Laplace Transform and Differential Equations

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UNIVERSITY OF
KWAZULU-NATAL
Department of Mathematics
at Howard College
Le
ture Notes in
Engineering Mathemati
s
LAPLACE TRANSFORM
and
DIFFERENTIAL
EQUATIONS
2009
Preliminaries 1
Chapter Zero
PRELIMINARIES
1. Improper Integrals
If
∫M
x0
f(x) dx exists for every M , and limM→∞
∫M
x0
f(x) dx also exists and is finite, then we
define ∫ ∞
x0
f(x) dx = lim
M→∞
∫ M
x0
f(x) dx.
We also say the integral converges. It may be shown that∣∣∣∣
∫ ∞
x0
f(x) dx
∣∣∣∣ ≤
∫ ∞
x0
|f(x)| dx;
if the integral on the right converges, then the integral on the left also converges and we say it
converges absolutely.
An absolutely convergent integral is always convergent, too; but the converse is not neces-
sarily true.
To establish whether an improper integral converges or diverges, one may sometimes use
the comparison test: suppose |f(x)| ≤ |g(x)| from some point onwards (say, for x > a): then
∫ ∞
a
|f(x)| dx =∞ =⇒
∫ ∞
a
|g(x)| dx =∞
(both integrals diverge). Vice-versa,
∫ ∞
a
|f(x)| dx <∞ ⇐=
∫ ∞
a
|g(x)| dx <∞
(both integrals converge absolutely). The comparison test is useful, for example, when it is
relatively easy to integrate f(x) but not g(x), or vice-versa.
2. Integration by Parts
The “product rule” for differentiation is usually written as (uv)′ = u′v+ uv′. By simple manip-
ulations, we get immediately
u′v = (uv)′ − uv′.
Integrating with respect to the independent variable (let us call it x), we get
∫
u′v dx = uv −
∫
uv′ dx. (1)
2 Preliminaries
Using definite integration instead, we get
∫ b
a
u′v dx =
[
u v
]b
a
−
∫ b
a
u v′ dx. (2)
Both formulas (1) and (2) are called “integration by parts”.
3. An Important Result
Observe that ∫ ∞
0
e−x dx =
[
− e−x
]∞
0
= −(0− 1) = 1.
Then consider
∫∞
0
xe−x dx: integrating by parts, we get
∫ ∞
0
xe−x dx =
[
− xe−x
]∞
0
−
∫ ∞
0
(− e−x) dx.
We need to find [
− xe−x
]∞
0
= lim
M→∞
[
−Me−M
]
−
[
− 0e−0
]
=
= lim
M→∞
−M
eM
.
The limit on the right is of the form ∞/∞, and de l’Hospital’s theorem is applicable: one gets
immediately that this limit is zero. Substituting back, it follows
∫ ∞
0
xe−x dx = 0 +
∫ ∞
0
e−x dx = 1.
In the same way, using integration by parts, we see that
∫ ∞
0
x2e−x dx = lim
M→∞
[
−M
2
eM
]
+
02
e0
−
∫ ∞
0
(
− 2xe−x
)
dx.
By de l’Hospital’s rule (applied twice)
lim
M→∞
[
−M
2
eM
]
= 0;
hence ∫ ∞
0
x2e−x dx = 0 + 2
∫ ∞
0
xe−x dx = 2 · 1.
More generally: if n is a positive integer, then applying de l’Hospital’s theorem n times, we may
show that
lim
M→∞
Mn
eM
= 0;
in other words, the exponential function eM always diverges faster than any (arbitrarily large)
power Mn, as M tends to infinity. This statement remains true even if eM is replaced by
eεM , where ε is positive and arbitrarily small. Exponential growth is inherently stronger than
polynomial growth of any degree.
Preliminaries 3
Now, using these results, and repeating integration by parts n times, it is easy to see that
∫ ∞
0
xne−x dx = 0 +
∫ ∞
0
nxn−1e−x dx =
= 0 +
∫ ∞
0
n(n− 1)xn−2e−x dx =
= 0 +
∫ ∞
0
n(n− 1)(n − 2)xn−3e−x dx = etc. etc.
= n! ,
where n! = 1 · 2 · 3 · · · n.
4. Euler’s Formula
Euler studied the properties of the quantity†
w = cosφ+ i sinφ,
where i2 = −1. It is easy to verify that the magnitude and argument of w are, respectively
|w| = 1, arg(w) = φ.
If two such expressions are multiplied together, one gets this identity:
(cos φ1 + i sinφ1)(cosφ2 + i sinφ2) = cos(φ1 + φ2) + i sin(φ1 + φ2)
(verify this). Following Euler, we define the exponential of an imaginary quantity in the following
way:
eiφ
def
= cosφ+ i sinφ.
It is easy to see that this definition extends to imaginary exponentials all properties established
for real exponentials: for example
e−iφ =
1
eiφ
, eiφ1eiφ2 = ei(φ1+φ2),
d eicφ
dφ
= iceicφ,
and so on. The following formulas are also very useful and should be memorized:
cosφ =
eiφ + e−iφ
2
sinφ =
eiφ − e−iφ
i2
.
As an exercise, use Euler’s formula to derive the identities
cosA cosB = 12 [cos(A+B) + cos(A−B)], sinA cosB = 12 [sin(A+B) + sin(A−B)],
which you saw in high school.
† Euler did not call it a complex quantity, as we should have done, because complex numbers
had not yet been properly defined. In a sense, he was ahead of his times.
4 The Laplace Transform
Chapter One
THE LAPLACE TRANSFORM
1. Introduction
◮Definition: Suppose f(t) is a given function of t, and the integral∫ ∞
0
e−stf(t) dt (3)
converges for some value of the parameter s. Then the integral (3) is called the Laplace Transform
of f(t) and is denoted either as L [f(t)] or F (s). ◭
◮Example 1 If f(t) = 3t+ 4,
L [3t+ 4] =
∫ ∞
0
(3t+ 4) e−st dt.
If s > 0, by substituting st = x (so that t = +∞ corresponds to x = +∞) we get immediately
L [3t+ 4] = 3
s2
∫ ∞
0
xe−x dx+
4
s
∫ ∞
0
e−x dx =
3
s2
+
4
s
. [s > 0]
On the other hand, if s ≤ 0, then t = +∞ is mapped into x = −∞; we get a divergent integral,
hence the Laplace transform of 3t+ 4 is not defined for negative s. ◭
◮Example 2 If f(t) = 7e2t, then
L [7e2t] = ∫ ∞
0
7e2te−st dt =
[
7e(2−s)t
2− s
]∞
0
=
7
s− 2 .
Note that in this example the Laplace transform is defined only for s > 2. ◭
◮Example 3 Find L [f(t)], if f(t) = (e7t + 3)2.
Solution: Expanding the square, we get
(
e7t + 3
)2
= e14t + 6e7t + 9, and hence
L [f(t)] =
∫ ∞
0
(
e14t + 6e7t + 9
)
e−st dt =
1
s− 14 +
6
s− 7 +
9
s
.
Note that in this example the Laplace transform is defined only for s > 14. ◭
The Laplace Transform 5
◮Example 4 Find L [f(t)], if f(t) = 2 for 4 ≤ t ≤ 7; f(t) = 0 everywhere else.
Solution: One has immediately
L [f(t)] =
∫ 7
4
2e−st dt =
[
2e−st
−s
]7
4
=
2e−4s − 2e−7s
s
.
Comment: A function like f(t) in this example, is called a transient because it “comes to life”,
so to speak, when t = 4, at which point it jumps to the value 2 (here we have a discontinuity);
when t is increased beyond the point t = 7 (another discontinuity) f(t) vanishes for good. ◭
Before we begin to study the properties of the Laplace transform, which will take a fair amount
of time, let us make some preliminary remarks.
• The integration variable is commonly called t, rather than x. This is because in most
applications, t physically represents time. There may be exceptions to this rule.
• The Laplace transform defines a correspondence between functions of t and functions of s.
Such a correspondence is a linear map: in other words, given two functions of t, f1 and f2,
and two constants c1 and c2, then
L [c1f1(t) + c2f2(t)] = c1L [f1(t)]+ c2L [f2(t)]. (linearity)
• We assume that s is real. This keeps the foregoing discussion as simple as possible; but be
warned, certain properties of the Laplace transform are easier to study if s is treated as a
complex variable.
• If ∫∞
0
e−stf(t) dt converges absolutely for a certain s, then it converges absolutely for all
values of s greater than that. This follows from the simple observation that if s > s0, then
|e−st| < |e−s0t| for all positive t.
• If two functions, say f(t)and g(t) are identical for t ≥ 0, then clearly they have the same
Laplace transform; their behavior for negative t is irrelevant. So, with no loss of generality,
we may assume f(t) = 0 identically for negative t.
• For the integral (3) to converge for some s, it is sufficient that |f(t)| be integrable to the
right of the origin, and do not diverge faster than an exponential as t → ∞. We have
already noted that all powers of t, and hence all polynomials, meet thisrequirement.
2. Notation. Uniqueness
The function f(t) appearing in (3) is sometimes called the direct function or pre-image; the
transform itself is usually called the image of f(t).
Following an established tradition, we shall use the same letter of the alphabet for the direct
function and for its transform, with the understanding that small letters such as f , g, x, y, ω
etc. will be used for direct functions, and the corresponding capital letters F , G, X, Y , Ω etc.
for their transforms.† Occasionally we shall also need Greek letters for dummy variables: in
this case, recall that τ (“tau”) and σ (“sigma”) are the Greek equivalent of small t and small s,
respectively.
Derivatives with respect to t will generally be denoted by dots (Newton’s notation), as it is
common in physics and engineering. Derivatives with respect to s will be denoted by the more
† Warning: some books use small letters for transforms and capital letters for direct functions,
which can be very confusing.
6 The Laplace Transform
usual primes. For example:
x˙ =
dx
dt
, y¨ =
d2y
dt2
;
(FG)′′ = F ′′G+ 2F ′G′ + FG′′ =
d2F
ds2
G+ 2
dF
ds
dG
ds
+ F
d2G
ds2
.
We conclude these preliminaries with a rather subtle point. It is obvious that if f(t) ≡ g(t),
then F (s) = G(s). But it is not obvious at all whether from F (s) = G(s) one may deduce that
f(t) = g(t) anywhere. In a sense, this is blatantly false, since f and g may take arbitrary values
for t < 0; and that is precisely the reason why we stipulated that direct functions be re-defined
to be identically zero for negative t.
The question of equality of the original functions, when the transforms are equal, is the
object of Lerch’s theorem,† which says that under very reasonable assumptions, if F (s) = G(s)
then f(t) = g(t) for all positive t, except at most a finite number of isolated points.
From the point of view of applications, Lerch’s theorem says that if F (s) = G(s) then “for
all practical purposes” f(t) = g(t).
3. Basic Transforms
The following transforms must be committed to memory.
L [ect] = 1
s− c [s > c] (4)
L [coshωt] = s
s2 − ω2 [s > |ω|] (5)
L [sinhωt] = ω
s2 − ω2 [s > |ω|] (6)
L [cosωt] = s
s2 + ω2
[s > 0] (7)
L [sinωt] = ω
s2 + ω2
[s > 0] (8)
L [tc] = c!
sc+1
. [s > 0, c > −1] (9)
The proof of (4) is straightforward:
L [ect] =
∫ ∞
0
ecte−st dt =
[
e(c−s)t
c− s
]∞
0
= 0− 1
c− s =
1
s− c ,
as long as s > c. The proofs for (5) and (6) are corollaries:
L [coshωt] = L [ 12(eωt + e−ωt)] = 12(s − ω) + 12(s + ω) = ss2 − ω2 ,
L [sinhωt] = L [ 1
2
(
eωt − e−ωt)] = 1
2(s − ω) −
1
2(s + ω)
=
ω
s2 − ω2 .
In order to prove (7) and (8), we may use Euler’s formula:
eiωt = cosωt+ i sinωt. (10)
† Matya´sˇ Lerch (1860–1922), Czech.
The Laplace Transform 7
We have then by (4):
L [eiωt] = 1
s− iω =
s+ iω
s2 + ω2
.
Now (by definition) L [cosωt] and L [sinωt] are certainly real: hence, separating real and imag-
inary part in the last expression, we get
L [cosωt] = Re
[
s+ iω
s2 + ω2
]
=
s
s2 + ω2
;
L [sinωt] = Im
[
s+ iω
s2 + ω2
]
=
ω
s2 + ω2
.
◮Example 5 Find the Laplace transform of sin3 10t.
Solution: First of all, we must put sin3 10t into a simple form, without powers. This may be done
using the identities you learnt in high school, but perhaps the best way is by Euler’s formula
(10), which gives
sinωt =
eiωt − e−iωt
i2
.
It follows
sin3 10t =
(
ei10t − e−i10t
i2
)3
;
expanding the cube, we get
sin3 ωt =
ei30t − 3ei10t + 3e−i10t − e−i30t
−i8 =
=
ei30t − e−i30t
−i4 · 2 − 3 ·
ei10t − e−i10t
−i4 · 2 =
= − 1
4
sin 30t+ 3
4
sin 10t.
From this identity, and formula (8), we get immediately that
L [ sin3 10t] = − 14 · 30s2 + 900 + 34 · 10s2 + 100 . ◭
◮Example 6 Find the Laplace transform of sin 2t cos 9t.
Solution: Simple manipulations (or Euler’s formula) give
sin 2t cos 9t = 12 sin 11t− 12 sin 7t;
it follows immediately that
L [sin 2t cos 9t] = 11/2
s2 + 121
− 7/2
s2 + 49
.
◭
The proof of (9) is bit more involved and requires a new concept, which is introduced in the
next section.
You will find that, in applications, the few basic transforms (4–9) listed above are often
all that one needs to know. Just like we seldom calculate derivatives as a limit of the form
8 The Laplace Transform
(
f(x + h) − f(x))/h, but rather through the rules of “differential calculus”, so there exists a
“Laplace calculus” which allows us to calculate many Laplace transforms without going back to
the definition (3).
4. The Factorial Function
Consider L [tn], n being a positive integer. We substitute st with x in the integral (3) (if s is
positive, this maps t = +∞ into x = +∞), and then integrate by parts n times:
L [tn] =
∫ ∞
0
tne−st dt =
1
sn+1
∫ ∞
0
xne−x dx =
n!
sn+1
. [s > 0]
Essentially, we have repeated the procedure outlined in section 0.3 . It works because we started
with an integer power tn; but what about, for example,
√
t, or more general powers of t? For
positive s, integrating by parts once, we get
L [√t ] = ∫ ∞
0
t1/2e−st dt =
1
s3/2
∫ ∞
0
x1/2e−x dx = 0− 1
s3/2
∫ ∞
0
1
2
x−1/2
(− e−x) dx.
The integral on the right converges, but cannot be done analytically; another round of integra-
tion by parts leads nowhere, because
∫∞
0
x−3/2e−x dx is divergent (convince yourself of this).
Integration by substitution is also useless; it has been shown that there is no change of variables
that resolves this integral as a combination of elementary functions (polynomials, sine/cosine
etc.).
However, such an integral certainly
exists. As the picture shows,
∫ ∞
0
x1/2e−x dx = A,
where A is the area A below the graph
of
f(x) = x1/2e−x
and above the x axis, in the whole first
quadrant. x
f(x)
A
 2 4 6 8 0
 0
 0.2
 0.4
It’s easy to show that even though this region is unbounded, its area is finite. In other
words, A may be calculated with arbitrary precision (using a computer program, why not?). It
may therefore be used to extend the definition of factorial that you learnt in first year.
◮Definition: If c is a real number greater than −1, we define the generalized factorial of c by
means of the expression
c! =
∫ ∞
0
xce−x dx. (11)
If c is a positive integer, the integral on the right may be done by parts and is found to be equal
to the product 1 · 2 · 3 · · · c. ◭
This is all we need to conclude our proof of (9): for s > 0 we have immediately:
L [tc] =
∫ ∞
0
tce−st dt =
1
sc+1
∫ ∞
0
xce−x dx =
c!
sc+1
, [c > −1]
The Laplace Transform 9
where c! is defined in (11). As an exercise, convince yourself that
∫∞
0
xce−x dx =∞ if c ≤ −1.
Factorials possess a simple and useful recursive property: if c > −1, then
(c+ 1)! = (c+ 1) · c! (12)
If c is a positive integer, this follows immediately from the “old” definition of factorial. If c is
an arbitrary real number greater than −1, then integrating by parts we get
(c+ 1)! =
∫ ∞
0
xc+1e−x dx =
[
− xc+1e−x
]∞
0
+
∫ ∞
0
(c+ 1)xce−x dx.
The integrated part is zero: by assumption c+ 1 > 0, hence
lim
x→0
xc+1 = 0;
also (see section 0.3),
lim
x→∞
xc+1e−x = 0.
It follows
(c+ 1)! = 0 +
∫ ∞
0
(c+ 1)xce−x dx = (c+ 1) · c! ,
which is (12), as required to prove.
An important consequence of (12) is that once a table of generalized factorials has been
calculated for any interval of unit length, it may be extended outside such an interval using
ordinary multiplications.
◮Example 7 Calculate 5.3273! , assuming that a table of c! is available only for 0 ≤ c ≤ 1.
Solution: By (12), we may write 5.3273! = 5.3273 . 4.3273!. Repeating this procedure four more
times,we get
5.3273! = 5.3273 . 4.3273 . 3.3273 . 2.3273 . 1.3273 . 0.3273!
and at this point we look up 0.3273! in the table. There, we find that 0.3273! ≈ 0.89371; therefore
5.3273! ≈ 211.7553, which lies between 5! = 120 and 6! = 720. ◭
We told you a half-lie. You probably thought that the only case when c! may be calculated in an
easy way, is if c is a positive integer or zero. Whether this statement is true or false, depends on
what you mean by “easy”. As it happens, using the Laplace transform it is possible to calculate
c! when c is half-odd; we shall come back to this in chapter 2, where it is shown that
(− 1
2
)! =
√
π.
From this, one gets immediately, using (12):
1
2 ! =
1
2 · (− 12 )! = 12
√
π
3
2 ! =
3
2 · 12 ! = 34
√
π
5
2
! = 5
2
· 3
2
! = 15
8
√
π
7
2 ! =
7
2 · 52 ! = 10516
√
π
9
2 ! = etc.
10 The Laplace Transform
Finally, a word of warning. Most books still introduce generalized factorials by means of Legen-
dre’s so-called “gamma function”; that is, they write Γ(c+1) for c! . That extra +1 is there only
for historical reasons, and is as necessary as the human appendix.† Today, as the best computer
programs (such as maxima, maple, mathematica, etc.) accept both notations, there is no
need for us to persist with the bulkier gamma notation. Most important, the exclamation mark
is a good reminder of the link with “ordinary” factorials.
5. Basic Properties; Multiplication by s
Before we begin to study the analytic properties of the Laplace transform, we must make some
assumptions on the function f(t) that appears in the definition (3). Clearly, it is not necessary
that f(t) be continuous for the integral in (3) to converge: the function in example 4, for instance,
is discontinuous. On the other hand, it is easy to produce functions with no Laplace transform
for any value of s: for example, f(t) = et
2
or f(t) = 1/t.
There is no special term in the mathematical literature describing “a function that admits
a Laplace transform”, but we can easily do without it. However, we introduce now a couple of
definitions that will help us through our discussion.
◮Definition: A function f(t) is said to be sectionally continuous if it is continuous everywhere
except at a finite number of discontinuities, and at each point of discontinuity both the limit
from the right and the limit from the left, of f(t), exist and are finite. ◭
◮Example 8 Let f(t) = t for t ∈ [0, 1]; f(t) = t2 for t ∈ (1,∞). Then f(t) is continuous
for every t; f˙(t) is only sectionally continuous, because f˙(t) = 1 in (0, 1), f˙(t) = 2t in (1,∞),
limt→1− f˙(t) = 1, limt→1+ f˙(t) = 2. At t = 1, which is a point of discontinuity, the limit from
the left and the limit from the right of f˙ are different, but both are finite. ◭
◮Definition: A function f(t) is of exponential order if there exist two positive constants c and
B such that |f(t)| < B ect, for all t from a certain point onwards. ◭
These definitions are broad enough to include most functions of practical relevance. Clearly, if
a function is sectionally continuous and of exponential order, then the integral in (3) converges,
and hence its Laplace transform exists—for sufficiently large s, possibly.
It also follows that, if f(t) is of exponential order, and s is large enough, then
lim
t→∞
|f(t)|e−st = 0
(i.e., this is true for all s greater than some constant γ).
Suppose now that f(t), f˙(t) are of exponential order; f(t) is continuous and f˙(t) sectionally
continuous. By definition,
L [f˙(t)] =
∫ ∞
0
(df/dt) e−st dt.
Integrating by parts, we get
L [f˙(t)] =
[
f(t) e−st
]∞
0
−
∫ ∞
0
f(t)(−se−st ) dt.
† Quoted from G. Arfken, Mathematical Methods for Physicists, Wiley (USA) 1970.
The Laplace Transform 11
This may also be written:
L [f˙(t)] = lim
t→∞
f(t) e−st − lim
t→0+
f(t) e−st + s
∫ ∞
0
f(t) e−st dt.
Since we assume that f(t) is of exponential order, the first of the limits above is zero (for suffi-
ciently large s). The second limit is equal to limt→0+ f(t), i.e., the limit of f(t) as t approaches
zero from the right—remember, by definition f(t) ≡ 0 if t is negative. We write
lim
t→0+
f(t) = f0.
It follows immediately that
L [f˙(t)] = −f0 + sF (s). (13)
◮Example 9 Find the function x(t) such that 5x˙+ 3x = 0 and x0 = 2.
Solution: We take the Laplace transform of the equation. We get
L [5x˙+ 3x] = 0,
where L [x] = X(s), and L [x˙] = −x0 + sX(s) = −2 + sX(s). It follows
5(−2 + sX) + 3X = 0,
which yields
X =
10
5s + 3
=
2
s+ 3/5
.
Now, by inspection, we recognize that
1
s+ 3/5
= L [e−3t/5];
Therefore, it follows immediately that
X = L [2e−3t/5],
and finally (by Lerch’s theorem) x = 2e−3t/5. ◭
Transforms of higher-order derivatives may be calculated by repeated application of (13). For
example, if f and f˙ are continuous, and f¨ is sectionally continuous, then
L [f¨(t)] = −f˙0 + sL [f˙(t)],
where f˙0 denotes limt→0+ f˙(t). Substituting for L [f˙(t)] its expression (13), we get
L [f¨(t)] = −f˙0 + s(−f0 + sF (s)) =
= −f˙0 − sf0 + s2F (s). (14)
12 The Laplace Transform
Similarly,
L [ ˙¨f(t)] = −f¨0 + sL [f¨(t)]) =
= −f¨0 − sf˙0 − s2f0 + s3F (s), (15)
where f¨0 denotes limt→0+ f¨(t). Naturally, here, we must assume that f , f˙ and f¨ are continuous,
and the third derivative of f(t) is sectionally continuous.
Formulas for higher-order derivatives may be calculated in the same way, if needed. How-
ever, in each case, the assumptions must be extended: the highest derivative of f(t) must be
sectionally continuous, and all lower-order derivatives (including f) must be continuous. In
typical engineering applications, these assumptions are generally satisfied.
◮Example 10 Calculate L [ cos2 t] with and without (13).
Solution: Without using (13), one may find L [ cos2 t] using the identity cos2 t = 12 (cos 2t + 1).
This gives
L [cos2 t] = 12L [cos 2t] + 12L [1]
= 12 ·
(
s
s2 + 4
+
1
s
)
.
In order to use (13), we first note that f0 = 1 and f˙ = −2 sin t cos t = − sin 2t. It then follows
that
L [f˙ ] = − 2
s2 + 4
,
and, by (13), that
L [f˙ ] = sL [cos2 t]− 1.
By comparison, one gets that
sL [cos2 t]− 1 = − 2
s2 + 4
,
which finally yields
L [cos2 t] = 1
s
− 2
s(s2 + 4)
.
Since
1
2 ·
(
s
s2 + 4
+
1
s
)
=
s2 + 2
s(s2 + 4)
=
1
s
− 2
s(s2 + 4)
,
the two results are identical. ◭
This example works because cos2 t is continuous throughout, and so are its derivatives of any
order. But the continuity requirements are crucial, as the following example shows.
◮Example 11 Find F (s), if f(t) = t for 0 ≤ t < 2, and f(t) = 0 everywhere else.
Solution: Before we start: if you apply (13), you get the wrong answer. Let us see why. The
graph of f(t) is pictured on the right; note the discontinuity at t = 2. It follows that
The Laplace Transform 13
f˙ =
{
1 for 0 ≤ t < 2
0 everywhere else.
Hence, by (3),
L [f˙ ] = ∫ 2
0
1 · e−st dt = 1− e
−2s
s
.
t
 f(t)
20
Since f0 = 0, if you applied (13) you would get that L [f ] = (1 − e−2s)/s2. But this is wrong!
Working from (3), and integrating by parts, we get
L [f ] =
∫ 2
0
te−st dt =
[
− te
−st
s
]2
0
+
1
s
∫ 2
0
e−st dt = −2e
−2s
s
+
1− e−2s
s2
.
This is the correct answer. Formula (13) may not be used because f(t) is not continuous at
t = 2. ◭
Next example is very similar, except for a small detail: f(t) is continuous and f˙(t) is sectionally
continuous. Therefore (13) is applicable.
◮Example 12 Find F (s) if f(t) = 3t for 0 ≤ t ≤ 2; f(t) = 6 for t > 2.
Solution: As the picture shows, f(t) is continuous. Observe that
f˙ =
{
3 if 0 ≤ t ≤ 2
0 everywhere else.
Also, f˙(t) is sectionally continuous. By (3), we find that
L [f˙] =
∫ 2
0
3e−st dt =
3(1− e−2s)
s
.
t20
 f(t)
6
Since f is continuous and f˙ sectionally continuous, (13) is applicable. Therefore,
L [f˙ ] = −f0 + sF (s) = 3(1 − e
−2s)
s
.
But f0 = 0, hence
F (s) =
3(1− e−2s)
s2
.
On the other hand, one may calculate the transform from (3), integrating by parts:
F (s) =
∫ 2
0
3te−st dt +
∫ ∞
2
6e−st dt
=
3
s2
− 3(2s + 1)e
−2s
s2
+
6e−2s
s
.
It’s easy to see that the final result is the same. ◭
14 The Laplace Transform
6. Division by s
It follows from the fundamental theorem of calculus that
d
dt
∫ t
0
f(τ) dτ = f(t).
The integral on the left-hand side is an anti-derivative of f(t)†. We now call it g:
g(t)
def
=
∫ t
0
f(τ) dτ,
and observe that
g˙(t) = f(t), g(0+) = 0.
Taking the Laplace transform of the identity
L [g˙(t)] = L [f(t)]
and using (13), we get
−g0 + sG(s) = F (s).
But g0 = 0. Hence,
G(s) =
F (s)
s
,
which may finally be expanded as
L
[ ∫ t
0
f(τ) dτ
]
=
F (s)
s
. (16)
Naturally, this formula may be iterated any number of times. For example, if we integrate
between zero and t the function g defined at the beginning of this section, we get another function
of t, let us call it h, such that h¨(t) = f(t), and h(0) = h˙(0) = 0; therefore L [h(t)] = F (s)/s2.
Then, integrating h between zero and t and reasoning in the same way, we obtain yet
another function of t, and its Laplace transform will be F (s)/s3. And so on: the process may
be repeated as many times as needed.
◮Example 13 Find the function x(t) such that 2x˙+ 4x = 2 and x0 = 1.
Solution: We take the Laplace transform of the equation. We get
L [2x˙+ 4x] = L [2],
where L [x] = X(s), L [x˙] = −x0 + sX(s) = −1 + sX(s), and L [2] = 2/s. It follows
2(−1 + sX) + 4X = 2
s
,
† Be careful: the dummy variable of integration must not be t, because it appears as a limit
of integration.
The Laplace Transform 15
which—solving for X—yields
X =
1
2s + 4
(
2 +
2
s
)
=
1
s+ 2
+
1
s(s+ 2)
.
Now, by inspection, we recognize that
1
s+ 2
= L [e−2t];
the second term may be simplified by rule (16):
1
s(s+ 2)
= L
[∫ t
0
e−2τ dτ
]
= L [1
2
− 1
2
e−2t
]
.
Combining the two terms, it follows immediately that
X = L [e−2t]+ L [ 12 − 12e−2t] = L [ 12 + 12e−2t],
and finally (by Lerch’s theorem) x = 12 +
1
2e
−2t. ◭
◮Example 14 Since
tn =
∫ t
0
nτn−1 dt,
we get immediately
L [tn] = nL [t
n−1]
s
.
In this way, starting from
L [1] =
∫ ∞
0
e−st dt =
1
s
,
we get
L [t] = 1
s2
, L [t2] = 2!
s3
, L [t3] = 3!
s4
, L [t4] = 4!
s5
,
and so on. We recover (9), though for integer powers only. ◭
These transforms may also be understood in terms of convolution, a more advanced technique
that we’ll study is section 2.7.
7. Multiplication by t
It may be shown that if the integral (3) converges absolutely, then F (s) may be differentiated
any number of times with respect to s. For example,
F ′(s) =
d
ds
∫ ∞
0
f(t) e−st dt =
∫ ∞
0
∂
∂s
[
f(t) e−st
]
dt = −
∫ ∞
0
f(t) te−st dt.
Repeating this process n times, we obtain
L [tnf(t)] = (−1)n d
nF (s)
dsn
. (17)
◮Example 15 Find L [t cosh 4t].
Solution: Since L [cosh 4t] = s/(s2 − 16), then by (17)
L [t cosh 4t] = − d
ds
s
s2 − 16 = −
1
s2 − 16 +
2s2
(s2 − 16)2 =
s2 + 16
(s2 − 16)2 .
As an exercise, calculate
∫∞
0
t cosh 4t e−st dt (integrating by parts) and verify that you get the
same answer. ◭
16 The Laplace Transform
◮Example 16 Find L [t4ect].
Solution: We have from (4): L [ect] = 1/(s − c). Hence, by (17):
L [t4ect] =
(
− d
ds
)4 1
s− c =
4!
(s− c)5 . ◭
◮Example 17 Find L [teiωt] and hence L [t cosωt], L [t sinωt].
Solution: From (4) we get L [eiωt] = 1/(s − iω). It follows
L [teiωt] = − d
ds
1
s− iω =
1
(s− iω)2 =
(s+ iω)2
(s2 + ω2)2
=
s2 + i2sω − ω2
(s2 + ω2)2
.
Separating real and imaginary parts, it follows immediately
L [t cos ωt] = s
2 − ω2
(s2 + ω2)2
;
L [t sinωt] = 2sω
(s2 + ω2)2
.
As an exercise, calculate from first principles
∫∞
0
e−stt cosωt dt or
∫∞
0
e−stt sinωt dt, and verify
that you get the same results. ◭
◮Example 18 Find the function x(t) such that 2x˙+ x = t3e−t/2, and x0 = 0.
Solution: Taking the Laplace transform of the equation, we get
2L [x˙] + L [x] = L [t3e−t/2].
writing L [x] = X and using (13) we get immediately
L [x˙] = −x0 + sX = sX.
Then, we note that
L [e−t/2] = 1
s+ 1/2
=⇒ L [t3e−t/2] = − d3
dt3
(
1
s+ 1/2
)
=
3!
(s+ 1/2)4
;
this step (multiplication by t3) follows from (17). Substituting these expressions into the pre-
ceding equation yields:
2sX +X =
6
(s+ 1/2)4
.
Hence
X =
6
(2s+ 1)(s + 1/2)4
=
3
(s + 1/2)5
.
Now, we observe that
d4
dt4
(
1
s+ 1/2
)
=
4!
(s+ 1/2)5
;
The Laplace Transform 17
therefore,
3
(s+ 1/2)5
=
3
4!
d4
dt4
(
1
s+ 1/2
)
= 18 L
[
t4e−t/2
]
;
this step (multiplication by t4) also follows from (17). Hence,
X = L [ 18 t4e−t/2],
and finally, by Lerch’s theorem,
x = 1
8
t4e−t/2
is the required function. ◭
8. Division by t
If f(t) is of exponential order, then certainly
lim
s→∞
∫ ∞
0
|f(t)| e−st dt = 0.
In other words, if a function of s does not approach zero as s → ∞, then it is not a Laplace
transform, in the ordinary sense of the word.†
Given a function f(t) with a Laplace transform F (s), we introduce
G(s)
def
=
∫ ∞
s
F (σ) dσ.
Note that σ is a dummy variable; s is the lower limit of integration and we treat it as a parameter.
Convince yourself that this integral is defined in such a way that if s→∞, then G(s) tends to
zero. Therefore, G(s) may be the Laplace transform of some function g(t).
By the fundamental theorem of calculus, G(s) is an anti-derivative of −F (s):
G′(s) = −F (s) = −L [f(t)].
On the other hand, it follows from (17) that
G′(s) = L [−tg(t)].
By comparison, we get f(t) = tg(t), which may be written
g(t) =
f(t)
t
.
Finally, tranforming again, we get G(s) = L [f(t)/t], which may be expanded as
L
[
f(t)
t
]
=
∫ ∞
s
F (σ) dσ. (18)
† This comment does not apply to “generalized functions”, which you’ll find in more advanced
courses. But generalized functions are not functions in the usual sense of the word.
18 The Laplace Transform
◮Example 19 Find L [sin t/t].
Solution: From (8) we get L [sin t] = 1/(s2 + 1). Hence, it follows from (18) that
L
[
sin t
t
]
=
∫ ∞
s
dσ
σ2 + 1
=
[
arctan σ
]∞
s
= 12π − arctan s = arccot s.
Comment: By definition, L [sin t/t] = ∫∞
0
(sin t/t) e−st dt. So, we get the interesting result
∫ ∞
0
sin t
t
e−st dt = 12π − arctan s. [s > 0]
For example, for s = 1 we get
∫∞
0
(e−t sin t/t) dt = 12π − arctan 1 = 14π. Letting s→ 0+ we get∫ ∞
0
sin t
t
dt = 1
2
π,
which is a famous example of a convergent improper integral that is not absolutely convergent.
Note also that these integrals may not be done using elementary calculus, as the anti-derivatives
may not be written in terms of elementary functions. We find the surprising result that definite
integration is elementary while indefinite integration is not. ◭
◮Example 20 Using the result of example 19, find G(s), given that g(t) =
∫ t
0
(sin τ/τ) dτ .
Solution: In example 19 we established that
L
[
sin t
t
]
= arccot s.
Hence, applying (16) we get immediately that
L
[ ∫ t
0
sin τ
τ
dτ
]
=
arccot s
s
.
Comment: the function g(t) =
∫ t
0
(sin τ/τ) dτ is called integral sine, denoted si(t), and is used
in many engineering applications. ◭◮Example 21 Find L [(et − 1)/t].
Solution: From (4) and (9) we get, respectively, L [et] = 1/(s− 1) and L [1] = 1/s. It follows
that
L
[
et − 1
t
]
=
∫ ∞
s
(
1
σ − 1 −
1
σ
)
dσ =
=
[
ln
(
σ − 1
σ
)]∞
s
= ln 1− ln s− 1
s
= ln
s
s− 1 .
Again, although the indefinite integral
∫∞
0
(
e−st(et−1)/t) dt cannot be done in closed form, the
Laplace transform has been found by means of (18). ◭
In principle, rule (18) may be iterated: division of f(t) by t2 corresponds to integrating F (s)
twice, and so on. In practice, examples of this kind are likely to be fairly long. You may find
The Laplace Transform 19
one at the end of this chapter (see section 1.11). Next example shows how double integration
may, in some cases, be avoided.
◮Example 22 Find the Laplace transform of sin2 t/t and hence the Laplace transform of
sin2 t/t2.
Solution: The first half of this problem is very simple: from the identity
sin2 t = 12 (1− cos 2t)
we get immediately that
L [ sin2 t] = 1
2
(
1
s
− s
s2 + 4
)
,
and hence that
L
[
sin2 t
t
]
=
1
2
∫ ∞
s
(
1
σ
− σ
σ2 + 4
)
dσ = 1
2
ln
√
s2 + 4
s
.
Naturally, for this step we applied equation (18). For the second part, we could use (18) again,
but the following shortcut is better. If we define
f(t) =
sin2 t
t
:
then the preceding result may be written
F (s) = 1
2
ln
√
s2 + 4
s
.
Differentiating f(t) we get that
f˙ =
2 sin t cos t
t
− sin
2 t
t2
=
sin 2t
t
− sin
2 t
t2
;
Note that f(0+) = 0. It follows that
sin2 t
t2
=
sin 2t
t
− f˙
and hence, by (13), that
L
[
sin2 t
t2
]
= L
[
sin 2t
t
]
−
(
s F (s)− f(0+)
)
= L
[
sin 2t
t
]
− 1
2
s ln
√
s2 + 4
s
.
The first term on the right-hand side is almost identical to what we found in example 19:
L
[
sin 2t
t
]
= arccot(s/2).
So, finally, we get:
L
[
sin2 t
t2
]
= arccot(s/2)− 1
2
s ln
√
s2 + 4
s
. ◭
20 The Laplace Transform
OVERVIEW
The rules discussed in sections 5–8 may, at first, seem hard to grasp. They are not: let us put
them all together in a table.
f(t) F (s)
f˙(t) sF (s)− f0∫ t
0
f(τ) dτ
F (s)
s
t f(t) −F ′(s)
f(t)
t
∫ ∞
s
F (σ) dσ
The first line is simply a reminder that we use lower-case letter to denote functions of t, and the
corresponding upper-case letters to denote their transforms.
You should be able to spot the thread linking the other formulas: Differentiation with
respect to one variable, t or s, is associated with multiplication by the other; integration with
respect to one variable is associated with division by the other.
We are slightly over-simplifying, now; some fine points must also be borne in mind. For
example, integration with respect to s ends at infinity, whereas integration with respect to t
starts at zero. The rule for L [f˙ ] must account for the possibility that f(t) be discontinuous at
t = 0, hence f0 appears in the second line; on the other hand, the rule for F
′(s) has a minus
sign, which should not be overlooked.
However, the common thread (in italics above) is easy to remember. Using the six basic
transforms (4–9) as building blocks and combining the rules above, one may derive a wide variety
of transforms, with no direct call for integral calculus. We now complement this table with two
more basic rules: the shift rules.
9. Shifting s
Consider a function f(t) with Laplace transform F (s). Let α be a constant. By definition,
L [e−αtf(t)] = ∫ ∞
0
f(t) e−αt e−st dt.
It follows immediately
L [e−αt f(t)] = ∫ ∞
0
f(t) e(−s−α)t dt =
= F (s + α). (19)
◮Example 23 Find L [e−4t√t ].
Solution: By (9) we get that
[√
t
]
= (1/2)!
/
s3/2. Therefore, L [e−4t√t ] = (1/2)!/(s+4)3/2. ◭
◮Example 24 Find L [e−5t cos 3t].
Solution: Since L [cos 3t] = s/(s2 + 9), we get immediately that
L [e−5t cos 3t] = s+ 5
(s + 5)2 + 9
=
s+ 5
s2 + 10s + 34
.
◭
The Laplace Transform 21
◮Example 25 Find L [t3e2t].
Solution: Since L [t3] = 6/s4, we get immediately by (19) that
L [t3e2t] = 6
(s− 2)4 .
Note, however, that one may also start from L [e2t] = 1/(s − 2) and apply (17) three times:
L [t3e2t] = − d3
ds3
(
1
s− 2
)
=
6
(s− 2)4 . ◭
◮Example 26 Find L [sin 2t cosh 3t].
Solution: By definition,
sin 2t cosh 3t = sin 2t ·
(
e3t + e−3t
2
)
=
= 12e
3t sin 2t+ 12e
−3t sin 2t.
It follows immediately that
L [sin 2t cosh 3t] = 1
(s− 3)2 + 4 +
1
(s+ 3)2 + 4 ◭
◮Example 27 Find F (s), if f(t) = t−1
∫ t
0
e−3τ sin τ dτ .
Solution: We observe that f(t) is obtained by performing three operations on the function sin t,
which are:
(i) multiplication by e−3t,
(ii) integration with respect to t,
(iii) division by t.
The corresponding operations on F (s) are:
(i) shift by 3 units to the left
(ii) division by s,
(iii) integration with respect to s.
Proceeding along these lines, we get:
L [sin t] = 1
s2 + 1
,
L [e−3t sin t] = 1
(s+ 3)2 + 1
,(i)
L
[∫ t
0
e−3τ sin τ dτ
]
=
1
s[(s+ 3)2 + 1]
,(ii)
and finally:
(iii) F (s) = L
[
1
t
∫ t
0
e−3τ sin τ dτ
]
=
∫ ∞
s
dσ
σ[(σ + 3)2 + 1]
.
22 The Laplace Transform
A simple substitution yields∫ ∞
s
dσ
σ[(σ + 3)2 + 1]
=
∫ ∞
s+3
dx
(x− 3)(x2 + 1) .
It is possible to show that
1
(x− 3)(x2 + 1) =
1
10
x− 3 −
1
10 (x+ 3)
x2 + 1
,
as you can easily verify. The expansion above may be done by the methods you learnt in first
year, but we’ll come back to this subject in the next chapter.
So, integrating, we get:
F (s) =
1
10
∫ ∞
s+3
[
1
x− 3 −
x+ 3
x2 + 1
]
dx =
=
1
10
[
ln |x− 3| − 12 ln |x2 + 1| − 3 arctan x
]∞
s+3
=
=
1
10
[
ln
∣∣∣∣ x− 3√x2 + 1
∣∣∣∣− 3 arctan x
]∞
s+3
=
=
1
10
[
ln 1− 3 · π
2
− ln s√
(s+ 3)2 + 1
+ 3arctan(s+ 3)
]
.
Simplifying, we find that
F (s) = 110 ln
√
(s+ 3)2 + 1
s
− 310 arccot(s+ 3). ◭
10. Shifting t
We have discussed the the effects of a shift of the s axis. A shift of the t axis may be studied in
a similar way.
This may be useful, for instance, if the function f(t) is defined by different formulas over
different pieces of the t axis. In such a case, the following approach may be useful. Suppose
f(t) =
{
0 if t < T ,
φ(t− T ) if t ≥ T ,
where φ(t) is a function of t having Laplace transform Φ(s). Then, by definition,
F (s) =
∫ T
0
f(t) e−st dt +
∫ ∞
T
f(t) e−st dt =
= 0 +
∫ ∞
T
φ(t− T ) e−st dt.
Substituting t = T + τ , we get
F (s) =
∫ ∞
0
φ(τ)e−sT−sτ dτ =
= e−sT Φ(s). (20)
The Laplace Transform 23
◮Example 28 Find L [f(t)], if f(t) =
{
0 if t < 3,
sin 2t if t ≥ 3.
Solution: We may not apply rule (20) as long as f(t) is written in this way, because the depen-
dence on t− 3 is not explicit. To make it so, we manipulate the function slightly:
sin 2t = sin 2(t− 3 + 3) = sin 2(t− 3) · cos 6 + cos 2(t− 3) · sin 6.
Then we define a function
φ(t)
def
= sin 2t cos 6 + cos 2t sin 6,
which has the Laplace transform
Φ(s) =
2 cos 6
s2 + 4
+
s sin 6
s2 + 4
.
Since obviously
sin 2t = φ(t− 3),
i.e.,
f(t) =
{
0 if t < 3,
φ(t− 3) if t ≥ 3,
we are in a position to apply (20). It follows that
F (s) =
e−3s(2 cos 6 + s sin 6)
s2 + 4
.
◭
In most applications, rule (20) is combined with a simple and useful mathematical tool: the step
function.
◮Definition: The function u(t), which is defined as
u(t) =
{
0 for negative t
1
2 for t = 0
1 for positive t
is called the unit step function, or also Heaviside’s function. ◭
Do not worry about the definition of u(t) for t = 0: it is purely conventional and doesnot affect
the Laplace transform in any way. It finds its place in more advanced topics.
Heaviside’s function jumps from 0 to 1 when its argument is increased across zero. So, for
example, what is u(t − 74)? Since t − 74 < 0 when t < 74, and t − 74 > 0 when t > 74, we
see that u(t− 74) jumps from 0 to 1 as t is increased across the point t = 74. By rule (20), the
Laplace transform of this function would be simply
F (s) = e−74s · L [1] = e
−74s
s
.
Very often two step functions are combined to form a function that is zero for t up to a certain
point, and becomes zero again from another point onwards.
24 The Laplace Transform
◮Example 29 Consider the function
f(t) = u(t− 1)− u(t− 4),
pictured on the right: until t = 1 both u’s are
zero, so their difference is zero. After t = 4,
both u’s are = 1, hence their difference is zero
again. Between t = 1 and t = 4, only u(t−1)
is equal to one, while u(t − 4) is still zero:
hence, u(t− 1)− u(t− 4) = 1 there.
t
1
−1
1 4
 f(t)
The graph of this function is a rectangle with unit height and a length of three units. Its
Laplace transform is
F (s) = L [u(t− 1)] − L [u(t− 4)] = e
−s − e−4s
s
.
◭
Things get interesting when we multiply a given function of t by u(t − a) − u(t − b): we get a
new function that coincides with the old one between t = a and t = b, but is identically zero
everywhere else. Such a function is called a transient.
◮Example 30 Find F (s), if f(t) = sin 5t for t between 1 and 4; f(t) ≡ 0 everywhere else.
Solution: The graph of sin 5t extends, of course, from −∞ to +∞. However, if sin 5t is multiplied
by u(t − 1) − u(t − 4), the part of the graph lying outside the interval (1, 4) is “wiped off”, so
to speak, while the part between 1 and 4 is not affected.
The graph of f(t), pictured on the
right, jumps from 0 to sin 5 at t = 1, and
then from sin 20 to 0 at t = 4. We may
write
f(t) = sin 5t · [u(t− 1)− u(t− 4)] =
= sin 5t · u(t− 1)− sin 5t · u(t− 4).
Proceeding like in example 28, we write
f(t) = sin 5(t− 1 + 1) · u(t− 1)−
− sin 5(t− 4 + 4) · u(t− 4).
t
1
4
 f(t)
Simplifying, we get:
f(t) =
[
sin 5(t− 1) cos 5 + cos 5(t− 1) sin 5] · u(t− 1)−
− [ sin 5(t− 4) cos 20 + cos 5(t− 4) sin 20] · u(t− 4),
and finally, by (20):
F (s) =
[
5 cos 5
s2 + 25
+
s sin 5
s2 + 25
]
e−s −
[
5 cos 20
s2 + 25
+
s sin 20
s2 + 25
]
e−4s.
◭
The Laplace Transform 25
It should be noted that problems of this type may always be done from first principles. However,
with some practice you’ll find that the method of this section is usually better. For instance,
the last example may also be done by calculating
F (s) =
∫ 4
1
e−st sin 5t dt
using integration by parts or Euler’s formula. Do it, as an exercise.
◮Example 31 Find F (s), if f(t) =
∣∣1/2− t∣∣.
Solution: First of all, recall that, by definition, |A| = −A whenever A is negative. Hence,
since the expression 1/2 − t changes sign for t = 1/2, then |1/2 − t| = t − 1/2 if t > 1/2.
Also, recall that f(t) ≡ 0 by definition if t is negative.
Therefore,
f(t) =


0 if t is negative,
1/2 − t if 0 < t < 1/2,
t− 1/2 if t > 1/2,
as shown in the picture. Using Heaviside’s unit step func-
tion, we may write
f(t) = (1/2− t)
[
u(t)− u(t− 1/2)
]
+ (t− 1/2)u(t− 1/2).
/1 2
/1 2
 f(t)
t
Simplifying, we get immediately:
f(t) = (1/2− t)u(t) + 2(t− 1/2)u(t− 1/2).
Considering the transforms
L [1/2] =
1/2
s
, L [t] = 1
s2
,
and applying (20), we get immediately
F (s) =
1/2
s
− 1
s2
+
2e−s/2
s2
.
As an exercise, calculate
∫ 1/2
0
(1/2 − t)e−st dt +
∫∞
1/2
(t − 1/2)e−st dt and verify that you get the
same answer. ◭
◮Example 32 Find F (s) if f(t) = 3t for 0 ≤ t ≤ 2; f(t) = 6 for t > 2 (this is the same as
example 12).
Solution: We note that
f(t) = 3t
[
u(t)− u(t− 2)]+ 6u(t− 2) =
= 3t u(t) − 3(t− 2 + 2)u(t− 2) + 6u(t− 2) =
= 3t u(t) − 3(t− 2)u(t − 2).
26 The Laplace Transform
It follows immediately that
F (s) =
3
s2
− 3e
−2s
s2
:
same result as in example 12. ◭
◮Example 33 Find F (s) if f(t) = 3t− t2 for 0 < t < 3, f(t) ≡ 0 everywhere else.
Solution: Write
f(t) = (3t− t2) · (u(t)− u(t− 3)).
To get an idea of the graph of f(t), imagine that you
take the parabola y = 3t−t2 and remove everything
that lies below the t axis, as shown in the picture.
Now, we have f(t) = (3t−t2)·u(t)−(3t−t2)·u(t−3).
The first term may be left as it is, but we must do
some groundwork on the second term, if we are to
use the shift theorem (20). Hence, we write
t
 f(t)
 1
 2
 1 2 3 0
(3t− t2) · u(t− 3) = [3(t− 3 + 3)− (t− 3 + 3)2] · u(t− 3) =
=
[
3(t− 3) + 9− (t− 3)2 − 6(t− 3)− 9] · u(t− 3) =
=
[− 3(t− 3)− (t− 3)2] · u(t− 3).
In this way, the second term depends on t is only through the expression t− 3. And since
L [− 3t− t2] = − 3
s2
− 2
s3
,
we get by (20) that
L [(− 3(t− 3)− (t− 3)2) · u(t− 3)] = −( 3
s2
+
2
s3
)
e−3s.
So much for the second term. The first term requires no rearrangements, and we get immediately:
L [(3t− t2) · u(t)] = ( 3
s2
− 2
s3
)
· e0s;
obviously e0s = 1. So, finally, we get:
F (s) =
3
s2
− 2
s3
+
(
3
s2
+
2
s3
)
e−3s.
◭
SHIFT RULES
Formulas (20) and (19) are often called shift properties. They are summarized in the following
table.
f(t) F (s)
f(t− T )u(t− T ) e−sTF (s)
e−atf(t) F (s + a)
The Laplace Transform 27
11. Additional Examples
As a rule, the linearity property
L [f1 + f2] = L [f1] + L [f2]
holds only if L [f1] and L [f2] exist each one on its own. It may happen, though, that the Laplace
transform of f1 + f2 does exist, even if f1 and f2 do not have a Laplace transform.
◮Example 34 Find F (s), if f(t) = (et − 1) · t−3/2.
Solution: We would like to use the shift theorem (19). However,
∫ ∞
0
ett−3/2e−st dt−
∫ ∞
0
t−3/2e−st dt = ∞−∞
which is clearly meaningless. So, F (s) may not be “split” as we did in several previous examples;
see for instance example 26.
On the other hand, it’s easy to see that F (s) exists, and we are going to find it in two steps.
First of all we write
g(t) =
et − 1
t1/2
, f(t) =
et − 1
t3/2
=
g(t)
t
,
and consider G(s). This is easy, because L [ett−1/2] and L [t−1/2] exist separately:
L [t−1/2] = (−1/2)!
s1/2
, L [ett−1/2] = (−1/2)!
(s− 1)1/2 ;
the second one comes via the shift theorem (19). Therefore,
G(s) =
(−1/2)!
(s− 1)1/2 −
(−1/2)!
s1/2
.
Finally, we go back to f(t) and use (18)—division by t corresponds to integration by s:
F (s) =
∫ ∞
s
[
(−1/2)!
(σ − 1)1/2 −
(−1/2)!
σ1/2
]
dσ =
= (−1/2)!
[
(σ − 1)1/2
1/2
− σ
1/2
1/2
]∞
s
We should show that, in the equation above, the last expression in square brackets goes to zero
as σ tends to infinity; this is a good revision example in 1st-year calculus. Writing
(σ − 1)1/2
1/2
− σ
1/2
1/2
=
(σ − 1)1/2 − σ1/2
1/2
· (σ − 1)
1/2 + σ1/2
(σ − 1)1/2 + σ1/2 ,
and simplifying the numerator, we get:
(σ − 1)1/2
1/2
− σ
1/2
1/2
=
−2
(σ − 1)1/2 + σ1/2 .
28 The Laplace Transform
It’s now clear that the right-hand side goes to zero as σ →∞, as required. So, finally:
F (s) = 0 − (−1/2)!
[
(σ − 1)1/2
1/2
− σ
1/2
1/2
]
σ=s
=
= 2(−1/2)!
[
s1/2 − (s− 1)1/2] .
In section 2.7 we’ll see that (−1/2)! = √π. See also the comments at the end of section 1.4. ◭
◮Example 35 Find L [(2e3t − 3e2t + 1) · t−5/2].
Solution: This problem is very similar to the preceding one, so we’ll look only at the main points.
We define
h(t) =
2e3t − 3e2t + 1
t1/2
, g(t) =
2e3t − 3e2t + 1
t3/2
=
h(t)
t
, f(t) =
2e3t − 3e2t + 1
t5/2
=
g(t)
t
.
We seek F (s). Neither f(t) nor g(t) may by split as we did previously; however,for h(t) it’s
correct to write
H(s) = L [2e3t · t−1/2] − L [3e2t · t−1/2] + L [t−1/2] =
=
2(−1/2)!
(s− 3)1/2 −
3(−1/2)!
(s− 2)1/2 +
(−1/2)!
s1/2
.
Using (18)—division by t corresponds to integration by s, we get:
G(s) =
∫ ∞
s
H(σ) dσ = (−1/2)!
[
2(σ − 3)1/2
1/2
− 3(σ − 2)
1/2
1/2
+
σ1/2
1/2
]∞
s
.
Proceeding like in the previous example, it’s possible to see that the expression in square brackets
goes to zero as σ tends to infinity. Therefore,
G(s) = 2(−1/2)!
[− 2(s − 3)1/2 + 3(s− 2)1/2 − s1/2].
One more application of (18) yields
F (s) =
∫ ∞
s
G(σ) dσ = 2(−1/2)!
[−2(σ − 3)3/2
3/2
+
3(σ − 2)3/2
3/2
− σ
3/2
3/2
]∞
s
.
Once again, it’s possible to show that the expression is square brackets goes to zero as σ tends
to infinity. So, the final answer is
F (s) = 4
3
(−1/2)!
[
2(s− 3)3/2 − 3(s − 2)3/2 + s3/2] .
As an exercise, fill in the details of this example. ◭
◮Example 36 Find F (s), if f(t) = te|t−1|.
Solution: First of all, note that
f(t) =
{
0 if t is negative, by definition;
te1−t if 0 < t < 1,
tet−1 if t is greater than 1.
The Laplace Transform 29
Using Heaviside’s unit step function, we write
f(t) = te1−t
[
u(t)− u(t− 1)]+ tet−1 u(t− 1).
Simplifying, we get:
f(t) = te1−t u(t) + t
(
et−1 − e1−t)u(t− 1) =
= ete−t u(t) + 2t sinh(t− 1) · u(t− 1) =
= ete−t u(t) + 2[1 + (t− 1)] sinh(t− 1) · u(t− 1).
Considering the transforms
L [te−t] = 1
(s+ 1)2
, L [sinh t] = 1
s2 − 1 , L [t sinh t] = −
d
ds
(
1
s2 − 1
)
=
2s
(s2 − 1)2 ,
and applying (20), we get immediately
F (s) =
e
(s + 1)2
+ 2e−s
[
s
s2 − 1 +
2s
(s2 − 1)2
]
.
As an exercise, find F (s) from first principles, i.e., calculating
∫ 1
0
te1−te−st dt+
∫∞
1
tet−1e−st dt
and verify that you get the same answer. Compare the amount of work required. ◭
◮Example 37 Find the Laplace transform of f(t) = e4t
∫ t
0
e−3z
∫ z
0
e−2y
∫ y
0
e5x cos x dx dy dz.
Solution: Begin with
L [cos t] = s
s2 + 1
.
Applying (19) we get:
L [e5t cos t] = s− 5
(s− 5)2 + 1 .
Applying (16) we get:
L
[ ∫ t
0
e5x cos x dx
]
=
s− 5
s [(s− 5)2 + 1] .
Applying (19) we get:
L
[
e−2t
∫ t
0
e5x cos x dx
]
=
s− 3
(s+ 2) [(s − 3)2 + 1] .
Applying (16) we get:
L
[ ∫ t
0
e−2y
∫ y
0
e5x cosx dx dy
]
=
s− 3
s(s+ 2) [(s − 3)2 + 1] .
Applying (19) we get:
L
[
e−3t
∫ t
0
e−2y
∫ y
0
e5x cos x dx dy
]
=
s
(s+ 3)(s + 5) [s2 + 1]
.
Applying (16), we get:
L
[ ∫ t
0
e−3z
∫ z
0
e−2y
∫ y
0
e5x cos x dx dy dz
]
=
1
(s+ 3)(s + 5) [s2 + 1]
.
Finally, applying (19), we get:
F (s) = L
[
e4t
∫ t
0
e−3z
∫ z
0
e−2y
∫ y
0
e5x cos x dx dy dz
]
=
1
(s2 − 1) [(s − 4)2 + 1] . ◭
30 Tutorial Problems The Laplace Transform
PROBLEMS
Transforms of Elementary Functions
1. Find the Laplace transform of each of the following functions. In each case, specify the
values of s for which the integral (3) converges.
(a) 2e4t
(b) 3e−2t
(c) 5t− 3
(d) 2t2 − e−t
(e) 3 cos 5t
(f) 10 sin 6t
(g) 6 sin 2t− 5 cos 2t
(h) (t2 + 1)2
(i) (sin t− cos t)2
(j) 3 cosh 5t− 4 sinh 5t
(k) (5e2t − 3)2
(l) 4 cos2 2t
2. Find the following Laplace transforms.
(a) L [cosh2 4t]
(b) L [3t4 − 2t3 + 4e−3t − 2 sin 5t+ 3cos 2t]
(c) L [cosh3 t]
(d) L [sin5 t]
(e) L [sin 3t cos 2t]
(f) L [cos 2t cos 5t].
3. Using the identity c! = (c + 1)!/(c + 1) [see (12)], show that lim
c→−1+
c! =∞. Use a good
computer package to observe this result numerically.
Multiplication by s
4. Let f(t) =
{
et if 0 < t < 1,
0 if t > 1,
and g(t) =
{
et if 0 < t < 1,
e if t > 1.
Find F (s), G(s) and show that rule (13) holds for g but not for f . Why is it so?
5. Given f(t) =
{
2t if 0 ≤ t ≤ 1
t if t > 1,
find (a) F (s), (b) L [f˙(t)].
Does formula (13) hold? Explain.
6. Given f(t) =
{
t2 if 0 ≤ t ≤ 1
0 if t > 1,
find (a) F (s), (b) L [f¨(t)].
Does formula (14) hold? Explain.
Division by s
7. Calculate L [ ∫ t
0
(τ3 − 5τ + sinh 2τ) dτ]: (a) By doing the integral first, (b) Using for-
mula (16). Verify that you get the same answer.
8. Calculate L [ ∫ t
0
(2 sin τ cos 7τ) dτ
]
applying formula (16).
Multiplication by t
9. Find the following Laplace transforms using formula (17).
(a) L [t3e−3t]
(b) L [(t+ 2)2et]
(c) L [t(3 sin 2t− 2 cos 2t)]
(d) L [t2 sin t]
(e) L [t cosh 3t]
(f) L [t sinh 2t]
(g) L [(t− 1)(t− 2) sin 3t]
(h) L [t3 cos t] .
10. Applying (17), calculate: (a)
∫∞
0
t e−3t sin t dt, (b)
∫∞
0
t2 e−t cos t dt.
The Laplace Transform Tutorial Problems 31
Division by t
11. Find the following Laplace transforms; read example 22 before attempting (d).
(a) L [(1 − e−t)/t]
(b) L [sinh2 t/t]
(c) L [(cosh t− cos t)/t]
(d) L [sinh2 t/t2].
12. Calculate (a) L
[ ∫ t
0
1− e−τ
τ
dτ
]
, (b) L
[
t
∫ t
0
sin τ
τ
dτ
]
.
Shifting
13. Find the following Laplace transforms.
(a) L [e−t cos 2t]
(b) L [2e3t sin 4t]
(c) L [e2t(3 sin 4t− 4 cos 4t)]
(d) L [e−t(3 sinh 2t− 5 cosh 2t)]
(e) L [e−t sin2 t]
(f) L [(1 + te−t)3]
(g) L [t2 u(t− 5)]
(h) L [e−2tt2u(t− 1)].
14. Find:
(a) L
[
sinh t
3
√
t
]
(b) L
[
e−3t sin 2t
t
]
15. Find F (s), given:
(a) f(t) =
{
2t if 0 ≤ t < 5,
0 everywhere else.
(b) f(t) =
{
sinπt if 0 < t < 1,
0 everywhere else.
16. Find F (s), given:
(a) f(t) = |t2 − 4| (b) f(t) = |2− et|
17. Find L [ |t2 − 7t+ 12| ].
ANSWERS
1 (a) 2/(s − 4), s > 4
(b) 3/(s + 2), s > −2
(c) (5− 3s)/s2, s > 0
(d) (4 + 4s− s3)/s3(s+ 1), s > 0
(e) 3s/(s2 + 25), s > 0
(f) 60/(s2 + 36), s > 0
(g) (12 − 5s)/(s2 + 4), s > 0
(h) (s4 + 4s2 + 24)/s5, s > 0
(i) (s2 − 2s+ 4)/s(s2 + 4), s > 0
(j) (3s − 20)/(s2 − 25), s > 5
(k) 25/(s− 4)− 30/(s− 2)+ 9/s, s > 4
(l) 2/s + 2s/(s2 + 16), s > 0.
2 (a) (s2 − 32)/s(s2 − 64)
(b) 72/s5 − 12/s4 + 4/(s + 3)− 10/(s2 + 25) + 3s/(s2 + 4)
(c) [s/(s2 − 9) + 3s/(s2 − 1)]/4
(d) [5/(s2 + 25)− 15/(s2 + 9) + 10/(s2 + 1)]/16
(e) [5/(s2 + 25) + 1/(s2 + 1)]/2
(f) [s/(s2 + 49) + s/(s2 + 9)]/2.
3 Using the open-source package maxima one gets that (−0.9)! ≈ 9.51, (−0.999)! ≈ 999.42
and (−0.999999)! ≈ 106. As c gets closer to −1, c! becomes practically equal 1/(c + 1) [up to
the 6th significant digit].
4 F =
1− e1−s
s− 1 , G =
1− e1−s
s− 1 +
e1−s
s
. Rule (13) does not hold for f because f is
discontinuous at t = 1, whereas g is continuous throughout and g˙ is sectionally continuous.
5 (a) F (s) = 2/s2 − e−s(1/s + 1/s2); (b) L [f˙ ] = 2/s − e−s/s.
Formula (13) does not hold because f is not continuous.
32 Tutorial Problems The Laplace Transform
6 (a) F (s) = 2/s3 − e−s(2/s3 + 2/s2 + 1/s); (b) L [f¨ ] = 2(1− e−s)/s.
Formula (14) does not hold because f and f˙ are not continuous.
8
8
s
( 1
s2 + 64
)
− 6
s
( 1
s2 + 36
)
.
9 (a) 6/(s + 3)4
(b) (4s2 − 4s+ 2)/(s − 1)3
(c) (8 + 12s− 2s2)/(s2 +4)2
(d) (6s2 − 2)/(s2 + 1)3
(e) (s2 + 9)/(s2 − 9)2
(f) 4s/(s2 − 4)2
(g) (6s4 − 18s3 + 126s2 − 162s + 432)/(s2 + 9)3
(h) (6s4 − 36s2 + 6)/(s2 + 1)4.
10 (a) 3/50, (b) −1/2.
11 (a) ln
s+ 1
s
(b) 12 ln
s√
s2 − 4
(c) 1
2
ln
s2 + 1
s2 − 1
(d) arcoth(s/2)− 12s ln
s√
s2 − 4
12 (a)
1
s
ln
(
1 +
1
s
)
, (b) (arccot s)/s2 + 1/s(s2 + 1).
13 (a) (s+ 1)/(s2 + 2s + 5)
(b) 8/(s2 − 6s+ 25)
(c) 4(5− s)/(s2 − 4s+ 20)
(d) (1− 5s)/(s2 + 2s− 3)
(e) 2/(s + 1)(s2 + 2s+ 5)
(f) 1/s+3/(s+1)2+6/(s+2)3+6/(s+3)4
(g) (2/s3 + 10/s2 + 25/s) e−5s
(h)
[
2/(s+2)3+2/(s+2)2+1/(s+2)
]
e−2−s
14 (a) F (s) =
(−1/3)!
2(s− 1)2/3 −
(−1/3)!
2(s+ 1)2/3
, (b) F (s) = arccot(s+ 3)/2.
15 (a) F (s) =
2
s2
− (2 + 10s) e
−5s
s2
, (b) F (s) =
π(1 + e−s)
(s2 + π2)
.16 (a) F (s) =
4
s
− 2
s3
+
[
8
s2
+
4
s3
]
e−2s, (b) F (s) =
2
s
− 1
s− 1 +
[
4
s− 1 −
4
s
]
2−s.
17
(
2
s3
− 7
s2
+
12
s
)
− 2
(
2
s3
− 1
s2
)
e−3s + 2
(
2
s3
+
1
s2
)
e−4s.
The Inverse Transform 33
Chapter Two
THE INVERSE TRANSFORM
1. Introduction
◮Definition: If F (s) is the Laplace transform of f(t), then f(t) is called the inverse Laplace
transform of F (s), and is denoted
f(t) = L−1 [F (s)].
◭
There is a formula (Mellin’s formula) expressing L−1 [F ] as a definite integral, analogous to
the integral (3) which defines L [f ]. However, Mellin’s formula requires some knowledge of the
theory of complex variables, which at this stage you don’t have.
On the other hand, it is still possible to find the inverse Laplace transform in a wide range
of cases, using only the basic transforms (4)–(9), the rules described in chapter 1, and some
skill. And since skill comes only through practice, the best advice, at this point, is that you go
through as many examples as possible, then more, and then a few more.
Laplace calculus uses a variety of techniques, some of which are routine, some tricky: it is
similar, in this respect, to integral calculus. You must build your own personal library of such
tricks. Moreover, as you’ll get good, you’ll begin to spot other ways of doing the examples in
these notes, and you’ll want to know whether your method is better than the one presented
here. Try your way, then; be bold. Compare the methods; find points going for/against each
one.
◮Example 38 Find L−1 [(s − 2)4/s6].
Solution: Expand the numerator by the binomial formula:
(s− 2)4
s6
=
s4 − 8s3 + 24s2 − 32s + 16
s6
=
1
s2
− 8
s3
+
24
s4
− 32
s5
+
16
s6
.
Then, by (9), it follows immediately:
L−1
[
(s− 2)4
s6
]
= t− 8
2!
t2 +
24
3!
t3 − 32
4!
t4 +
16
5!
t5 =
= t− 4t2 + 4t3 − 4
3
t4 +
2
15
t5. ◭
◮Example 39 Find L−1 [1/(s − a)b+1].
Solution: Since
b!
sb+1
= L [tb],
34 The Inverse Transform
we apply the s-shift property (19): we get immediately
L−1
[
1
(s− a)b+1
]
=
tb eat
b!
.
Comment: if b is an integer, we may proceed differently, noting that
b!
(s− a)b+1 = (−1)
b d
b
dsb
[
1
s− a
]
=
(−d
ds
)b
L [eat];
then it follows, by rule (13) [multiplication by t]:
1
(s− a)b+1 = L
[
tbeat
b!
]
.
However, the first method is more general because it works even when b is not integer. ◭
◮Example 40 Find f(t), if F (s) = (3s − 5)/(s − 1)4.
Solution: Write
3s− 5
(s− 1)4 =
3(s − 1 + 1)− 5
(s− 1)4 =
3
(s− 1)3 −
2
(s − 1)4 .
Now, proceeding like in the previous example, we get immediately
F (s) = 32L
[
t2 et
]− 13L [t3 et],
and finally f(t) = 16 t
2et
(
9− 2t). ◭
◮Example 41 Find L−1 [(s − 16)/(s2 − 2s− 24)].
Solution: Write s2 − 2s − 24 = (s − 1)2 − 25 (this step is called “completing the square”). It
follows
s− 16
s2 − 2s− 24 =
s− 1
(s− 1)2 − 25 −
15
(s− 1)2 − 25 .
But
s
s2 − 25 = L [cosh 5t]
5
s2 − 25 = L [sinh 5t];
hence, by the s-shift property (19), it follows
s− 1
(s− 1)2 − 25 = L [e
t cosh 5t]
15
(s− 1)2 − 25 = L [3e
t sinh 5t].
and finally
L−1
[
s− 16
s2 − 2s− 24
]
= et
(
cosh 5t− 3 sinh 5t). ◭
The Inverse Transform 35
◮Example 42 Find f(t), if F (s) = (2s3 + s2 + 2s+ 2)/(s5 + 2s4 + s3).
Solution: Observe that s3 may be factored out in the denominator. Therefore,
2s3 + s2 + 2s + 2
s5 + 2s4 + s3
=
2s3 + s2 + 2s + 2
s3(s2 + 2s+ 2)
.
Now, we split the numerator:
2s3 + s2 + 2s+ 2
s3(s2 + 2s+ 2)
=
2s3
s3(s2 + 2s+ 2)
+
s2 + 2s+ 2
s3(s2 + 2s+ 2)
=
2
s2 + 2s + 2
+
1
s3
.
Finally, we observe that s2 + 2s+ 2 = (s + 1)2 + 1. So,
F (s) =
2
(s + 1)2 + 1
+
1
s3
,
and finally f(t) = 2 e−t sin t+ 12 t
2. ◭
◮Example 43 Find f(t), if F (s) = s/
√
(s+ 4)5.
Solution 1: Rewrite F (s):
F (s) =
s+ 4− 4√
(s+ 4)5
=
1√
(s + 4)3
− 4√
(s+ 4)5
.
The inverse-transform of the right-hand side may be found by applying the shift theorem (19),
which reduces it to a pair of basic transforms of the form (9):
L−1
[
1√
(s + 4)3
− 4√
(s+ 4)5
]
= e−4tL−1
[
1
s3/2
− 4
s5/2
]
=
e−4t t1/2
1
2 !
− 4e
−4t t3/2
3
2 !
.
Solution 2: Multiplication by s corresponds to differentiation by t. Therefore, using (13), we
get:
L−1
[
s√
(s + 4)5
]
=
d
dt
(
L−1
[
1√
(s+ 4)5
])
=
d
dt
(
e−4t t3/2
3
2 !
)
=
−4e−4t t3/2
3
2 !
+
3
2
e−4t t1/2
3
2 !
.
Note that this step relies also on the obvious fact that limt→0+ e−4t t3/2 = 0 (Why?). After
simplifications, this result becomes identical to the one obtained before. ◭
2. Partial Fractions: Heaviside’s Method
Expansion in partial fractions is a well known technique and is usually taught in a first-year
calculus course. Here, we shall only discuss some of its main features with a view to applications.
Suppose
F (s) =
N(s)
D(s)
,
36 The Inverse Transform
where both the numerator and the denominator are polynomials in s, and†
degree (N) < degree (D).
Also, suppose initially that D(s) may be written as the product of m linear factors, i.e.,
D(s) = (s− r1)(s− r2) · · · (s− rm). (21)
This means that D(s) has only simple roots and there are m of them, where m = degree (D); it
also means that when (21) is expanded out in terms of powers of s, the coefficient of the leading
term (which is sm) is exactly 1.
Under these assumptions, one may write
N(s)
D(s)
=
c1
s− r1 +
c2
s− r2 + · · ·+
cm
s− rm , (22)
and the constants c1 . . . cm are found by Heaviside’s “cover-up” method.† Here is how Heaviside’s
method works: multiply both sides of (22) by (s− rk); we get
(s− rk)N(s)
D(s)
= c1
s− rk
s− r1 + c2
s− rk
s− r2 + · · ·+ ck
s− rk
s− rk + · · ·+ cm
s− rk
s− rm .
Now simplify the k-th fraction on the right-hand side, which is clearly equal to 1. The resulting
formula is true for every s; in particular, if s = rk, the right-hand side is
R. H. S. = c1 · 0 + c2 · 0 + · · ·+ ck · 1 + · · ·+ cm · 0 = ck.
The left-hand side may be simplified too, because by (21),
L. H. S. =
(s− rk)N(s)
(s− r1)(s− r2) · · · (s− rk) · · · (s − rm) .
Simplifying the common factor (s − rk) is tantamount to “covering up” the same factor in the
denominator of the original fraction (hence the slang name): therefore
N(rk)
(rk − r1)(rk − r2) · · · (rk − rk−1) (rk − rk+1) · · · (rk − rm) = ck. (23)
◮Example 44 Expand (3s+ 7)/(s − 3)(s + 1) in partial fractions.
Solution: Write
3s+ 7
(s − 3)(s + 1) =
A
s− 3 +
B
s+ 1
.
Apply (23). Covering up s − 3 on the left-hand side and letting s = 3, we get A = 4. Covering
up s+ 1 and letting s = −1 we get B = −1. So, finally,
3s+ 7
(s − 3)(s + 1) =
4
s− 3 −
1
s+ 1
.
◭
† If the degree of N is not less than the degree of D, then one may always divide N by D by
long division, eventually getting N/D = Q+R/D, where Q is the quotient of the division and
R, the remainder, is of a degree less than D.
† Oliver Heaviside (1850–1925), English.
The Inverse Transform 37
◮Example 45 Solve the differential equation x¨ − 2x˙ − 24x = 0, given x(0+) = 1, and
x˙(0+) = −14.
Solution: The transformed equation is
14− s+ s2X − 2(−1 + sX)− 24X = 0,
or
(s2 − 2s − 24)X = s− 16.
Hence, X = (s − 16)/(s2 − 2s− 24).
This is precisely the transform of example 41, so we could use the result found there and
relax. However, we need to practice partial fractions. Therefore, we first find the roots of the
denominator, i.e., we solve
s2 − 2s − 24 = 0,
and get s = 6 and s = −4. Then we write
X =
s− 16
(s− 6)(s + 4) =
c1
s− 6 +
c2s+ 4
.
Now, “covering up” the factor (s− 6) in the expression in the middle and setting s = 6, we get
c1 =
6− 16
6 + 4
= −1.
Similarly, “covering up” s+ 4 and setting s = −4, we get
c2 =
−4− 16
−4− 6 = 2.
It follows
X = − 1
s− 6 +
2
s+ 4
,
and finally
x = −e6t + 2e−4t.
Is this the same answer found in example 41? Yes it is:
et cosh 5t− 3et sinh 5t = 12
(
et+5t + et−5t
)− 32(et+5t − et−5t) = −e6t + 2e−4t,
as expected. ◭
◮Example 46 Expand (s3 − 2s2 + 3s− 5)/s(s − 1)(s − 2)(s + 3).
Solution: Identify the roots of D(s): they are equal to 0, 1, 2, −3. Prepare for expansion:
s3 − 2s2 + 3s− 5
s(s− 1)(s − 2)(s + 3) =
c1
s
+
c2
s− 1 +
c3
s− 2 +
c4
s+ 3
.
38 The Inverse Transform
By Heaviside’s method, we get:
cover up s: c1 =
03 − 2 · 02 + 3 · 0− 5
(0− 1)(0 − 2)(0 + 3) =
−5
6
cover up s− 1: c2 = 1
3 − 2 · 12 + 3 · 1− 5
1(1− 2)(1 + 3) =
−3
−4
cover up s− 2: c3 = 2
3 − 2 · 22 + 3 · 2− 5
2(2− 1)(2 + 3) =
1
10
cover up s+ 3: c4 =
(−3)3 − 2 · (−3)2 + 3 · (−3)− 5
(−3)(−3 − 1)(−3− 2) =
−59
−60 .
It follows that
s3 − 2s2 + 3s− 5
s(s− 1)(s − 2)(s + 3) = −
5
6s
+
3
4(s − 1) +
1
10(s − 2) +
59
60(s + 3)
.
◭
One more comment, very useful: go back to (22). If we let s → ∞, we certainly get 0 = 0,
because the degree of D(s) is assumed to be greater than the degree of N(s) by at least one
unit.† However, if we multiply both sides by s and then let s → ∞, we find that the left-hand
side may approach a finite limit, as well as zero; and on the right-hand side we find m limits
which may all be done by inspection. This procedure, called “testing the transform at infinity”
may be used
(i) as a quick numerical check on the calculations, or
(ii) to find a coefficient, when all but one have been computed.
◮Example 47 Go back to the last example. Consider the expansion
s3 − 2s2 + 3s− 5
s(s− 1)(s − 2)(s + 3) =
c1
s
+
c2
s− 1 +
c3
s− 2 +
c4
s+ 3
.
and multiply both sides by s: we get
s(s3 − 2s2 + 3s − 5)
s(s− 1)(s − 2)(s + 3) = c1
s
s
+ c2
s
s− 1 + c3
s
s− 2 + c4
s
s+ 3
.
Note that:
L. H. S. =
s4 + lower powers of s
s4 + lower powers of s
:
hence
lim
s→∞[L. H. S.] = 1.
By inspection,
lim
s→∞[R. H. S.] = c1 + c2 + c3 + c4.
And indeed
1 = −5
6
+
3
4
+
1
10
+
59
60
,
† As a rule, before embarking on a partial fractions expansion, you should always verify that
this is the case.
The Inverse Transform 39
as expected. In this way we have checked our calculations. ◭
3. Partial Fractions: Multiple Roots.
When D(s) has one multiple root, or more, then Heaviside’s method does not yield all coeffi-
cients. However, it still works fine for all the simple roots; it also produces immediately one
coefficient for each multiple root.
Recall that by “testing at infinity” one may find one more coefficient, in addition to the
ones found by the standard “cover up” method. If only one coefficient is missing, this is enough
to complete an expansion.
◮Example 48 Expand (3s− 2)/(s + 5)(s − 1)2.
Solution: We seek an expansion of the form
3s− 2
(s+ 5)(s − 1)2 =
c
s+ 5
+
b1
s− 1 +
b2
(s− 1)2 .
Multiplying both sides of this equation by s+ 5 and setting s = −5, we find
c =
3 · (−5)− 2
(−5− 1)2 = −
17
36
.
Similarly, multiplying both sides by (s − 1)2 and setting s = 1 we find
b2 =
3 · 1− 2
1 + 5
=
1
6
.
Now only b1 remains to be found. Testing the transform at infinity, we see that
sN(s)
D(s)
=
3s2 + lower powers of s
s3 + lower powers of s
,
hence sN(s)/D(s) tends to 0 as s→∞. Therefore
0 = −17
36
+ b1,
and finally b1 = 17/36. ◭
◮Example 49 Expand (s3 + 11)/(s − 1)2(s+ 2)2.
Solution: Write
s3 + 11
(s− 1)2(s+ 2)2 =
c1
s− 1 +
c2
(s − 1)2 +
b1
s+ 2
+
b2
(s+ 2)2
.
We find c2 by multiplying both sides of this equation by (s− 1)2 and setting s = 1:
c2 =
1 + 11
(1 + 2)2
=
4
3
.
40 The Inverse Transform
Similarly, multiplying both sides by (s + 2)2 and setting s = −2, we find b2:
b2 =
−8 + 11
(−2− 1)2 =
1
3
.
At this point we have
s3 + 11
(s− 1)2(s+ 2)2 =
c1
s− 1 +
4
3(s − 1)2 +
b1
s+ 2
+
1
3(s + 2)2
. (24)
Multiplying both sides by s and letting s→∞ we get
1 = c1 + b1.
We need one more bit of information; we get it by testing one numerical value of s in (24): any
value not used so far would work, but s = 0 seems to be easy enough. We get
11
4
=
c1
−1 +
4
3
+
b1
2
+
1
12
,
or
4
3
= −c1 + 1
2
b1,
hence b1 = 14/9, c1 = −5/9. ◭
Broadly speaking, finding the coefficients in the presence of multiple roots requires more work,
be it with pencil and paper, or computer time. Several “generalized Heaviside’s methods” have
been devised to handle multiple roots, but none of them ultimately avoids a fair amount of
tedious calculations. A good discussion may be found in G. Doetsch, Guide to the applications
of the Laplace and Z-transforms, van Nostrand (1971).
One method that’s easy to remember but not particularly fast, consists of moving terms with
known coefficients from the right-hand side to the left-hand side, rearranging and simplifying.
It is best described by an example.
◮Example 50 Find f(t), given that F (s) = 16/(s2 − 3s+ 2)s4.
Solution: We expand F (s) in partial fractions. We write
16
(s2 − 3s+ 2)s4 =
16
(s − 1)(s − 2)s4 =
A
s− 2 +
B
s− 1 +
C1
s
+
C2
s2
+
C3
s3
+
C4
s4
.
By the “cover-up” method we get:
A =
16
24
= 1,
B =
16
−1 = −16,
C4 =
16
(−1)(−2) = 8.
By testing the transform at infinity, we get
0 = 1− 16 + C1,
The Inverse Transform 41
hence C1 = 15. So far we have established that
16
(s2 − 3s+ 2)s4 =
1
s− 2 −
16
s− 1 +
15
s
+
C2
s2
+
C3
s3
+
8
s4
.
Now, we move the term 8/s4 across to the left-hand side. It follows that
16
(s2 − 3s+ 2)s4 −
8
s4
=
1
s− 2 −
16
s− 1 +
15
s
+
C2
s2
+
C3
s3
.
Simplify the left-hand side:
16
(s2 − 3s+ 2)s4 −
8
s4
=
−8s2 + 24s
(s2 − 3s+ 2)s4 =
−8s+ 24
(s2 − 3s+ 2)s3 .
Substite back: it follows that
−8s+ 24
(s2 − 3s+ 2)s3 =
1
s− 2 −
16
s− 1 +
15
s
+
C2
s2
+
C3
s3
,
and C3 may be now found by Heaviside’s method:
C3 =
24
2
= 12.
Repeat the procedure. Move the term 12/s3 to the left-hand side: it follows that
−8s+ 24
(s2 − 3s + 2)s3 −
12
s3
=
1
s− 2 −
16
s− 1 +
15
s
+
C2
s2
.
Simplify the left-hand side:
−8s+ 24
(s2 − 3s + 2)s3 −
12
s3
=
−12s2 + 28s
(s2 − 3s+ 2)s3 =
−12s + 28
(s2 − 3s+ 2)s2 .
Sobstitute back: it follows that
−12s+ 28
(s2 − 3s+ 2)s2 =
1
s− 2 −
16
s− 1 +
15
s
+
C2
s2
,
and C2 may be found by Heaviside’s method:
C2 =
28
2
= 14.
All the coefficients have been found:
16
(s2 − 3s+ 2)s4 =
1
s− 2 −
16
s− 1 +
15
s
+
14
s2
+
12
s3
+
8
s4
.
So, finally, f(t) = e2t − 16et + 15 + 14t+ 6t2 + 43 t3. ◭
42 The Inverse Transform
Another method, which at first seems simple, is to plug in as many values of s as there are missing
coefficients. These values may be freely chosen, as long as none of them coincides with a root
of D(s). In this way one gets a system of n linear equations with n unknowns. The drawback
of this method is that the amount of work needed to solve a system of n linear equations grows
(for large n) with speed of n3; it can be rather laborious even for n = 3.
Yet another approach to the previous example is to apply rule (16) [division by s is associated
with integration with respect to t]. We’ll come back to this idea in section 5.
The so-called calculus of residues is probably much better than any of the methods describedso far, but since it requires some knowledge of the theory of complex variables, we leave it for a
more advanced course.
4. Partial Fractions: Complex Roots
In principle, if D(s) has complex roots, the methods described in the sections 2–3 are still
applicable. The only difference is that the coefficients of the expansion are, in general, complex.
Complex arithmetic is inherently more time-consuming than real arithmetic.† In applications,
real trigonometric functions are generally preferable to complex exponentials, though a good
case may sometimes be made for using the latter.
In engineering applications N(s) and D(s) are virtually certain to be real polynomials. It
may be shown, then, that the complex roots of D(s), if present, come in complex conjugate pairs,
and the corresponding coefficients in the partial fractions expansion are also complex conjugate.
To fix the ideas, consider a simple case where D(s) is real and has a complex root a + ib
with multiplicity 1. Then a− ib is also a root, and the expansion has the form
N(s)
D(s)
=
c+ id
s− a− ib +
c− id
s− a+ ib + (other terms).
Grouping terms on the right-hand side, we get
N(s)
D(s)
=
Bs+ C
(s− a)2 + b2 + (other terms),
where B and C are real, and may be expressed in terms of the “old” constants a,b, c and d
(convince yourself of this). However there is no need to find c and d, since one may find B and
C directly. The following examples illustrate this method.
◮Example 51 Expand (2s2 − 2s+ 1)/(s − 1)(s2 + 4) in partial fractions.
Solution: Both following expansions are possible:
2s2 − 2s + 1
(s − 1)(s2 + 4) =
c1
s− 1 +
c2 − id2
(s− i2) +
c2 + id2
(s+ i2)
,
=
A
s− 1 +
Bs+ C
s2 + 4
,
but the first requires complex arithmetic, the second one does not.
Let us do the latter. We find immediately
A =
2− 2 + 1
1 + 4
=
1
5
† Both when done by a computer and when done by pencil and paper.
The Inverse Transform 43
(by Heaviside’s method: nothing new here). To find B and C, we may either continue with the
“cover-up” method, or use a couple of shortcuts.
Method 1: By the “cover-up” method. Multiply both sides by s2 + 4 and simplify: it follows
(2s2 − 2s+ 1)(s2 + 4)
(s− 1)(s2 + 4) =
A(s2 + 4)
s− 1 +
(Bs+ C)(s2 + 4)
s2 + 4
2s2 − 2s+ 1
s− 1 =
A(s2 + 4)
s− 1 +Bs+ C.
If we now set s = ±i2 , we make s2 + 4 = 0. Either root may be used; at the end the final
results will be the same. For instance, setting s = i2 we get
−8− i4 + 1
i2− 1 = 0 + i2B + C =⇒
−1 + i18
5
= i2B + C.
Hence, separating real and imaginary part, we find B = 9/5 and C = −1/5. So, finally,
2s2 − 2s + 1
(s − 1)(s2 + 4) =
1/5
s− 1 +
(9s − 1)/5
s2 + 4
.
Method 2: Use the test at infinity. Multiply both sides of the expansion by s and let s → ∞.
We get
2s3 + · · ·
s3 + · · · =
s/5
s− 1 +
Bs2 + Cs
s2 + 4
,
and hence (in the limit of s→∞)
2 = 1/5 +B =⇒ B = 9/5.
At this point, only C is missing. We find it by inspection, by substituting for s an arbitrary
value that hasn’t yet been used. For example, s = 0, why not? We get immediately
1
−4 =
1/5
−1 +
C
4
=⇒ C = −1/5,
and finally
2s2 − 2s + 1
(s − 1)(s2 + 4) =
1/5
s− 1 +
(9s − 1)/5
s2 + 4
.
We get the same expansion, as expected. ◭
◮Example 52 Expand (s2 − 16s+ 23)/(s2 − 2s+ 5)(s − 1)(s − 3) in partial fractions.
Solution: The quadratic factor in the denominator is the product of two complex-conjugate
pairs:
s2 − 2s+ 5 = (s− 1 + i2)(s − 1− i2).
Hence, D(s) has two complex roots and two real roots, namely 1+ i2, 1− i2, 1 and 3. We seek
an expansion of the form
s2 − 16s + 23
(s2 − 2s+ 5)(s − 1)(s − 3) =
A
s− 1 +
B
s− 3 +
Cs+D
s2 − 2s + 5
44 The Inverse Transform
Coefficients A and B are found immediately by Heaviside’s method:
A =
1− 16 + 23
(1− 2 + 5)(1 − 3) = −1
B =
9− 48 + 23
(9− 6 + 5)(3 − 1) = −1
To find C and D, we have again the choice of two simple methods.
Method 1: Continuing with the “cover-up” method, we multiply both sides of the expansion by
s2 − 2s+ 5,
(s2 − 16s+ 23)(s2 − 2s + 5)
(s2 − 2s+ 5)(s − 1)(s − 3) = A
s2 − 2s+ 5
s− 1 +B
s2 − 2s+ 5
s− 3 +
(Cs+D)(s2 − 2s+ 5)
s2 − 2s+ 5 ,
simplify
s2 − 16s + 23
(s − 1)(s − 3) = A
s2 − 2s+ 5
s− 1 +B
s2 − 2s+ 5
s− 3 + Cs+D,
and set s = 1 + i2 (recall that s2 − 2s+ 5 = 0 if s = 1± i2). We get
(1 + i2)2 − 16(1 + i2) + 23
(1 + i2− 1)(1 + i2− 3) = 0 + 0 + C(1 + i2) +D,
hence
4− i28
−4− i4 = C +D + i2C,
which yields
1− i7
−1− i =
(1− i7)(−1 + i)
2
= 3 + i4 =
= C +D + i2C.
Separating the imaginary part we get
i4 = i2C,
hence C = 2, and finally D = 3− C = 1.
Method 2: By the test at infinity. Note that F (s) is of second degree in the numerator, fourth
degree in the denominator, hence sF (s)→ 0 as s tends to infinity. On the other hand,
s ·
(
A
s− 1 +
B
s− 3 +
Cs+D
s2 − 2s+ 5
)
→ A+B +C as s→∞.
Since we already found that A = −1 and B = −1, we get immediately that C = 2. We still
miss D, but when only one coefficient is missing, we may find it by inspection, substituting for
s any number that hasn’t been used. Again, s = 0 is the obvious choice. We get immediately
23
5(−1)(−3) = 1 +
1
3
+
D
5
,
and hence D = 1. ◭
The Inverse Transform 45
We get the impression that the second method is slightly faster than Heaviside’s “cover-up”
method. The problem is, it may be used only once. For a partial-fractions expansion with two
complex roots or more, you must use the cover-up method at least once, and finish the job by
the method described above.
◮Example 53 Expand (s2 − 4s− 10)/(s2 − 2s + 10)(s2 + 4) in partial fractions.
Solution: The denominator has two pairs of complex conjugate roots, namely s = 1 ± i3 and
s = ±i2. We seek an expansion of the form
s2 − 4s− 10
(s2 − 2s + 10)(s2 + 4) =
As+B
s2 − 2s+ 10 +
Cs+D
s2 + 4
.
Multiplying both sides of this equation by s2 + 4 and setting s = i2 we get
(i2)2 − 4(i2) − 10
(i2)2 − 2(i2) + 10 = C(i2) +D,
or −14− i8
6− i4 = i2C +D.
Simplifying the left-hand side, we get
−14− i8
6− i4 =
−7− i4
3− i2 =
(−7− i4)(3 + i2)
13
= −1− i2,
and finally (comparing real and imaginary part) C = −1, D = −1. Now we “test at infinity”:
lim
s→∞
s · s
2 − 4s− 10
(s2 − 2s + 10)(s2 + 4) = 0,
lim
s→∞
s ·
(
As+B
s2 − 2s+ 10 +
Cs+D
s2 + 4
)
= A+ C.
But we already know that C = −1, hence A = 1. Finally, substituting s = 0 (any real number
would do), we get
−10
(10)(4)
=
B
10
+
D
4
:
having found before that D = −1, we deduce that B = 0. ◭
◮Example 54 Expand (s3 + 1)/(s − 1)(s2 + 1)2 in partial fractions.
Solution: We seek an expansion of the form
s3 + 1
(s− 1)(s2 + 1)2 =
A
s− 1 +
Bs+ C
s2 + 1
+
Ds+ E
(s2 + 1)2
.
By Heaviside’s method,
1 + 1
(1 + 1)2
= A+ 0 + 0 =⇒ A = 1/2.
46 The Inverse Transform
Continuing with Heaviside’s method, we multiply through by (s2 + 1)2 and simplify: we get
s3 + 1
(s− 1) =
A(s2 + 1)2
s− 1 + (Bs+ C)(s
2 + 1) +Ds+ E.
Substituting s = i we get
−i+ 1
i− 1 = 0 + 0 + iD + E.
The left-hand side of this equation is equal to −1, hence it follows immediately that D = 0 and
E = −1. Testing at infinity we get the equation
s4 + · · ·
s5 + · · · =
As
s− 1 +
Bs2 + Cs
s2 + 1
+
Ds2 +Es
s4 + · · · ;
as s→∞ this yields 0 = A+B. Having already found that A = 1/2, we deduce that B = −1/2.
At this point, only C remains to be found: so, we substitute s = 0 in the expansion. We obtain
−1 = −A+ C + E,
where A = 1/2 and E = −1. It follows that C = 1/2. ◭
5. Manipulations of the Transform
In some applications it may happen that it is much easier to find L−1 [s F (s)], for a given F (s),
than L−1 [F (s)] . In these cases L−1 [F (s)] may be found using property (16) [division