EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 2, Article Number 5986
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Modified Laplace-type Transform and Its Properties

İnci Ege

Department of Mathematics, Faculty of Science, University of Aydın Adnan Menderes,
Aydın, Türkiye

Abstract. In this article, we introduce the Modified Laplace-type transform, develop convergence
properties, and obtain fundamental formulas of some elementary functions such as power functions,
sine, cosine, hyperbolic sine, hyperbolic cosine, and exponential functions. We derive translation
theorems and a scale-preserving theorem and also show the relationship between the modified
Laplace type transform and the modified degenerate Gamma function. This integral transform
is applied to solve linear ordinary differential equations with constant coefficients and a Volterra
integral equation of the second kind.

2020 Mathematics Subject Classifications: 44A10, 44A20, 34A25, 44A05

Key Words and Phrases: Laplace transform, Laplace-type integral transform, modified degen-
erate Gamma function, integral equation

1. Introduction

A transformation is a mathematical technique that changes one function into another.
An integral transform maps a function from its original function space into another func-
tion space by using integration as a tool to solve differential and integral equations. Its
motivation comes from some classes of problems that are difficult to solve in their original
representations. An integral transform takes a function from its original domain into an-
other, which may make solving the equation much easier than in the original domain. The
transformed function can generally be mapped back to the original function space using
the inverse transform. An integral transform T is of the form

(Tv)(t) =

∫ x2

x1

v(x)k(x, t) dx,

where v is the input function, Tv is the output function and k(x, t) is the kernel of the
transform.

Numerous useful integral transforms have been defined; see for example [1–20]. Each
is specified by a choice of the kernel function k of two variables. Perhaps the most well-
known integral transform is the Laplace transform. Besides mathematics, it is utilized

DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.5986

Email address: iege@adu.edu.tr (İ. Ege)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 2 of 20

in other fields of study wildly as engineering, physics, astronomy, etc. For example, it is
used to solve differential equations occurring in the analysis of electronic circuits. Laplace
transform is defined by

F (u) = L{f(t)} =

∫ ∞

0
e−utf(t) dt (1)

provided that the integral converges [10].
There are many integral transforms in the Laplace class and most of them have been

named after the mathematicians who introduced them. Some of these are the Sumudu
transform [21], the Natural transform [22], the Elzaki transform [23], the Aboodh trans-
form [24], the ZZ transform [25] and the Polynomial integral transform [26]. The Sumudu
[21] and Elzaki transforms [23] are defined respectively as

S(u) = S{f(t)} =
1

u

∫ ∞

0
e−

t
u f(t) dt, (2)

and

E(u) = E{f(t)} = u

∫ ∞

0
e−

t
u f(t) dt. (3)

Recently, the Laplace-type integral transform is introduced by Kim in [27] as

Fα(u) = Gα{f(t)} = uα
∫ ∞

0
e−

t
u f(t) dt, (4)

where f(t) be an integrable function on [0.∞), u > 0 and α ∈ Z.
The modified degenerate Gamma function defined in [28] as

Γ∗
λ(x) =

∫ ∞

0
tx−1(1 + λ)−

t
λ dt, λ ∈ (0, 1) and Re(x) > 0, (5)

and satisfies the properties that

Γ∗
λ(x+ 1) =

λx

ln(1 + λ)
Γ∗
λ(x),

Γ∗
λ(n+ 1) =

λn+1n!

lnn+1(1 + λ)
, n = 1, 2, . . . . (6)

Degenerate versions of existing integral transforms have also been studied in the last
few years. For example, Kim and Kim introduced the degenerate Laplace transform [29]
and Upadhyaya gives further results for the degenerate Laplace transform [30–32]. Cam-
pos et al. defined degenerate Laplace-type integral transform and gave its properties [3].
Also, Duran defined the degenerate Sumudu transform [33] and Kalavathi et al. defined
the degenerate Elzaki transform [34].

Motivated by the above-mentioned research, in this paper we proposed the modi-
fied Laplace-type transform. The mentioned transform is indicated by the operator G∗

α,λ



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 3 of 20

through this research. We define the modified Laplace-type transform and provide some
of their properties and relations, and derive the modified Laplace-type transform of some
functions such as power functions, sine, cosine, hyperbolic sine, hyperbolic cosine, ex-
ponential function, and function derivatives. Moreover, we attain relations between the
Laplace-type transform and the modified Laplace-type transform and also give a rela-
tion between the modified Laplace-type transform and the modified degenerate Gamma
function. Also, we give some operational properties of modified Laplace-type transform.

2. The main results

In this section, we present the modified Laplace-type transform G∗
α,λ, give sufficient

conditions for the existence, and calculate the modified Laplace-type transform of some
frequently used functions.

Definition 1. Let λ ∈ (0,∞), α ∈ Z, u > 0 and f(t) be an integrable function defined for
all t ≥ 0. Then the integral

G∗
α,λ{f(t)} = uα

∫ ∞

0
(1 + λ)−

t
uλ f(t) dt (7)

is said to be the modified Laplace-type transform G∗
α,λ of f(t) provided that the integral in

(7) exists.
Since the function G∗

α,λ{f(t)} is depend on the variable u, it can be denoted as F ∗
α,λ(u).

We note that
lim
λ→0

G∗
α,λ{f(t)} = Gα{f(t)}, (8)

lim
λ→0
α=1

G∗
α,λ{f(t)} = E{f(t)} (9)

and
lim
λ→0
α=−1

G∗
α,λ{f(t)} = S{f(t)}. (10)

Now we give sufficient conditions for the existence of the new integral transform.

Theorem 1. (Existence property of G∗
α,λ) Let f(t) be a piecewise-continious function on

every finite interval [0, a] and of exponential order as t goes to infinity with |f(t)| ≤ Mekt

for t > L, where and k, L,M are constants and M > 0. Then G∗
α,λ{f(t)} exists for

1
u > kλ

ln(1+λ) .

Proof. We can write

uα
∫ ∞

0
(1 + λ)−

t
uλ f(t) dt = uα

∫ L

0
(1 + λ)−

t
uλ f(t) dt+ uα

∫ ∞

L
(1 + λ)−

t
uλ f(t) dt. (11)



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 4 of 20

Since the function f(t) is continuous on the interval [0, L], there exist an M > 0 such that
|f(t)| ≤ M . It gives that

uα
∫ L

0
(1 + λ)−

t
uλ f(t) dt ≤ Muα

∫ L

0
(1 + λ)−

t
uλ dt =

Mλuα+1

ln(1 + λ)
[1− (1 + λ)−

L
uλ ] < ∞.

Hence the first integral on the right-hand side of equation (11) exists. Also, since

|(1 + λ)−
t
uλ f(t)| ≤ M(1 + λ)−

t
uλ ekt = Me−

t
uλ

ln(1+λ)ekt = Me
t
(
k− ln(1+λ)

uλ

)

we have ∣∣∣∣uα ∫ ∞

L
(1 + λ)−

t
uλ f(t) dt

∣∣∣∣ ≤ uα
∫ ∞

L
e−

t
uλ

ln(1+λ) |f(t)| dt

= Muα lim
h→∞

∫ h

L
e
t
(
k− ln(1+λ)

uλ

)
dt =

Mλuα+1

ln(1 + λ)− kuλ
e
L
(
k− ln(1+λ)

uλ

)
< ∞

for k < ln(1+λ)
uλ . Then we proved that the integral

uα
∫ ∞

0
(1 + λ)−

t
uλ f(t) dt (12)

exists, and the result follows.

Theorem 2. The modified Laplace-type transform of the function f(t) = tn, n = 1, 2, . . .
is given by

G∗
α,λ{tn} =

n!λn+1uα+n+1

lnn+1(1 + λ)
.

Proof. By considering the Definition 1 for f(t) = tn, leads to

G∗
α,λ{tn} = uα

∫ ∞

0
(1 + λ)−

t
uλ tn dt = uα lim

h→∞

∫ h

0
(1 + λ)−

t
uλ tn dt.

Using integration by parts, we have

G∗
α,λ{tn} = uα lim

h→∞

[
−tnuλ(1 + λ)

−t
uλ

ln(1 + λ)

∣∣∣∣h
0

+
nuλ

ln(1 + λ)

∫ h

0
tn−1(1 + λ)

−t
uλ dt

]
.

Since for 1
uλ > 0 we have limh→ ∞

hnuλ(1 + λ)
−h
uλ

ln(1 + λ)
= 0, then

G∗
α,λ{tn} =

nuλ

ln(1 + λ)
G∗

α,λ{tn−1},

and so

G∗
α,λ{tn−1} =

(n− 1)uλ

ln(1 + λ)
G∗

α,λ{tn−2}.



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 5 of 20

Hence, we get

G∗
α,λ{tn} =

n(n− 1)u2λ2

ln2(1 + λ)
G∗

α,λ{tn−2}.

Continiuting this process we get

G∗
α,λ{tn} =

n(n− 1) . . . 2unλn

lnn(1 + λ)
G∗

α,λ{1}.

Also by taking f(t) = 1 in the Definition 1, we have

G∗
α,λ{1} = uα

∫ ∞

0
(1 + λ)−

t
uλ dt = uα lim

h→∞

∫ h

0
(1 + λ)−

t
uλ dt =

λuα+1

ln(1 + λ)
,

and the result follows.

Note that, from the Theorem 2 we have

lim
λ→0

G∗
α,λ{tn} = lim

λ→0

n!λn+1uα+n+1

lnn+1(1 + λ)
= n!uα+n+1 = Gα{tn}

for n = 0, 1, 2, . . ..
Also note that, by using the equation (6) we give the relation between the modified Laplace-
type transform and the modified degenerate Gamma function Γ∗

λ for λ ∈ (0, 1) as

G∗
α,λ{tn} = Γ∗

λ(n+ 1)uα+n+1.

Theorem 3. The modified Laplace-type transform of the function f(t) = eat, is given by

G∗
α,λ{eat} =

λuα+1

ln(1 + λ)− aλu
for u <

ln(1 + λ)

aλ
and a ∈ R.

Proof. Using the equation (7) for f(t) = eat and writing (1 + λ)−
t
uλ = e

−t ln(1+λ)
λu we

obtain

G∗
α,λ{eat} = uα

∫ ∞

0
(1 + λ)−

t
uλ eat dt = uα lim

h→∞

∫ h

0
e
t
(
a− ln(1+λ)

λu

)
dt

= lim
h→∞

λuα+1

aλu− ln(1 + λ)

[
e
h
(
a− ln(1+λ)

λu

)
− 1

]
=

λuα+1

ln(1 + λ)− aλu

for a < ln(1+λ)
λu , and the proof is completed.

Note that, from the Theorem 3 we have

lim
λ→0

G∗
α,λ{eat} = lim

λ→0

λuα+1

ln(1 + λ)− aλu
=

uα+1

1− au
= Gα{eat}.

Theorem 4. The modified Laplace-type transform of the function f(t) = sin at is given
by

G∗
α,λ{sin at} =

aλ2uα+2

ln2(1 + λ) + a2λ2u2
. (13)



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 6 of 20

Proof. By writing sin at = eiat−e−iat

2i , we obtain

G∗
α,λ{sin at} = uα

∫ ∞

0
(1 + λ)−

t
uλ sin at dt = uα

∫ ∞

0
e

−t ln(1+λ)
λu

(
eiat − e−iat

2i

)
dt

=
uα

2i
lim
h→∞

∫ h

0

[
e
−t

(
ln(1+λ)

λu
−ia

)
− e

−t
(
ia+

ln(1+λ)
λu

)]
dt

=
uα

2i
lim
h→∞

[
λu

iaλu− ln(1 + λ)
e
−h

(
ln(1+λ)

λu
−ia

)
+

λu

ln(1 + λ)− iaλu

]
−uα

2i
lim
h→∞

[
−λu

iaλu+ ln(1 + λ)
e
−h

(
ia+

ln(1+λ)
λu

)
+

λu

ln(1 + λ) + iaλu

]

=
λuα+1

2i

[
1

ln(1 + λ)− iaλu
− 1

ln(1 + λ) + iaλu

]
=

aλ2uα+2

ln2(1 + λ) + a2λ2u2

for ia− ln(1+λ)
λu < 0 and ia+ ln(1+λ)

λu > 0. Also since ln(1+λ)
λ > 0, we get the result.

Note that, from the Theorem 4 we have

lim
λ→0

G∗
α,λ{sin at} = lim

λ→0
=

aλ2uα+2

ln2(1 + λ) + a2λ2u2
=

auα+2

1 + a2u2
= Gα{sin at}.

Now, we give the transform of the mathematical construct Dirac’s delta function δ
defined as

δ(t) =

{
+∞, t = 0

0, otherwise
,

∫ ∞

−∞
δ(t) dt = 1.

Theorem 5. The modified Laplace-type transform of the Dirac’s delta function is

G∗
α,λ{δ(t− a)} = uα(1 + λ)−

a
λu , a ≥ 0.

Proof. Let

fh(t− a) =

{
1
h , a ≤ t ≤ a+ h

0, otherwise.

Then we have,

G∗
α,λ{fh(t− a)} = uα

∫ ∞

0
(1 + λ)−

t
uλ fh(t− a) dt =

uα

h

∫ a+h

a
(1 + λ)−

t
uλ dt

= − λuα+1

ln(1 + λ)h
(1 + λ)−

a
uλ

[
(1 + λ)−

h
uλ − 1

]
.

Since δ(t− a) is the limit of fh as h → 0 we get

G∗
α,λ{δ(t− a)} = lim

h→0
G∗

α,λ{fh(t− a)} = uα(1 + λ)−
a
λu

by L’hospital rule, and the result follows.

Note that, from the theorem 5 we have

lim
λ→0

G∗
α,λ{δ(t− a)} = lim

λ→0
uα(1 + λ)−

a
λu = uαe−

a
u = Gα{δ(t− a)}.



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 7 of 20

3. Some operational properties

In this section, we present some useful operational properties of the modified Laplace-
type transform. Firstly, the modified Laplace-type transform satisfies the following linear-
ity property.

Theorem 6. (Linearity) Let i = 1, 2, . . . n. If fi(t) is a function whose modified Laplace-
type integral transform exists, then for any constant αi we have

G∗
α,λ

{
n∑

i=1

αifi(t)

}
=

n∑
i=1

αiG
∗
α,λ{fi(t)}. (14)

Proof. Let fi(t) be any function whose modified Laplace-type integral transform exists
for i = 1, 2, . . . n. Then

G∗
α,λ

{
n∑

i=1

αifi(t)

}
= uα

∫ ∞

0

[
(1 + λ)−

t
uλ

n∑
i=1

αifi(t)

]
dt

=

n∑
i=1

αiu
α

∫ ∞

0
(1 + λ)−

t
uλ fi(t) dt =

n∑
i=1

αiG
∗
α,λ{fi(t)},

and the result follows.

Theorem 7. The modified Laplace-type transform of the function f(t) = sinh at is given
by

G∗
α,λ{sinh at} =

aλ2uα+2

ln2(1 + λ)− a2λ2u2
for u <

ln(1 + λ)

aλ
.

Proof. By using the equation sinh at =
eat − e−at

2
and the linearity properity (14) we

get

G∗
α,λ{sinh at} =

1

2

[
G∗

α,λ{eat} −G∗
α,λ{e−at}

]
.

Now, using the Theorem 3 the result follows.

Note that,

lim
λ→0

G∗
α,λ{sinh at} = lim

λ→0

aλ2uα+2

ln2(1 + λ)− a2λ2u2
=

auα+2

1− a2u2
= Gα{sinh at}.

Theorem 8. (Transform of Derivatives) If f(t), f
′
(t), . . . f (n−1)(t) are continious func-

tions on the interval [0, L] and of exponential order as t goes to infinity for t > L while
f (n)(t) is piecewise-continious on the interval [0, L], then

G∗
α,λ{f (n)(t)} =

lnn(1 + λ)

λnun
G∗

α,λ{f(t)} − uα
lnn−1(1 + λ)

λn−1un−1
f(0)

−uα
lnn−2(1 + λ)

λn−2un−2
f

′
(0)− . . .− uαf (n−1)(0), (15)

where f (n)(t) = dn

dtn f(t) and n = 1, 2, 3, . . ..



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 8 of 20

Proof. Using the Definition (1) and the integration by parts, we have

G∗
α,λ{f

′
(t)} = uα

∫ ∞

0
(1 + λ)−

t
uλ f

′
(t) dt = uα lim

h→∞

∫ h

0
(1 + λ)−

t
uλ f

′
(t) dt

= uα lim
h→∞

[
(1 + λ)−

t
uλ f(t)

∣∣∣∣h
0

+
ln(1 + λ)

λu

∫ h

0
(1 + λ)−

t
uλ f(t) dt

]
(16)

= −uαf(0) +
ln(1 + λ)

λu
G∗

α,λ{f(t)},

and the equation (15) follows for n = 1.
Now for proving the equation (15), we use induction method on n. The equation (15) is
true for n = 1. By using the equation (16) we can write

G∗
α,λ{f (k+1)(t)} = G∗

α,λ{
d

dt
f (k)(t)} =

ln(1 + λ)

λu
G∗

α,λ{f (k)(t)} − uαfn(0)

=
ln(1 + λ)

λu

[
lnk(1 + λ)

λkuk
G∗

α,λ{f(t)} − uα
lnk−1(1 + λ)

λk−1uk−1
f(0)− . . .− uαf (k−1)(0)

]
− uαf (k)(0)

=
lnk+1(1 + λ)

λkuk
G∗

α,λ{f(t)} − uα
lnk(1 + λ)

λkuk
f(0)− . . .− uα

ln(1 + λ)

λu
f (k−1)(0)− uαf (k)(0).

This proves that the equation (15) is true for n = k + 1 and the equation (15) follows.

Note that, by using the Theorem 8 we have

lim
λ→0

G∗
α,λ{f (n)(t)}

= lim
λ→0

[
lnn(1 + λ)

λnun
G∗

α,λ{f(t)} − uα
lnn−1(1 + λ)

λn−1un−1
f(0)− . . .− uαf (n−1)(0)

]
=

1

un
Gα{f(t)} −

1

un−1
f(0)uα − 1

un−2
f

′
(0)uα − . . .− uαf (n−1)(0)

= Gα{f (n)(t)}.

for n = 1, 2, 3, . . ..

Corollary 1. The modified Laplace-type transform of the function f(t) = cos at is given
by

G∗
α,λ{cos at} =

λ ln(1 + λ)uα+1

ln2(1 + λ) + a2λ2u2
. (17)

Proof. Let f(t) = 1
a sin at. Then f

′
(t) = cos at and f(0) = 0. Now using the linearity

property (14) and the equation (16) we get

G∗
α,λ{cos at} =

ln(1 + λ)

aλu
and G∗

α,λ{sin at} =
λ ln(1 + λ)uα+1

ln2(1 + λ) + a2λ2u2
.



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 9 of 20

Corollary 2. The modified Laplace-type transform of the function f(t) = cosh at is given
by

G∗
α,λ{cosh at} =

λ ln(1 + λ)uα+1

ln2(1 + λ)− a2λ2u2
for u <

ln(1 + λ)

aλ
.

Proof. With the similar proof of the Corollary 1 we get the result.

Note that, from the Corollaries 1 and 2 we have

lim
λ→0

G∗
α,λ{cos at} = lim

λ→0

λ ln(1 + λ)uα+1

ln2(1 + λ) + a2λ2u2
=

uα+1

1 + a2u2
= Gα{cos at},

lim
λ→0

G∗
α,λ{cosh at} = lim

λ→0

λ ln(1 + λ)uα+1

ln2(1 + λ)− a2λ2u2
=

uα+1

1− a2u2
= Gα{cosh at}.

Let Fα,λ be the modified Laplace-type transform and Fα be the the Laplace-type
integral transform of f(t), G∗

α,λ{f(t)} = F ∗
α,λ(u) and Lα{f(t)} = Fα(u). Besides the

relation limλ→0G
∗
α,λ{f(t)} = Gα{f(t)} we give the following equation relation between

the modified Laplace-type transform and the Laplace-type integral transform. Since

G∗
α,λ{f(t)} = uα

∫ ∞

0
(1 + λ)−

t
uλ f(t) dt = uα

∫ ∞

0
e

−t ln(1+λ)
λu f(t) dt

=

(
ln(1 + λ)

λ

)α

Fα

(
λu

ln(1 + λ)

)
,

this implies that

F ∗
α,λ(u) =

(
ln(1 + λ)

λ

)α

Fα

(
λu

ln(1 + λ)

)
. (18)

Now replace u by
ln(1 + λ)

λ
u in the equation (18) we have the relation

Fα(u) =

(
λ

ln(1 + λ)

)α

F ∗
α,λ

(
ln(1 + λ)

λ
u

)
.

Example 1. Suppose we want to find G∗
α,λ{cos at} using the relation by the equation (18).

Let G∗
α,λ{f(t)} = F ∗

α,λ(u) and Lα{f(t)} = Fα(u). Since

Lα{cos at} = Fα(u) =
uα+1

1 + a2u2
,

we get

G∗
α,λ{cos at} =

(
ln(1 + λ)

λ

)α

Fα

(
λu

ln(1 + λ)

)
=

(
ln(1 + λ)

λ

)α( λu

ln(1 + λ)

)α+1 1

1 +
a2λ2u2

ln2(1 + λ)

=
λ ln(1 + λ)uα+1

ln2(1 + λ) + a2λ2u2
.



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 10 of 20

Theorem 9. (The First Translation) Let Gα{f(t)} = Fα(u). Then

G∗
α,λ{eatf(t)} =

(
ln(1 + λ)− aλu

λ

)α

Fα

(
λu

ln(1 + λ)− aλu

)
for u <

ln(1 + λ)

aλ
.

Proof. By using the equation (7) we have

G∗
α,λ{eatf(t)} = uα

∫ ∞

0
(1 + λ)−

t
uλ eatf(t) dt = uα

∫ ∞

0
e
−t

(
ln(1+λ)

λu
−a

)
f(t) dt

= uα
(

λu

ln(1 + λ)− aλu

)α( ln(1 + λ)− aλu

λu

)α ∫ ∞

0
e

−
t

λu

ln(1 + λ)− aλu f(t) dt

=

(
ln(1 + λ)− aλu

λu

)α

Fα

{
λu

ln(1 + λ)− aλu

}
for

λu

ln(1 + λ)− aλu
> 0, and the result follows.

Note that, by using the Theorem 9 we have

lim
λ→0

G∗
α,λ{eatf(t)} = lim

λ→0

[(
ln(1 + λ)− aλu

λ

)α

Fα

(
λu

ln(1 + λ)− aλu

)]
= (1− au)α Fα

{
u

1− au

}
= Gα{eatf(t)},

and since

Gα{sin bt} =
buα+2

1 + b2u2
= Fα(u),

Gα{cos bt} =
uα+1

1 + b2u2
= Fα(u)

in [27], we get the following results:

G∗
α,λ{eat sin bt} =

bλ2uα+2

(ln(1 + λ)− aλu)2 + b2λ2u2
,

G∗
α,λ{eat cos bt} =

λuα+1 (ln(1 + λ)− aλu)

(ln(1 + λ)− aλu)2 + b2λ2u2
.

Theorem 10. (The Second Translation) Let G∗
α,λ{f(t)} = F ∗

α,λ(u). Then for a ≥ 0 we
have

G∗
α,λ{f(t− a)H(t− a)} = (1 + λ)

−a
λu F ∗

α,λ(u),

where H(t) is the Heaviside function, which is defined by H(t) = 1 if t ≥ 0 and H(t) = 0
if t < 0.

In particular, the modified Laplace-type transform of the Heaviside function is

G∗
α,λ{H(t− a)} =

λ

ln(1 + λ)
(1 + λ)

−a
λu uα+1.



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 11 of 20

Proof. We have

G∗
α,λ{f(t− a)H(t− a)} = uα

∫ ∞

0
(1 + λ)−

t
uλ f(t− a)H(t− a) dt

= uα
∫ ∞

a
(1 + λ)−

t
uλ f(t− a) dt.

Now, using the change of variable t− a = y we get

G∗
α,λ{f(t− a)H(t− a)} = uα

∫ ∞

0
(1 + λ)−

(a+y)
uλ f(y) dy

= uα(1 + λ)−
a
uλ

∫ ∞

0
(1 + λ)−

y
uλ f(y) dy = (1 + λ)−

a
uλF ∗

α,λ(u)

and

G∗
α,λ{H(t− a)} = uα

∫ ∞

0
(1 + λ)−

t
uλH(t− a) dt

= uα
∫ ∞

a
(1 + λ)−

t
λu dt =

λ

ln(1 + λ)
(1 + λ)−

a
λuuα+1.

Note that by using the Theorem 10 we have

lim
λ→0

G∗
α,λ{f(t− a)H(t− a)} = lim

λ→0
(1 + λ)

−a
λu F ∗

α,λ(u)

= e−
a
uFα(u) = Gα{f(t− a)H(t− a)}, (19)

and

lim
λ→0

G∗
α,λ{H(t− a)} = lim

λ→0

λ

ln(1 + λ)
(1 + λ)

−a
λu uα+1 = e−

a
uuα+1 = Gα{H(t− a)}. (20)

Theorem 11. (Transform of an Integral) Let f(t) be a piecewise-continious function for
t ≥ 0 and integrable. Then if G∗

α,λ{f(t)} = F ∗
α,λ(u) we have

G∗
α,λ

{∫ t

0
f(s) ds

}
=

λu

ln(1 + λ)
F ∗
α,λ(u). (21)

Proof. Let g(t) =
∫ t
0 f(s) ds. Then by using the Theorem 8 we get

G∗
α,λ{f(t)} = G∗

α,λ{g
′
(t)} =

ln(1 + λ)

λu
G∗

α,λ{g(t)} − uαg(0) =
ln(1 + λ)

λu
G∗

α,λ{g(t)},

since g(0) = 0. Hence,

F ∗
α,λ(u) =

ln(1 + λ)

λu
G∗

α,λ{g(t)},

and the result follows.

Note that by using the Theorem 11 we have

lim
λ→0

G∗
α,λ

{∫ t

0
f(s) ds

}
= lim

λ→0

λu

ln(1 + λ)
F ∗
α,λ(u) = uF (u) = Gα

{∫ t

0
f(s) ds

}
.



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 12 of 20

Theorem 12. (Change of Scale) Let G∗
α,λ{f(t)} = F ∗

α,λ(u). Then

G∗
α,λ{f(at)} =

1

aα+1
F ∗
α,λ(au) for a > 0. (22)

Proof. Let at = w. Then

G∗
α,λ{f(at)} = uα

∫ ∞

0
(1 + λ)−

t
uλ f(at) dt =

uα

a

∫ ∞

0
(1 + λ)

− w
λ(au) f(w) dw

=
uα

a

(au)α

(au)α

∫ ∞

0
(1 + λ)

− w
λ(au) f(w) dw =

1

aα+1
F ∗
α,λ(au)

for a > 0, and the result follows.

Note that by using the Theorem 12 we have

lim
λ→0

G∗
α,λ{f(at)} = lim

λ→0

1

aα+1
F ∗
α,λ(au) =

1

aα+1
Fα(au).

Theorem 13. Let G∗
α,λ{f(t)} = F ∗

α,λ(u) and g(t) = f(t − a) for t ≥ a and g(t) = 0 for
0 ≤ t < a, a > 0. Then

G∗
α,λ{g(t)} = (1 + λ)−

a
uλF ∗

α,λ(u).

Proof. Let us write

G∗
α,λ{g(t)} = uα

∫ ∞

0
(1 + λ)−

t
uλ g(t) dt = uα

∫ ∞

a
(1 + λ)−

t
uλ f(t− a) dt.

Now let t− a = y. Then we find

G∗
α,λ{g(t)} = uα

∫ ∞

0
(1 + λ)−

y+a
uλ f(y) dy = (1 + λ)−

a
uλuα

∫ ∞

0
(1 + λ)−

y
uλ f(y) dy,

and the result follows.

Now, we give an example to illustrate the last theorem.

Example 2. For

g(t) =

{
sin t, t ≥ π

2

0, 0 ≤ t < π
2

we want to find G∗
α,λ{g(t)}.

Since cos
(
t− π

2

)
= sin t, we have

g(t) =

{
cos
(
t− π

2

)
, t ≥ π

2

0, 0 ≤ t < π
2 .



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 13 of 20

Then by the Theorem 13 for f(t) = cos t we have G∗
α,λ{g(t)} = (1 + λ)−

π
2uλF ∗

α,λ(u). Also,
since

G∗
α,λ{cos t} = F ∗

α,λ(u) =
λ ln(1 + λ)uα+1

ln2(1 + λ) + λ2u2
,

by the equation (17) we have

G∗
α,λ{g(t)} = (1 + λ)−

π
2uλ

λ ln(1 + λ)uα+1

ln2(1 + λ) + λ2u2
.

Theorem 14. (Multiplication of Powers of the Variable) Let G∗
α,λ{f(t)} = F ∗

α,λ(u). Then

G∗
α,λ{tnf(t)} =

λn

lnn(1 + λ)
u2n

dn

dun
F ∗
α,λ(u)−

λn

lnn(1 + λ)

(
n

1

)
(α− (n− 1))u2n−1 dn−1

dun−1
F ∗
α,λ(u)

+
λn

lnn(1 + λ)

(
n

2

)
(α− (n− 1))(α− (n− 2))u2n−2 dn−2

dun−2
F ∗
α,λ(u)

− λn

lnn(1 + λ)

(
n
3

)
(α− (n− 1))(α− (n− 2))(α− (n− 3))u2n−3 dn−3

dun−3
F ∗
α,λ(u) + . . .

+(−1)n
λn

lnn(1 + λ)
(α− (n− 1))(α− (n− 2)) . . . αunF ∗

α,λ(u) (23)

for n = 1, 2, 3, . . ..

Proof. We use the induction method on n for proving the equation (23). Since

d

du
F ∗
α,λ(u) = αuα−1

∫ ∞

0
(1 + λ)−

t
uλ f(t) dt+

uα−2 ln(1 + λ)

λ

∫ ∞

0
(1 + λ)−

t
uλ tf(t) dt

=
α

u
F ∗
α,λ(u) +

ln(1 + λ)

λu2
G∗

α,λ{tf(t)},

the equation (23) is true for n = 1. Now suppose that the equation (23) is valid for n.
Since

G∗
α,λ{tnf(t)} = uα

∫ ∞

0
(1 + λ)−

t
uλ tnf(t) dt

we have
d

du
G∗

α,λ{tnf(t)} =
α

u
G∗

α,λ{tnf(t)}+
ln(1 + λ)

λu2
G∗

α,λ{tn+1f(t)}.

Then

G∗
α,λ{tn+1f(t)} =

λu2

ln(1 + λ)

d

du
G∗

α,λ{tnf(t)} −
λαu

ln(1 + λ)
G∗

α,λ{tnf(t)}. (24)

Now, by the inductive hypothesis, we get



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 14 of 20

G∗
α,λ{tn+1f(t)} =

λn+1

lnn+1(1 + λ)

[
u2n+1 dn+1

dun+1
F ∗
α,λ(u)−

(
n+ 1

1

)
(α− n)u2n+1 dn

dun
F ∗
α,λ(u)

]

+
λn+1

lnn+1(1 + λ)

[(
n+ 1

2

)
(α− n) (α− (n− 1))u2n

dn−1

dun−1
F ∗
α,λ(u) + . . .

]
+

λn+1

lnn+1(1 + λ)

[
(−1)n+1(α− n) (α− (n− 1)) . . . αun+1F ∗

α,λ(u)
]
.

Hence the equation (23) is valid for n+ 1, and the result follows.

Example 3. We want to find G∗
α,λ{tet}.

By the Theorem 14 we can write

G∗
α,λ{tet} =

λu2

ln(1 + λ)

d

du
F ∗
α,λ(u)−

λαu

ln(1 + λ)
F ∗
α,λ(u),

where F ∗
α,λ(u) = G∗

α,λ{et}. Since G∗
α,λ{et} =

λuα+1

ln(1 + λ)− λu
for u <

ln(1 + λ)

λ
, then

G∗
α,λ{tet} = λu2

ln(1+λ)

d

du

(
λuα+1

ln(1 + λ)− λu

)
− λαu

ln(1+λ)

λuα+1

ln(1 + λ)− λu
=

λ2uα+2

(ln(1 + λ)− λu)2
.

Note that, we can find the same result for G∗
α,λ{tet} by using the Theorem 9 as the

following: Since

G∗
α,λ{ett} =

(
ln(1 + λ)− λu

λ

)α

Fα

(
λu

ln(1 + λ)− λu

)
where Fα(u) = Gα{t} and Gα{t} = uα+2 we get

G∗
α,λ{tet} =

λ2uα+2

(ln(1 + λ)− λu)2
.

Theorem 15. (Convolution) Let G∗
α,λ{f(t)} = F ∗

α,λ(u), and G∗
α,λ{g(t)} = G∗

α,λ(u). Then
the modified Laplace-type transform of the convolution is given as

G∗
α,λ{(f ∗ g)(t)} =

1

uα
F ∗
α,λ(u)G

∗
α,λ(u), (25)

where f ∗ g is the convolution of two functions defined by

(f ∗ g)(t) =
∫ t

0
f(x)g(t− x) dx. (26)

Proof. By the equations (7) and (26) we have

G∗
α,λ{(f ∗ g)(t)} = uα

∫ ∞

0
(1 + λ)−

t
uλ

(∫ t

0
f(x)g(t− x) dx

)
dt



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 15 of 20

= uα
∫ ∞

0

∫ ∞

x
(1 + λ)−

t
uλ f(x)g(t− x) dt dx.

Now putting t− x = w we have

G∗
α,λ{(f ∗ g)(t)} = uα

∫ ∞

0

∫ ∞

0
(1 + λ)−

w+x
uλ f(x)g(w) dw dx

=
1

uα

(
uα
∫ ∞

0
(1 + λ)−

w
uλ g(w)

)(
uα
∫ ∞

0
(1 + λ)−

x
uλ f(x) dx

)
,

and the result follows.

Note that, the modified Laplace-type transform preserves the associative property
concerning the convolution operator:

G∗
α,λ{((f ∗ g) ∗ h)(t)} = G∗

α,λ{(f ∗ (g ∗ h))(t)}.

Let G∗
α,λ{f(t)} = F ∗

α,λ(u). Then f(t) is called as the inverse Laplace-type transform

of F ∗
α,λ(u) and defined by G∗−1

α,λ {F ∗
α,λ(u)} = f(t).

Also note that, the inverse modified Laplace-type transform is linear. Namely, let αi ∈ R,
G∗

α,λ{fi(t)} = F ∗
i,α,λ(u) for i = 1, 2, . . .. Then

G∗−1

α,λ

{
n∑

i=1

αiF
∗
i,α,λ(u)

}
=

n∑
i=1

αiG
∗−1

α,λ

{
F ∗
i,α,λ(u)

}
.

4. Applications

In this section, we give examples to illustrate the use of the mentioned transform in
solving certain initial value problems described by ordinary differential equations and a
Volterra integral equation of the second kind.

Example 4. Consider the first order differential equation

dx

dt
+ x = 0 with the condition x(0) = 1. (27)

Applying the modified Laplace-type transform to both sides of the equation (27) and using
the linearity property we get

−uαx(0) +
ln(1 + λ)

λu
G∗

α,λ{x(t)}+G∗
α,λ{x(t)} = 0.

Putting x(0) = 1 we have

G∗
α,λ{x(t)} =

λuα+1

ln(1 + λ) + λu
.

Now, applying the inverse modified Laplace-type transform, and using the Theorem 3
gives the solution x(t) = e−t.



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 16 of 20

Example 5. Consider the first order differential equation

dx

dt
+ x = 3t with the condition x(0) = 1. (28)

Applying the modified Laplace-type transform we have

−uαx(0) +
ln(1 + λ)

λu
G∗

α,λ{x(t)}+ 2G∗
α,λ{x(t)} =

3λ2uα+2

ln2(1 + λ)
.

Then by using initial condition and partial fraction we get

G∗
α,λ{x(t)} =

3λ3uα+3

ln2(1 + λ) [ln(1 + λ) + λu]
+

λuα+1

ln(1 + λ) + λu

= − 3λuα+1

ln(1 + λ)
+

3λ2uα+2

ln2(1 + λ)
+

4λuα+1

ln(1 + λ) + λu
.

Taking the inverse modified Laplace-type transform of the last equation and using the
Theorems 2 and 3 leads to the solution x(t) = −3 + 3t+ 4e−t.

Now, we use the modified Laplace-type transform for solving a Volterra integral equa-
tion of the second kind.

Example 6. Consider the integral equation

x(t) = t2 +

∫ t

0
x(v) sin(t− v) dv. (29)

By the equation (26) we write the equation (29) as

x(t) = t2 + (x ∗ sin)(t).

Operating the modified Laplace-type transform on both sides to the last equation and
using the Convolution Theorem 15 we have

G∗
α,λ{x(t)} = G∗

α,λ{t2}+G∗
α,λ{(x ∗ sin)(t)} = G∗

α,λ{t2}+
1

uα
G∗

α,λ{x(t)}G∗
α,λ{sin t}.

Then,

G∗
α,λ{x(t)}

(
1−

G∗
α,λ{sin t}

uα

)
= G∗

α,λ{t2}.

Now using the Theorems 2 and 4 we get

G∗
α,λ{x(t)} =

2λ3uα+3

ln3(1 + λ)
uα

1

uα − λ2uα+2

ln2(1 + λ) + λ2u2



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 17 of 20

=
2λ3uα+3

ln3(1 + λ)

ln2(1 + λ) + λ2u2

ln2(1 + λ)
=

2λ3uα+3

ln3(1 + λ)

2λ5uα+5

ln5(1 + λ)
.

Lastly, taking the inverse modified Laplace-type transform of the last equation leads to

the solution x(t) = t2 +
1

12
t4.

Table 1: The modified Laplace-type transform of some elementary functions.

f(t) = G∗−1

α,λ {F ∗
α,λ(u)} F ∗

α,λ(u) = G∗
α,λ{f(t)}

1
λuα+1

ln(1 + λ)

t
λ2 uα+2

ln2(1 + λ)

tn, (n = 1, 2, . . .)
n!λn+1uα+n+1

lnn+1(1 + λ)

eat
λuα+1

ln(1 + λ)− aλu

sin at
aλ2uα+2

ln2(1 + λ) + a2λ2u2

cos at
λ ln(1 + λ)uα+1

ln2(1 + λ) + a2λ2u2

sinh at
aλ2uα+2

ln2(1 + λ)− a2λ2u2

cosh at
λ ln(1 + λ)uα+1

ln2(1 + λ)− a2λ2u2

eat sin bt
bλ2uα+2

(ln(1 + λ)− aλu)2 + b2λ2u2

eat cos bt
λuα+1 (ln(1 + λ)− aλu)

(ln(1 + λ)− aλu)2 + b2λ2u2



İ. Ege / Eur. J. Pure Appl. Math, 18 (2) (2025), 5986 18 of 20

5. Conclusion

In the presented paper, we have considered the concept of the modified Laplace-type
transform and give relations between some integral transforms in the Laplace class such as
the Laplace-type, Sumudu, and Elzaki transforms. We have proved some properties and
derived the modified Laplace-type transform of power functions, sine, cosine, hyperbolic
sine, hyperbolic cosine, exponential function, and function derivatives. We give some
operational properties such as linearity, translations, and scale preserving. Besides these,
we have also examined the relation between the modified Laplace-type transform and
the modified degenerate Gamma function. We have applied the new transform to solve
some ordinary differential equations and a Volterra integral equation. In the future, the
presented transform can be used in many scopes, such as solving various complicated
problems by developing a mathematical model using some differential equations.

Acknowledgements

The author expresses gratitude to the anonymous reviewers for their valuable feedback
and constructive suggestions, which have enhanced the quality of this manuscript.

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