# Fractional Hermite–Jensen–Mercer Integral Inequalities with respect to Another Function and Application

Authors: Saad Ihsan Butt; Muhammad Umar; Khuram Ali Khan; Artion Kashuri; Homan Emadifar
Journal: Complexity
 (2021)

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## Abstract

In this paper, authors prove new variants of Hermite–Jensen–Mercer type inequalities usingψ–Riemann–Liouville fractional integrals with respect to another function via convexity. We establish generalized identities involving ψ–Riemann–Liouville fractional integral pertaining first and twice differentiable convex function λ, and these will be used to derive novel estimates for some fractional Hermite–Jensen–Mercer type inequalities. Some known results are recaptured from our results as special cases. Finally, an application from our results using the modified Bessel function of the first kind is established as well.


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## Body

## 1. Introduction and Preliminaries

The theory of fractional integrals and derivatives has occurred in many fields and directions such as partial differential equations, difference equations, probability, and stochastic processes (see [1–6]). Behind it, the theory of convex functions with integral inequalities is also useful.Definition 1.
A functionλ:J⊂ℜ⟶ℜ is said to be convex on J if(1)λ1−ζy1+ζy2≤1−ζλy1+ζλy2,holds for every y1,y2∈J and ζ∈0,1.
One of the best-known inequalities for convex functions is the following Hermite–Hadamard’s inequality: ifλ:J⊆ℜ⟶ℜ is a convex function in J, where y1,y2∈J and y1<y2, then(2)λy1+y22≤1y2−y1∫y1y2λζdζ≤λy1+λy22H−H.
It is also known as classicalH−H inequality. A number of mathematicians in the field of applied and pure mathematics have dedicated their efforts to extend, generalize, counterpart, and refine Hermite–Hadamard’s inequality H−H for different classes of convex functions. For more recent results obtained on inequality H−H, we refer the reader to references [7–10].
Letσi∈0,1 be nonnegative weights such that ∑i=1nσi=1. The Jensen inequality states that, if λ is convex function on y1,y2, then(3)λ∑i=1nσiℓi≤∑i=1nσiλℓi,holds for all ℓi∈y1,y2 and all i=1,2,…,n, see [11].
In the literature, Jensen’s inequality and Hermite–Hadamard’s inequality are highly familiar results pertaining convex functions. One of the well-known and most significant inequalities in mathematical analysis is Jensen’s and related inequalities. Jensen’s inequality for differentiable convex functions plays a significant role in the field of inequalities as several other inequalities can be seen as special cases of it. It is used in order to make claims regarding the function while just a little is known or is needed to be known about the distribution. Furthermore, this inequality has been used in various areas of sciences and technology to solve several problems, such as engineering, mathematical statistics, financial economics, and computer science. Some recent results can be seen in [12–14].
Jensen’s inequality has a following variant gave by Mercer (see [15]).Theorem 1.
Letλ be a convex function on y1,y2, then(4)λy1+y2−∑i=1nσiℓi≤λy1+λy2−∑i=1nσiλℓi,holds for all ℓi∈y1,y2 and all σi∈0,1,i=1,2,…,n.Jensen–Mercer’s type inequality is a topic of supreme interest as it gives more information with explicit boundary conditions. It is quite effective for applications in operator analysis in higher dimensions [16–18]. Moradi et al. established some new improvements and generalization of Jensen–Mercer’s type inequalities [19]. Recently, in [20], Adil et al. gave applications of Jensen–Mercer’s inequality in information theory. They computed new estimates for Csiszár and related divergences. Taking into consideration the wonderful packages of Jensen’s and associated inequalities in diverse fields of mathematics and engineering sciences, their generalizations and upgrades were a subject of an excellent hobby for the researchers in the last few years as obvious from a massive variety of investigation on it (see [21–24]).In [25], Vanterler da Costa Sousa and Capelas de Oliveira introduced ψ–fractional integrals and ψ–Hilfer fractional derivative with respect to another function. They also studied Gronwall inequalities using ψ–Hilfer operator (see [26]).Definition 2 (see [23]).
Suppose thaty1,y2−∞≤y1<y2≤∞ and α>0. Also let ψ be an increasing and positive monotone function on y1,y2, having a continuous derivative ψ′ on y1,y2. Then, the left-sided and right-sided ψ–Riemann–Liouville fractional integrals of a function λ with respect to another function ψ on y1,y2 are defined as follows:(5)Iy1+α:ψλℓ=1Γα∫y1ℓψ′ζψℓ−ψζα−1λζdζ,y1<ℓ,Iy2−α:ψλℓ=1Γα∫ℓy2ψ′ζψζ−ψℓα−1λζdζ,ℓ<y2,respectively.
If we chooseψζ=ζ and ψζ=lnζ, then we get, respectively, Riemann–Liouville and Hadamard fractional integrals.
Motivated by previous results, we will establish several new Hermite–Hadamard–Mercer type inequalities involvingψ–Riemann–Liouville fractional integrals (i.e., Riemann–Liouville fractional integral of any function with respect to another function). Moreover, our results recover several known results. Finally, an application using the modified Bessel function of the first kind will be established as well.


## 2. Hermite–Jensen–Mercer Type Inequalities

Throughout the paper, the following assumption will be used in the sequel.A1: Let 0≤y1<y2, λ:y1,y2⟶ℜ be a positive function and λ∈L1y1,y2. Also suppose, ψ⋅ is an increasing and positive monotone function on y1,y2, having a continuous derivative ψ′ on y1,y2 and α>0.Theorem 2.
IfA1 is satisfied and λ is a convex function on y1,y2, then(6)λy1+y2−ℓ1+ℓ22≤λy1+λy2−Γα+12ℓ2−ℓ1α×Iψ−1ℓ1+α:ψλ∘ψψ−1ℓ2+Iψ−1ℓ2−α:ψλ∘ψψ−1ℓ1≤λy1+λy2−λℓ1+ℓ22,(7)λy1+y2−ℓ1+ℓ22≤2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ2≤λy1+λy2−λℓ1+λℓ22,for all ℓ1,ℓ2∈y1,y2, where Γ⋅ is the gamma function.Proof.
Using Jensen–Mercer’s inequality, we have(8)λy1+y2−x+z2≤λy1+λy2−λx+λz2,for all x,z∈y1,y2.
Now, by change of variablesx=ζℓ1+1−ζℓ2 and z=1−ζℓ1+ζℓ2, for all ℓ1,ℓ2∈y1,y2 and ζ∈0,1 in (8), we get(9)λy1+y2−ℓ1+ℓ22≤λy1+λy2−λζℓ1+1−ζℓ2+λ1−ζℓ1+ζℓ22.
Multiplying the above inequality byζα−1 on both sides and integrating with respect to ζ on 0,1, we obtain(10)1αλy1+y2−ℓ1+ℓ22≤1αλy1+λy2−12∫01ζα−1λζℓ1+1−ζℓ2+λ1−ζℓ1+ζℓ2dζ,where(11)α2∫01ζα−1λζℓ1+1−ζy2+λ1−ζy1+ζy2dζ=α2∫01ζα−1λζℓ1+1−ζℓ2dζ+α2∫01ζα−1λ1−ζℓ1+ζℓ2dζ.
Now, letζ=ψγ−ℓ1/ℓ2−ℓ1, then dζ=ψ′γdγ/ℓ2−ℓ1. Using the above equality, we obtain(12)=α2∫ψ−1ℓ1ψ−1ℓ2ℓ2−ψγℓ2−ℓ1α−1λψγψ′γℓ2−ℓ1dγ+α2∫ψ−1ℓ1Ψ−1ℓ2ψγ−ℓ1ℓ2−ℓ1α−1λψγψ′γℓ2−ℓ1dγ=Γα+12ℓ2−ℓ1α1Γα∫ψ−1ℓ1ψ−1ℓ2ψ′γℓ2−ψγα−1λ∘ψγdγ+Γα+12ℓ2−ℓ1α1Γα∫ψ−1ℓ1ψ−1ℓ2ψ′γψγ−ℓ1α−1λ∘ψγdγ.
So, the final form will be of this type as follows:(13)=Γα+12ℓ2−ℓ1α×Iψ−1ℓ1+α:ψλ∘ψψ−1ℓ2+Iψ−1ℓ2−α:ψλ∘ψψ−1ℓ1,and so the first inequality of (6) is proved.
Regarding the second inequality of (6), since λ is convex function, then for ζ∈0,1, we have(14)λℓ1+ℓ22=λζℓ1+1−ζℓ2+1−ζℓ1+ζℓ22≤λζℓ1+1−ζℓ2+λ1−ζℓ1+ζℓ22.
Multiplying the above inequality byζα−1 on both sides and integrating with respect to ζ on 0,1, we get(15)1αλℓ1+ℓ22≤12∫01ζα−1λζℓ1+1−ζℓ2+λ1−ζℓ1+ζℓ2.
Letψγ=ζℓ1+1−ζℓ2 and ψβ=1−ζℓ1+ζℓ2. Then, we have(16)λℓ1+ℓ22≤Γα+12ℓ2−ℓ1αIψ−1ℓ1+α:ψλ∘ψψ−1ℓ2+Iψ−1ℓ2−α:ψλ∘ψψ−1ℓ1.
Multiplying by−1, we will get(17)−Γα+12ℓ2−ℓ1αIψ−1ℓ1+α:ψλ∘ψψ−1ℓ2+Iψ−1ℓ2−α:ψλ∘ψψ−1ℓ1≤−λℓ1+ℓ22.
Addingλy1+λy2 both sides in (17), we obtain our second inequality of (6).
To prove the first inequality of (7) by using the convexity of λ, we have(18)2λy1+y2−x+z2≤λy1+y2−x+λy1+y2−z,for all x,z∈y1,y2. By change of variables x=ζℓ1+1−ζℓ2 and z=1−ζℓ1+ζℓ2,ζ∈0,1, we get(19)2λy1+y2−ℓ1+ℓ22≤λy1+y2−ζℓ1+1−ζℓ2+λy1+y2−1−ζℓ1+ζℓ2.
Multiplying the above inequality byζα−1 on both sides and integrating with respect to ζ over 0,1, we get(20)2αλy1+y2−ℓ1+ℓ22≤∫01ζα−1λy1+y2−ζℓ1+1−ζℓ2+λy1+y2−1−ζℓ1+ζℓ2dζ.
Hence, by change of variables, we obtain(21)λy1+y2−ℓ1+ℓ22≤2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ2+α:ψλ∘Ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ2,and so the first inequality of (7) is proved.
About the second inequality of (7), since λ is convex function, then for ζ∈0,1, we obtain(22)λy1+y2−ζℓ1+1−ζℓ2≤λy1+λy2−ζλℓ1+1−ζλℓ2,(23)λy1+y2−1−ζℓ1+ζℓ2≤λy1+λy2−1−ζλℓ1+ζλℓ2.
By adding inequalities (22) and (23), we have(24)λy1+y2−ζℓ1+1−ζℓ2+λy1+y2−1−ζℓ1+ζℓ2≤2λy1+λy2−λℓ1+λℓ2.
Multiplying the above inequality byζα−1 on both sides and integrating with respect to ζ on 0,1, we get(25)2αΓαℓ2−ℓ1αIψ−1y1+y2−ℓ2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ2≤2λy1+λy2−λℓ1+λℓ2⋅1α.
Multiplying byα/2, we will get(26)2α−1Γα+1ℓ2−ℓ1αIψ−1y1+y2−ℓ2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ2≤λy1+λy2−λℓ1+λℓ22.
From inequalities (21) and (26), we get the desired double inequality (7).Remark 1.
Takingψγ=γ in Theorem 2, we will get Theorem 2 proved in [27].Remark 2.
Takingψγ=γ and α=1 in Theorem 2, we will obtain Theorem 2 proved by Kian and Moslehian in [28].Theorem 3.
IfA1 is satisfied and λ is a convex function on y1,y2, then(27)λy1+y2−ℓ1+ℓ22≤2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ1+ℓ2/2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ∘ψψ−1y1+y2−ℓ2≤λy1+λy2−λℓ1+λℓ22,for all ℓ1,ℓ2∈y1,y2.Proof.
About the first inequality (27) by using the convexity of λ, we have(28)2λy1+y2−x+z2≤λy1+y2−x+λy1+y2−z,for all x,z∈y1,y2. By change of variables x=ζ/2ℓ1+2−ζ/2ℓ2 and z=2−ζ/2ℓ1+ζ/2ℓ2,ζ∈0,1, we get(29)2λy1+y2−ℓ1+ℓ22≤λy1+y2−ζ2ℓ1+2−ζ2ℓ2+λy1+y2−2−ζ2ℓ1+ζ2ℓ2.
Multiplying the above inequality byζα−1 on both sides and integrating with respect to ζ over 0,1, we have(30)2αλy1+y2−ℓ1+ℓ22≤∫01ζα−1λy1+y2−ζ2ℓ1+2−ζ2ℓ2+λy1+y2−2−ζ2ℓ1+ζ2ℓ2dζ.
Hence, by change of variables, we obtain(31)λy1+y2−ℓ1+ℓ22≤2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ1+ℓ2/2+α:ψλ∘Ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ∘ψψ−1y1+y2−ℓ2,which proved the first inequality of (27).
Regarding the second inequality of (27), since λ is convex function, then for ζ∈0,1, we have(32)λy1+y2−ζ2ℓ1+2−ζ2ℓ2≤λy1+λy2−ζ2λℓ1+2−ζ2λℓ2,(33)λy1+y2−2−ζ2ℓ1+ζ2ℓ2≤λy1+λy2−2−ζ2λℓ1+ζ2λℓ2.
By adding inequalities (32) and (33), we get(34)λy1+y2−ζ2ℓ1+2−ζ2ℓ2+λy1+y2−2−ζ2ℓ1+ζ2ℓ2≤2λy1+λy2−λℓ1+λℓ2.
Multiplying the above inequality byζα−1 on both sides and integrating with respect to ζ on 0,1, we obtain(35)∫01ζα−1λy1+y2−ζ2ℓ1+2−ζ2ℓ2+λy1+y2−2−ζ2ℓ1+ζ2ℓ2dζ≤2λy1+λy2−λℓ1+λℓ2∫01ζα−1dζ.
Then, we have the following inequality:(36)2αΓαℓ2−ℓ1αIψ−1y1+y2−ℓ1+ℓ2/2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ∘ψψ−1y1+y2−ℓ2≤2λy1+λy2−λℓ1+λℓ2⋅1α.
Multiplying byα/2, we will get(37)2α−1Γα+1ℓ2−ℓ1αIψ−1y1+y2−ℓ1+ℓ2/2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ∘ψψ−1y1+y2−ℓ2≤λy1+λy2−λℓ1+λℓ22.
From inequalities (31) and (37), we get the desired double inequality (27).Remark 3.
Takingψγ=γ in Theorem 3, we will get Theorem 3 proved in [27].Remark 4.
Takingψγ=γ and α=1 in Theorem 3, we will obtain Theorem 2.1 proved by Kian and Moslehian in [28].Theorem 4.
IfA1 is satisfied and λ is a convex function on y1,y2, then(38)λy1+y2−ℓ1+ℓ22≤2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ2+α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22+Iψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22≤λy1+λy2−λℓ1+λℓ22,for all ℓ1,ℓ2∈y1,y2.Proof.
Regarding the first part of inequality (38) by using the convexity of λ, we have(39)2λy1+y2−x+z2≤λy1+y2−x+λy1+y2−z,for all x,z∈y1,y2. By change of variables x=1+ζ/2ℓ1+1−ζ/2ℓ2 and z=1−ζ/2ℓ1+1+ζ/2ℓ2,ζ∈0,1, we get(40)2λy1+y2−ℓ1+ℓ22≤λy1+y2−1+ζ2ℓ1+1−ζ2ℓ2+λy1+y2−1−ζ2ℓ1+1+ζ2ℓ2.
Multiplying the above inequality byζα−1 on both sides and integrating with respect to ζ∈0,1, we obtain(41)2αλy1+y2−ℓ1+ℓ22≤∫01ζα−1λy1+y2−1+ζ2ℓ1+1−ζ2ℓ2+λy1+y2−1−ζ2ℓ1+1+ζ2ℓ2dζ.
Hence, by change of variables, we have(42)λy1+y2−ℓ1+ℓ22≤2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ2+α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22+Iψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22,which concludes the first inequality of (38).
About the second inequality of (38), since λ is convex function, then for ζ∈0,1, we get(43)λy1+y2−1+ζ2ℓ1+1−ζ2ℓ2≤λy1+λy2−1+ζ2λℓ1+1−ζ2λℓ2,(44)λy1+y2−1−ζ2ℓ1+1+ζ2ℓ2≤λy1+λy2−1−ζ2λℓ1+1+ζ2λℓ2.
By adding inequalities (43) and (44), we have(45)λy1+y2−1+ζ2ℓ1+1−ζ2ℓ2+λy1+y2−1−ζ2ℓ1+1+ζ2ℓ2≤2λy1+λy2−λℓ1+λℓ2.
Multiplying the above inequality byζα−1 on both sides and integrating with respect to ζ over 0,1, we obtain(46)∫01ζα−1λy1+y2−1+ζ2ℓ1+1−ζ2ℓ2+λy1+y2−1−ζ2ℓ1+1+ζ2ℓ2dζ≤2λy1+λy2−λℓ1+λℓ2∫01ζα−1dζ.
Then, we have the following inequality:(47)2αΓαℓ2−ℓ1α×Iψ−1y1+y2−ℓ2+α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22+Iψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22≤2λy1+λy2−λℓ1+λℓ2⋅1α.
Multiplying byα/2, we will get(48)2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ2+α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22+Iψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22≤λy1+λy2−λℓ1+λℓ22.
So, the second inequality of (38) holds.Remark 5.
Takingψγ=γ in Theorem 4, we will get Theorem 2 proved in [29].Remark 6.
Takingψγ=γ and α=1 in Theorem 4, we will obtain Theorem 2.1 proved by Kian and Moslehian in [28].


## 3. New Generalized Identities and Their Integral Inequalities

In this section, the following lemmas will play a basic role in our next results.Lemma 1.
IfA1 is satisfied and λ:y1,y2⟶ℜ is a differentiable function on L1y1,y2, then(49)λy1+y2−ℓ1+λy1+y2−ℓ22−Γα+12ℓ2−ℓ1α×Iψ−1y1+y2−ℓ2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ2=12ℓ2−ℓ1α×∫ψ−1y1+y2−ℓ2ψ−1y1+y2−ℓ1ψγ−y1+y2−ℓ2α−y1+y2−ℓ1−ψγα×λ′∘ψγψ′γdγ,for all ℓ1,ℓ2∈y1,y2.Proof.
It suffices to note that(50)I=λy1+y2−ℓ1−λy1+y2−ℓ22−I1+I2,where(51)I1=Γα+12ℓ2−ℓ1αIΨ−1y1+y2−ℓ2+α:ψλ∘ψψ−1y1+y2−ℓ1=α2ℓ2−ℓ1α∫ψ−1y1+y2−ℓ2ψ−1y1+y2−ℓ1ψ′γy1+y2−ℓ1−ψγα−1λ∘ψγdγ=−12ℓ2−ℓ1α∫ψ−1y1+y2−ℓ1ψ−1y1+y2−ℓ2dy1+y2−ℓ1−ψγαλ∘ψγ=12ℓ2−ℓ1αλy1+y2−ℓ2ℓ2−ℓ1α+∫ψ−1y1+y2−ℓ2ψ−1y1+y2−ℓ1ψ′γy1+y2−ℓ1−ψγαλ′∘ψγdγ,(52)I2=Γα+12ℓ2−ℓ1αIψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ2=α2ℓ2−ℓ1α∫ψ−1y1+y2−ℓ2ψ−1y1+y2−ℓ1ψ′γ−y1+y2−ℓ2+ψγα−1λ∘ψγdγ=12ℓ2−ℓ1α∫ψ−1y1+y2−ℓ2ψ−1y1+y2−ℓ1d−y1+y2−ℓ2+ψγαλ∘ψγ=12ℓ2−ℓ1αλy1+y2−ℓ1ℓ2−ℓ1α−∫ψ−1y1+y2−ℓ2ψ−1y1+y2−ℓ1ψ′γ−y1+y2−ℓ2+ψγαλ′∘ψγdγ.
Substituting (51) and (52) in (50), we get the desired equality (49).Remark 7.
Forℓ1=y1 and ℓ2=y2 in Lemma 1, we will get Lemma 3.1 proved in [30].Theorem 5.
IfA1 is satisfied and λ′ is a convex function on y1,y2, then(53)|λy1+y2−ℓ1+λy1+y2−ℓ22−Γα+12ℓ2−ℓ1αIψ−1y1+y2−ℓ2+α:ψλy1+y2−ℓ1+Iψ−1y1+y2−ℓ1−α:ψλy1+y2−ℓ2|≤ℓ2−ℓ1α+11−12αλ′y1+λ′y2−λ′ℓ1+λ′ℓ22.for all ℓ1,ℓ2∈y1,y2.Proof.
Here, we will use Lemma1, properties of modulus, and Jensen–Mercer’s inequality.
For everyγ∈ψ−1y1+y2−ℓ2,ψ−1y1+y2−ℓ1, we have y1+y2−ℓ2<ψγ<y1+y2−ℓ1.
Letζ=y1+y2−ℓ1−ψγ/ℓ2−ℓ1, and then ψγ=y1+y2−ζℓ1+1−ζℓ2. So, we get(54)|λy1+y2−ℓ1+λy1+y2−ℓ22−Γα+12ℓ2−ℓ1αIy1+y2−ℓ2+α:ψλy1+y2−ℓ1+Iy1+y2−ℓ1−α:ψλy1+y2−ℓ2|≤12ℓ2−ℓ1α∫ψ−1y1+y2−ℓ2ψ−1y1+y2−ℓ1ψγ−y1+y2−ℓ2α−y1+y2−ℓ1−ψγα×λ′∘ψγψ′γdγ=ℓ2−ℓ12∫01ζα−1−ζαλ′y1+y2−ζℓ1+1−ζℓ2dζ≤ℓ2−ℓ12∫01ζα−1−ζαλ′y1+λ′y2−ζλ′ℓ1+1−ζλ′ℓ2dζ=ℓ2−ℓ12I1+I2,where(55)I1=∫01/21−ζα−ζαλ′y1+λ′y2−ζλ′ℓ1+1−ζλ′ℓ2dζ=λ′y1+λ′y21α+1−2−αα+1−λ′ℓ11α+1α+2−2−α−1α+1+λ′ℓ21α+2−2−α−1α+1,(56)I2=∫1/21ζα−1−ζαλ′y1+λ′y2−ζλ′ℓ1+1−ζλ′ℓ2dζ=λ′y1+λ′y21α+1−2−αα+1−λ′ℓ11α+2−2−α−1α+1+λ′ℓ21α+1α+2−2−α−1α+1.
Substituting (55) and (56) in (54), we get (53).Remark 8.
Forℓ1=y1 and ℓ2=y2 in Theorem 5, we will get Theorem 3.4 proved in [30].Remark 9.
Takingψγ=γ in Theorem 5, we will get Theorem 4 proved in [27].Lemma 2.
IfA1 is satisfied and λ:y1,y2⟶ℜ is a differentiable function on L1y1,y2, then(57)λy1+y2−ℓ1+λy1+y2−ℓ22−2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ2+α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22+Iψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22=ℓ2−ℓ14∫01ζαλ′y1+y2−1+ζ2ℓ1+1−ζ2ℓ2dζ−∫01ζαλ′y1+y2−1−ζ2ℓ1+1+ζ2ℓ2dζ.Proof.
It suffices to note that(58)I=ℓ2−ℓ14I1−I2,where(59)I1=∫01ζαλ′y1+y2−1+ζ2ℓ1+1−ζ2ℓ2dζ=2ℓ2−ℓ1λy1+y2−ℓ1−2αℓ2−ℓ1∫01ζα−1λy1+y2−1+ζ2ℓ1+1−ζ2ℓ2dζ=2ℓ2−ℓ1λy1+y2−ℓ1−2α+1Γα+1ℓ2−ℓ1α+1Iψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22,(60)I2=∫01ζαλy1+y2−1−ζ2ℓ1+1+ζ2ℓ2dζ=−2ℓ2−ℓ1λy1+y2−ℓ2−2αℓ2−ℓ1∫01ζα−1λy1+y2−1−ζ2ℓ1+1+ζ2ℓ2dζ=−2ℓ2−ℓ1λy1+y2−ℓ2−2α+1Γα+1ℓ2−ℓ1α+1Iψ−1y1+y2−ℓ2+α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22.
Substituting (59) and (60) in (58), we get (57).Remark 10.
Takingψγ=γ in Lemma 2, we will get Lemma 1 proved in [29].Theorem 6.
IfA1 is satisfied and λ′ is a convex function on y1,y2, then(61)|λy1+y2−ℓ1+λy1+y2−ℓ22−2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ2+α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22+Iψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22|≤ℓ2−ℓ14α+2supξ∈y1,y2λ″ξ,for all ℓ1,ℓ2∈y1,y2.Proof.
From Lemma2 and using mean value theorem for λ′, we have(62)λy1+y2−ℓ1+λy1+y2−ℓ22−2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ2+α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22+Iψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22=ℓ2−ℓ124∫01ζα+1λ″ξdζ,where ξ∈y1,y2. This leads us to(63)|λy1+y2−ℓ1+λy1+y2−ℓ22−2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ2+α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22+Iψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22|≤ℓ2−ℓ124∫01ζα+1λ″ξdζ≤ℓ2−ℓ124supξ∈y1,y2λ″ξ∫01ζα+1dζ=ℓ2−ℓ124α+2supξ∈y1,y2λ″ξ.Remark 11.
Forψγ=γ in Theorem 6, we will get Theorem 3 proved in [29].Theorem 7.
IfA1 is satisfied and λ′ is a convex function on y1,y2, then(64)|λy1+y2−ℓ1+λy1+y2−ℓ22−2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ2+α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22+Iψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22|≤ℓ2−ℓ12α+1λ′y1+λ′y2−λ′ℓ1+λ′ℓ22,for all ℓ1,ℓ2∈y1,y2.Proof.
By using Lemma2, properties of modulus, and Jensen–Mercer inequality, we have(65)|λy1+y2−ℓ1+λy1+y2−ℓ22−2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ2+α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22+Iψ−1y1+y2−ℓ1−α:ψλ∘ψψ−1y1+y2−ℓ1+ℓ22|≤ℓ2−ℓ14∫01ζαλ′y1+y2−1+ζ2ℓ1+1−ζ2ℓ2dζ+∫01ζαλ′y1+y2−1−ζ2ℓ1+1+ζ2ℓ2dζ≤ℓ2−ℓ14∫01ζαλ′y1+λ′y2−1+ζ2λ′ℓ1+1−ζ2λ′ℓ2dζ+∫01ζαλ′y1+λ′y2−1−ζ2λ′ℓ1+1+ζ2λ′ℓ2dζ,and after integration, we get required result.Remark 12.
Forψγ=γ in Theorem 7, we will get Theorem 4 proved in [29].Lemma 3.
IfA1 is satisfied and λ:y1,y2⟶ℜ is a differentiable function on L1y1,y2, then(66)λy1+y2−ℓ1+ℓ22−2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ1+ℓ2/2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ∘ψψ−1y1+y2−ℓ2=ℓ2−ℓ14∫01ζαλ′y1+y2−ζ2ℓ1+2−ζ2ℓ2dζ−∫01ζαλ′y1+y2−2−ζ2ℓ1+ζ2ℓ2dζ.Proof.
See the proof of Lemma2.Remark 13.
Forψγ=γ, ℓ1=y1, and ℓ2=y1 in Lemma 3, we will get Lemma 3 proved in [31].Remark 14.
Forψγ=γ in Lemma 3, we will get Lemma 2 proved in [27].Theorem 8.
IfA1 is satisfied and λ′ is a convex function on y1,y2, then(67)|λy1+y2−ℓ1+ℓ22−2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ1+ℓ2/2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ∘ψψ−1y1+y2−ℓ2|≤ℓ2−ℓ12α+1λ′y1+λ′y2−λ′ℓ1+λ′ℓ22,for all ℓ1,ℓ2∈y1,y2.Proof.
By using Lemma3, properties of modulus, and Jensen–Mercer inequality, we have(68)|λy1+y2−ℓ1+ℓ22−2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ1+ℓ2/2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ∘ψψ−1y1+y2−ℓ2|≤ℓ2−ℓ14∫01ζαλ′y1+y2−ζ2ℓ1+2−ζ2ℓ2dζ+∫01ζαλ′y1+y2−2−ζ2ℓ1+ζ2ℓ2dζ≤ℓ2−ℓ14∫01ζαλ′y1+λ′y2−ζ2λ′ℓ1+2−ζ2λ′ℓ2dζ+∫01ζαλ′y1+λ′y2−2−ζ2λ′ℓ1+ζ2λ′ℓ2dζ.
After integration, we get required result.Remark 15.
Takingψγ=γ in Theorem 8, we will get Theorem 5 proved in [27].Theorem 9.
IfA1 is satisfied and λ′q is convex function, then(69)|λy1+y2−ℓ1+ℓ22−2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ1+ℓ2/2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ∘ψψ−1y1+y2−ℓ2|≤ℓ2−ℓ141pα+11/pλ′y1q+λ′y2q−14λ′ℓ1q+34λ′ℓ2q1/q+λ′y1q+λ′y2q−34λ′ℓ1q+14λ′ℓ2q1/q,where q>1 and 1/p+1/q=1 for all ℓ1,ℓ2∈y1,y2.Proof.
Applying Lemma3, Hölder and Jensen–Mercer inequalities, the fact that λ′q is convex function, and properties of modulus, we have(70)|λy1+y2−ℓ1+ℓ22−2α−1Γα+1ℓ2−ℓ1α×Iψ−1y1+y2−ℓ1+ℓ2/2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ°ψψ−1y1+y2−ℓ2|≤ℓ2−ℓ14∫01ζαλ′y1+y2−ζ2ℓ1+2−ζ2ℓ2dζ+ℓ2−ℓ14∫01ζαλ′y1+y2−2−ζ2ℓ1+ζ2ℓ2dζ≤ℓ2−ℓ14∫01ζpαdζ1/p∫01λ′y1+y2−ζ2ℓ1+2−ζ2ℓ2qdζ1/q+ℓ2−ℓ14∫01ζpαdζ1/p∫01λ′y1+y2−2−ζ2ℓ1+ζ2λ′ℓ2qdζ1/q≤ℓ2−ℓ141pα+11/pλ′y1q+λ′y2q−14λ′ℓ1q+34λ′ℓ2q1/q+ℓ2−ℓ141pα+11/pλ′y1q+λ′y2q−34λ′ℓ1q+14λ′ℓ2q1/q.
After further simplifications, we get required result.Remark 16.
Forψγ=γ in Theorem 9, we will get Theorem 6 proved in [27].Lemma 4.
IfA1 is satisfied and λ:y1,y2⟶ℜ is a twice differentiable function on L1y1,y2, then(71)2α−1Γα+1ℓ2−ℓ1αIψ−1y1+y2−ℓ1+ℓ2/2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ∘ψψ−1y1+y2−ℓ2−λy1+y2−ℓ1+ℓ22=ℓ2−ℓ128α+1∫011−ζα+1λ″y1+y2−1+ζ2ℓ1+1−ζ2ℓ2dζ+∫011−ζα+1λ″y1+y2−1−ζ2ℓ1+1+ζ2ℓ2dζ.Proof.
It suffices to note that(72)I=ℓ2−ℓ128α+1I1+I2,where(73)I1=∫011−ζα+1λ″y1+y2−1+ζ2ℓ1+1−ζ2ℓ2dζ=−2ℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22+2α+1ℓ2−ℓ1∫011−ζαλ′y1+y2−1+ζ2ℓ1+1−ζ2ℓ2dζ=−2ℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22−4α+1ℓ2−ℓ12λy1+y2−ℓ1+ℓ22+4αα+1ℓ2−ℓ12∫011−ζα−1λy1+y2−1+ζ2ℓ1+1−ζ2ℓ2dζ=−2ℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22−4α+1ℓ2−ℓ12λy1+y2−ℓ1+ℓ22+2α+2Γα+2ℓ2−ℓ1α+2Iψ−1y1+y2−ℓ1+ℓ2/2+α:ψλ∘ψψ−1y1+y2−ℓ1,(74)I2=∫011−ζα+1λ″y1+y2−1−ζ2ℓ1+1+ζ2ℓ2dζ=2ℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22−2α+1ℓ2−ℓ1∫011−ζαλ′y1+y2−1−ζ2ℓ1+1+ζ2ℓ2dζ=2ℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22−4α+1ℓ2−ℓ12λy1+y2−ℓ1+ℓ22+4αα+1ℓ2−ℓ12∫011−ζα−1λy1+y2−1−ζ2ℓ1+1+ζ2ℓ2dζ=2ℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22−4α+1ℓ2−ℓ12λy1+y2−ℓ1+ℓ22+2α+2Γα+2ℓ2−ℓ1α+2Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ∘ψψ−1y1+y2−ℓ2.
Substituting (73) and (74) in (72), we get (71).Corollary 1.
If we setℓ1=y1 and ℓ2=y2, we get(75)2α−1Γα+1y2−y1αIψ−1y1+y2/2+α:ψλ∘ψψ−1y2+Iψ−1y1+y2/2−α:ψλ∘ψψ−1y1−λy1+y22=y2−y128α+1∫011−ζα+1λ″1+ζ2y1+1−ζ2y2dζ+∫011−ζα+1λ″1−ζ2y1+1+ζ2y2dζ.Remark 17.
If we setψγ=γ in Lemma 4, we get Lemma 2 of [29].
Moreover, if we setℓ1=y1 and ℓ2=y2, we obtain Lemma 1 of [32].Remark 18.
Forψγ=γ, α=1, ℓ1=y1, and ℓ2=y2 in Lemma 4, it reduces to Lemma 2 proved in [32].Theorem 10.
IfA1 is satisfied and λ″ is a convex function on y1,y2, then(76)|2α−1Γα+1ℓ2−ℓ1αIψ−1y1+y2−ℓ1+ℓ2/2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ∘ψψ−1y1+y2−ℓ2−λy1+y2−ℓ1+ℓ22|≤ℓ2−ℓ124α+1α+2λ″y1+λ″y2−λ″ℓ1+λ″ℓ22.Proof.
By using Lemma4, properties of modulus, and Jensen–Mercer inequality, we have(77)|2α−1Γα+1ℓ2−ℓ1αIψ−1y1+y2−ℓ1+ℓ2/2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ∘ψψ−1y1+y2−ℓ2−λy1+y2−ℓ1+ℓ22|≤ℓ2−ℓ128α+1∫011−ζα+1λ″y1+y2−1+ζ2ℓ1+1−ζ2ℓ2dζ+∫011−ζα+1λ″y1+y2−1−ζ2ℓ1+1+ζ2ℓ2dζ≤ℓ2−ℓ128α+1∫011−ζα+1λ″y1+λ″y2−1+ζ2λ″ℓ1+1−ζ2λ″ℓ2dζ+∫011−ζα+1λ″y1+λ″y2−1−ζ2λ″ℓ1+1+ζ2λ″ℓ2dζ,and after integration, we get required result.Corollary 2.
If we setℓ1=y1 and ℓ2=y2 in Theorem 10, we get(78)|2α−1Γα+1y2−y1αIψ−1y1+y2/2+α:ψλ∘ψψ−1y2+Iψ−1y1+y2/2−α:ψλ∘ψψ−1y1−λy1+y22|≤y2−y124α+1α+2λ″y1+λ″y22.Remark 19.
If we setψγ=γ in Theorem 10, we obtain Theorem 5 of [29].
Moreover, if we setℓ1=y1 and ℓ2=y2, we get Theorem 5 of [32].Corollary 3.
If we setψγ=γ, ℓ1=y1, ℓ2=y2, and α=1 in Theorem 10, we get Proposition 1 of [33]:(79)1y2−y1∫y1y2λxdx−λy1+y22≤y2−y1224λ″y1+λ″y22.Theorem 11.
IfA1 is satisfied and λ″q is convex function, then(80)|2α−1Γα+1ℓ2−ℓ1αIψ−1y1+y2−ℓ1+ℓ2/2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ∘ψψ−1y1+y2−ℓ2−λy1+y2−ℓ1+ℓ22|≤ℓ2−ℓ128α+11pα+1+11/pλ″y1q+λ″y2q−3λ″ℓ1q+λ″ℓ2q41/q+λ″y1q+λ″y2q−λ″ℓ1q+3λ″ℓ2q41/q,where q>1 and 1/p+1/q=1 for all ℓ1,ℓ2∈y1,y2.Proof.
From Lemma4, Hölder and Jensen–Mercer inequalities, the fact that λ″q is convex function, and properties of modulus, we have(81)|2α−1Γα+1ℓ2−ℓ1αIψ−1y1+y2−ℓ1+ℓ22+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ∘ψψ−1y1+y2−ℓ2−λy1+y2−ℓ1+ℓ22|≤ℓ2−ℓ128α+1∫011−ζα+1λ″y1+y2−1+ζ2ℓ1+1−ζ2ℓ2dζ+∫011−ζα+1λ″y1+y2−1−ζ2ℓ1+1+ζ2ℓ2dζ≤ℓ2−ℓ128α+1∫011−ζpα+1dζ1/p∫01λ″y1+y2−1+ζ2ℓ1+1−ζ2ℓ2qdζ1/q+∫011−ζpα+1dζ1/p∫01λ″y1+y2−1−ζ2ℓ1+1+ζ2ℓ2qdζ1/q≤ℓ2−ℓ128α+1∫011−ζpα+1dζ1/p×∫01λ″y1q+λ″y2q−1+ζ2λ″ℓ1q−1−ζ2λ″ℓ2qdζ1/q+∫01λ″y1q+λ″y2q−1−ζ2λ″ℓ1q−1+ζ2λ″ℓ2qdζ1/q=ℓ2−ℓ128α+11pα+1+11/pλ″y1q+λ″y2q−3λ″ℓ1q+λ″ℓ2q41/q+λ″y1q+λ″y2q−λ″ℓ1q+3λ″ℓ2q4q1/q.Remark 20.
If we setψγ=γ in Theorem 11, we get Theorem 6 of [29].Lemma 5.
IfA1 satisfied and λ:y1,y2⟶ℜ is a twice differentiable function on L1y1,y2, then(82)2α−1Γα+1ℓ2−ℓ1αIψ−1y1+y2−ℓ1+ℓ2/2+α:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ∘ψψ−1y1+y2−ℓ2−λy1+y2−ℓ1+ℓ22=ℓ2−ℓ128α+1∫01ζα+1λ″y1+y2−2−ζ2ℓ1+ζ2ℓ2dζ+∫01ζα+1λ″y1+y2−ζ2ℓ1+2−ζ2ℓ2dζ.Proof.
It suffices to note that(83)I=ℓ2−ℓ128α+1I1+I2,where(84)I1=∫01ζα+1λ″y1+y2−2−ζ2ℓ1+ζ2ℓ2dζ=−2ℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22+2α+1ℓ2−ℓ1∫01ζαλ′y1+y2−2−ζ2ℓ1+ζ2ℓ2dζ=−2ℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22−4α+1ℓ2−ℓ12λy1+y2−ℓ1+ℓ22+4αα+1ℓ2−ℓ12∫01ζα−1λy1+y2−2−ζ2ℓ1+ζ2ℓ2dζ=−2ℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22−4α+1ℓ2−ℓ12λy1+y2−ℓ1+ℓ22+2α+2Γα+2ℓ2−ℓ1α+2Iψ−1y1+y2−ℓ1+ℓ2/2+α:ψλ∘ψψ−1y1+y2−ℓ1,(85)I2=∫01ζα+1λ″y1+y2−ζ2ℓ1+2−ζ2ℓ2dζ=2ℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22−2α+1ℓ2−ℓ1∫01ζαλ′y1+y2−ζ2ℓ1+2−ζ2ℓ2dζ=2ℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22−4α+1ℓ2−ℓ12λy1+y2−ℓ1+ℓ22+4αα+1ℓ2−ℓ12∫01ζα−1λy1+y2−ζ2ℓ1+2−ζ2ℓ2dζ=2ℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22−4α+1ℓ2−ℓ12λy1+y2−ℓ1+ℓ22+2α+2Γα+2ℓ2−ℓ1α+2Iψ−1y1+y2−ℓ1+ℓ2/2−α:ψλ∘ψψ−1y1+y2−ℓ2.
Substituting (84) and (85) in (83), we get (82).Corollary 4.
If we setℓ1=y1 and ℓ2=y2, we get(86)2α−1Γα+1y2−y1αIψ−1y1+y2/2+α:ψλ∘ψψ−1y2+Iψ−1y1+y2/2−α:ψλ∘ψψ−1y1−λy1+y22=y2−y128α+1∫01ζα+1λ″2−ζ2y1+ζ2y2dζ+∫01ζα+1λ″ζ2y1+2−ζ2y2dζ.Corollary 5.
If we setψγ=γ, we get(87)2α−1Γα+1ℓ2−ℓ1αIy1+y2−ℓ1+ℓ2/2+αλy1+y2−ℓ1+Iy1+y2−ℓ1+ℓ2/2−αλy1+y2−ℓ2−λy1+y2−ℓ1+ℓ22=ℓ2−ℓ128α+1∫01ζα+1λ″y1+y2−2−ζ2ℓ1+ζ2ℓ2dζ+∫01ζα+1λ″y1+y2−ζ2ℓ1+2−ζ2ℓ2dζ.
Moreover, if we setℓ1=y1 and ℓ2=y2, we get(88)2α−1Γα+1y2−y1αIy1+y2/2+αλy2+Iy1+y2/2−αλy1−λy1+y22=y2−y128α+1∫01ζα+1λ″2−ζ2y1+ζ2y2dζ+∫01ζα+1λ″ζ2y1+2−ζ2y2dζ.Remark 21.
By using Lemma5, we can get the same results of Theorems 10 and 11, so we omit their proof here.Lemma 6.
IfA1 is satisfied and λ:y1,y2⟶ℜ is a twice differentiable function on L1y1,y2, then(89)2α−2Γαℓ2−ℓ1α−1Iψ−1y1+y2−ℓ1+ℓ2/2+α−1:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α−1:ψλ∘ψψ−1y1+y2−ℓ2−λy1+y2−ℓ1+ℓ22=ℓ2−ℓ12α⋅22−α∫01/2ζαλ″y1+y2−ζℓ2+1−ζℓ1dζ+∫1/211−ζαλ″y1+y2−ζℓ2+1−ζℓ1dζ.Proof.
It suffices to note that(90)I=ℓ2−ℓ12α⋅21−αI1+I2,where(91)I1=∫01/2ζαλ″y1+y2−ζℓ2+1−ζℓ1dζ=−12αℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22+αℓ2−ℓ1∫01/2ζα−1λ′y1+y2−1−ζℓ1+ζℓ2dζ=−12αℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22−α2α−1ℓ2−ℓ12λy1+y2−ℓ1+ℓ22+αα−1ℓ2−ℓ12∫01/2ζα−2λy1+y2−1−ζℓ1+ζℓ2dζ=−12αℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22−α2α−1ℓ2−ℓ12λy1+y2−ℓ1+ℓ22+Γα+1ℓ2−ℓ1α+1Iψ−1y1+y2−ℓ1+ℓ2/2+α−1:ψλ∘ψψ−1y1+y2−ℓ1,(92)I2=∫1/211−ζαλ″y1+y2−ζℓ2+1−ζℓ1dζ=12αℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22−αℓ2−ℓ1∫1/211−ζα−1λ′y1+y2−1−ζℓ1+ζℓ2dζ=12αℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22+α2α−1ℓ2−ℓ12λy1+y2−ℓ1+ℓ22+αα−1ℓ2−ℓ12∫1/211−ζα−2λy1+y2−1−ζℓ1+ζℓ2dζ=12αℓ2−ℓ1λ′y1+y2−ℓ1+ℓ22−α2α−1ℓ2−ℓ12λy1+y2−ℓ1+ℓ22+Γα+1ℓ2−ℓ1α+1Iψ−1y1+y2−ℓ1+ℓ2/2−α−1:ψλ∘ψψ−1y1+y2−ℓ2.
Substituting (91) and (92) in (90), we get (89).Corollary 6.
If we setℓ1=y1 and ℓ2=y2 in Lemma 6, we get(93)2α−2Γαy2−y1α−1Iψ−1y1+y2/2+α:ψλ∘ψψ−1y2+Iψ−1y1+y2/2−α:ψλ∘ψψ−1y1−λy1+y22=y2−y12α⋅22−α∫01/2ζαλ″ζy1+1−ζy2dζ+∫1/211−ζαλ″ζy1+1−ζy2dζ.Corollary 7.
If we setψγ=γ in Lemma 6, we get(94)2α−2Γαℓ2−ℓ1α−1Jy1+y2−ℓ1+ℓ2/2+αλy1+y2−ℓ1+Jy1+y2−ℓ1+ℓ2/2−αλy1+y2−ℓ2−λy1+y2−ℓ1+ℓ22=ℓ2−ℓ12α⋅22−α∫01/2ζαλ″y1+y2−ζℓ2+1−ζℓ1dζ+∫1/211−ζαλ″y1+y2−ζℓ2+1−ζℓ1dζ.
Moreover, if we setℓ1=y1 and ℓ2=y2, we obtain Lemma 2.1 of [34] for m=1.Theorem 12.
IfA1 is satisfied and λ″ is a convex function on y1,y2, then(95)|2α−2Γαℓ2−ℓ1α−1Iψ−1y1+y2−ℓ1+ℓ2/2+α−1:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α−1:ψλ∘ψψ−1y1+y2−ℓ2−λy1+y2−ℓ1+ℓ22|≤ℓ2−ℓ124αα+1λ″y1+λ″y2−λ″ℓ1+λ″ℓ22.Proof.
By using Lemma6, properties of modulus, and Jensen–Mercer inequality, we have(96)|2α−2Γαℓ2−ℓ1α−1Iψ−1y1+y2−ℓ1+ℓ2/2+α−1:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α−1:ψλ∘ψψ−1y1+y2−ℓ2−λy1+y2−ℓ1+ℓ22|≤ℓ2−ℓ12α⋅22−α∫01/2ζαλ″y1+y2−1−ζℓ1+ζℓ2dζ+∫1/211−ζαλ″y1+y2−1−ζℓ1+ζℓ2dζ≤ℓ2−ℓ12α⋅22−α∫01/2ζαλ″y1+λ″y2−1−ζλ″ℓ1+ζλ″ℓ2dζ+∫1/211−ζαλ″y1+λ″y2−1−ζλ″ℓ1+ζλ″ℓ2dζ,and after integration, we get required result.Corollary 8.
If we setℓ1=y1 and ℓ2=y2 in Theorem 12, we get(97)|2α−2Γαy2−y1α−1Iψ−1y1+y2/2+α−1:ψλ∘ψψ−1y2+Iψ−1y1+y2/2−α−1:ψλ∘ψψ−1y1−λy1+y22|≤y2−y128αα+1λ″y1+λ′′y2.Corollary 9.
If we setψγ=γ in Theorem 12, we get(98)|2α−2Γαℓ2−ℓ1α−1Jy1+y2−ℓ1+ℓ2/2+α−1λy1+y2−ℓ1+Jy1+y2−ℓ1+ℓ2/2−α−1λy1+y2−ℓ2−λy1+y2−ℓ1+ℓ22|≤ℓ2−ℓ124αα+1λ″y1+λ″y2−λ″ℓ1+λ″ℓ22.Remark 22.
If we setψγ=γ, ℓ1=y1, and ℓ2=y2 in Theorem 12, we get Theorem 2.1 of [34].
Moreover, if we setα=2, we obtain Proposition 1 of [33].Theorem 13.
IfA1 is satisfied and λ″q is convex function, then(99)|2α−2Γαℓ2−ℓ1α−1Iψ−1y1+y2−ℓ1+ℓ2/2+α−1:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α−1:ψλ∘ψψ−1y1+y2−ℓ2−λy1+y2−ℓ1+ℓ22|≤ℓ2−ℓ12α⋅22−α12pα+1pα+11/pλ″y1q+λ″y2q2−3λ″ℓ1q+λ″ℓ2q81/q+λ″y1q+λ″y2q2−λ″ℓ1q+3λ″ℓ2q81/q,where q>1 and 1/p+1/q=1 for all ℓ1,ℓ2∈y1,y2.Proof.
Applying Lemma6, Hölder and Jensen–Mercer inequalities, the fact that λ″q is convex function, and properties of modulus, we have(100)|2α−2Γαℓ2−ℓ1α−1Iψ−1y1+y2−ℓ1+ℓ2/2+α−1:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α−1:ψλ∘ψψ−1y1+y2−ℓ2−λy1+y2−ℓ1+ℓ22|≤ℓ2−ℓ12α⋅22−α∫01/2ζαλ″y1+y2−1−ζℓ1+ζℓ2dζ+∫1/211−ζαλ″y1+y2−1−ζℓ1+ζℓ2dζ≤ℓ2−ℓ12α⋅22−α∫01/2ζpαdζ1/p∫01/2λ″y1+y2−1−ζℓ1+ζℓ2qdζ1/q+∫1/211−ζpαdζ1/p∫1/21λ″y1+y2−1−ζℓ1+ζℓ2qdζ1/q≤ℓ2−ℓ12α⋅22−α12pα+1pα+11/p×∫01/2λ″y1q+λ″y2q−1−ζλ″ℓ1q−ζλ″ℓ2qdζ1/q+∫1/21λ″y1q+λ″y2q−1−ζλ″ℓ1q−ζλ″ℓ2qdζ1/q=ℓ2−ℓ12α⋅22−α12pα+1pα+11/pλ″y1q+λ″y2q2−3λ″ℓ1q+λ″ℓ2q81/q+λ″y1q+λ″y2q2−λ″ℓ1q+3λ″ℓ2q81/q.Corollary 10.
If we setℓ1=y1 and ℓ2=y2 in Theorem 12, we get(101)|2α−2Γαy2−y1α−1Iψ−1y1+y2/2+α−1:ψλ∘ψψ−1y2+Iψ−1y1+y2/2−α−1:ψλ∘ψψ−1y1−λy1+y22|≤ℓ2−ℓ12α⋅22−α12pα+1pα+11/p|λ′′y1|q+3|λ′′y2|q81/q+3λ″y1q+λ″y2q81/q.Corollary 11.
If we setψγ=γ in Theorem 12, we get(102)|2α−2Γαℓ2−ℓ1α−1Jy1+y2−ℓ1+ℓ2/2+α−1λy1+y2−ℓ1+Jy1+y2−ℓ1+ℓ2/2−α−1λy1+y2−ℓ2−λy1+y2−ℓ1+ℓ22|≤ℓ2−ℓ12α⋅22−α12pα+1pα+11/pλ″y1q+λ″y2q2−3λ″ℓ1q+λ″ℓ2q81/q+λ″y1q+λ″y2q2−λ″ℓ1q+3λ″ℓ2q81/q.Theorem 14.
IfA1 is satisfied and λ″q is convex function, then(103)|2α−2Γαℓ2−ℓ1α−1Iψ−1y1+y2−ℓ1+ℓ2/2+α−1:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α−1:ψλ∘ψψ−1y1+y2−ℓ2−λy1+y2−ℓ1+ℓ22|≤ℓ2−ℓ12α⋅22−α12α+1α+11−1/q×λ″y1q+λ″y2q2α+1α+1−α+3λ″ℓ1q2α+2α+1α+2−λ″ℓ2q2α+2α+21/q+λ″y1q+λ″y2q2α+1α+1−λ″ℓ1q2α+2α+2−α+3λ″ℓ2q2α+2α+1α+21/q,where q≥1 for all ℓ1,ℓ2∈y1,y2.Proof.
From Lemma2, power-mean and Jensen–Mercer inequalities, the fact that λ″q is convex function, and properties of modulus, we have(104)|2α−2Γαℓ2−ℓ1α−1Iψ−1y1+y2−ℓ1+ℓ2/2+α−1:ψλ∘ψψ−1y1+y2−ℓ1+Iψ−1y1+y2−ℓ1+ℓ2/2−α−1:ψλ∘ψψ−1y1+y2−ℓ2−λy1+y2−ℓ1+ℓ22|≤ℓ2−ℓ12α⋅22−α∫01/2ζαλ″y1+y2−1−ζℓ1+ζℓ2dζ+∫1/211−ζαλ″y1+y2−1−ζℓ1+ζℓ2dζ≤ℓ2−ℓ12α⋅22−α∫01/2ζαdζ1−1/q×λ″y1q+λ″y2q∫01/2ζαdζ−λ″ℓ1q∫01/2ζα1−ζdζ−λ″ℓ2q∫01/2ζα+1dζ1/q+∫1/211−ζαdζ1−1/q×λ″y1q+λ″y2q∫1/211−ζαdζ−λ″ℓ1q∫1/21ζ1−ζαdζ−λ″ℓ2q∫1/211−ζα+1dζ1/q=ℓ2−ℓ12α⋅22−α12α+1α+11−1/q×λ″y1q+λ″y2q2α+1α+1−α+3λ″ℓ1q2α+2α+1α+2−λ″ℓ2q2α+2α+21/q+λ″y1q+λ″y2q2α+1α+1−λ″ℓ1q2α+2α+2−α+3λ″ℓ2q2α+2α+1α+21/q.Remark 23.
If we setψγ=γ, ℓ1=y1, and ℓ2=y2 in Theorem 14, we get Theorem 2.2 of [34] for m=s=1.
Moreover, if we setα=2, we obtain Proposition 5 of [33].


## 4. Application

In this last section, we will give an application of our results using modified Bessel function of the first kind.Let the functionℑp:ℜ⟶1,+∞ be defined by(105)ℑpℓ1=2pΓp+1ℓ1−vIpℓ1,p>1,ℓ1∈ℜ.For this, we recall the modified Bessel function of the first kindℑp which is defined as follows [35]:(106)ℑpℓ1=∑n≥0ℓ1/2p+2nn!Γp+n+1.The first and then th order derivative formula of ℑp is, respectively, given by the following [36]:(107)ℑp′ℓ1=ℓ12p+1ℑp+1ℓ1,(108)dnℑpℓ1dℓ1n=2n−2pπℓ1p−nΓp+1×F32p+12,p+22;p+1−n2,p+2−n2,p+1;ℓ124,where F32⋅,⋅;⋅,⋅,⋅;⋅ is the hypergeometric function defined by the following [36]:(109)F32p+12,p+22;p+1−n2,p+2−n2,p+1;ℓ124=∑k=0∞p+1/2kp+2/2kp−1/2kp−2/2kp+1k⋅ℓ12k4k⋅k!,and for some parameter v, the Pochhammer symbol vk is defined as(110)v0=1,vk=vv+1⋯v+k+1,k=1,2,3,….Proposition 1.
Let0<y1<y2 be real numbers and p>−1, then(111)ℑpy2−ℑpy1y2−y1−y1+y24p+1ℑp+1y1+y22≤y2−y124823−2pπΓp+1×y1p−3⋅F32p+12,p+22;p−22,p−12,p+1;y124+y2p−3⋅F32p+12,p+22;p−22,p−12,p+1;y224.≤y2−y124823−2pπΓp+1×y1p−3⋅F32p+12,p+22;p−22,p−12,p+1;y124+y2p−3⋅F32p+12,p+22;p−22,p−12,p+1;y224.Proof.
Letλℓ1=ℑp′ℓ1. Note that the function ℓ1⟶ℑ‴ℓ1 is convex on the interval 0,+∞ for each p>−1. Using Corollary 3 and relations (107) and (108), we obtain the desired inequality (111).Remark 24.
Using the same technique like Proposition1, we can obtain some new interesting inequalities pertaining modified Bessel function of the first kind or the well-known q–digamma function for q∈0,1 from our generic results. We omit here their proofs, and the details are left to the interested reader.


## 5. Conclusion

In this article, some new Hermite–Jensen–Mercer type inequalities involvingψ–Riemann–Liouville fractional integrals are found. Several ψ–Riemann–Liouville fractional integral inequalities using identities as auxiliary results are provided, and the known results are recaptured as special cases as well. Finally, the efficiency of our results is showed with an application via modified Bessel function of the first kind. We hope that current work using our idea and technique will attract the attention of researchers working in mathematical analysis and other related fields in pure and applied sciences.



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