EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 3, Article Number 6467
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

A Novel of Ω-Proportional Fractional Integrals of a
Function with Respect to Another Function

Jamshed Nasir1, Haitham Qawaqneh2, Hassen Aydi3,4,∗

1 Department of Mathematics and Statistics, Virtual University of Pakistan,
Lahore Campus, 54000, Pakistan
2 Al-Zaytoonah University of Jordan, Amman 11733, Jordan
3 Institut Supérieur d’Informatique et des Technologies de Communication,
Université de Sousse, H. Sousse 4000, Tunisia
4 Department of Mathematics and Applied Mathematics, Sefako Makgatho Health Sciences
University, Ga-Rankuwa, South Africa

Abstract. This paper explores a key topic in fractional calculus, which is the sophisticated idea
of proportional fractional integrals with regard to another function. Our focus is on synchronous,
monotonic, and bounded functions. We investigate the mathematical features and theoretical
underpinnings of these integrals. The paper sheds fresh information on the behavior and uses
of fractional integrals by concentrating on these particular types of functions, underscoring their
potential for modeling intricate systems and processes. The findings provide new approaches for
future study and useful applications, expanding our grasp of fractional calculus.

2020 Mathematics Subject Classifications: 26D15, 26D51, 26D07, 26D10

Key Words and Phrases: Proportional fractional integral, Ω-proportional fractional integral of
another function, Synchronous functions, Monotone function

1. Introduction

Integral inequalities are fundamental tools in mathematical analysis, as they provide
valuable insights into the behavior of a function’s integral—especially when exact eval-
uation is difficult or impossible. Common examples include Holder’s and Minkowski’s
inequalities, both of which are closely related to Lp-spaces and norms. These inequalities
play a key role in the study of function sequences and the stability of solutions to differ-
ential equations across various fields. By establishing upper and lower bounds, integral
inequalities are also vital in solving optimization problems, see ([1]-[8]). The study of
differential equations, functional analysis, and probability theory all depend on integral

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i3.6467

Email addresses: jamshed@vu.edu.pk (J. Nasir),
h.alqawaqneh@zuj.edu.jo (H. Qawaqneh), hassen.aydi@isima.rnu.tn (H. Aydi)

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



J. Nasir, H. Qawaqneh, H. Aydi / Eur. J. Pure Appl. Math, 18 (3) (2025), 6467 2 of 16

inequalities, which are basic tools in mathematical analysis that provide integrals bound-
aries. These inequalities, which frequently involve requirements on monotonicity, convex-
ity, or other functional features, establish links between integrals of functions. Hölder’s,
Minkowski’s, and Gronwall’s inequalities are classical examples that are essential for es-
timating solutions and demonstrating the existence and uniqueness of a variety of math-
ematical problems. In addition to helping with theoretical research, integral inequalities
have many uses in fields like economics, engineering, and physics where integral expressions
naturally occur in modeling and analysis. For more details, see ([9]-[17]).

Differentiation and integration are extended to non-integer (fractional) orders in frac-
tional calculus, a generalization of classical calculus. A more flexible and precise modeling
of complex systems with memory and hereditary qualities is made possible by fractional
calculus, which permits operations of arbitrary order in contrast to classical calculus,
which works with integer-order derivatives and integrals. This discipline has received a
lot of interest lately because of its applicability in a number of fields, including biological
systems, control theory, viscoelasticity, anomalous diffusion, and signal processing. The
Riemann–Liouville, Caputo, and Grunwald–Letnikov derivatives are among the concepts
that form the mathematical basis of fractional calculus, and each is appropriate for a
particular kind of issue. With the use of these instruments, fractional calculus offers a
strong foundation for explaining dynamic phenomena that traditional models are unable
to effectively represent. Mathematicians including Leibniz, Liouville, Riemann, and oth-
ers investigated the idea of extending fractional calculus. The fractional derivative, which
has multiple definitions (Riemann-Liouville, Caputo), is appropriate for a certain set of
features and applications see ([18]-[21]).

Definition 1. [22] Consider f ∈ L[a, b]. The left-right-sided Riemann-Liouville (R–L)
fractional integrals of order ξ > 0 are defined by

aJ
ξf(τ) =

1

Γ(ξ)

∫ τ

a
(τ − µ)ξ−1f(µ)dµ, a < τ (1)

and

Jξbf(τ) =
1

Γ(ξ)

∫ b

τ
(µ− τ)ξ−1f(µ)dµ, τ < b, (2)

where the Gamma function is defined as Γ(ξ) =
∫∞
0 e−uuξ−1du.

This integral is motivated by the reputed and well known Cauchy formula as follows:∫ x

a
dτ1

∫ τ1

a
dτ2...

∫ τn−1

a
f (taun) dτn =

1

Γ (n)

∫ x

a
(−τ)n−1 f (τ) dτ. (3)

Definition 2. [23, 24] Suppose (a, b) is a finite interval of real line ℜ and ℜ(ξ) > 0.
Also that suppose Ω(x) is an increasing and positive monotone function on (a, b), having a
continuous derivative Ω′(x) on (a, b). The left-right sided fractional integrals of a function
f with respect to another function Ω on [a, b] are defined by

(Jξ
a+,Ω

f)(τ) =
1

Γ(ξ)

∫ τ

a
(Ω(τ)− Ω(µ))ξ−1Ω′(µ)f(µ)dµ, a < τ (4)



J. Nasir, H. Qawaqneh, H. Aydi / Eur. J. Pure Appl. Math, 18 (3) (2025), 6467 3 of 16

and

(Jξ
b−,Ω

f)(τ) =
1

Γ(ξ)

∫ b

τ
(Ω(µ)− Ω(τ))ξ−1Ω′(µ)f(µ)dµ, τ < b. (5)

From (4) and (5),

(Jξ
a+,Ω

f)(τ) = (Jξ
b−,Ω

f)(τ) = 0 (6)

If we choose Ω(x) = x in the integral formulas (4) and (5), we have

Jξ
a+,Ω

= Jξ
a+

and Jξ
b−,Ω

= Jξ
b− . (7)

If a = 0 in (4), we can write

(Jξ
0+,Ω

f)(τ) =
1

Γ(ξ)

∫ τ

0
(Ω(τ)− Ω(µ))ξ−1Ω′(µ)f(µ)dµ, 0 < τ (8)

(Jξ
0+,Ω

f)(τ) = f(τ).

For the convenience of establishing the results, we give the semi-group property:

Jξ
a+,Ω

Jβ
a+,Ω

f(τ) = Jξ+β
a+,Ω

f(τ), ξ ≥ 0, β ≥ 0,

which gives the commutative property holding as

Jξ
a+,Ω

Jβ
a+,Ω

f(τ) = Jβ
a+,Ω

Jξ
a+,Ω

f(τ).

Definition 3. (Modified conformable derivatives) For γ ∈ [0, 1], let the functions x0, x1 :
[0, 1]×ℜ → [0,+∞) be continuous such that for all t ∈ ℜ, we have

lim
ϱ→0+

x1 (γ, κ) = 1, lim
γ→0+

x0 (γ, κ) = 0, lim
γ→1−

x1 (γ, κ) = 0, lim
γ→1−

x0 (γ, κ) = 1,

and
x1 (γ, κ) ̸= 0, γ ∈ [0, 1], x0 (γ, κ) ̸= 0, γ ∈ (0, 1].

Then the modified conformable differential operator of order γ is defined by

Dγf(κ) = x1 (γ, κ) f(κ) + x0 (γ, κ) f
′(κ). (9)

The derivative given in (9) is said to be proportional derivative. For more details, see the
literature [25]-[26].



J. Nasir, H. Qawaqneh, H. Aydi / Eur. J. Pure Appl. Math, 18 (3) (2025), 6467 4 of 16

Definition 4. [27] For γ > 0 and ξ ∈ C, Re(ξ) > 0, the left and right proportional
fractional integrals of f are respectively defined as

(
aI

ξ,γf
)
(κ) =

1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(κ−τ)
(κ− τ)ξ−1 f (τ) dτ (10)

and (
Iξ,γb f

)
(κ) =

1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(τ−κ)
(τ − κ)ξ−1 f (τ) dτ. (11)

Remark 1. If we choose γ = 1 in the integral formulas (10) and (11), we find (1) and
(2).

The fractional proportional derivative of a function with respect to another function
is as follows:

Definition 5. [28] For γ ∈ (0, 1], ξ ∈ C, such that for all κ ∈ ℜ, ℜ(ξ) > 0,Ω ∈
C[a, b] where Ω′ > 0, we define left and right fractional integrals of f with respect to
Ω by

(
aI

ξ,γ,Ωf
)
(κ) =

1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(µ))
(Ω (κ)− Ω (µ))ξ−1Ω′(µ)f (µ) dµ (12)

and (
Iξ,γ,Ωb f

)
(κ) =

1

γξΓ (ξ)

∫ b

κ
e

γ−1
γ

(Ω(µ)−Ω(κ))
(Ω (µ)− Ω (κ))ξ−1Ω′(µ)f (µ) dµ. (13)

Remark 2. If we choose Ω(y) = y in the integral formulas (12) and (13), we find (10)
and (11).

Remark 3. If we choose Ω(y) = y and γ = 1 in the integral formulas (12) and (13), we
find (1) and (2).

Remark 4. If we choose γ = 1 in the integral formulas (12) and (13), we find (4) and
(5).

In fractional calculus, the proportional fractional integrals with respect to another
function are an advanced topic. With a specific emphasis on synchronous, monotonic
and bounded functions, it entails integrating a function using a fractional order that
is proportionate to another function. In contrast to monotonic functions, which either
continuously rise or decrease, synchronous functions change jointly in a predictable way.



J. Nasir, H. Qawaqneh, H. Aydi / Eur. J. Pure Appl. Math, 18 (3) (2025), 6467 5 of 16

2. Main Results

The development of new integral inequalities involving the Ω-proportional fractional
integral of a function with respect to another function marks a significant advancement
in the theory of fractional calculus and its applications. These inequalities are formulated
within a generalized integral framework where the integration process is governed by a
proportionality function Ω, and the integration is carried out with respect to another
function rather than the independent variable. Such an approach allows for a more flexible
and context-sensitive analysis of functions, especially those exhibiting memory effects,
scaling behavior, or singularities.

In this section, we prove some Ω−proportional fractional integrals of a function with
respect to another function by synchronous and monotonic functions on [0,+∞).

Theorem 1. Suppose that f and g are two synchronous functions on [0,+∞), then for
κ > a, ξ ∈ C, ℜ(ξ) > 0, γ ∈ (0, 1], the following Ω−proportional holds:

[
(
aI

ξ,γ,Ωfg
)
(κ)] . [aI

ξ,γ,Ω(1)] ≥ [
(
aI

ξ,γ,Ωf
)
(κ)] . [

(
aI

ξ,γ,Ωg
)
(κ)]. (14)

Proof. If f and g are synchronous functions, we have

[f (τ)− f (ϱ)][g (τ)− g (ϱ)] ≥ 0. (15)

From (15), it can be written as

f (τ) g (τ) + f (ϱ) g (ϱ) ≥ f (τ) g (ϱ) + f (ϱ) g (τ) . (16)

Multiplying both sides of (16) with 1
γξΓ(ξ)

e
γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ), τ ∈

(a, κ) with respect to τ, we obtain

f (τ) g (τ)
1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ)+

f (ϱ) g (ϱ)
1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ)

≥ f (τ) g (ϱ)
1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ)

+ f (ϱ) g (τ)
1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) . (17)

Integrating the inequality (17) at (a, κ) with respect to τ, we have

1

γξΓ (ξ)

∫ t

a
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1 f (τ) g (τ) Ω′ (τ) dτ+



J. Nasir, H. Qawaqneh, H. Aydi / Eur. J. Pure Appl. Math, 18 (3) (2025), 6467 6 of 16

1

γξΓ (ξ)

∫ t

a
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1 f (ϱ) g (ϱ) Ω′ (τ) dτ

≥ 1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1 f (τ) g (ϱ) Ω′ (τ) dτ

+
1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1 f (ϱ) g (τ) Ω′ (τ) dτ.

[
(
aI

ξ,γ,Ωfg
)
(κ)] + f (ϱ) g (ϱ) [aI

ξ,γ,Ω(1)] ≥ g (ϱ) [
(
aI

ξ,γ,Ωf
)
(κ)] + f (ϱ) [

(
aI

ξ,γ,Ωg
)
(κ)].

(18)

Multiplying both sides of (18) with 1
γξΓ(ξ)

e
γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))ξ−1Ω′ (ϱ), ϱ ∈

(a, κ) with respect to ϱ, we obtain

[
(
aI

ξ,γ,Ωfg
)
(κ)]

1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))ξ−1Ω′ (ϱ)+

f (ϱ) g (ϱ) [aI
ξ,γ,Ω(1)]

1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))ξ−1Ω′ (ϱ)

≥ g (ϱ) [
(
aI

ξ,γ,Ωf
)
(κ)]

1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))ξ−1Ω′ (ϱ)

+ f (ϱ) [
(
aI

ξ,γ,Ωg
)
(κ)]

1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))ξ−1Ω′ (ϱ) . (19)

Integrating inequality (19) at (a, κ) with respect to ϱ, we have

[
(
aI

ξ,γ,Ωfg
)
(κ)]

1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))ξ−1Ω′ (ϱ) dϱ +

[aI
ξ,γ,Ω(1)]

1

γξΓ (ξ)

∫ t

a
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))ξ−1 f (ϱ) g (ϱ) Ω′ (ϱ) dϱ

≥ [
(
aI

ξ,γ,Ωf
)
(κ)]

1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))ξ−1 g (ϱ) Ω′ (ϱ) dϱ

+ [
(
aI

ξ,γ,Ωg
)
(κ)]

1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))ξ−1 f (ϱ) Ω′ (ϱ) dϱ.

Therefore, the inequality can be written as

[
(
aI

ξ,γ,Ωfg
)
(κ)] . [aI

ξ,γ,Ω(1)] ≥ [
(
aI

ξ,γ,Ωf
)
(κ)] . [

(
aI

ξ,γ,Ωg
)
(κ)].

This completes the proof.



J. Nasir, H. Qawaqneh, H. Aydi / Eur. J. Pure Appl. Math, 18 (3) (2025), 6467 7 of 16

Theorem 2. Suppose that f and g are two synchronous functions on [0,+∞), then for
κ > a, ξ ∈ C, ξ > 0, β > 0, γ ∈ (0, 1], the following Ω−proportional holds:

[
(
aI

ξ,γ,Ωfg
)
(κ)] . [aI

β,γ,Ω(1)] + [
(
aI

β,γ,Ωfg
)
(κ)][aI

ξ,γ,Ω(1)]

≥ [
(
aI

ξ,γ,Ωf
)
(κ)] .[

(
aI

β,γ,Ωg
)
(κ) + [

(
aI

β,γ,Ωf
)
(κ)].[

(
aI

ξ,γ,Ωg
)
(κ)]. (20)

Proof. Let f and g be synchronous functions on [0,+∞). For all τ, ϱ ≥ 0, multiplying

both sides of (18) with 1
γβΓ(β)

e
γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1Ω′ (ϱ), ϱ ∈ (a, κ) with re-

spect to ϱ, we obtain

[
(
aI

ξ,γ,Ωfg
)
(κ)]

1

γβΓ (β)
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1Ω′ (ϱ)

+ f (ϱ) g (ϱ) [aI
ξ,γ,Ω(1)].

1

γβΓ (β)
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1Ω′ (ϱ)

≥ g (ϱ) [
(
aI

ξ,γ,Ωf
)
(κ)]

1

γβΓ (β)
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1Ω′ (ϱ)

+ f (ϱ) [
(
aI

ξ,γ,Ωg
)
(κ)].

1

γβΓ (β)
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1Ω′ (ϱ) . (21)

Integrating the inequality (21) at (a, κ) with respect to ϱ, then we have

[
(
aI

ξ,γ,Ωfg
)
(κ)]

1

γβΓ (β)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1Ω′ (ϱ) dϱ

+ [aI
ξ,γ,Ω(1)].

1

γβΓ (β)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1 f (ϱ) g (ϱ) Ω′ (ϱ) dϱ

≥ [
(
aI

ξ,γ,Ωf
)
(κ)]

1

γβΓ (β)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1 g (ϱ) Ω′ (ϱ) dϱ

+ [
(
aI

ξ,γ,Ωg
)
(κ)].

1

γβΓ (β)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1 f (ϱ) Ω′ (ϱ) dϱ. (22)

Then we have

[
(
aI

ξ,γ,Ωfg
)
(κ)] . [aI

β,γ,Ω(1)] + [
(
aI

β,γ,Ωfg
)
(κ)][aI

ξ,γ,Ω(1)]

≥ [
(
aI

ξ,γ,Ωf
)
(κ)] .[

(
aI

β,γ,Ωg
)
(κ) + [

(
aI

β,γ,Ωf
)
(κ)].[

(
aI

ξ,γ,Ωg
)
(κ)].

This completes the proof.

Remark 5. It is obvious that if we let ξ = β in Theorem 2, it reduces to Theorem 1.

Theorem 3. Suppose that f, g and θ are three monotone functions defined on [0,+∞),
satisfying the following inequality

[f (τ)− f (ϱ)][g (τ)− g (ϱ)][θ (τ)− θ (ϱ)] ≥ 0,



J. Nasir, H. Qawaqneh, H. Aydi / Eur. J. Pure Appl. Math, 18 (3) (2025), 6467 8 of 16

then for all τ, ϱ ∈ [a, κ] κ > a, ξ > 0, β > 0, γ ∈ (0, 1], the Ω−proportional holds:

[
(
aI

ξ,γ,Ωfgθ
)
(κ)] [aI

β,γ,Ω(1)] − [aI
ξ,γ,Ω(1)][

(
aI

β,γ,Ωfgθ
)
(κ)]

≥ [
(
aI

ξ,γ,Ωfθ
)
(κ)] . [

(
aI

β,γ,Ωg
)
(κ)] + [

(
aI

ξ,γ,Ωgθ
)
(κ)] . [

(
aI

β,γ,Ωf
)
(κ)]

− [
(
aI

ξ,γ,Ωθ
)
(κ)] . [

(
aI

β,γ,Ωfg
)
(κ)] + [

(
aI

ξ,γ,Ωfg
)
(κ)] . [

(
aI

β,γ,Ωθ
)
(κ)]

+ [
(
aI

ξ,γ,Ωf
)
(κ)] . [

(
aI

β,γ,Ωgθ
)
(κ)]− [

(
aI

ξ,γ,Ωg
)
(κ)] . [

(
aI

β,γ,Ωfθ
)
(κ)]. (23)

Proof. Since f, g and θ are three monotonic functions defined on [0,+∞), then for all
τ, ϱ ≥ 0, we have

[f (τ)− f (ϱ)][g (τ)− g (ϱ)][θ (τ)− θ (ϱ)] ≥ 0. (24)

From (24), it can be written as

f (τ) g (τ) θ (τ)− f (ϱ) g (ϱ) θ (ϱ)− f (τ) g (ϱ) θ (τ)− f (ϱ) g (τ) θ (τ)

+ f (ϱ) g (ϱ) θ (τ)− f (τ) g (τ) θ (ϱ)− f (τ) g (ϱ) θ (ϱ) + f (ϱ) g (τ) θ (ϱ) ≥ 0. (25)

Multiplying both sides of (25) with 1
γξΓ(ξ)

e
γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ), τ ∈

(a, κ) with respect to τ, we obtain

1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (τ) g (τ) θ (τ)

− 1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (ϱ) g (ϱ) θ (ϱ)

− 1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (τ) g (ϱ) θ (τ)

− 1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (ϱ) g (τ) θ (τ)

+
1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (ϱ) g (ϱ) θ (τ)

− 1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (τ) g (τ) θ (ϱ)

− 1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (τ) g (ϱ) θ (ϱ)

+
1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (ϱ) g (τ) θ (ϱ) ≥ 0. (26)

Integrating the inequality (26) at (a, κ) with respect to τ, we have

1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (τ) g (τ) θ (τ) dτ



J. Nasir, H. Qawaqneh, H. Aydi / Eur. J. Pure Appl. Math, 18 (3) (2025), 6467 9 of 16

− 1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (ϱ) g (ϱ) θ (ϱ) dτ

− 1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (τ) g (ϱ) θ (τ) dτ

− 1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (ϱ) g (τ) θ (τ) dτ

+
1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (ϱ) g (ϱ) θ (τ) dτ

− 1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (τ) g (τ) θ (ϱ) dτ

− 1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (τ) g (ϱ) θ (ϱ) dτ

+
1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (ϱ) g (τ) θ (ϱ) dτ ≥ 0. (27)

That is,

[
(
aI

ξ,γ,Ωfgθ
)
(κ)]− f (ϱ) g (ϱ) θ (ϱ) .[aI

ξ,γ,Ω(1)] ≥ g (ϱ) [
(
aI

ξ,γ,Ωfθ
)
(κ)]

+ f (ϱ) [
(
aI

ξ,γ,Ωgθ
)
(κ)] − f (ϱ) g (ϱ) [

(
aI

ξ,γ,Ωθ
)
(κ)] + θ (ϱ) [

(
aI

ξ,γ,Ωfg
)
(κ)]

+ g (ϱ) θ (ϱ) [
(
aI

ξ,γ,Ωf
)
(κ)] − f (ϱ) θ (ϱ) [

(
aI

ξ,γ,Ωg
)
(κ)]. (28)

Multiplying both sides of (28) with 1
γβΓ(β)

e
γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1Ω′ (ϱ), ϱ ∈

(a, κ) with respect to ϱ, we obtain

[
(
aI

ξ,γ,Ωfgθ
)
(κ)]

∫ κ

a

1

γβΓ (β)
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1Ω′ (ϱ)

− [aI
ξ,γ,Ω(1)]

∫ κ

a

1

γβΓ (β)
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1Ω′ (ϱ) f (ϱ) g (ϱ) θ (ϱ) dϱ

≥ [
(
aI

ξ,γ,Ωfθ
)
(κ)]

∫ κ

a

1

γβΓ (β)
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1Ω′ (ϱ) g (ϱ) dϱ

+ [
(
aI

ξ,γ,Ωgθ
)
(κ)]

∫ κ

a

1

γβΓ (β)
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1Ω′ (ϱ) f (ϱ) dϱ

− [
(
aI

ξ,γ,Ωθ
)
(κ)]

∫ κ

a

1

γβΓ (β)
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1Ω′ (ϱ) f (ϱ) g (ϱ) dϱ

+ [
(
aI

ξ,γ,Ωfg
)
(κ)]

∫ κ

a

1

γβΓ (β)
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1Ω′ (ϱ) θ (ϱ) dϱ

+ [
(
aI

ξ,γ,Ωf
)
(κ)]

∫ κ

a

1

γβΓ (β)
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1Ω′ (ϱ) g (ϱ) θ (ϱ) dϱ

− [
(
aI

ξ,γ,Ωg
)
(κ)]

∫ κ

a

1

γβΓ (β)
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))β−1Ω′ (ϱ) f (ϱ) θ (ϱ) dϱ. (29)



J. Nasir, H. Qawaqneh, H. Aydi / Eur. J. Pure Appl. Math, 18 (3) (2025), 6467 10 of 16

That is,

[
(
aI

ξ,γ,Ωfgθ
)
(κ)] [aI

β,γ,Ω(1)] − [aI
ξ,γ,Ω(1)][

(
aI

β,γ,Ωfgθ
)
(κ)]

≥ [
(
aI

ξ,γ,Ωfθ
)
(κ)] . [

(
aI

β,γ,Ωg
)
(κ)] + [

(
aI

ξ,γ,Ωgθ
)
(κ)] . [

(
aI

β,γ,Ωf
)
(κ)]

− [
(
aI

ξ,γ,Ωθ
)
(κ)] . [

(
aI

β,γ,Ωfg
)
(κ)] + [

(
aI

ξ,γ,Ωfg
)
(κ)] . [

(
aI

β,γ,Ωθ
)
(κ)]

+ [
(
aI

ξ,γ,Ωf
)
(κ)] . [

(
aI

β,γ,Ωgθ
)
(κ)]− [

(
aI

ξ,γ,Ωg
)
(κ)] . [

(
aI

β,γ,Ωfθ
)
(κ)]. (30)

This completes the proof.

3. Inequalities involving Ω−proportional fractional integrals of a
function with respect to another function by bounded functions

In this section, we prove some Ω−proportional fractional integrals of a function with
respect to another function by bounded functions.

Theorem 4. Suppose that f is an integrable function on [a, b], then for κ > a, ξ ∈ C,
ℜ(ξ) > 0, γ ∈ (0, 1],Φ1,Φ2 ∈ [a, b] and Φ1 ≤ f ≤ Φ2, the following Ω−proportional holds:(

aI
ξ,γ,ΩΦ2

)
(κ)] [

(
aI

ξ,γ,hf
)
(κ)] + [

(
aI

ξ,γ,hf
)
(κ)] [

(
aI

ξ,γ,hΦ1

)
(κ)]

≥ [
(
aI

ξ,γ,hΦ2

)
(κ)][

(
aI

ξ,γ,hΦ1

)
(κ)] + [

(
aI

ξ,γ,hf
)
(κ)]2. (31)

Proof. If τ ,ϱ ∈ (a, κ), then we have

[Φ2 (τ)− f (τ)][f (ϱ)− Φ1 (ϱ)] ≥ 0. (32)

From (32), it can be written as

Φ2 (τ) f (ϱ) + f (τ) Φ1 (ϱ) ≥ Φ2 (τ) Φ1 (ϱ) + f (ϱ) f (τ) . (33)

Multiplying both sides of (16) with 1
γξΓ(ξ)

e
γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ), τ ∈

(a, κ) with respect to τ, we obtain

Φ2 (τ) f (ϱ)
1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ)+

f (τ) Φ1 (ϱ)
1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ)

≥ Φ2 (τ) Φ1 (ϱ))
1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ)



J. Nasir, H. Qawaqneh, H. Aydi / Eur. J. Pure Appl. Math, 18 (3) (2025), 6467 11 of 16

+ f (ϱ) f (τ)
1

γξΓ (ξ)
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) . (34)

Integrating the inequality (34) at (a, κ) with respect to τ, we have

f (ϱ)
1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) Φ2 (τ) dτ+

Φ1 (ϱ)
1

γξΓ (ξ)

∫ t

a
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (τ) dτ

≥ Φ1 (ϱ))
1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) Φ2 (τ) dτ

+ f (ϱ)
1

γξΓ (ξ)

∫ t

a
e

γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ) f (τ) dτ.

That is,

f(ϱ)[
(
aI

ξ,γ,ΩΦ2

)
(κ)] + Φ1 (ϱ) [

(
aI

ξ,γ,Ωf
)
(κ)]

≥ Φ1 (ϱ) [
(
aI

ξ,γ,ΩΦ2

)
(κ)] + f (ϱ) [

(
aI

ξ,γ,Ωf
)
(κ)]. (35)

Multiplying both sides of (35) with 1
γξΓ(ξ)

e
γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))ξ−1Ω′ (ϱ), ϱ ∈

(a, κ) with respect to ϱ and integrating inequality at (a, κ) with respect to ϱ, then we
obtain

[
(
aI

ξ,γ,ΩΦ2

)
(κ)]

1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))ξ−1Ω′ (ϱ) f(ϱ)dϱ +

[
(
aI

ξ,γ,Ωf
)
(κ)]

1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))ξ−1Ω′ (ϱ) Φ1 (ϱ) dϱ

≥ [
(
aI

ξ,γ,ΩΦ2

)
(κ)]

1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))ξ−1Ω′ (ϱ) Φ1 (ϱ) dϱ +

[
(
aI

ξ,γ,Ωf
)
(κ)]

1

γξΓ (ξ)

∫ κ

a
e

γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))ξ−1Ω′ (ϱ) f (ϱ) dϱ. (36)

Thus, we get the inequality (31).

Theorem 5. Suppose that f is an integrable function on [a, b], then for κ > a, ξ ∈ C,
ℜ(ξ) > 0, γ ∈ (0, 1],Φ1,Φ2 ∈ [a, b] and Φ1 ≤ f ≤ Φ2, λ1, λ2 > 0 and 1

λ1
+ 1

λ2
= 1, the

following Ω−proportional holds:

1

λ1

[(
aI

ξ,γ,Ω (Φ2 − f)λ1

)
(κ)

] [(
aI

ξ,γ,h (1)
)]

+
1

λ2

[(
aI

ξ,γ,h (1)
)][(

aI
ξ,γ,Ω (f− Φ1)

λ2

)
(κ)

]



J. Nasir, H. Qawaqneh, H. Aydi / Eur. J. Pure Appl. Math, 18 (3) (2025), 6467 12 of 16

+

[(
aI

ξ,γ,hΦ2

)
(κ)

][(
aI

ξ,γ,hΦ1

)
(κ)

]
+

[(
aI

ξ,γ,hf
)
(κ)

]2
≥[(

aI
ξ,γ,hΦ2

)
(κ)

][(
aI

ξ,γ,hf
)
(κ)

]
+

[(
aI

ξ,γ,hf
)
(κ)

][(
aI

ξ,γ,hΦ1

)
(κ)

]
(37)

Proof. With the help of well-known Young’s inequality (see [16]), one has

1

λ1
uλ1 +

1

λ2
vλ2 ≥ uv; where u, v ≥ 0. (38)

By setting the requirement u = Φ2 − f and v = f− Φ1, we have

1

λ1
[Φ2 − f]λ1 +

1

λ2
[f− Φ1]

λ2 ≥ [Φ2 − f][f− Φ1]; where u, v ≥ 0. (39)

Multiplying both sides of (39) with 1
γξΓ(ξ)

e
γ−1
γ

(Ω(κ)−Ω(τ))
(Ω (κ)− Ω (τ))ξ−1Ω′ (τ), τ ∈

(a, κ) with respect to τ, and integrating with respect to τ ∈ (a, κ), we obtain

1

λ1

[(
aI

ξ,γ,Ω (Φ2 − f)λ1

)
(κ)

]
+

1

λ2

[(
aI

ξ,γ,h (1)
)][(

(f (ϱ)− Φ1 (ϱ))
λ2

)
(κ)

]
+Φ1 (ϱ) [

(
aI

ξ,γ,ΩΦ2

)
(κ)] + f (ϱ) [

(
aI

ξ,γ,Ωf
)
(κ)]

≥ f (ϱ) [
(
aI

ξ,γ,ΩΦ2

)
(κ)] + Φ1 (ϱ) [

(
aI

ξ,γ,Ωf
)
(κ)] (40)

Multiplying both sides of (40) with 1
γξΓ(ξ)

e
γ−1
γ

(Ω(κ)−Ω(ϱ))
(Ω (κ)− Ω (ϱ))ξ−1Ω′ (ϱ), ϱ ∈

(a, κ) with respect to ϱ and integrating inequality at (a, κ) with respect to ϱ, then af-
ter getting the simplification, we get

1

λ1

[(
aI

ξ,γ,Ω (Φ2 − f)λ1

)
(κ)

] [(
aI

ξ,γ,h (1)
)]

+
1

λ2

[(
aI

ξ,γ,h (1)
)][(

aI
ξ,γ,Ω (f− Φ1)

λ2

)
(κ)

]
+

[(
aI

ξ,γ,hΦ2

)
(κ)

][(
aI

ξ,γ,hΦ1

)
(κ)

]
+

[(
aI

ξ,γ,hf
)
(κ)

]2
≥[(

aI
ξ,γ,hΦ2

)
(κ)

][(
aI

ξ,γ,hf
)
(κ)

]
+

[(
aI

ξ,γ,hf
)
(κ)

][(
aI

ξ,γ,hΦ1

)
(κ)

]
.

This completes the proof.



J. Nasir, H. Qawaqneh, H. Aydi / Eur. J. Pure Appl. Math, 18 (3) (2025), 6467 13 of 16

4. Special cases

Here, we aim at present some new generalizations via proportional fractional integrals
with respect to another function, which are the new estimates of the main consequences.

Corollary 1. Under the assumptions of Theorem 1, the following inequality holds:

[
(
aI

ξ,γfg
)
(κ)] . [aI

ξ,γ(1)] ≥ [
(
aI

ξ,γf
)
(κ)] . [

(
aI

ξ,γg
)
(κ)].

Proof. Letting Ω(x) = x in Theorem 1 yields the proof of Corollary 1.

Corollary 2. Under the assumptions of Theorem 2, the following inequality holds:

[
(
aI

ξ,γfg
)
(κ)] . [aI

β,γ(1)] + [
(
aI

β,γfg
)
(κ)][aI

ξ,γ(1)]

≥ [
(
aI

ξ,γf
)
(κ)] .[

(
aI

β,γg
)
(κ) + [

(
aI

β,γf
)
(κ)].[

(
aI

ξ,γg
)
(κ)].

Proof. Letting Ω(x) = x in Theorem 2 yields the proof of Corollary 2.

Corollary 3. Under the assumptions of Theorem 2, the following inequality holds:

[
(
aI

γ,Ωfg
)
(κ)] . [aI

γ,Ω(1)] + [
(
aI

γ,Ωfg
)
(κ)][aI

γ,Ω(1)]

≥ [
(
aI

γ,Ωf
)
(κ)] .[

(
aI

γ,Ωg
)
(κ) + [

(
aI

γ,Ωf
)
(κ)].[

(
aI

γ,Ωg
)
(κ)].

Proof. Letting ξ = 1 = β in Theorem 2 yields the proof of Corollary 3.

5. Conclusion

The Ω-proportional fractional integral of a function with respect to another function,
in conclusion, provides a strong and cohesive framework that greatly expands the current
and classical fractional integral operators. This innovative method expands the versa-
tility and usefulness of fractional calculus in simulating intricate, nonlocal, and memory-
dependent events by introducing the proportionality function Ω and permitting integration
with respect to another function. In addition to generalizing well-known theorems, the
recently developed integral inequalities within this framework provide new opportunities
for theoretical investigation and real-world applications. These contributions provide the
foundation for future study and advancement in the field by offering strong mathematical
tools for the analysis of fractional differential equations and the creation of more precise
models in a variety of scientific and technical fields.

Authors’ Contributions

All authors contribute equally in this paper.



J. Nasir, H. Qawaqneh, H. Aydi / Eur. J. Pure Appl. Math, 18 (3) (2025), 6467 14 of 16

Conflict of interest

The authors declare that they have no conflict of interest.

Acknowledgements

The authors acknowledge the financial support from Al-Zaytoonah University of Jor-
dan, Amman 11733, Jordan.

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