# Fejér–Pachpatte–Mercer-Type Inequalities for Harmonically Convex Functions Involving Exponential Function in Kernel

Authors: Saad Ihsan Butt; Saba Yousaf; Khuram Ali Khan; Rostin Matendo Mabela; Abdullah M. Alsharif
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

In the present study, fractional variants of Hermite–Hadamard, Hermite–Hadamard–Fejér, and Pachpatte inequalities are studied by employing Mercer concept. Firstly, new Hermite–Hadamard–Mercer-type inequalities are presented for harmonically convex functions involving fractional integral operators with exponential kernel. Then, weighted Hadamard–Fejér–Mercer-type inequalities involving exponential function as kernel are proved. Finally, Pachpatte–Mercer-type inequalities for products of harmonically convex functions via fractional integral operators with exponential kernel are constructed.


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## Body

## 1. Introduction

Integral inequalities have been widely used in various sciences, including mathematical sciences, applied sciences, differential equations, and functional analysis. In the last two decades, these inequalities have gained attention from researchers. In most mathematical analysis areas, many types of integral inequalities are used. They are very important in approximation theory and numerical analysis, which estimate the error’s approximation. Integral inequalities are useful tools in the study of different classes of differential equations and integral equations. They are today employed not only in mathematics but also in physics, computer science, and biology.Convexity has several uses in business, medicine, industry, and art that have a significant impact on our daily lives. One of the most important applications of the convex function is the formulation of inequalities. Many equalities and inequalities have been defined for convex functions, but Jensen’s inequality and the Hermite–Hadamard integral inequality are the most notable results [1–3]. The following notion of convex function plays a significant role in optimization theory and in other fields of sciences.Definition 1.
A functionΩ:I⊂ℝ⟶ℝ is called a convex function on I, if(1)Ωηa1+1−ηa2≤ηΩa1+1−ηΩa2,for all a1,a2∈I and η∈0,1, holds.
The following classical Jensen’s inequality is defined as generalization of convex functions.Theorem 1 (see [1]).
Suppose thatΩ is a convex function on a1,a2; then,(2)Ω∑ℓ=1nηℓϰℓ≤∑ℓ=1nηℓΩϰℓ,for all ϰℓ∈a1,a2 and ηℓ∈0,1, where ℓ=1,2,…,n with ∑ℓ=1nηℓ=1.Inequality (2) is a key to extract applications in information theory. It is very useful in computing optimal bounds for joint and conditional entropies and mutual information (for instance, see [4–6] and the references therein).Hermite–Hadamard inequalities may be considered as a refinement of the concept of convexity, and it is simply inferred from Jensen’s inequality as follows.Theorem 2 (see [2]).
IfΩ is a convex function on the interval a1,a2 with a1<a2, then(3)Ωa1+a22≤1a2−a1∫a1a2Ωϰdϰ≤Ωa1+Ωa22.It has been applied in several branches such as finance, engineering, and science (see [2]). In recent years, Hermite–Hadamard inequalities for convex functions have gotten a lot of attention, and as a result, there have been a lot of refinements and generalizations.In 2003, Mercer presented a variant of Jensen’s inequality as follows.Theorem 3 (see [7]).
Suppose thatΩ is a convex function on the interval θ1,θ2; then,(4)Ωθ1+θ2−∑ℓ=1nηℓϰℓ≤Ωθ1+Ωθ2−∑ℓ=1nηℓΩϰℓ,for all ϰℓ∈θ1,θ2 and ηℓ∈0,1, where ℓ=1,2,…,n with ∑ℓ=1nηℓ=1.Over the years, the Jensen–Mercer inequality became a topic of foremost interest for many scholars as they have investigated and studied in various ways including bringing it to a higher dimension and acquiring it for convex operators along with its several refinements, operator variants for superquadratic functions, improvements, and many generalizations with applications in information theory (see [8–10]).In [11], Íşcan gave the definition of harmonic convexity as follows.Definition 2.
A functionΩ:I⊂ℝ\0⟶ℝ is said to be harmonically convex on I, if(5)Ωa1a2ηa1+1−ηa2≤ηΩa2+1−ηΩa1,holds for all a1,a2∈I and η∈0,1.
In [11], Íşcan for the first time introduced the Hermite–Hadamard inequality for harmonically convex function along with the following identity.Theorem 4 (see [11]).
Suppose thatΩ:I⊆0,∞⟶ℝ is a harmonically convex function and a1,a2∈I with a1<a2. If Ω∈La1,a2, then(6)Ω2a1a2a1+a2≤a1a2a2−a1∫a1a2Ωϰϰ2dϰ≤Ωa1+Ωa22.Lemma 1 (see [11]).
IfΩ:I⊆0,∞⟶ℝ is a differentiable function on Io and a1,a2∈I with a1<a2 and Ω′∈La1,a2, then(7)Ωa1+Ωa22−a1a2a2−a1∫a1a2Ωϰϰ2dϰ=a1a2a2−a12∫011−2ηηa2+1−ηa12Ω′a1a2ηa2+1−ηa1dη.The most prominent inequalities connected to the integral mean of a harmonically convex function are the Hermite–Hadamard inequalities or their weighted versions which are called Hermite–Hadamard–Fejér inequalities for harmonically convex functions. In [12], Chen and Wu established the Hermite–Hadamard–Fejér inequality for harmonically convex functions.Theorem 5 (see [12]).
LetΩ:I⊆0,∞⟶ℝ be a harmonically convex function and a1,a2∈I with a1<a2. If Ω∈La1,a2 and ω:a1,a2⊆0,∞⟶ℝ is non-negative, integrable, and harmonically symmetric with respect to 2a1a2/a1+a2, that is, ωϰ=ω1/1/a1+1/a2−1/ϰ, then(8)Ω2a1a2a1+a2∫a1a2ωϰϰ2dϰ≤∫a1a2Ωxωϰϰ2dϰ≤Ωa1+Ωa22∫a1a2ωϰϰ2dϰ.In [13], Chen and Wu obtained two Hermite–Hadamard-type inequalities for products of harmonically convex functions as follows.Theorem 6 (see [13]).
LetΩ,ω:a1,a2⊆0,∞⟶0,∞, a1,a2∈0,∞, be functions such that Ω,ω,Ωω∈La1,a2. If Ω and ω are harmonically convex on a1,a2, then(9)a1a2a2−a1∫a1a2Ωϰωϰϰ2dϰ≤13Ma1,a2+16Na1,a2,(10)2Ω2a1a2a1+a2ω2a1a2a1+a2≤a1a2a2−a1∫a1a2Ωϰωϰϰ2dϰ+16Ma1,a2+13Na1,a2,where Ma1,a2=Ωa1ωa1+Ωa2ωa2 and Na1,a2=Ωa1ωa2+Ωa2ωa1.Fractional calculus is an extension of classical calculus. Fractional calculus is now a well-known technique in engineering science, with wide range of applications in material modeling. A growing number of degree considerations have recently been given to fractional calculus and its numerous applications. Fractional integral/derivative operators are extremely important in the development of fractional calculus. Fractional differential equations and dynamical frameworks were established a few decades ago as key tools for exhibiting a variety of phenomena in various branches of pure and applied sciences. Many physical problems can be modeled using fractional differential equations, including heat equations, wave equations, Poisson equations, and Laplace equations, biological populations, fluid mechanics, thermodynamics, viscoelasticity, vibration, advection-diffusion, groundwater flow with memory, and signal processing [14, 15]. Several studies have shown that fractional operators can accurately explain complex long memory and multiscale phenomena in materials that are difficult to capture using standard mathematical methods including classical differential calculus. The significance of fractional calculus can be more understandable, and several works involving fractional calculus have been done.Several well-known inequalities and related results can be generalized and extended via fractional integral operators (see [16–19] and the references therein).Definition 3 (see [15]).
LetΩ∈La1,a2. The Riemann–Liouville fractional integrals Ja1+αΩ and Ja2−αΩ of order α>0 with a1≥0 are defined by(11)Ja1+αΩϰ=1Γα∫a1ϰϰ−uα−1Ωudu,ϰ>a1,Ja2−αΩϰ=1Γα∫ϰa2u−ϰα−1Ωudu,ϰ<a2,respectively. Here, Γα is the gamma function defined by Γα=∫0∞e−uuα−1du and Ja1+0Ωϰ  =  Ja2−0Ωϰ=Ωϰ.
In [20], Iscan and Wu for the first time introduced Hermite–Hadamard-type inequalities for harmonically convex functions for Riemann–Liouville fractional integral operators along with the following integral identity.Theorem 7 (see [20]).
LetΩ:a1,a2⊆0,∞⟶ℝ be a function with a1<a2 and Ω∈La1,a2. If Ω is a harmonically convex function, then(12)Ω2a1a2a1+a2≤Γα+12a1a2a2−a1αJ1/a1−αΩ∘g1a2+I1/a2+αΩ∘g1a1≤Ωa1+Ωa22,where gu=1/u.Lemma 2 (see [20]).
Suppose thatΩ:a1,a2⊆0,∞⟶ℝ is a differentiable function on Io and Ω′∈La1,a2; then,(13)Ωa1+Ωa22−Γα+12a1a2a2−a1αJ1/a1−αΩ∘g1a2+I1/a2+αΩ∘g1a1=a1a2a2−a12∫01ηα−1−ηαηa1+1−ηa22Ω′a1a2ηa1+1−ηa2dη,where gu=1/u.In [21], İşcan and Kunt represented the Hermite–Hadamard–Fejér-type inequality for harmonically convex functions in Riemann–Liouville fractional integral forms as follows.Theorem 8 (see [21]).
LetΩ:I⊆0,∞⟶ℝ be a harmonically convex function and a1,a2∈I with a1<a2. If Ω∈La1,a2 and ω:a1,a2⊆0,∞⟶ℝ is nonnegative, integrable, and harmonically symmetric with respect to 2a1a2/a1+a2, that is, ωϰ=ω1/1/a1+1/a2−1/ϰ, then(14)Ω2a1a2a1+a2J1/a1−αω∘g1a2+J1/a2+αω∘g1a1≤J1/a1−αΩω∘g1a2+J1/a2+αΩω∘g1a1≤Ωa1+Ωa22J1/a1−αω∘g1a2+J1/a2+αω∘g1a1,where α>0gu=1/u, u∈1/a2,1/a1.The importance of Hadamard-type inequalities is due to their roles in various fields of modern mathematics such as numerical analysis, probability, mathematical analysis, and related fields [2, 22]. Many researchers generalize and extend their studies to Hermite–Hadamard, Hermite–Hadamard–Fejér, and Pachpatte-type inequalities involving fractional integrals for various classes of convex functions (see [20, 21, 23, 24] and the references therein).Recently (from 2020 to 2021), some new kinds of fractional treatment of Hermite–Jensen–Mercer-type inequalities for a variety of fractional integral operators were presented in [25–27]. All these results were investigated for convex functions or s-convex functions, and many applications to special functions like Bessel and q-digamma functions were obtained.Since there a is massive literature about the development of fractional Mercer integral inequalities involving convex functions but still there exist many gaps to be filled for fractional integral inequalities for other classes of convex functions. Therefore, the basic aim of this paper is to present three new Hadamard–Mercer-type inequalities for harmonically convex functions using fractional integral operators with exponential kernel. We also give fractional Mercer integral inequalities for product of two harmonically convex functions. We hope that the new techniques formulated in this paper are more energizing than the accessible one.Ahmad et al. [23] gave the definition of two new fractional integral operators with an exponential kernel.Definition 4.
LetΩ∈La1,a2. The fractional integral operators Ia1αΩϰ and Ia2αΩϰ of order α∈0,1 are, respectively, defined by(15)Ia1αΩϰ=1α∫a1ϰexp−1−ααϰ−uΩudu,ϰ>a1,Ia2αΩϰ=1α∫ϰa2exp−1−ααu−ϰΩudu,ϰ<a2.Remark 1.
Ifα=1, then(16)limα⟶1Ia1αΩϰ=∫a1ϰΩudu,limα⟶1Ia2αΩϰ=∫ϰa2Ωudu.
For the convenience of expression, throughout the paper, we set(17)ρ=1−ααa2−a1a1a2.


## 2. Hermite–Hadamard–Mercer-Type Inequalities for Harmonically Convex Function

Theorem 9.
Suppose thatΩ:θ1,θ2⊆0,∞⟶ℝ is a positive function with 0≤θ1<θ2. If Ω is a harmonically convex function on θ1,θ2 and Ω∈Lθ1,θ2, then(18)Ω11/θ1+1/θ2−a1+a2/2a1a2≤Ωθ1+Ωθ2−1−α21−exp−ρI1/a1αΩ∘g1a2+I1/a2αΩ∘g1a1≤Ωθ1+Ωθ2−Ω2a1a2a1+a2,(19)Ω11/θ1+1/θ2−a1+a2/2a1a2≤1−α21−exp−ρI1/θ1+1/θ2−1/a1αΩ∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αΩ∘g1θ1+1θ2−1a1≤12Ω11/θ1+1/θ2−1/a1+Ω11/θ1+1/θ2−1/a2≤Ωθ1+Ωθ2−Ωa1+Ωa22,for all a1,a2∈θ1,θ2, α>0, gu=1/u, u∈1/θ2,1/θ1 and ρ is defined in (17).Proof.
Using the Jensen–Mercer inequality for harmonically convex function, we have(20)Ω11/θ1+1/θ2−a1¯+a2¯/2a1¯a1¯≤Ωθ1+Ωθ2−Ωa1¯+Ωa2¯2,for all a1¯, a2¯∈θ1,θ2. By changing of the variables a1¯=a1a2/ηa1+1−ηa2, a2¯=a1a2/ηa2+1−ηa1 for all a1, a2∈θ1,θ2 and η∈0,1 in (20), we obtain(21)Ω11/θ1+1/θ2−a1+a2/2a1a2≤Ωθ1+Ωθ2−12Ωa1a2ηa1+1−ηa2+Ωa1a2ηa2+1−ηa1.
Multiplying byexp−ρη on both sides of (21) and then integrating with respect to η over 0,1, we have(22)1−exp−ρρΩ11/θ1+1/θ2−a1+a2/2a1a2≤1−exp−ρρΩa1+Ωa2−12∫01exp−ρηΩa1a2ηa1+1−ηa2+Ωa1a2ηa2+1−ηa1dη=1−exp−ρρΩθ1+Ωθ2−a1a22a2−a1∫1/a21/a1exp−1−αα1a1−uΩ1udu+∫1/a21/a1exp−1−ααu−1a2Ω1udu=1−exp−ρρΩθ1+Ωθ2−αa1a22a2−a1I1/a2αΩ∘g1a1+I1/a1αΩ∘g1a2.
Multiplying byρ/1−exp−ρ on both sides of above equation and putting the value of ρ which is given in (17), we get(23)Ω11/θ1+1/θ2−a1+a2/2a1a2≤Ωθ1+Ωθ2−1−α21−exp−ρI1/a1αΩ∘g1a2+I1/a2αΩ∘g1a1.
Thus, the first inequality of (18) is proved. Now we prove the second inequality in (18); since Ω is a harmonically convex function, then, for η∈0,1, it yields(24)Ω2a1a2a1+a2=Ω2η/a1+1−η/a2+1−η/a1+η/a2≤12Ωa1a2ηa2+1−ηa1+Ωa1a2ηa1+1−ηa2.
Multiplying byexp−ρη on both sides of (24) and then integrating with respect to η over 0,1, we have(25)1−exp−ρρΩ2a1a2a1+a2≤12∫01exp−ρηΩa1a2ηa2+1−ηa1+Ωa1a2ηa1+1−ηa2dη=αa1a22a2−a1I1/a1αΩ∘g1a2+I1/a2αΩ∘g1a1.
Then,(26)−Ω2a1a2a1+a2≥−1−α21−exp−ρI1/a1αΩ∘g1a2+I1/a2αΩ∘g1a1.
AddingΩa1+Ωa2 to both sides of (26), we find the second inequality of (18).
Now, we prove inequality (19). Since Ω is a harmonically convex function, then, we have that for any a1,a2∈θ1,θ2,(27)Ω2a1a2a1+a2=Ω11/2η/a1+1−η/a2+1−η/a1+η/a2≤12Ω1η/a1+1−η/a2+Ω11−η/a1+η/a2≤Ωa1+Ωa22.
Replacinga1 and a2 by 1/1/θ1+1/θ2−1/a1 and 1/1/θ1+1/θ2−1/a2, respectively, in (27), we get(28)Ω11/θ1+1/θ2−a1+a2/2a1a2≤12Ω11/θ1+1/θ2−η/a1+1−η/a2+Ω11/θ1+1/θ2−1−η/a1+η/a2≤12Ω11/θ1+1/θ2−1/a1+Ω11/θ1+1/θ2−1/a2.
Multiplying byexp−ρη on both sides of (28) and then integrating with respect to η over 0,1, we obtain(29)1−exp−ρρΩ11/θ1+1/θ2−a1+a2/2a1a2≤12∫01exp−ρηΩ11/θ1+1/θ2−η/a1+1−η/a2dη+∫01exp−ρηΩ11/θ1+1/θ2−1−η/a1+η/a2dη≤1−exp−ρ2ρΩ11/θ1+1/θ2−1/a1+Ω11/θ1+1/θ2−1/a2.
It is obvious that(30)12∫01exp−ρηΩ11/θ1+1/θ2−η/a1+1−η/a2dη+∫01exp−ρηΩ11/θ1+1/θ2−1−η/a1+η/a2dη=a1a22a2−a1∫1/θ1+1/θ2−1/a11/θ1+1/θ2−1/a2exp−1−αα1θ1+1θ2−1a2−uΩ1udu+∫1/θ1+1/θ2−1/a11/θ1+1/θ2−1/a2exp−1−ααu−1θ1+1θ2−1a1Ω1udu=αa1a22a2−a1I1/θ1+1/θ2−1/a1αΩ∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αΩ∘g1θ1+1θ2−1a1.
Using the Jensen–Mercer inequality for harmonically convex function, we conclude that(31)Ω11/θ1+1/θ2−a1+a2/2a1a2≤1−α21−exp−ρI1/θ1+1/θ2−1/a1αΩ∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αΩ∘g1θ1+1θ2−1a1≤12Ω11/θ1+1/θ1−1/a1+Ω11/θ1+1/θ1−1/a2≤Ωθ1+Ωθ2−Ωa1+Ωa22.
So, inequality (19) is proved.Remark 2.
If we takea1=θ1 and a2=θ2 in Theorem 9, then we have Theorem 2.1 in [24].Remark 3.
Forα⟶1, we have(32)limα⟶11−α21−exp−ρ=a1a22a2−a1.
Under the assumptions of Theorem9 with α=1, one has(33)Ω11/θ1+1/θ2−a1+a2/2a1a2≤Ωθ1+Ωθ2−∫01Ωa1a2ηa1+1−ηa2dη≤Ωθ1+Ωθ2−Ω2a1a2a1+a2,(34)Ω11/θ1+1/θ2−a1+a2/2a1a2≤a1a2a2−a1∫a1a21η2Ω11/θ1+1/θ2−1/ηdη≤Ωθ1+Ωθ2−Ωa1+Ωa22,for all a1, a2∈θ1,θ2. Inequalities (33) and (34) were proved by Baloch et al. in [28, Theorem 3.5] and [29, Theorem 2.1].Remark 4.
Ifα⟶1, a1=θ1, and a2=θ2 in Theorem 9, then we have Hermite–Hadamard inequality (6) for harmonically convex function which was proved by Íşcan in [11].Lemma 3.
LetΩ:θ1,θ2⊆0,∞⟶ℝ be a differentiable function on θ1,θ2 with θ1<θ2. If Ω′∈Lθ1,θ2, then(35)12Ω11/θ1+1/θ2−1/a1+Ω11/θ1+1/θ2−1/a2−1−α21−exp−ρI1θ1+1θ2−1a1αΩ°g1θ1+1θ2−1a2+I1θ1+1θ2−1a2αΩ°g1θ1+1θ2−1a1=a2−a12a1a21−exp−ρ∫01exp−ρ1−η1/θ1+1/θ2−η/a1+1−η/a22Ω′11/θ1+1/θ2−η/a1+1−η/a2dη−∫01exp−ρη1/θ1+1/θ2−η/a1+1−η/a22Ω′11/θ1+1/θ2−η/a1+1−η/a2dη,for all a1,a2∈θ1,θ2, α>0, and gu=1/u, u∈1/θ2,1/θ1.Proof.
LetAη=1/θ1+1/θ2−η/a1+1−η/a2. It suffices to note that(36)12Ω11/θ1+1/θ2−1/a1+Ω11/θ1+1/θ2−1/a2−1−α21−exp−ρI1/θ1+1/θ1−1/a1αΩ∘g1θ1+1θ2−1a2+I1/θ1+1/θ1−1/a2αΩ∘g1θ1+1θ2−1a1=a2−a12a1a21−exp−ρ∫01exp−ρ1−ηAη2Ω′11/θ1+1/θ2−η/a1+1−η/a2dη−∫01exp−ρηAη2Ω′11/θ1+1/θ2−η/a1+1−η/a2dη=I1−I2.
By integrating by part, we have(37)I1=121−exp−ρexp−ρ1−ηΩ1Aη01−ρ∫01exp−ρηΩ1Aηdη=121−exp−ρΩ11/θ1+1/θ2−1/a1−exp−ρΩ11/θ1+1/θ2−1/a2−ρa1a2a2−a1∫1/θ1+1/θ2−1/a11/θ1+1/θ2−1/a2exp−1−ααu−1θ1+1θ2−1a1Ω1udu=121−exp−ρΩ11/θ1+1/θ2−1/a1−exp−ρΩ11/θ1+1/θ2−1/a2−ραa1a2a2−a1I1/θ1+1/θ2−1/a2αΩ∘g1θ1+1θ2−1a1,and similarly we get(38)I2=121−exp−ρexp−ρηΩ1Aη01−−ρ∫01exp−ρηΩ1Aηdη=121−exp−ρexp−ρΩ11/θ1+1/θ1−1/a1−Ω11/θ1+1/θ2−1/a2+ρa1a2a2−a1∫1/θ1+1/θ1−1/a11/θ1+1/θ1−1/a2exp−1−αα1θ1+1θ2−1a2−uΩ1udu=121−exp−ρexp−ρΩ11/θ1+1/θ2−1/a1−Ω11/θ1+1/θ2−1/a2+ραa1a2a2−a1I1/θ1+1/θ1−1/a1αΩ∘g1θ1+1θ2−1a2.
Using (37) and (40) in (36), we get equality (35).Remark 5.
From Lemma2 with α⟶1, a1=θ1, and a2=θ2, we indeed have Lemma 1 which was proved by Íşcan in [11].Theorem 10.
IfΩ:I=θ1,θ2⊆0,∞⟶ℝ is a differentiable function on θ1,θ2 with θ1<θ2 and Ω′∈Lθ1,θ2. If Ω′ is a harmonically convex on θ1,θ2, then(39)|12Ω11/θ1+1/θ2−1/a1+Ω11/θ1+1/θ2−1/a2−1−α21−exp−ρI1/θ1+1/θ2−1/a1αΩ∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αΩ∘g1θ1+1θ2−1a1|≤a2−a12a1a21−exp−ρ∫01/2exp−ρη−exp−ρ1−ηημ12+1−ημ22Ω′11/θ1+1/θ2−1/a1+1−ημ12+ημ22Ω′11/θ1+1/θ2−1/a2dη,where(40)μ1=1θ1+1θ2−ηa1+1−ηa2,(41)μ2=1θ1+1θ2−1−ηa1+ηa2,for all a1,a2∈θ1,θ2, α>0, and gu=1/u, u∈1/θ2,1/θ1.Proof.
SinceΩ′ is a harmonically convex on θ1,θ2, using Lemma 2, we can obtain(42)|12Ω11/θ1+1/θ2−1/a1+Ω11/θ1+1/θ2−1/a2−1−α21−exp−ρI1/θ1+1/θ2−1/a1αΩ∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αΩ∘g1θ1+1θ2−1a1|≤a2−a12a1a21−exp−ρ∫01exp−ρη−exp−ρ1−η1/θ1+1/θ2−η/a1+1−η/a22Ω′11/θ1+1/θ2−η/a1+1−η/a2dη≤a2−a12a1a21−exp−ρ∫01/2exp−ρη−exp−ρ1−η1/θ1+1/θ2−η/a1+1−η/a22ηΩ′11/θ1+1/θ2−1/a1+1−ηΩ′11/θ1+1/θ2−1/a2dη+∫1/21exp−ρ1−η−exp−ρη1/θ1+1/θ2−η/a1+1−η/a22ηΩ′11/θ1+1/θ2−1/a1+1−ηΩ′11/θ1+1/θ2−1/a2dη≤a2−a12a1a21−exp−ρ∫01/2exp−ρη−exp−ρ1−η1/θ1+1/θ2−η/a1+1−η/a22ηΩ′11/θ1+1/θ2−1/a1+1−ηΩ′11/θ1+1/θ2−1/a2dη+∫01/2exp−ρη−exp−ρ1−η1/θ1+1/θ2−1−η/a1+η/a221−ηΩ′11/θ1+1/θ2−1/a1+ηΩ′11/θ1+1/θ2−1/a2dη=a2−a12a1a21−exp−ρ∫01/2exp−ρη−exp−ρ1−ηημ12+1−ημ22Ω′11/θ1+1/θ2−1/a1+1−ημ12+ημ22Ω′11/θ1+1/θ2−1/a2dη.
This completes the proof.Remark 6.
Forα⟶1, we have(43)limα⟶11−α21−exp−ρ=a1a22a2−a1,limα⟶1exp−ρη−exp−ρ1−η21−exp−ρ=1−2η2.
Ifα⟶1, a1=θ1, and a2=θ2 in Theorem 10, then we get Theorem 2.6 which was proved by İşcan in [11].


## 3. Fejér–Hadamard–Mercer-Type Inequality for Harmonically Convex Function

Theorem 11.
IfΩ:I⊆0,∞⟶ℝ is a harmonically convex function for θ1,θ2∈I with θ1<θ2. such that If Ω∈Lθ1,θ2 and ω:θ1,θ2⊆0,∞⟶ℝ is nonnegative, integrable, and harmonically symmetric with respect to 2θ1θ2/θ1+θ2, that is, ωϰ=ω1/1/θ1+1/θ2−1/ϰ, then(44)Ω11/θ1+1/θ2−a1+a2/2a1a2I1/θ1+1/θ2−1/a1αω∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αω∘g1θ1+1θ2−1a1≤I1/θ1+1/θ2−1/a1αω∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αΩω∘g1θ1+1θ2−1a1≤12Ω11/θ1+1/θ2−1/a1+Ω11/θ1+1/θ2−1/a2I1/θ1+1/θ2−1/a1αω∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αω∘g1θ1+1θ2−1a1,for all a1,a2∈θ1,θ2, α>0, and gu=1/u, u∈1/θ2,1/θ1.Proof.
IfΩ is a harmonically convex function on θ1,θ2, then for all a1,a2∈θ1,θ2,(45)Ω2a1a2a1+a2=Ω11/2η/a1+1−η/a2+1−η/a1+η/a2≤12Ω1η/a1+1−η/a2+Ω11−η/a1+η/a2≤Ωa1+Ωa22.
Replacinga1 and a2 by 1/1/θ1+1/θ2−1/a1 and 1/1/θ1+1/θ2−1/a2, respectively, we get(46)Ω11/θ1+1/θ2−a1+a2/2a1a2≤12Ω11/θ1+1/θ2−η/a1+1−η/a2+Ω11/θ1+1/θ2−1−η/a1+η/a2≤Ω11/θ1+1/θ2−1/a1+Ω11/θ1+1/θ2−1/a2.
Multiplying by(47)exp−ρηω11/θ1+1/θ2−1−η/a1+η/a2,on both sides of (46) and then integrating with respect to η over 0,1, we obtain(48)2Ω11/θ1+1/θ2−a1+a2/2a1a2∫01exp−ρηω11/θ1+1/θ2−1−η/a1+η/a2dη≤∫01exp−ρηΩ11/θ1+1/θ2−η/a1+1−η/a2ω11/θ1+1/θ2−1−η/a1+η/a2dη+∫01exp−ρηΩ11/θ1+1/θ2−1−η/a1+η/a2ω11/θ1+1/θ2−1−η/a1+η/a2dη≤Ω11/θ1+1/θ2−1/a1+Ω11/θ1+1/θ2−1/a2∫01exp−ρηω11/θ1+1/θ2−1−η/a1+η/a2dη,where(49)∫01exp−ρηΩ11/θ1+1/θ2−η/a1+1−η/a2ω11/θ1+1/θ2−1−η/a1+η/a2dη+∫01exp−ρηΩ11/θ1+1/θ2−1−η/a1+η/a2ω11/θ1+1/θ2−1−η/a1+η/a2dη=a1a2a2−a1∫1/θ1+1/θ2−1/a11/θ1+1/θ2−1/a2exp−1−ααu−1θ1+1θ2−1a1Ω11/θ1+1/θ2−1/a1+1/θ1+1/θ2−1/a2−uω1udu+∫1/θ1+1/θ2−1/a11/θ1+1/θ2−1/a2exp−1−ααu−1θ1+1θ2−1a1Ω1uω1udu=a1a2a2−a1∫1/θ1+1/θ2−1/a11/θ1+1/θ2−1/a2exp−1−αα1θ1+1θ2−1a2−uΩ1uω11/θ1+1/θ2−1/a1+1/θ1+1/θ2−1/a2−udu+∫1/θ1+1/θ2−1/a11/θ1+1/θ2−1/a2exp−1−ααu−1/θ1+1/θ2−1/a1Ω1uω1udu=αa1a2a2−a1I1/θ1+1/θ2−1/a1αΩω∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αΩω∘g1θ1+1θ2−1a1.
That is,(50)2Ω11/θ1+1/θ2−a1+a2/2a1a2∫01exp−ρηω11/θ1+1/θ2−1−η/a1+η/a2dη≤αa1a2a2−a1I1/θ1+1/θ2−1/a1αΩω∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αΩω∘g1θ1+1θ2−1a1≤Ω11/θ1+1/θ2−1/a1+Ω11/θ1+1/θ2−1/a2∫01exp−ρηω11/θ1+1/θ2−1−η/a1+η/a2dη.
Sinceω is symmetric with respect to 2θ1θ2/θ1+θ2, we have(51)I1/θ1+1/θ2−1/a2αω∘g1θ1+1θ2−1a1=I1/θ1+1/θ2−1/a1αω∘g1θ1+1θ2−1a2=12I1/θ1+1/θ2−1/a1αω∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αω∘g1θ1+1θ2−1a1.
Therefore, we have(52)Ω11/θ1+1/θ2−a1+a2/2a1a2I1/θ1+1/θ2−1/a1αω∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αω∘g1θ1+1θ2−1a1≤I1/θ1+1/θ2−1/a1αΩω∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αΩω∘g1θ1+1θ2−1a1≤12Ω11/θ1+1/θ1−1/a1+Ω11/θ1+1/θ1−1/a2I1/θ1+1/θ2−1/a1αω∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αω∘g1θ1+1θ2−1a1.
Thus, the proof of Theorem11 is complete.Remark 7.
If we takea1=θ1 and a2=θ2 in Theorem 11, we will get Theorem 3.2 in [24].Remark 8.
Under the assumptions of Theorem11 with α=1, we have(53)Ω11/θ1+1/θ2−a1+a2/2a1a2∫1/θ1+1/θ2−1/a11/θ1+1/θ2−1/a2ωϰϰ2dϰ≤∫1/θ1+1/θ2−1/a11/θ1+1/θ2−1/a2Ωϰϰ2ωϰdϰ≤12Ω11/θ1+1/θ2−1/a1+Ω11/θ1+1/θ2−1/a2∫1/θ1+1/θ2−1/a11/θ1+1/θ2−1/a2ωϰϰ2dϰ.Remark 9.
Ifα=1, a1=θ1, and a2=θ2 in Theorem 11, then we can get Hermite–Hadamard–Fejér inequality (8) for harmonically convex function which was proved by Chen and Wu in [12].


## 4. Pachpatte–Mercer-Type Inequality for Harmonically Convex Function

Theorem 12.
LetΩ,ω:θ1,θ2⊆0,∞⟶ℝ be functions such that Ω,ω,Ωω∈Lθ1,θ2. If Ω and ω are harmonically convex on θ1,θ2, then(54)αa1a22a2−a1I1/θ1+1/θ2−1/a1αΩω∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αΩω∘g1θ1+1θ2−1a1≤ρ2−2ρ+4−ρ2+2ρ+4exp−ρ2ρ3Mθ1,θ2,a1,a2+ρ−2+exp−ρρ+2ρ3Nθ1,θ2,a1,a2,(55)2Ω11/θ1+1/θ2−a1+a2/2a1a2ω11/θ1+1/θ2−a1+a2/2a1a2≤1−α21−exp−ρI1/θ1+1/θ2−1/a1αΩω∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αΩω∘g1θ1+1θ2−1a1+ρ−2+exp−ρρ+2ρ21−exp−ρMθ1,θ2,a1,a2+ρ2−2ρ+4−ρ2+2ρ+4exp−ρ2ρ21−exp−ρNθ1,θ2,a1,a2,where(56)Mθ1,θ2,a1,a2=Ω11/θ1+1/θ2−1/a1ω11/θ1+1/θ2−1/a1+Ω11/θ1+1/θ2−1/a2ω11/θ1+1/θ2−1/a2,Nθ1,θ2,a1,a2=Ω11/θ1+1/θ2−1/a1ω11/θ1+1/θ2−1/a2+Ω11/θ1+1/θ2−1/a2ω11/θ1+1/θ2−1/a1,for all a1,a2∈θ1,θ2, α>0, and gu=1/u, u∈1/θ2,1/θ1.Proof.
SinceΩ and ω are harmonically convex on θ1,θ2, then for all η∈0,1, we have(57)Ω11/θ1+1/θ2−η/a1+1−η/a2ω11/θ1+1/θ2−η/a1+1−η/a2≤η2Ω11/θ1+1/θ2−1/a1ω11/θ1+1/θ2−1/a1+1−η2Ω11/θ1+1/θ2−1/a2ω11/θ1+1/θ2−1/a2+η1−ηΩ11/θ1+1/θ2−1/a1ω11/θ1+1/θ2−1/a2+Ω11/θ1+1/θ2−1/a2ω11/θ1+1/θ2−1/a1,Ω11/θ1+1/θ2−1−η/a1+η/a2ω11/θ1+1/θ2−1−η/a1+η/a2≤η2Ω11/θ1+1/θ2−1/a2ω11/θ1+1/θ2−1/a2+1−η2Ω11/θ1+1/θ2−1/a1ω11/θ1+1/θ2−1/a1+η1−ηΩ11/θ1+1/θ2−1/a1ω11/θ1+1/θ2−1/a2+Ω11/θ1+1/θ2−1/a2ω11/θ1+1/θ2−1/a2.
Adding these inequalities, we have(58)Ω11/θ1+1/θ2−η/a1+1−η/a2ω11/θ1+1/θ2−η/a1+1−η/a2+Ω11/θ1+1/θ2−1−η/a1+η/a2ω11/θ1+1/θ2−1−η/a1+η/a2≤2η2−2η+1Mθ1,θ2,a1,a2+2η1−ηNθ1,θ2,a1,a2.
Multiplying byexp−ρη on both sides of (58) and then integrating with respect to η∈0,1, we have(59)∫01exp−ρηΩ11/θ1+1/θ2−η/a1+1−η/a2ω11/θ1+1/θ2−η/a1+1−η/a2dη+∫01exp−ρηΩ11/θ1+1/θ2−1−η/a1+η/a2ω11/θ1+1/θ2−1−η/a1+η/a2dη≤Mθ1,θ2,a1,a2∫01exp−ρη2η2−2η+1dη+Nθ1,θ2,a1,a2∫01exp−ρη2η1−ηdη=ρ2−2ρ+4−ρ2+2ρ+4exp−ρρ3Mθ1,θ2,a1,a2+2ρ−2+exp−ρρ+2ρ3Nθ1,θ2,a1,a2.
Thus, we have(60)αa1a22a2−a1I1/θ1+1/θ2−1/a1αΩω∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αΩω∘g1θ1+1θ2−1a1≤ρ2−2ρ+4−ρ2+2ρ+4exp−ρ2ρ3Mθ1,θ2,a1,a2+ρ−2+exp−ρρ+2ρ3Nθ1,θ2,a1,a2.
Thus, inequality (54) is proved. Now we prove inequality (55). By using harmonic convexity of the function Ω on θ1,θ2, we have(61)Ω2a1a2a1+a2ω2a1a2a1+a2=Ω11/2η/a1+1−η/a2+1−η/a1+η/a2ω11/2η/a1+1−η/a2+1−η/a1+η/a2≤12Ω1η/a1+1−η/a2+Ω11−η/a1+η/a212ω1η/a1+1−η/a2+ω11−η/a1+η/a2.
Replacinga1 and a2 by 1/1/θ1+1/θ2−1/a1 and 1/1/θ1+1/θ2−1/a2, respectively, we get(62)Ω11/θ1+1/θ2−a1+a2/2a1a2ω11/θ1+1/θ2−a1+a2/2a1a2≤12Ω11/θ1+1/θ2−η/a1+1−η/a2+Ω11/θ1+1/θ2−1−η/a1+η/a212ω11/θ1+1/θ2−η/a1+1−η/a2+ω11/θ1+1/θ2−1−η/a1+η/a2≤14Ω11/θ1+1/θ2−η/a1+1−η/a2ω11/θ1+1/θ2−η/a1+1−η/a2+14Ω11/θ1+1/θ2−1−η/a1+η/a2ω11/θ1+1/θ2−1−η/a1+η/a2+η1−η2Mθ1,θ2,a1,a2+2η2−2η+14Nθ1,θ2,a1,a2.
Thus,(63)Ω11/θ1+1/θ2−a1+a2/2a1a2ω11/θ1+1/θ2−a1+a2/2a1a2≤14Ω11/θ1+1/θ2−η/a1+1−η/a2ω11/θ1+1/θ2−η/a1+1−η/a2+14Ω11/θ1+1/θ2−1−η/a1+η/a2ω11/θ1+1/θ2−1−η/a1+η/a2+η1−η2Mθ1,θ2,a1,a2+2η2−2η+14Nθ1,θ2,a1,a2.
Multiplying byexp−ρη on both sides of (63) and then integrating with respect to η∈0,1, we obtain(64)1−exp−ρρΩ11/θ1+1/θ2−a1+a2/2a1a2ω11/θ1+1/θ2−a1+a2/2a1a2≤14∫01exp−ρηΩ11/θ1+1/θ2−η/a1+1−η/a2ω11/θ1+1/θ2−η/a1+1−η/a2dη+14∫01exp−ρηΩ11/θ1+1/θ2−1−η/a1+η/a2ω11/θ1+1/θ2−1−η/a1+η/a2dη+Mθ1,θ2,a1,a2∫01exp−ρηη1−η2dη+Nθ1,θ2,a1,a2∫01exp−ρη2η2−2η+14dη=αa1a24a2−a1I1/θ1+1/θ2−1/a1αΩω∘g1θ1+1θ2−1a2+I1/θ1+1/θ2−1/a2αΩω∘g1θ1+1θ2−1a1+ρ−2+exp−ρρ+22ρ3Mθ1,θ2,a1,a2+ρ2−2ρ+4−ρ2+2ρ+4exp−ρ4ρ3Nθ1,θ2,a1,a2.
Thus,(65)Ω11/θ1+1/θ2−a1+a2/2a1a2ω11/θ1+1/θ2−a1+a2/2a1a2≤1−α41−exp−ρI1/θ1+1/θ1−1/a1αΩω∘g1θ1+1θ2−1a2+I1/θ1+1/θ1−1/a2αΩω∘g1θ1+1θ2−1a1+ρ−2+exp−ρρ+22ρ21−exp−ρMθ1,θ2,a1,a2+ρ2−2ρ+4−ρ2+2ρ+4exp−ρ4ρ21−exp−ρNθ1,θ2,a1,a2.Remark 10.
If we takea1=θ1 and a2=θ2 in Theorem 12, then we get Theorem 12 in [24].Remark 11.
Forα⟶1, we have(66)limα⟶11−α21−exp−ρ=a1a22a2−a1,limα⟶1ρ−2+exp−ρρ+2ρ3=16,limα⟶1ρ2−2ρ+4−ρ2+2ρ+4exp−ρ2ρ21−exp−ρ=13.
Under the assumptions of Theorem12 with α=1, we have(67)a1a2a2−a1∫1/θ1+1/θ2−1/a11/θ1+1/θ2−1/a2Ωϰϰ2ωϰdϰ≤Mθ1,θ2,a1,a23+Nθ1,θ2,a1,a26,2Ω11/θ1+1/θ2−a1+a2/2a1a2ω11/θ1+1/θ2−a1+a2/2a1a2≤a1a2a2−a1∫1/θ1+1/θ2−1/a11/θ1+1/θ2−1/a2Ωϰϰ2ωϰdϰ+Mθ1,θ2,a1,a26+Nθ1,θ2,a1,a23.Remark 12.
Ifα=1, a1=θ1, and a2=θ2 in Theorem 12, then we obtain inequalities (9) and (10) for harmonically convex functions proved by Chen and Wu in [13].


## 5. Concluding Remarks and Future Directions

In this study, we introduce for the first time the unified variants of Hermite–Hadamard, Fejér–Hadamard, and Pachpatte–Mercer-type inequalities for harmonically convex functions for fractional integral operators with the exponential kernel. New integral identity involving fractional integral operators with exponential kernel is developed. A compact analysis of newly obtained results and their connections is explained in Remarks2–12. As special cases, we get Hermite–Hadamard, Fejér–Hadamard, and Pachpatte–Mercer-type inequalities for classical calculus with explicit boundary values. Some particular cases reflect the related existing results. One of the direct impact and utilization of the results extracted in this paper is to obtain inequalities involving following new fractional integral operators containing Mittag-Leffler nonsingular kernels:(68)Ia1αΩϰ=1α∫a1ϰEα,1−1−ααϰ−uαΩudu,ϰ>a1,Ia2αΩϰ=1α∫ϰa2Eα,1−1−ααu−ϰαΩudu,ϰ<a2,for α∈0,1 and Ω∈La1,a2, where Eα,νξ is a Mittag-Leffler-type function:(69)Eα,νξ=∑ι=0∞ξιΓαι+ν.These integral inequalities may be helpful in the circumstances where upper and lower bounds matter for fractional integral operators involving nonlocal kernels. It is natural to investigate such results for other general convexities like harmonicallyh−convex functions introduced by Noor et al. in [30]. Also, it is interesting to construct such inequalities over fractal domains where we may get optimal and sharp local fractional integral inequalities involving Mittag-Leffler kernel.



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