A Generalization of Integral Transform 


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 11, No. 4, 2018, 1130-1142
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

A Generalization of Integral Transform

Benedict Barnes1,∗, C. Sebil1, A. Quaye1

1 Department of Mathematics, Kwame Nkrumah University of Science and Technology,
Kumasi, Ghana

Abstract. In this paper, the generalization of integral transform (GIT) of the function G{f(t)}
is introduced for solving both the differential and interodifferential equations. This transform
generalizes the integral transforms which use exponential functions as their kernels and the integral
transform with polynomial function as a kernel. The generalized integral transform converts the
differential equation into us domain (the transformed variables) and reconverts the result by its
inverse operator. In particular, if u = 1, then the generalized integral transform coincides with
the Laplace transform and this result can be written in another form as the polynomial integral
transform.

2010 Mathematics Subject Classifications: 44B53, 44B54

Key Words and Phrases: Generalization of integral transform, kernel, differential equation

1. Introduction

The role of integral transforms has been increasingly recognized in the scientific world.
Most of the problems in fluid mechanics, circuit design, heat transfer, population of species,
are all continuous processes, which are modelled as either differential or integrodifferen-
tial equations with known initial or boundary conditions. Searching for the solutions of
differential equations is a pertinent issue which concerns every scientist. Over the years,
the researchers across the globe have come out with some methods for solving differential
equations. The type and order of the differential equation determines the method that
has to be selected to find the solution of the equation. The classical methods such as
the separation of variables solves only separable differential equations. In similar vein,
the use of integrating factor method solves a linear differential equation in an appropriate
functional space. In addition, the differential equation has to be written in the standard
form before searching for an appropriate integrating factor which transforms the differen-
tial equation into a separable form, from which the solution is obtained. This tedious and
cumbersome method of searching for a solution of a differential equation is heartbreaking
and undesirable.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v11i4.3330

Email addresses: ewiekwamina@gmail.com bbarnes.cos@knust.edu.gh ( B. Barnes)

http://www.ejpam.com 1130 c© 2018 EJPAM All rights reserved.



B. Barnes, C. Sebil, A. Quaye / Eur. J. Pure Appl. Math, 11 (4) (2018), 1130-1142 1131

In order to overcome the shortcomings of the classical methods for solving differential
equations Laplace transform was introduced which converts the differential equation into
an algebraic equation, which is then transformed the result back by means of its inverse
operator to obtain the desired solution in a suitable functional space, for example, see [1].
The authors in [2] extended the Laplace transform to double Laplace transform. In [3], the
author established the convolution property of the double Laplace transform and applied
this property to solve the homogeneous linear partial differential equations. The authors in
[4] applied the double Laplace transform to solve nonhomogeneous linear partial differen-
tial equations. Currently, integral transform method has become reliable method for solv-
ing differential equations, integral equations and integro-differential equations. Sumudu
transform was observed by [5]. The authors in [6], also introduced natural transform for
solving differential equations. In [7], the authors extended the natural transform from one
dimension to two dimensions and also, compared the double transform with Laplace and
Sumudu transforms. The author in [8] introduced a polynomial integral transform PIT.
In his work, the properties of the PIT were established and applied these properties for
solving both the linear ordinary differential equations and linear partial differential equa-
tions. In [9], the authors established the relationships of the PIT among some integral
transforms by equating the domain element that appears in the function f(t) in PIT with
other domain elements in f(t) of other integral transforms.

It is not enough to introduce integral transform for solving differential equations with-
out establishing its relationships with other integral transforms. By and large, the integral
transformation of the operator of an unknown function into an algebraic equation in the
same functional space is paramount and inevitably, the generalization of integral trans-
form (GIT) is introduced in this paper. This method uses exponential function as its
kernel, which converts a linear differential equations to an algebraic equation in terms of
us and finally, transforms the resulting algebraic equation by means of its inverse operator
to obtain the solution of the differential equation.

This paper is organized as follows. The section one contains the introduction; the
GIT, its properties and applications are captured in section two and the last section of
this paper contains the conclusion.

2. Main Results

In this section, the generalization of integral transform is introduced. We definitions
of some integral transforms that will enable us to achieve our results.

Definition 1 (Polynomial Integral Transform). Let f(x) be a function defined for x ≥ 0.
Then the integral

B(f(x)) = B(s) =

∫ ∞
1

f(lnx)x−(s+1)dx, (1)

is the polynomial integral transform of f(x) for all x ∈ [1,∞), see [8].



B. Barnes, C. Sebil, A. Quaye / Eur. J. Pure Appl. Math, 11 (4) (2018), 1130-1142 1132

Definition 2 (Natural Transform). Let f(t) be a function defined for t ≥ 1. Then the
integral

N{f(t)} =

∫ ∞
0

f(ut)e−stdt, (2)

is the natural transform of f(t) for t ∈ [0,∞). see [6].

Definition 3 (Sumudu Transform). Let f(t) be a function defined for t ≥ 0. Then the
integral

S{f(t)} =

∫ ∞
0

f(ut)e−tdt, (3)

is the Sumudu transform of f(t) for all t ∈ [0,∞). see [5].

Definition 4 (Fourier Transform). Let f(t) be a function defined for t ≥ 0. Then the
integral

F{f(t)} =

∫ ∞
0

f(t)e−istdt (4)

is the Fourier transform of f(t) for all t ∈ C, see [3].

2.1. The Derivation of the Generalization of Integral Transform

In this subsection, the proof of the GIT is provided in theorem 1 below.

Theorem 1. Let f(t) be a function defined for t ≥ 0. Then the integral

G{f(t)} = G(s) = u

∫ ∞
0

f(ut)e−ustdt,

is the generalized integral transform of f(t) for all t ∈ [0,∞).

Proof : Equating the kernels in the polynomial integral transform and natural trans-
form in equations (1) and (2), we obtain

x−(s+1) = e−st

⇒ lnx−(s+1) = ln e−st

⇒ −(s+ 1) lnx = −st

⇒ x = e
st

(s+1) (5)

⇒ dx =
s

(s+ 1)
e

st
(s+1)dt

Substituting equation (5) into equation (1) yields

G(f(t)) =
s

(s+ 1)

∫ ∞
0

f
( st

s+ 1

)
e
− s2t

(s+1)dt

⇒ G(f(t)) = u

∫ ∞
0

f(ut)e−sutdt, (6)

where, u = s
(s+1) .



B. Barnes, C. Sebil, A. Quaye / Eur. J. Pure Appl. Math, 11 (4) (2018), 1130-1142 1133

2.2. The Sufficient Condition for Existence of a Generalization of Integral
Transform

Theorem 2. Let f(t) be a piecewise continuous function on the interval [0,∞) and of
exponential order k for t > T , then G{f(t)} exists for s > k.

Proof : Setting f(t) ≤M |ekt|. We see that:

G{f(t)} ≤ u

∞∫
0

e−ustf(ut)dt

‖G{f(t)}‖ ≤

∥∥∥∥∥∥u
∞∫
0

e−ustf(ut)dt

∥∥∥∥∥∥
‖G{f(t)}‖ ≤ ‖u‖

∥∥∥∥∥∥
∞∫
0

e−ustf(ut)dt

∥∥∥∥∥∥
‖G{f(t)}‖ ≤ ‖u‖

∞∫
0

∣∣e−ustf(ut)
∣∣ dt

‖G{f(t)}‖ ≤ ‖u‖
∞∫
0

e−ustMekutdt, ∀t > T

‖G{f(t)}‖ ≤ u

∫ T

0
Mekute−ustdt+ u

∫ ∞
T

Mekute−ustdt

= uM

∫ T

0
e−(s−k)utdt+ uM lim

m→∞

∫ m

T
e−(s−k)utdt

= uM

[
− 1

(s− k)u
e−(s−k)ut

]T
0

+ uM lim
m→∞

[
− 1

(s− k)u
e−(s−k)ut

]m
T

‖G{f(t)}‖ ≤ uM

[
− 1

(s− k)u
e−(s−k)uT +

1

(s− k)u
e−(s−k)u(0)

]
+ uM lim

m→∞

[
− 1

(s− k)u
e−(s−k)um +

1

(s− k)u
e−(s−k)uT

]
‖G{f(t)}‖ ≤ 1

(s− k)u
, u > 0, s > k.

We observed that G{f(t)} exists for u > 0, s > k for some t > T .

2.3. Properties of a Generalization of Integral Transform

Corollary 1. (Linearity property of GIT) Let f(t) and g(t) be functions defined for t ≥ 0.
Then

G{αf(t) + βg(t)} = αG{f(t)}+ βG{g(t)},



B. Barnes, C. Sebil, A. Quaye / Eur. J. Pure Appl. Math, 11 (4) (2018), 1130-1142 1134

where α and β are scalars.

Proof : We see from the definition of the generalized integral transform that:

G{αf(t) + βg(t)} = u

∫ ∞
0

(
αf(ut) + βg(ut)

)
e−sutdt

⇒ G{αf(t) + βg(t)} = u

∫ ∞
0

αf(ut)e−sutdt+ u

∫ ∞
0

βg(ut)e−sutdt

⇒ G{αf(t) + βg(t)} = αu

∫ ∞
0

f(ut)e−sutdt+ βu

∫ ∞
0

g(ut)e−sutdt

⇒ G{αf(t) + βg(t)} = αG{f(t)}+ βG{g(t)}.

Corollary 2. (First shifting theorem of a function using the GIT) If G{f(t)} = G(s),
then

G{eαxf(t)} = G(s− α), for s > 1.

Proof : Setting G{f(t)} = G(s) = u
∫∞
0 f(ut)e−ustdt, then

G{eαtf(t)} = u

∫ ∞
0

eαutf(ut)e−ustdt

G{eαtf(t)} = u

∫ ∞
0

f(ut)e−(s−α)utdt

G{eαtf(t)} = G(s− α).

Corollary 3. (Second shifting theorem of a function using the GIT).
Setting

Hc(t) =

{
0, 0 ≤ t < c
1, t ≥ c

be a unit step function. Then

G{Hcf(t− c)} = e−csG(s).

Proof : We see the definition of the GIT that:

G{Hc(t)f(t− c)} = u

∫ ∞
0

Hc(t)f(ut− c)e−ustdt

⇒ G{Hc(t)f(t− c)} = u lim
m→∞

∫ m

0
1.f(ut− c)e−ustdt

⇒ G{Hc(t)f(t− c)} = lim
m→∞

∫ um−c

0
f(v)e−(s−c)vdv

G{Hcf(t− c)} = e−csG(s).



B. Barnes, C. Sebil, A. Quaye / Eur. J. Pure Appl. Math, 11 (4) (2018), 1130-1142 1135

2.4. The GIT of Derivative

In this section, we derive an expression for the derivative of function which shall enable
us to search for the solutions of ordinary differential equations. The result is in theorem
3 below.

Theorem 3. If y(t), y′(t), . . . , y(n−1)(t) are continuous on [0,∞) and are of exponential
order and if f (n)(t) is piecewise continuous on [0,∞), then

G{f (n)(t)} = unsnY (s)− unsn−1y(0)− un−1sn−2y′(0)− . . .− uy(n−1)(0),

where G(s) = G{f(t)}.

Proof : By induction, we consider the function y(t). We see from theorem (2.5) that:

G{y(t)} = Y (s) = u

∞∫
0

e−usty(ut)dt

Considering G
{
dy
dt

}
. Thus,

G

{
dy

dt

}
= u

∞∫
0

e−ust
dy(ut)

dt
dt

G

{
dy

dt

}
= u lim

m→∞

m∫
0

e−ust
dy(ut)

dt
dt.

Using the integration by part yields

G

{
dy

dt

}
= u

 lim
m→∞

[
e−usty(ut)

]m
0

+ us lim
m→∞

m∫
0

e−usty(ut)dt


G

{
dy

dt

}
= u

 lim
m→∞

[
e−usmy(um)− e−us(0)y(0)

]
+ us lim

m→∞

m∫
0

e−usty(ut)dt


G

{
dy

dt

}
= u

 lim
m→∞

e−usmy(um)− lim
m→∞

y(0) + us

∞∫
0

e−usty(ut)dt


G

{
dy

dt

}
= u

0− y(0) + us

∞∫
0

e−usty(ut)dt


G

{
dy

dt

}
= −uy(0) + us

u ∞∫
0

e−usty(ut)dt





B. Barnes, C. Sebil, A. Quaye / Eur. J. Pure Appl. Math, 11 (4) (2018), 1130-1142 1136

G

{
dy

dt

}
= usY (s)− uy(0)

Considering G
{
d2y
dt2

}
, we have:

G

{
d2y

dt2

}
= u

∞∫
0

e−ust
d2y(ut)

dt2
dt

G

{
d2y

dt2

}
= u lim

m→∞

m∫
0

e−ust
d2y(ut)

dt2
dt

Using the integration by parts yields

G

{
d2y

dt2

}
= u

 lim
m→∞

[
e−ust

dy(ut)

dt

]m
0

+ us lim
m→∞

m∫
0

e−ust
dy(ut)

dt
dt


= u

 lim
m→∞

[
e−usm

dy(um)

dt
− e−us×0dy(0)

dt

]
+ s

u ∞∫
0

e−ust
dy(ut)

dt
dt


= −uy′(0) + us(usY (s)− uy(0))

= −uy′(0) + u2s2Y (s)− u2sy(0)

G

{
d2y

dt2

}
= u2s2Y (s)− u2sy(0)− uy′(0)

Considering G
{
d3y
dt3

}
, we have:

G

{
d3y

dt3

}
= u

∞∫
0

e−ust
d3y(ut)

dt3
dt

G

{
d3y

dt3

}
= u lim

m→∞

m∫
0

e−ust
d3y(ut)

dt3
dt.

Using the integration by parts, we obatin

G

{
d3y

dt3

}
= u

 lim
m→∞

[
e−ust

d2y(ut)

dt2

]m
0

+ us lim
m→∞

m∫
0

e−ust
d2y(ut)

dt2
dt


G

{
d3y

dt3

}
= −uy′′(0) + us(u2s2Y (s)− u2y(0)− u2sy′(0))

G

{
d3y

dt3

}
= −uy′′(0) + u3s3Y (s)− u3s2y(0)− u2sy′(0)



B. Barnes, C. Sebil, A. Quaye / Eur. J. Pure Appl. Math, 11 (4) (2018), 1130-1142 1137

G

{
d3y

dt3

}
= u3s3Y (s)− u3s2y(0)− u2sy′(0)− uy′′(0)

...

G

{
dny

dtn

}
= unsnY (s)− unsn−1y(0)− un−1sn−2y′(0)− . . .− uy(n−1)(0)

Corollary 4. (Convolution Theorem for GIT) If f(t) and g(t) are piecewise continuous
on [0,∞) and of exponential order, then

G{f ∗ g} = G1{f(t)}G2{g(t)}
G{f ∗ g} = G1(s)G2(s).

Proof : Setting G1{f(t)} = G1(s) = u
∞∫
0

e−usεf(uε)dε and G2{g(t)} = G2(s) =

u
∞∫
0

e−usτg(uτ)dτ

G1(s)G2(s) =

u ∞∫
0

e−usεf(uε)dε

u ∞∫
0

e−usτg(uτ)dτ


G1(s)G2(s) = u2

∞∫
0

∞∫
0

e−us(ε+τ)f(uε)g(uτ)dεdτ

G1(s)G2(s) = u2
∞∫
0

f(uτ)dτ

∞∫
0

g(uε)e−us(ε+τ)dε (7)

t = ε+ τ
dt = dε

}
(8)

Substituting equation(8) into equation(7) yields

G1(s)G2(s) = u2
∞∫
0

f(uτ)dτ

∞∫
0

e−ustg(u(t− τ))dt

G1(s)G2(s) = u2
∞∫
0

e−ustdt

t∫
0

f(uτ)g(u(t− τ))dτ

= u

∞∫
0

e−ust

u
t∫

0

f(uτ)g(u(t− τ))dτ

 dt

G1(s)G2(s) = G{f ∗ g}.

Corollary 5. (Commutativity of two functions using the GIT) The convolution of func-
tions f(t) and g(t) commute.



B. Barnes, C. Sebil, A. Quaye / Eur. J. Pure Appl. Math, 11 (4) (2018), 1130-1142 1138

Proof : By the convolution of f(t) and g(t), we have:

f(t) ∗ g(t) = u

∫ ∞
0

f(uτ)g(u(t− τ))e−ustdτ, (9)

Setting t− τ = v
dτ = −dv.

}
(10)

Substituting equation(10) into equation(9) yields

f(t) ∗ g(t) = −u
∫ −∞
t

f(u(t− v))g(uv)e−us(t−v)dv

f(t) ∗ g(t) = u

∫ ∞
0

g(uv)f(u(t− v))e−usvdv

f(t) ∗ g(t) = g(t) ∗ f(t)

This completes the prove.

2.5. Illustration of the GIT

In this section, we show some of the areas where GIT can be used to solve problems.

Example 1.
dy

dt
+ 3y(t) = 2, y(0) = 0

Taking the GIT of both sides,

G

{
dy

dt
+ 3y(t)

}
= G{2}

G

{
dy

dt

}
+G{3y(t)} = G{2}

usG(s)− uy(0) + 3G(s) =
2

s

usG(s) + 3G(s) =
2

s
+ uy(0)

G(s) =
2

s(us+ 3)

G(s) =
2

3s
− 2

3(us+ 3)
.

Taking the inverse GIT of both sides of the above equation yields

G−1{G(s)} = G−1
{

2

3s
− 2

3(us+ 3)

}
.



B. Barnes, C. Sebil, A. Quaye / Eur. J. Pure Appl. Math, 11 (4) (2018), 1130-1142 1139

By the linearity of the inverse GIT, we get

y(t) = G−1
{

2

3s

}
−G−1

{
2

3(us+ 3)

}
y(t) =

2

3

(
1− e−3t

)
Example 2.

y′′(t) + 16y(t) = cos(4t), y(0) = 0, y′(0) = 1.

Taking the GIT of both sides, we get

G
{
y′′ + 16y

}
= G{cos(4t)}

G
{
y′′
}

+ 16G{y} = G{cos(4t)}

u2s2Y (s)− u2sy(0)− uy′(0) + 16Y (s) =
s

s2 + 16

Y (s) =
us2 + s+ 16

(s2 + 16)(u2s2 + 16)
.

Taking the inverse generalized integral transform of both sides of the above equation yields

G−1{Y (s)} = G−1
{

us2 + s+ 16

(s2 + 16)(u2s2 + 16)

}
y(t) =

1

4
sin(4t) +

1

8
t sin(4t)

Example 3.

E(t) = L
di

dt
+Ri(t) +

1

C

t∫
0

i(τ)dτ (1)

Setting L = 0.1h, R = 2, C = 0. If, E(t) = 120t− 120φ(t− 1), i(0) = 0, we obtain

0.1
di

dt
+ 2i(t) +

1

0.1

t∫
0

i(τ)dτ = 120t− 120φ(t− 1)

=⇒ 0.1
di

dt
+ 2i(t) + 10

t∫
0

i(τ)dτ = 120t− 120φ(t− 1) (2)

Finding the GIT of the terms on both sides of the above equation yields

G

0.1
di

dt
+ 2i(t) +

1

0.1

t∫
0

i(τ)dτ = 120t− 120φ(t− 1)

 = G{120t− 120φ(t− 1)}



B. Barnes, C. Sebil, A. Quaye / Eur. J. Pure Appl. Math, 11 (4) (2018), 1130-1142 1140

By the linearly property of GIT, we get

0.1G

{
di

dt

}
+ 2G{i(t)}+ 10G


t∫

0

i(τ)dτ

 = 120G{t} − 120G{tφ(t− 1)}

=⇒ 0.1(usI(s)− ui(0)) + 2I(s) +
10

s
I(s) = 120

[
1

s2
− 1

s2
e−s − 1

s
e−s
]

Finding the inverse generalized integral transform on both sides of the above equation.
Thus,

G−1{I(s)} = G−1

{
120

[
1
s2
− 1

s2
e−s − 1

se
−s][

0.1us+ 2 + 10
s

] }
i(t) = 12[1− φ(t− 1)]− 12[e−10t − e−10(t−1)φ(t− 1)]−

120te−10t − 1080(t− 1)e−10(t−1φ(t− 1).

=⇒ i(t) =


12− 12e−10t − 120te−10t, 0 ≤ t ≤ 1

−12e−10t − 12e−10(t−1) − 1080(t− 1)e−10(t−1), ∀t ≥ 1

2.6. The Complex Generalized Integral Transform

In this section of the paper, we provide the complex form of the generalized integral
transform. Thus, the variable f(t) is transformed in a complex domain.

Corollary 6. Let f(t) be a function defined for t ≥ 0. Then the integral

G{f(t)} = G(is) = iu

∫ ∞
0

f(ut)e−iustdt,

is the complex generalized integral transform of f(t) for all t ∈ C+.

Proof : Equating the kernels in equations (1) and (8) yields

x−(s+1) = e−t

⇒ lnx−(s+1) = ln e−t

⇒ −(s+ 1) lnx = −t

⇒ x = e
ist

(s+1) (13)

⇒ dx =
is

(s+ 1)
e

ist
(s+1)dt.

Substituting equation (9) into equation (1) yields

G(f(t)) =
is

(s+ 1)

∫ ∞
0

f
( ist

s+ 1

)
e
− is2t

(s+1)dt



REFERENCES 1141

G(f(t)) = iu

∫ ∞
0

f(iut)e−iustdt, (14)

where u = s
(s+1) .

3. Conclusion

We observed that, if y(t), y′(t), . . . , y(n−1)(t) are continuous on [0,∞) and one of ex-
ponential order and f (n)(t) is piecewise continuous on [0,∞), then

G{fn(t)} = unsnY (s)− unsn−1y(0)− un−1sn−2y′(0)− . . .− uy(n−1)(0).

Undoubtedly, a generalization of an integral transform becomes the Laplace transform for
u = 1. Thus,

L(f(t)) =

∫ ∞
0

f(t)e−stdt.

The GIT has some properties like the expression for derivative which is unique as compared
to other integral transforms. The transformed variable u plays the role of dilating the
transformed domain to obtain the desired result. On the other hand, the the transformed
domain can be contracted by the transformed variable u.

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