α
2
+

π
ln(b/a)

2
< α
2
+

2π
ln(b/a)

2
< α
2
+

3π
ln(b/a)

2
< ···
There is a least eigenvalue
λ
1

= α
2
+

π
ln(b/a)

2
,
but there is no greatest eigenvalue, (λ
n
→ ∞ as n → ∞).
• For each eigenvalue, we found one unique, (to within a multiplicative constant), eigenfunction φ
n
. We were able
to choose the eigenfunctions to be real-valued. The eigenfunction
φ
n
= x
−α
sin

nπ
ln(x/a)
ln(b/a)

.
has exactly n − 1 zeros in the open interval a < x < b.
1454
• The eigenfunctions are orthogonal with respect to the weighting function σ(x) = x

2α−1
.

b
a
φ
n
(x)φ
m
(x)σ(x) dx =

b
a
x
−α
sin

nπ
ln(x/a)
ln(b/a)

x
−α
sin

mπ
ln(x/a)
ln(b/a)

x

2α−1
dx
=

b
a
sin

nπ
ln(x/a)
ln(b/a)

sin

mπ
ln(x/a)
ln(b/a)

1
x
dx
=
ln(b/a)
π

π
0
sin(nx) sin(mx) dx
=
ln(b/a)

2π

π
0
(cos((n − m)x) − cos((n + m)x)) dx
= 0 if n = m
• The eigenfunctions are complete. Any piecewise continuous function f(x) defined on a ≤ x ≤ b can be expanded
in a series of eigenfunctions
f(x) ∼
∞

n=1
c
n
φ
n
(x),
where
c
n
=

b
a
f(x)φ
n
(x)σ(x) dx

b
a

φ
2
n
(x)σ(x) dx
.
The sum converges to
1
2
(f(x
−
) + f(x
+
)). (We do not prove this property.)
1455
• The eigenvalues can be rel ated to the eigenfunctions with the Rayleigh quotient.
λ
n
=

−pφ
n
dφ
n
dx

b
a
+

b

a

p

dφ
n
dx

2
− qφ
2
n

dx

b
a
φ
2
n
σ dx
=

b
a

x
2α+1

x

−α−1

nπ
ln(b/a)
cos

nπ
ln(x/a)
ln(b/a)

− α sin

nπ
ln(x/a)
ln(b/a)

2

dx

b
a

x
−α
sin

nπ
ln(x/a)
ln(b/a)


2
x
2α−1
dx
=

b
a


nπ
ln(b/a)

2
cos
2
(·) − 2α
nπ
ln(b/a)
cos (·) sin (·) + α
2
sin
2
(·)

x
−1
dx


b
a
sin
2

nπ
ln(x/a)
ln(b/a)

x
−1
dx
=

π
0


nπ
ln(b/a)

2
cos
2
(x) − 2α
nπ
ln(b/a)
cos(x) sin(x) + α
2
sin

2
(x)

dx

π
0
sin
2
(x) dx
= α
2
+

nπ
ln(b/a)

2
Now we expand a function f(x) in a series of the eigenfunctions.
f(x) ∼
∞

n=1
c
n
x
−α
sin

nπ

ln(x/a)
ln(b/a)

,
where
c
n
=

b
a
f(x)φ
n
(x)σ(x) dx

b
a
φ
2
n
(x)σ(x) dx
=
2n
ln(b/a)

b
a
f(x)x
α−1
sin


nπ
ln(x/a)
ln(b/a)

dx
1456
Solution 29.4
y

− y

+ λy = 0, y(0) = y(1) = 0.
The factor that will p ut this equation in Sturm-Liouville form is
F (x) = exp


x
−1 dx

=
e
−x
.
The differential equation becomes
d
dx

e
−x

y


+ λ
e
−x
y = 0.
Thus we see that the eigenfunctions will be orthogonal with respect to the weighting function σ =
e
−x
.
Substituting y =
e
αx
into the differential equation yields
α
2
− α + λ = 0
α =
1 ±
√
1 − 4λ
2
α =
1
2
±

1/4 − λ.
If λ < 1/4 then the solutions to the differential equation are exponential and only the trivial solution satisfi es the

boundary conditions.
If λ = 1/4 then the solution is y = c
1
e
x/2
+c
2
x
e
x/2
and again only the trivial solution satisfies the boundary
conditions.
Now consider the case that λ > 1/4.
α =
1
2
± ı

λ − 1/4
The solutions are
e
x/2
cos(

λ − 1/4 x),
e
x/2
sin(

λ − 1/4 x).

The left boundary condition gives us
y = c
e
x/2
sin(

λ − 1/4 x).
1457
The right boundary condition demands that

λ − 1/4 = nπ, n = 1, 2, . . .
Thus we see that the eigenvalues and eigenfunctions are
λ
n
=
1
4
+ (nπ)
2
, y
n
=
e
x/2
sin(nπx).
If f(x) is a piecewise continuous function then we can expand it in a series of the eigenfunctions.
f(x) =
∞

n=1

a
n
e
x/2
sin(nπx)
The coefficients are
a
n
=

1
0
f(x)
e
−x
e
x/2
sin(nπx) dx

1
0
e
−x
(
e
x/2
sin(nπx))
2
dx
=


1
0
f(x)
e
−x/2
sin(nπx) dx

1
0
sin
2
(nπx) dx
= 2

1
0
f(x)
e
−x/2
sin(nπx) dx.
Solution 29.5
Consider the eigenvalue problem
y

+ λy = 0 y(0) = 0 y(1) + y

(1) = 0.
Since this is a Sturm-Liouville problem, there are only real eigenvalues. By the Rayleigh quotient, the eigenvalu es are
λ =

−φ
dφ
dx



1
0
+

1
0


dφ
dx

2

dx

1
0
φ
2
dx
,
1458
λ =
φ

2
(1) +

1
0


dφ
dx

2

dx

1
0
φ
2
dx
.
This demonstrates that there are only positive eigenvalues. The general solu tion of the differential equation for positive,
real λ is
y = c
1
cos

√
λx

+ c

2
sin

√
λx

.
The solution that satisfies the left boundary condition is
y = c sin

√
λx

.
For nontrivial solutions we must have
sin

√
λ

+
√
λ cos

√
λ

= 0
√
λ = −tan


√
λ

.
The positive solutions of this equation are eigenvalues with corresponding eigenfunctions sin

√
λx

. In Figure 29.1
we plot the functions x and −tan(x) and draw vertical lines at x = (n − 1/2)π, n ∈ N.
From this we see that there are an infinite number of eigenvalues, λ
1
< λ
2
< λ
3
< ···. In the limit as n → ∞,
λ
n
→ (n − 1/2)π. The limit is approached from above.
Now consider the eigenvalue problem
y

+ y = µy y(0) = 0 y(1) + y

(1) = 0.
From above we see that the eigenvalues satisfy


1 − µ = −tan


1 − µ

and that there are an infinite number of eigenvalu es. For large n, µ
n
≈ 1 − (n − 1/2)π. The eigenfunctions are
φ
n
= sin


1 − µ
n
x

.
1459
Figure 29.1: x and −tan(x).
To solve the inhomogeneous problem, we expand the solution and the inhomogeneity in a series of the eigenfunctions.
f =
∞

n=1
f
n
φ
n
, f

n
=

1
0
f(x)φ
n
(x) dx

1
0
φ
2
n
(x) dx
y =
∞

n=1
y
n
φ
n
We substitite the expansions into the diffe rential equation to determine the coefficients.
y

+ y = f
∞

n=1

µ
n
y
n
φ
n
=
∞

n=1
f
n
φ
n
y =
∞

n=1
f
n
µ
n
sin


1 − µ
n
x

1460

Solution 29.6
Consider the eigenvalue problem
y

+ y = µy y(0) = 0 y(1) + y

(1) = 0.
From Exercise 29.5 we see that the eigenvalues satisfy

1 − µ = −tan


1 − µ

and that there are an infinite number of eigenvalu es. For large n, µ
n
≈ 1 − (n − 1/2)π. The eigenfunctions are
φ
n
= sin


1 − µ
n
x

.
To solve the inhomogeneous problem, we expand the solution and the inhomogeneity in a series of the eigenfunctions.
f =
∞


n=1
f
n
φ
n
, f
n
=

1
0
f(x)φ
n
(x) dx

1
0
φ
2
n
(x) dx
y =
∞

n=1
y
n
φ
n

We substitite the expansions into the diffe rential equation to determine the coefficients.
y

+ y = f
∞

n=1
µ
n
y
n
φ
n
=
∞

n=1
f
n
φ
n
y =
∞

n=1
f
n
µ
n
sin



1 − µ
n
x

Solution 29.7
First consider λ = 0. The general solution is
y = c
1
+ c
2
x.
1461
y = cx satisfies the boundary conditions. Thus λ = 0 is an eigenvalue .
Now consider negative real λ. The general solution is
y = c
1
cosh

√
−λx

+ c
2
sinh

√
−λx


.
The solution that satisfies the left boundary condition is
y = c sinh

√
−λx

.
For nontrivial solutions of the boundary value problem, there must be negative real solutions of
√
−λ − sinh

√
−λ

= 0.
Since x = sinh x has no nonzero real solutions, this equation has no solutions for negative real λ. There are no negative
real eigenvalues.
Finally consider positive real λ. The general solution is
y = c
1
cos

√
λx

+ c
2
sin


√
λx

.
The solution that satisfies the left boundary condition is
y = c sin

√
λx

.
For nontrivial solutions of the boundary value problem, there must be positive real solutions of
√
λ − sin

√
λ

= 0.
Since x = sin x has no nonzero real solutions, this equation has no solutions for positive real λ. There are no positive
real eigenvalues.
There is only one real eigenvalue, λ = 0, with corresponding eigenfunction φ = x.
1462
The difficulty with the boundary conditions, y(0) = 0, y

(0) − y(1) = 0 is that the problem is not self-adjoint. We
demonstrate this by showing that the problem does not satisfy Green’s identity. Let u and v be two functions that
satisfy the boundary conditions, but not necessarily the differential equation.
u, L[v] − L[u], v = u, v


 − u

, v
= [uv

]
1
0
− u

, v

 − u

, v

 − [u

v]
1
0
+ u

, v

 − u

, v



= u(1)v

(1) − u

(1)v(1)
Green’s identity is not satisfied,
u, L[v] − L[u], v = 0;
The problem is not self-adjoint.
Solution 29.8
First we write the equation in formally self-adjoint form,
L[y] ≡ (xy

)

= −λxy, |y(0)| < ∞, y(1) = 0.
Let λ be an eigenvalue with corresponding eigenfunction φ. We derive the Rayleigh quotient for λ.
φ, L[φ] = φ, −λxφ
φ, (xφ

)

 = −λφ, xφ
[φxφ

]
1
0
− φ

, xφ


 = −λφ, xφ
We apply the boundary conditions and solve for λ.
λ =
φ

, xφ


φ, xφ
The Bessel equation of the first kind and order zero satisfies the problem,
y

+
1
x
y

+ y = 0, |y(0)| < ∞, y(r) = 0,
1463
where r is a positive root of J
0
(x). We make the change of variables ξ = x/r, u(ξ) = y(x) to obtain the problem
1
r
2
u

+
1

rξ
1
r
u

+ u = 0, |u(0)| < ∞, u(1) = 0,
u

+
1
ξ
u

+ r
2
u = 0, |u(0)| < ∞, u(1) = 0.
Now r
2
is the eigenvalue of the problem for u(ξ). From the Rayleigh quotient, the minimum eigenvalue obeys the
inequality
r
2
≤
φ

, xφ


φ, xφ
,

where φ is any test function that satisfies the boundary conditions. Taking φ = 1 −x we obtain,
r
2
≤

1
0
(−1)x(−1) dx

1
0
(1 − x)x(1 − x) dx
= 6,
r ≤
√
6
Thus the smallest zero of J
0
(x) is less than or equal to
√
6 ≈ 2.4494. (The smallest zero of J
0
(x) is approximately
2.40483.)
Solution 29.9
We assume that 0 < l < π.
Recall that the solution of a second order differential equation with piecewise continuous coefficient functions is
piecewise C
2
. This means that the solution is C

2
except for a finite number of points where it is C
1
.
First consid er the case λ = 0. A set of linearly independent solutions of the differential equation is {1, z}. The
solution which satisfies y(0) = 0 is y
1
= c
1
z. The solution which satisfies y(π) = 0 is y
2
= c
2
(π − z). There is a
solution for the problem if there are there are values of c
1
and c
2
such that y
1
and y
2
have the same position and slope
at z = l.
y
1
(l) = y
2
(l), y


1
(l) = y

2
(l)
c
1
l = c
2
(π − l), c
1
= −c
2
1464
Since there is only the trivial solution, c
1
= c
2
= 0, λ = 0 is not an eigenvalue.
Now consider λ = 0. For 0 ≤ z ≤ l a set of linearly independent solutions is

cos(
√
aλz), sin(
√
aλz)

.
The solution which satisfies y(0) = 0 is
y

1
= c
1
sin(
√
aλz).
For l < z ≤ π a set of linearly independent solutions is

cos(
√
bλz), sin(
√
bλz)

.
The solution which satisfies y(π) = 0 is
y
2
= c
2
sin(
√
bλ(π − z)).
λ = 0 is an eigenvalue if there are nontrivial solutions of
y
1
(l) = y
2
(l), y


1
(l) = y

2
(l)
c
1
sin(
√
aλl) = c
2
sin(
√
bλ(π − l)), c
1
√
aλ cos(
√
aλl) = −c
2
√
bλ cos(
√
bλ(π − l))
We divide the second equation by

(λ) since λ = 0 and write this as a linear algebra problem.

sin(
√

aλl) −sin(
√
bλ(π − l))
√
a cos(
√
aλl)
√
b sin(
√
bλ(π − l))

c
1
c
2

=

0
0

This system of equations has nontrivial solutions if and only if the determinant of the matrix is zero.
√
b sin(
√
aλl) sin(
√
bλ(π − l)) +
√

a cos(
√
aλl) sin(
√
bλ(π − l)) = 0
We can use trigonometric identities to write this equation as
(
√
b −
√
a) sin

√
λ(l
√
a − (π − l)
√
b)

+ (
√
b +
√
a) sin

√
λ(l
√
a + (π − l)
√

b)

= 0
1465
Clearly this equation has an infinite number of solutions for real, positive λ. However, it is not clear that this equation
does not have non-real solutions. In order to prove that, we will show that the problem is self-adjoint. Before going on
to that we note that the eigenfunctions have the form
φ
n
(z) =

sin

√
aλ
n
z

0 ≤ z ≤ l
sin

√
bλ
n
(π − z)

l < z ≤ π.
Now we prove that the problem is self-adjoint. We consider the class of functions which are C
2
in (0 . . . π) excep t

at the interior point x = l where they are C
1
and which satisfy the boundary conditions y(0) = y(π) = 0. Note that
the differential operator is not defined at the point x = l. Thus Green’s identity,
u|q|Lv = Lu|q|v
is not well-defined. To remedy this we must define a new inner product. We choose
u|v ≡

l
0
uv dx +

π
l
uv dx.
This new inner product does not require differentiability at the point x = l.
The problem is self-adjoint if Green’s indentity is satisfied . Let u and v be elements of our class of functions. In
1466
addition to the boundary conditions, we will use the f act that u and v satisfy y(l
−
) = y(l
+
) and y

(l
−
) = y

(l
+

).
v|Lu =

l
0
vu

dx +

π
l
vu

dx
= [vu

]
l
0
−

l
0
v

u

dx + [vu

]

π
l
−

π
l
v

u

dx
= v(l)u

(l) −

l
0
v

u

dx − v(l)u

(l) −

π
l
v

u


dx
= −

l
0
v

u

dx −

π
l
v

u

dx
= −[v

u]
l
0
+

l
0
v


u dx − [v

u]
π
l
+

π
l
v

u dx
= −v

(l)u(l) +

l
0
v

u dx + v

(l)u(l) +

π
l
v

u dx
=


l
0
v

u dx +

π
l
v

u dx
= Lv|Lu
The problem is self-adjoint. Hence the eigenvalues are real. There are an infinite number of positive, real eigenvalues
λ
n
.
Solution 29.10
1. Let v be an eigenfunction with the eigenvalue λ. We start with the differential equation and then take the inner
product with v.
(pv

)

− (qv

)

+ rv = λsv
v, (pv


)

− (qv

)

+ rv = v, λsv
1467
We use integration by parts and utilize the homogeneous boundary conditions.
[v(pv

)

]
b
a
− v

, (pv

)

 − [vqv

]
b
a
+ v


, qv

 + v, rv = λv, sv
−[v

pv

]
b
a
+ v

, pv

 + v

, qv

 + v, rv = λv, sv
λ =
v

, pv

 + v

, qv

 + v, rv
v, sv

We see that if p, q, r, s ≥ 0 then the eigenvalues will be positive. (Of course we assume that p and s are not
identically zero.)
2. First we prove that this problem is self-adjoint. Let u and v be functions that satisfy the boundary conditions,
but do not necessarily satsify the differential equation.
v, L[u] − L[v], u = v, (pu

)

− (qu

)

+ ru−(pv

)

− (qv

)

+ rv, u
Following our work in part (a) we use integration by parts to move the derivatives.
= (v

, pu

 + v

, qu


 + v, ru) −(pv

, u

 + qv

, u

 + rv, u)
= 0
This problem satisfies Green’s identity,
v, L[u] − L[v], u = 0,
and is thus self-adjoint.
Let v
k
and v
m
be eigenfunctions corresp ondi ng to the distinct eigenvalues λ
k
and λ
m
. We start with Green’s
identity.
v
k
, L[v
m
] − L[v
k
], v

m
 = 0
v
k
, λ
m
sv
m
 − λ
k
sv
k
, v
m
 = 0
(λ
m
− λ
k
)v
k
, sv
m
 = 0
v
k
, sv
m
 = 0
The eigenfunctions are orthogonal with respect to the weighting function s.

1468
3. From part (a) we know that there are only positive eigenvalues. The general solution of the differential equation
is
φ = c
1
cos(λ
1/4
x) + c
2
cosh(λ
1/4
x) + c
3
sin(λ
1/4
x) + c
4
sinh(λ
1/4
x).
Applying the condition φ(0) = 0 we obtain
φ = c
1
(cos(λ
1/4
x) − cosh(λ
1/4
x)) + c
2
sin(λ

1/4
x) + c
3
sinh(λ
1/4
x).
The condition φ

(0) = 0 reduces this to
φ = c
1
sin(λ
1/4
x) + c
2
sinh(λ
1/4
x).
We substitute the solution into the two right boundary conditions.
c
1
sin(λ
1/4
) + c
2
sinh(λ
1/4
) = 0
−c
1

λ
1/2
sin(λ
1/4
) + c
2
λ
1/2
sinh(λ
1/4
) = 0
We see that sin(λ
1/4
) = 0. The eigenvalues and eigenfunctions are
λ
n
= (nπ)
4
, φ
n
= sin(nπx), n ∈ N.
1469
Chapter 30
Integrals and Convergence
Never try to teach a pig to sing. It wastes your tim e and annoys the pig.
-?
30.1 Uniform Convergence of Integrals
Consider the improper integral

∞

c
f(x, t) dt.
The integral is convergent to S(x) if, given any  > 0, there exists T (x, ) such that





τ
c
f(x, t) dt − S(x)




<  for all τ > T(x, ).
The sum is uniformly convergent if T is independ ent of x.
Similar to the Weierstrass M-test for infinite sums we have a uniform convergence test for integrals. If there exists
a continuous function M(t) such that |f(x, t)| ≤ M(t) and

∞
c
M(t) dt is convergent, then

∞
c
f(x, t) dt is uniformly
convergent.
1470
If


∞
c
f(x, t) dt is uniformly convergent, we have the following properties:
• If f(x, t) is continuous for x ∈ [a, b] and t ∈ [c, ∞) then for a < x
0
< b,
lim
x→x
0

∞
c
f(x, t) dt =

∞
c

lim
x→x
0
f(x, t)

dt.
• If a ≤ x
1
< x
2
≤ b then we can interchange the order of integration.


x
2
x
1


∞
c
f(x, t) dt

dx =

∞
c


x
2
x
1
f(x, t) dx

dt
• If
∂f
∂x
is continuous, then
d
dx


∞
c
f(x, t) dt =

∞
c
∂
∂x
f(x, t) dt.
30.2 The Riemann-Lebesgue Lemma
Result 30.2.1 If

b
a
|f(x)|dx exists, then

b
a
f(x) sin(λx) dx → 0 as λ → ∞.
Before we try to justify the Riemann-Lebesgue lemma, we will need a preliminary result. Let λ be a positive constant.





b
a
sin(λx) dx





=






−
1
λ
cos(λx)

b
a





≤
2
λ
.
1471
We will prove the Riemann-Lebesgue lemma for the case when f(x) has limited total fluctuation on the interval
(a, b). We can express f(x) as the di fference of two functions
f(x) = ψ
+

(x) − ψ
−
(x),
where ψ
+
and ψ
−
are positive, increasing, bounded functions.
From the mean value theorem for positive, increasing functions, there exists an x
0
, a ≤ x
0
≤ b, such that





b
a
ψ
+
(x) sin(λx) dx




=





ψ
+
(b)

b
x
0
sin(λx) dx




≤ |ψ
+
(b)|
2
λ
.
Similarly,





b
a
ψ
−

(x) sin(λx) dx




≤ |ψ
−
(b)|
2
λ
.
Thus





b
a
f(x) sin(λx) dx




≤
2
λ
(|ψ
+
(b)| + |ψ

−
(b)|)
→ 0 as λ → ∞.
30.3 Cauchy Principal Value
30.3.1 Integrals on an Infinite Domain
The improper integral

∞
−∞
f(x) dx is defined

∞
−∞
f(x) dx = lim
a→−∞

0
a
f(x) dx + lim
b→∞

b
0
f(x) dx,
1472
when these limits exist. The Cauchy principal value of the i ntegral is defined
PV

∞
−∞

f(x) dx = lim
a→∞

a
−a
f(x) dx.
The principal value may exist when the integral diverges.
Example 30.3.1

∞
−∞
x dx diverges, but
PV

∞
−∞
x dx = lim
a→∞

a
−a
x dx = lim
a→∞
(0) = 0.
If the improper integral converges, then the Cauchy principal value exists and is eq ual to the value of the in tegral.
The principal value of the integral of an odd function is zero. If the principal value of the integral of an even function
exists, then the integral converges.
30.3.2 Singular Functions
Let f(x) have a singularity at x = 0. Let a and b satisfy a < 0 < b. The integral of f(x) is defined


b
a
f(x) dx = lim

1
→0
−


1
a
f(x) dx + lim

2
→0
+

b

2
f(x) dx,
when the limits exist. The Cauchy principal value of the integral is defined
PV

b
a
f(x) dx = lim
→0
+



−
a
f(x) dx +

b

f(x) dx

,
when the limit exists.
Example 30.3.2 The integral

2
−1
1
x
dx
1473
diverges, but the principal value exists.
PV

2
−1
1
x
dx = lim
→0
+



−
−1
1
x
dx +

2

1
x
dx

= lim
→0
+

−

1

1
x
dx +

2

1
x
dx


=

2
1
1
x
dx
= log 2
1474
Chapter 31
The Laplace Transform
31.1 The Laplace Transform
The Laplace transform of the function f (t) is defined
L[f(t)] =

∞
0
e
−st
f(t) dt,
for all values of s for which the i ntegral exists. The Laplace transform of f(t) is a function of s which we will denote
ˆ
f(s).
1
A function f(t) is of exponential order α if there exist constants t
0
and M such that
|f(t)| < M
e

αt
, for all t > t
0
.
If

t
0
0
f(t) dt exists and f(t) is of exponential order α then the Laplace transform
ˆ
f(s) exists for (s) > α.
Here are a few examples of these concepts.
• sin t is of exponential order 0.
1
Denoting the Laplace transform of f(t) as F (s) is also common.
1475
• t
e
2t
is of exponential order α for any α > 2.
•
e
t
2
is not of exponential order α for any α.
• t
n
is of exponential order α for any α > 0.
• t

−2
does not have a Laplace transform as the integral diverges.
Example 31.1.1 Consider the Laplace transform of f(t) = 1. Since f(t) = 1 is of exponential order α for any α > 0,
the Laplace transform integral converges for (s) > 0.
ˆ
f(s) =

∞
0
e
−st
dt
=

−
1
s
e
−st

∞
0
=
1
s
Example 31.1.2 The function f(t) = t
e
t
is of exponential order α for any α > 1. We compute the Laplace transform
of this function.

ˆ
f(s) =

∞
0
e
−st
t
e
t
dt
=

∞
0
t
e
(1−s)t
dt
=

1
1 − s
t
e
(1−s)t

∞
0
−


∞
0
1
1 − s
e
(1−s)t
dt
= −

1
(1 − s)
2
e
(1−s)t

∞
0
=
1
(1 − s)
2
for (s) > 1.
1476
Example 31.1.3 Consider the Laplace transform of the Heaviside function,
H(t − c) =

0 for t < c
1 for t > c,
where c > 0.

L[H(t − c)] =

∞
0
e
−st
H(t − c) dt
=

∞
c
e
−st
dt
=

e
−st
−s

∞
c
=
e
−cs
s
for (s) > 0
Example 31.1.4 Next consider H(t − c)f(t − c).
L[H(t − c)f(t − c)] =


∞
0
e
−st
H(t − c)f(t − c) dt
=

∞
c
e
−st
f(t − c) dt
=

∞
0
e
−s(t+c)
f(t) dt
=
e
−cs
ˆ
f(s)
31.2 The Inverse Laplace Transform
The inverse Laplace transform in denoted
f(t) = L
−1
[
ˆ

f(s)].
1477