1
1.1. Introduction. The knowledge of “Integral Transforms” is an essential part of mathematical
background required by scientists and engineers. This is because the transform methods provide an
easy and effective means for the solutions of many problems arising in science and engineering.
For example,the Laplace transformation replaces a given function F (t) by another function f (s).
Then Laplace transformation converts an ordinary differential equation with some given initial
conditions into an algebraic equation in terms of f (s) Finally, using inverse Laplace transformation
we recover the original function F(t). Thus, the method of Laplace transformation is especially
useful for initial value problems, as it enables us to solve the problem without the trouble of finding
the general solution first and then evaluating the arbitrary constants. The use of Laplace transforms
provide a powerful technique of solving differential and integral equations.
1.2. Laplace transform. Definition. [Rohilkhand 2000, Meerut 2010]
Given a function F(t) defined for all real t 0, the Laplace transform of F (t) is a function of
a new variable s given by
L {F (t), s} = L {F (t)} = f (s) = F (s) = e st F(t ) dt ... (1)
0
The Laplace transform of F(t) is said to exist if the improper integral (1) converges for some
value of s, otherwise it does not exist.
Remark 1. Some authors use new variable p in place of s. Therefore, corresponding changes
may be done in proofs of articles and solutions of problems. Similarly, some authors use variable x
in place of t.
Remark 2. Some authors use f (t) in the place of F(t) and simultaneously use F(s) in place of
f (s) while defining Laplace transform.
Remark 3. L is called Laplace transformation operator.
Remark 4. The operation of multiplying F (t) by e–st and then integrating between the limits
0 to is known as Laplace transformation.
Remark 5. Improper integral. Write 0
e st F (t ) dt lim
0 e st F (t ) dt , where is a positive
real number. If the integral from 0 to exists for each > 0 and if the limit as exists, then
the improper integral is said to converge to that limiting value otherwise the integral is said to
diverge or fail to exist. Note that with regard to upper limit, the Laplace integral
0
e st F (t ) dt is to
be understand as an improper one.
1.3. Working rule to find Laplace transform of F (t), t 0 by using definition 1.2.
Step 1. By definition, L{F (t )} 0
e st F (t ) dt
Step 2. By definition of improper integral, we have
0
e st F (t ) dt lim
0 e st F (t ) dt
1.3
,1.4 The Laplace Transform
Step 3. Simplify e–st F (t) and evaluate 0
e st F (t ) dt .
Step 4. Evaluate lim
0 e st F (t ) dt
Step 5. From steps 1 and 2, we have
L{F (t )}
0
e st F (t ) dt lim
0 e st F (t ) dt
1.4. Piecewise (or sectionally) continuous function. Definition.
A function F(t) is called piecewise continuous or sectionally continuous in the closed interval
[a, b] if there exist a finite number of points t1, t2, . . . . . ., tn (a = t1 < t2 < t3 < < tn = b) such that
F(t) is continuous in each of open subin terval (t i, t i+1 ) and has finite right limit
F (ti + 0) and finite left limit F (ti + 1 – 0). Clearly F (t) need not necessarily be defined at the end
points of the subintervals.
O a
An example of a function F(t) which is sectionally continuous is shown in the above figure
by taking particular case n = 5. We note that F (t) is continuous in subinterval (t2, t3) and has finite
right and left limits F (t2 + 0) and F (t3 – 0) respectively. We also note that F (t) has discontinuities
at t2, t3 and t4.
1.5. Functions of exponential order. Definition. [Kanpur 2002]
If there exist a positive real constant m, a number and a finite number t such that
for all t t0, e t F ( t ) m or F (t ) m e t
we say that F (t) is a function of exponential order as t .
Equivalently, we also write F (t) = O (et), as t .
1.5 A. Solved examples based on functions of exponential order
Ex.1. Show that F (t) = t2 is of exponential order 3.
Sol. We have
t2 2t
lim e t F ( t ) = lim t
lim , by L Hospital s rule
t t e t e t
2
= tlim , by L Hospital s rule again
2 t
e
= 0, provided > 0
Hence F (t) = t is of exponential order. Again, | t2 | = t2 < e3t for all t > 0
2
Therefore t2 is a function of exponential order 3.
Ex.2. Show that F (t) = tn is of exponential order as t , n being any positive integer.
[Kanpur 2004]
,The Laplace Transform 1.5
Sol. Assuming that > 0, we have
n n 1
lim e t F ( t ) lim t lim nt , by L Hospital s rule lim n (n 1)t , by L Hospital s rule again
n 2
t t e t t e t
t 2 et
n (n 1)(n 2) 2 1t n n
lim , after repeated use of L Hospital s rule (n – 2) times more.
t n e t
=0
Hence tn is of exponential order , as t .
2
Ex.3. Show that the function F (t) = e t is not of exponential order as t .
2
Sol. We have, lim e t F (t ) = lim e t e t lim et (t ) for all values of .
t t t
Hence whatever be the value of , we cannot determine a real constant m such that
e t F (t ) m
Therefore, the given function is not of exponential order as t .
1.6. Functions of class A. Definition.
A function F (t) is said to belong class A if F (t) is of some exponential order as t and is
piecewise continuous over every finite interval of t 0.
1.7. Sufficient conditions for the existence of Laplace transform
Theorem. If F (t) is a function of class A, L {F (t)} exists.
or If F (t) is of some exponential order as t and is piecewise continuous over every finite
interval of t 0, then L {F (t)} exists.
or If F(t) is a function which is piecewise continuous on every finite interval in the range t 0
t
and satisfies | F (t ) | me for all t 0 and for constants and m, then the Laplace transform of
F(t) exists. [Osmania 2010; KU Kurukshetra 2004; Punjab 2005]
Proof. Since F (t) is of exponential order, say , we can find constants , m (> 0) and
t0 ( > 0) such that
| F (t) | < m et for t t0. ... (1)
t0
Now, L {F (t)} = e st F( t ) dt e st F(t ) dt e st F( t ) dt = I1 + I2, say ... (2)
0 0 t0
Since F (t) is piecewise continuous on every finite interval 0 t t0, I1 exists.
Again, | I2 | = t0
e st F(t ) dt t0
e st | F(t ) | dt m t0
e st e t dt , using (1)
e ( s ) t m e ( s )t 0
Thus, | I2 | < m . ...(3)
( s ) t0
s
Now, when s > , then e ( s )t0 0 as t . Hence (3) shows that
| I2 | is finite for all t0 > 0 when s > and hence I2 is also convergent. Then from (2), it follows that
L {F (t)} exists for all s > .
Remark: The conditions stated in the above theorem are sufficient to ensure the existence of
L {F (t)}. These are not necessary conditions for the existence of L{F (t)}. In other words, if the
, 1.6 The Laplace Transform
conditions of the above theorem are not satisfied, L{F(t)} may or may not exist as shown in the
following example: Consider the function F (t) = 1/ t .
As t 0, F (t) . Hence F (t) is not piecewise continuous on every finite interval in the
range t 0. Now, by definition, we have
2 2
L {1/ t }
0
e st (1/ t ) dt
s 0
e x dx, putting st = x and dt / t = 2dx/ s , s > 0
1/ 2
2
= , s 0. as
2
e x dx
s 2 s 0 2
Hence L {1/ t } exists for s > 0 even if 1/ t is not piecewise continuous in the range t 0.
1.8. Linearity property of Laplace transforms. If c1 and c2 be constants, then
L {c1 F1 (t) + c2 F2 (t)} = c1 L {F1(t)} + c2 L {F2 (t)}.
Proof. By definition, we have
L{c1 F1 (t) + c2 F2 (t)} = e st {c1 F1 ( t ) c2 F2 (t )}dt = c1 e st F1 ( t ) dt c2 e st F2 ( t ) dt
0 0 0
= c1L {F1 (t)} + c2 L {F2 (t)}, by definition.
1.9. Laplace transforms of some elementary functions
(i) To find Laplace transform of F(t) = 1.
[Kanpur 2003, Meerut 2004, M.S. Univ. T.N. 2007]
Sol. By definition of Laplace transform (see Art. 1.3), we have
e st
L{1} 0
e st (1) dt lim
0
e st dt lim
s
0
1 1 1 1
lim s 1 (0 1) , provided s > 0
s e s s
(ii) To find Laplace transform of the function F (t) = tn, n being any real number greater
than –1. [Ranchi 2010; Purvanchal 2006]
Sol. By definition, L {F (t)} = e st F (t ) dt .
0
L {tn} = e st t n dt e st t ( n 1) 1dt . ... (i)
0 0
From the properties of Gamma function , we know that
( m)
e ax x m 1dx
, if a > 0 and m > 0. ... (ii)
0 am
Replacing a by s, m by (n + 1) and x by t in (ii), we have
( n 1)
e st t ( n 1) 1dt , if s > 0 and n + 1 > 0
0 s n 1
(n 1)
(i) n
{t } = , s 0 and n 1.
s n 1
1.1. Introduction. The knowledge of “Integral Transforms” is an essential part of mathematical
background required by scientists and engineers. This is because the transform methods provide an
easy and effective means for the solutions of many problems arising in science and engineering.
For example,the Laplace transformation replaces a given function F (t) by another function f (s).
Then Laplace transformation converts an ordinary differential equation with some given initial
conditions into an algebraic equation in terms of f (s) Finally, using inverse Laplace transformation
we recover the original function F(t). Thus, the method of Laplace transformation is especially
useful for initial value problems, as it enables us to solve the problem without the trouble of finding
the general solution first and then evaluating the arbitrary constants. The use of Laplace transforms
provide a powerful technique of solving differential and integral equations.
1.2. Laplace transform. Definition. [Rohilkhand 2000, Meerut 2010]
Given a function F(t) defined for all real t 0, the Laplace transform of F (t) is a function of
a new variable s given by
L {F (t), s} = L {F (t)} = f (s) = F (s) = e st F(t ) dt ... (1)
0
The Laplace transform of F(t) is said to exist if the improper integral (1) converges for some
value of s, otherwise it does not exist.
Remark 1. Some authors use new variable p in place of s. Therefore, corresponding changes
may be done in proofs of articles and solutions of problems. Similarly, some authors use variable x
in place of t.
Remark 2. Some authors use f (t) in the place of F(t) and simultaneously use F(s) in place of
f (s) while defining Laplace transform.
Remark 3. L is called Laplace transformation operator.
Remark 4. The operation of multiplying F (t) by e–st and then integrating between the limits
0 to is known as Laplace transformation.
Remark 5. Improper integral. Write 0
e st F (t ) dt lim
0 e st F (t ) dt , where is a positive
real number. If the integral from 0 to exists for each > 0 and if the limit as exists, then
the improper integral is said to converge to that limiting value otherwise the integral is said to
diverge or fail to exist. Note that with regard to upper limit, the Laplace integral
0
e st F (t ) dt is to
be understand as an improper one.
1.3. Working rule to find Laplace transform of F (t), t 0 by using definition 1.2.
Step 1. By definition, L{F (t )} 0
e st F (t ) dt
Step 2. By definition of improper integral, we have
0
e st F (t ) dt lim
0 e st F (t ) dt
1.3
,1.4 The Laplace Transform
Step 3. Simplify e–st F (t) and evaluate 0
e st F (t ) dt .
Step 4. Evaluate lim
0 e st F (t ) dt
Step 5. From steps 1 and 2, we have
L{F (t )}
0
e st F (t ) dt lim
0 e st F (t ) dt
1.4. Piecewise (or sectionally) continuous function. Definition.
A function F(t) is called piecewise continuous or sectionally continuous in the closed interval
[a, b] if there exist a finite number of points t1, t2, . . . . . ., tn (a = t1 < t2 < t3 < < tn = b) such that
F(t) is continuous in each of open subin terval (t i, t i+1 ) and has finite right limit
F (ti + 0) and finite left limit F (ti + 1 – 0). Clearly F (t) need not necessarily be defined at the end
points of the subintervals.
O a
An example of a function F(t) which is sectionally continuous is shown in the above figure
by taking particular case n = 5. We note that F (t) is continuous in subinterval (t2, t3) and has finite
right and left limits F (t2 + 0) and F (t3 – 0) respectively. We also note that F (t) has discontinuities
at t2, t3 and t4.
1.5. Functions of exponential order. Definition. [Kanpur 2002]
If there exist a positive real constant m, a number and a finite number t such that
for all t t0, e t F ( t ) m or F (t ) m e t
we say that F (t) is a function of exponential order as t .
Equivalently, we also write F (t) = O (et), as t .
1.5 A. Solved examples based on functions of exponential order
Ex.1. Show that F (t) = t2 is of exponential order 3.
Sol. We have
t2 2t
lim e t F ( t ) = lim t
lim , by L Hospital s rule
t t e t e t
2
= tlim , by L Hospital s rule again
2 t
e
= 0, provided > 0
Hence F (t) = t is of exponential order. Again, | t2 | = t2 < e3t for all t > 0
2
Therefore t2 is a function of exponential order 3.
Ex.2. Show that F (t) = tn is of exponential order as t , n being any positive integer.
[Kanpur 2004]
,The Laplace Transform 1.5
Sol. Assuming that > 0, we have
n n 1
lim e t F ( t ) lim t lim nt , by L Hospital s rule lim n (n 1)t , by L Hospital s rule again
n 2
t t e t t e t
t 2 et
n (n 1)(n 2) 2 1t n n
lim , after repeated use of L Hospital s rule (n – 2) times more.
t n e t
=0
Hence tn is of exponential order , as t .
2
Ex.3. Show that the function F (t) = e t is not of exponential order as t .
2
Sol. We have, lim e t F (t ) = lim e t e t lim et (t ) for all values of .
t t t
Hence whatever be the value of , we cannot determine a real constant m such that
e t F (t ) m
Therefore, the given function is not of exponential order as t .
1.6. Functions of class A. Definition.
A function F (t) is said to belong class A if F (t) is of some exponential order as t and is
piecewise continuous over every finite interval of t 0.
1.7. Sufficient conditions for the existence of Laplace transform
Theorem. If F (t) is a function of class A, L {F (t)} exists.
or If F (t) is of some exponential order as t and is piecewise continuous over every finite
interval of t 0, then L {F (t)} exists.
or If F(t) is a function which is piecewise continuous on every finite interval in the range t 0
t
and satisfies | F (t ) | me for all t 0 and for constants and m, then the Laplace transform of
F(t) exists. [Osmania 2010; KU Kurukshetra 2004; Punjab 2005]
Proof. Since F (t) is of exponential order, say , we can find constants , m (> 0) and
t0 ( > 0) such that
| F (t) | < m et for t t0. ... (1)
t0
Now, L {F (t)} = e st F( t ) dt e st F(t ) dt e st F( t ) dt = I1 + I2, say ... (2)
0 0 t0
Since F (t) is piecewise continuous on every finite interval 0 t t0, I1 exists.
Again, | I2 | = t0
e st F(t ) dt t0
e st | F(t ) | dt m t0
e st e t dt , using (1)
e ( s ) t m e ( s )t 0
Thus, | I2 | < m . ...(3)
( s ) t0
s
Now, when s > , then e ( s )t0 0 as t . Hence (3) shows that
| I2 | is finite for all t0 > 0 when s > and hence I2 is also convergent. Then from (2), it follows that
L {F (t)} exists for all s > .
Remark: The conditions stated in the above theorem are sufficient to ensure the existence of
L {F (t)}. These are not necessary conditions for the existence of L{F (t)}. In other words, if the
, 1.6 The Laplace Transform
conditions of the above theorem are not satisfied, L{F(t)} may or may not exist as shown in the
following example: Consider the function F (t) = 1/ t .
As t 0, F (t) . Hence F (t) is not piecewise continuous on every finite interval in the
range t 0. Now, by definition, we have
2 2
L {1/ t }
0
e st (1/ t ) dt
s 0
e x dx, putting st = x and dt / t = 2dx/ s , s > 0
1/ 2
2
= , s 0. as
2
e x dx
s 2 s 0 2
Hence L {1/ t } exists for s > 0 even if 1/ t is not piecewise continuous in the range t 0.
1.8. Linearity property of Laplace transforms. If c1 and c2 be constants, then
L {c1 F1 (t) + c2 F2 (t)} = c1 L {F1(t)} + c2 L {F2 (t)}.
Proof. By definition, we have
L{c1 F1 (t) + c2 F2 (t)} = e st {c1 F1 ( t ) c2 F2 (t )}dt = c1 e st F1 ( t ) dt c2 e st F2 ( t ) dt
0 0 0
= c1L {F1 (t)} + c2 L {F2 (t)}, by definition.
1.9. Laplace transforms of some elementary functions
(i) To find Laplace transform of F(t) = 1.
[Kanpur 2003, Meerut 2004, M.S. Univ. T.N. 2007]
Sol. By definition of Laplace transform (see Art. 1.3), we have
e st
L{1} 0
e st (1) dt lim
0
e st dt lim
s
0
1 1 1 1
lim s 1 (0 1) , provided s > 0
s e s s
(ii) To find Laplace transform of the function F (t) = tn, n being any real number greater
than –1. [Ranchi 2010; Purvanchal 2006]
Sol. By definition, L {F (t)} = e st F (t ) dt .
0
L {tn} = e st t n dt e st t ( n 1) 1dt . ... (i)
0 0
From the properties of Gamma function , we know that
( m)
e ax x m 1dx
, if a > 0 and m > 0. ... (ii)
0 am
Replacing a by s, m by (n + 1) and x by t in (ii), we have
( n 1)
e st t ( n 1) 1dt , if s > 0 and n + 1 > 0
0 s n 1
(n 1)
(i) n
{t } = , s 0 and n 1.
s n 1
