# Some Generalized Fractional Integral Inequalities via Harmonic Convex Functions and Their Applications

Authors: Muhammad Uzair Awan; Nousheen Akhtar; Mustapha Raïssouli; Artion Kashuri; Muhammad Zakria Javed; Muhammad Aslam Noor
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

The aim of this paper is to derive from an auxiliary identity some new generalizations of fractional integral inequalities of Simpson, midpoint, and trapezoid using the class of harmonic convex functions. To show that our results are quite unifying, we discuss several new special cases. Finally, some applications regarding error estimations for the Simpson quadrature formula are presented to support our theoretical results.


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## Body

## 1. Introduction and Preliminaries

Fractional calculus was born on September 1695. Although the history of fractional calculus is very old, in recent years it has emerged as an interdisciplinary subject. Besides its great many applications in applied sciences, it has also played a significant role in modern analysis. Due to these facts, it received special attention by mathematicians, and as a result, a variety of new significant generalizations of the classical concepts of fractional calculus have been proposed in the literature (for details, see [1]). Since the emergence of fractional calculus, several researchers have started obtaining the fractional analogues of classical mathematical objects. For example, Sarikaya et al. [2] were the first to obtain some new fractional analogues of Hermite–Hadamard’s inequality. This idea opened a new direction of research for inequalities experts (for some more details, see [3–8]).We now recall some preliminary concepts that will be needed throughout this paper.LetS be a nonempty interval of ℝ. A function ϒ:S⟶ℝ is said to be convex, if(1)ϒ1−τk1+τk2≤1−τϒk1+τϒk2,∀k1,k2∈S,τ∈0,1.Recently, İşcan [9] introduced the class of harmonic convex functions asLetI⊂0,∞, be a nonempty interval. A function ϒ:I⟶ℝ, is said to be harmonic convex, if(2)ϒk1k2τk1+1−τk2≤1−τϒk1+τϒk2,∀k1,k2∈I,τ∈0,1.In order to explain more the concept of harmonic convex functions, we state the following remark, which may be of interest to the reader.Remark 1.
(i)
By virtue of the so-called arithmetic-harmonic mean inequality, namely,(3)k1!τk2≔k1k2τk1+1−τk2≤1−τk1+τk2:=k1∇τk2,it is easy to see that every convex increasing function onI is harmonic convex on I. However, a harmonic convex function is not necessary convex. For example, the concave function x↦logx is harmonic convex on 0,∞.(ii)
In terms of means, (2) can be simply written as(4)ϒk1!τk2≤ϒk1∇τϒk2,∀k1,k2∈I,τ∈0,1.With this, if we setΛx≔1/x for x>0, it is not hard to see that (2) is equivalent to(5)ϒ∘Λk1!τk2≤ϒ°Λk1∇τϒ°Λk2,∀k1,k2∈I,τ∈0,1.This means that,ϒ is harmonic convex on I if and only if ϒ°Λ is convex on I. It follows that every harmonic convex function on I is continuous on I.(iii)
According to (4), if ϒ is harmonic convex on I then we have(6)ϒk1!1−τ/2k2≤ϒk1∇1−τ/2ϒk2,∀k1,k2∈I,τ∈0,1,or, equivalently,(7)Υ2k1k21−τk1+1+τk2≤121+τϒk1+1−τϒk2,∀k1,k2∈I,τ∈0,1.
Now, let us recall some basic notions about fractional integrals. The Riemann–Liouville fractional integrals are defined asDefinition 1 (see [1]).
Letϒ∈L1k1,k2. The Riemann–Liouville fractional integrals Jk1+αϒ and Jk2−αϒ of order α>0 with k1≥0 are defined by(8)Jk1+αϒx=1Γα∫k1xx−τα−1ϒτdτ,x>k1,(9)Jk2−αϒx=1Γα∫xk2τ−xα−1ϒτdτ,x<k2,respectively, andΓαis the well-known Gamma function. Also, we defineJk1+0ϒx=Jk2−0ϒx=ϒx,
Thek-Riemann–Liouville fractional integrals are defined asDefinition 2 (see [10]).
LetΥ∈L1k1,k2. The k-Riemann–Liouville fractional integrals Jk1+,kαϒ and Jk2−,kαϒ of order α,k>0 with k1≥0 are given as follows:(10)Jk1+,kαϒx=1kΓkα∫k1xx−τα/k−1ϒτdτ,x>k1,(11)Jk2−,kαϒx=1kΓkα∫xk2τ−xα/k−1ϒτdτ,x<k2,respectively, and Γkα stands for the k-Gamma function.
In order to calculate integrals, we need hypergeometric functions. The integral form of the hypergeometric function is given asDefinition 3.
The hypergeometric functionℱ2k1,k2,k3,z has the following integral representation:(12)ℱ12k1,k2,k3,z=1βk2,k3−k2∫01τk2−11−τk3−k2−11−zτ−k1dτ,k3>k2>0,k1>0,where z<1 and β·,· refers to the Euler beta function.
Otherwise, Sarikaya and Ertugral [11] defined a new generalization of fractional integrals (which we call as Sarikaya fractional integral) as itemized in what follows.
Letφ:0,+∞⟶0,+∞ be a function satisfying the following conditions:(1)
∫01φτ/τdτ<+∞,(2)
φr/r2−φs/s2≤M3r−sφr/r2 for 1/2≤s/r≤2, where M1,M2, and M3 are independent of r,s>0.
Under the assumptions ofφ, the left-sided and the right-sided generalized fractional integrals are(13)Iφk1+k1,k2,k3,z=1βk2,k3−k2∫01τk2−11−τk3−k2−11−zτ−k1dτ,k3>k2>0,k1>0,ϒx=∫k1xφx−τx−τϒτdτ,x>k1,(14)Iφk2−ϒx=∫xk2φτ−xτ−xϒτdτ,x<k2.
Sarikaya’s fractional integrals are the generalization of some well-known fractional integrals like the Riemann–Liouville fractional integrals [1], k-Riemann–Liouville fractional integrals [10], Katugampola fractional integrals [12], conformable fractional integrals [7], etc.(1)
If we takeφτ=τ in operators (13) and (14), we have the classical Riemann integrals.(2)
If we chooseφτ=τα/Γα in operators (13) and (14), we get the Riemann–Liouville fractional integrals (see [1]).(3)
If we substituteφτ=τα/k/kΓkα in operators (13) and (14), we obtain the k-Riemann–Liouville fractional integrals (see [10]).(4)
If we takeφτ=τx−τα−1 in operators (13) and (14), we have the conformable fractional integrals, which were defined by Khalil et al. in [13].(5)
If we chooseφτ=τ/αexp−1−α/ατ for α∈0,1, in operators (13) and (14), we get the left-sided and the right-sided fractional integrals with the exponential kernel, which were defined in [11, 12].
To highlight our goal in this paper, let us recall the following results obtained by Dragomiret al. in [14]:Theorem 1.
Let us assume thatΥ:k1,k2⟶ℝ is a four times differentiable function on k1,k2, such that ϒ4∞≔supτ∈k1,k2Υ4τ<+∞ with k1<k2. Then, the following integral inequality holds:(15)16ϒk1+4ϒk1+k22+ϒk2−1k2−k1∫k1k2ϒτdτ≤12880k2−k14ϒ4∞.Just like the aforementioned inequalities, several integral inequalities related to Simpson’s integral inequality (15) have been found for convex functions (see [15–22]). But, our fundamental target in this paper is on another type of inequality, namely, the Simpson’s inequality for fractional integrals, by using the concept of harmonic convex functions.The remainder of this paper will be organized as follows: in Section2 we state an auxiliary lemma which will be a primordial tool for establishing our main results. Motivated by this lemma, we derive in Section 3 some new generalizations of fractional integral inequalities of Simpson, midpoint, and trapezoid type using a class of harmonic convex functions. To show that our results are quite unifying, we will discuss several new special cases. Finally, Section 4 displays some applications regarding error estimations of the Simpson quadrature formula as support for our theoretical results. We hope that the ideas and the techniques developed in this paper will inspire interested readers working in this field.


## 2. A New Auxiliary Result

In this section, we derive a key lemma that will help us in deriving our main results.Lemma 1.
Letφ:0,+∞⟶0,+∞ be as in the previous section. Let ϒ:k1,k2⟶ℝ be a differentiable function, with 0<k1<k2. If ϒ′ is integrable on k1,k2, then for ρ,σ≥0, we have(16)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k1+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2=2k1k2k2−k1Δ1∫01Δτ−Δ1ρ1+τk1+1−τk22ϒ′2k1k21+τk1+1−τk2dτ+∫01Δ1σ−Δτ1−τk1+1+τk22ϒ′2k1k21−τk1+1+τk2dτ,whereas before Λx≔1/x, and the function Δ:0,1⟶ℝ is defined by(17)Δτ≔Δφ,τk1,k2≔∫0τφk2−k1/2k2k1uudu.Proof.
First, we mention that the functionΔ defined by (17) is continuous on 0,1. By the standard rule of integration by parts, and using (13) and (14) with some algebraic operations, we have(18)∫01Δτ−Δ1ρ1+τk1+1−τk22ϒ′2k1k21+τk1+1−τk2dτ=12k1k2k2−k1Δ11−ρϒk2+ρϒ2k1k2k1+k2−Iφ1/k2+ϒ∘Λ2k1k2k1+k2,(19)∫01Δ1σ−Δτ1−τk1+1+τk22ϒ′2k1k21−τk1+1+τk2dτ=12k1k2k2−k1Δ11−σϒk1+σϒ2k1k2k1+k2−1/k1−Iφϒ∘Λ2k1k2k1+k2.
Adding equalities (18) and (19) and multiplying by the factor 2k1k2k2−k1/Δ1, we obtain the required result.Remark 2.
(i)
For the sake of simplicity, we write in the following1/k1− and 1/k2+ instead of 1/k1− and 1/k2+, respectively.(ii)
The previous lemma stems its importance in the fact that it includes a large class of examples and situations. Indeed, first, the parametersρ≥0 and σ≥0 could be chosen in an arbitrary manner. Secondly, as previously mentioned, φ belongs to a large class of functions. And thirdly, as explained in Remark 1, ϒ could be easily chosen as a harmonic convex function. Let us explain more these latter points in what follows.(i)
Taking special values forρ and σ in Lemma 1, we obtain the following:(1)
Ifρ=σ=2/3, then(20)16ϒk1+4ϒ2k1k2k1+k2+ϒk2−12Δ1Iφ1/k2+ϒ°Λ2k1k2k1+k2+Iφ1/k1+ϒ°Λ2k1k2k1+k2=2k1k2k2−k1Δ1∫01Δτ/2−Δ1/31+τk1+1−τk22ϒ′2k1k21+τk1+1−τk2dτ+∫01Δ1/3−Δτ/21−τk1+1+τk22ϒ′2k1k21−τk1+1+τk2dτ.(2)
Forρ=σ=1, we get(21)ϒ2k1k2k1+k2−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k2−ϒ∘Λ2k1k2k1+k2=k1k2k2−k1Δ1∫01Δτ−Δ11+τk1+1−τk22ϒ′2k1k21+τk1+1−τk2dτ+∫01Δ1−Δτ1−τk1+1+τk22ϒ′2k1k21−τk1+1+τk2dτ.(3)
Withρ=σ=0, we have(22)ϒk1+ϒk22−12Δ11k2+Iφϒ∘Λ2k1k2k1+k2+1k2+Iφϒ∘Λ2k1k2k1+k2=k1k2k2−k1Δ1∫01Δτ1+τk1+1−τk22ϒ′2k1k21+τk1+1−τk2dτ⋅∫01Δτ1−τk1+1+τk22−ϒ′2k1k21−τk1+1+τk2dτ.(ii)
We now discuss some other variants for Lemma1 when choosing special cases for φ as follows:(I)
If we takeφτ=τ, then(23)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2+2k1k2k2−k1∫1/k11/k2ϒ∘Λ1xdx=2k2k1k2−k1∫01τ−ρ1+τk1+1−τk22ϒ′2k1k21+τk1+1−τk2dτ+∫01σ−τ1−τk1+1+τk22ϒ′2k1k21−τk1+1+τk2dτ.(II)
If we chooseφτ=τα/Γα, then(24)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−2k1k2αΓα+1k2−k1αJ1/k2+αϒ°Λ2k1k2k1+k2+J1/k1−αϒ°Λ2k1k2k1+k2=2k2k1k2−k1∫01τα−ρ1+τk1+1−τk22ϒ′2k1k21+τk1+1−τk2dτ+∫01σ−τα1−τk1+1+τk22ϒ′2k1k21−τk1+1+τk2dτ.(III)
Withφτ=τα/k/kΓkα, we have(25)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−2k1k2α/kΓkα+kk2−k1α/kJ1/k2+,kαϒ∘Λ2k1k2k1+k2+J1/k1−,kαϒ∘Λ2k1k2k1+k2=2k2k1k2−k1∫01τα/k−ρ1+τk1+1−τk22ϒ′2k1k21+τk1+1−τk2dτ+∫01σ−τα/k1−τk1+1+τk22ϒ′2k1k21−τk1+1+τk2dτ.
Remark If we take φτ=τx−τα−1 or φτ=τ/αexp−1−α/ατ for α∈0,1, in Lemma 1, we can derive new identities regarding conformable fractional integrals and fractional integrals with the exponential kernel. We left to the reader the task of formulating these identities in a detailed manner.
We also state the following lemma which will be needed in the sequel.Lemma 2.
Letp1≥1 and let f,g be two continuous functions on 0,1. Then we have(26)∫01ftgtdt≤∫01ftdt1−1/p1×∫01ftgtp1dt1/p1.Proof.
It is based on the standard integral Hölder inequality when writing(27)fg=f1−1/p1f1/p1g.
The details are straightforward and therefore omitted here for the reader. □


## 3. Results and Discussions

### 3.1. Main Results

Our first main result in this section is as follows.Theorem 2.
Under the assumptions of Lemma1, if the function ϒ′ is harmonic convex on k1,k2, then the following inequality holds for Sarikaya fractional integrals(28)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤k1k2k2−k1Δ1ϒ′k1π1φρ+π2φσ+ϒ′k2π3φρ+π4φσ,where(29)π1φρ≔∫011−τΔτ−Δ1ρ1+τk1+1−τk22dτ,π2φσ≔∫011+τΔ1σ−Δτ1−τk1+1+τk22dτ,π3φρ≔∫011+τΔτ−Δ1ρ1+τk1+1−τk22dτ,(30)π4φσ≔∫011−τΔ1σ−Δτ1−τk1+1+τk22dτ.Proof.
Using Lemma1 and the properties of the modulus, we have(31)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1Δ1∫01ϒ′2k1k21+τk1+1−τk2dτ+∫01Δ1σ−Δτ1−τk1+1+τk22ϒ′2k1k21−τk1+1+τk2dτ
Since the functionϒ′ is harmonic convex on k1,k2, with the help of (7), we get(32)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤k1k2k2−k1Δ1×ϒ′k1∫011−τΔτ−Δ1ρ1+τk1+1−τk22dτ+∫011+τΔ1σ−Δτ1−τk1+1+τk22dτ+ϒ′k2∫011+τΔτ−Δ1ρ1−τk1+1+τk22dτ+∫011−τΔ1σ−Δτ1+τk1+1−τk22dτ=k1k2k2−k1Δ1ϒ′k1π1φρ+π2φσ+Υ′k2π3φρ+π4φσ.
This completes the proof.
We now state our second main result as recited in the following.Theorem 3.
Under the assumptions of Lemma1, if the function ϒ′p1 is harmonic convex on k1,k2 for some p1≥1, then(33)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1Δ1ψ11−1/p1ρϒ′k1p1π1φρ+ϒ′k2p1π3φρ21/p1+ψ21−1/p1σϒ′k1p1π2φσ+ϒ′k2p1π4φσ21/p1,where(34)ψ1ρ≔ψ1φρ≔∫01Δτ−Δ1ρ1+τk1+1−τk22dτ,ψ2σ≔ψ2φσ≔∫01Δ1σ−Δτ1−τk1+1+τk22dτ,and π1φρ,π2φσ,π3φρ,π4φσ are defined as in Theorem 2.Proof.
Using Lemma1, the properties of the modulus and Lemma 2, respectively, we have(35)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1Δ1∫01Δτ−Δ1ρdτ1+τk1+1−τk221−1/p1×∫01Δτ−Δ1ρ1+τk1+1−τk22ϒ′2k1k21+τk1+1−τk2p1dτ1/p1+∫01Δ1σ−Δτ1−τk1+1+τk221−1/p1×∫01Δ1σ−Δτ1−τk1+1+τk22ϒ′2k1k21−τk1+1+τk2p1dτ1/p1.
Since the functionϒ′p1 is harmonic convex on k1,k2, with the help of (7), we get(36)|1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/1/k1−ϒ∘Λ2k1k2k1+k2|≤k1k2k2−k1Δ1×ϒ′k1∫011−τΔτ−Δ1ρ1+τk1+1−τk22dτ+∫011+τΔ1σ−Δτ1−τk1+1+τk22dτ+|ϒ′k2|∫011+τΔτ−Δ1ρ1−τk1+1+τk22dτ+∫011−τΔ1σ−Δτ1+τk1+1−τk22dτ=k1k2k2−k1Δ1|ϒ′k1|π1φρ+π2φσ+|ϒ′k2|π3φρ+π4φσ.
This completes the proof.
We end this section by stating our third main result, which reads as follows:Theorem 4.
Under the assumptions of Lemma1, let us assume that the function |ϒ′|r1 is harmonic convex on k1,k2 for some r1>1. Let p1>1 be such that 1/p1+1/r1=1. Then we have(37)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ°Λ2k1k2k1+k2+Iφ1/k1−ϒ°Λ2k1k2k1+k2≤2k1k2k2−k1Δ1ϕ1/p1ρϒ′k1r1μ1+ϒ′k2r1μ221/r1+ϕ1/p1σϒ′k1r1μ3+ϒ′k2r1μ421/r1,where ϕ is defined, for t≥0, by(38)ϕt≔∫01Δτ−Δ1tp1dτ,μ1≔∫011−τ1+τk1+1−τk22r1dτ=k2−k1−2r122ℱ12r1,1,3,k2−k1k1+k2,(39)μ2≔∫011+τ1+τk1+1−τk22r1dτk2−k1−2r1ℱ122r1,1,2,k2−k1k1+k2+12−ℱ122r1,2,3,k2−k1k1+k2,(40)μ3≔∫011+τ1−τk1+1+τk22r1dτ=k1−k2−2r1ℱ122r1,1,2,k1−k2k1+k2+122ℱ12r1,2,3,k1−k2k1+k2,(41)μ4≔∫011−τ1−τk1+1+τk22r1dτ=k1−k2−2r12ℱ122r1,1,3,k1−k2k1+k2.Proof.
Using Lemma1, the properties of the modulus and Hölder’s inequality, we have(42)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1Δ1∫01Δτ−Δ1ρp1dτ1/p1×∫0111+τk1+1−τk22r1ϒ′2k1k21+τk1+1−τk2r1dτ1/r1+∫01Δ1σ−Δτp11/p1+∫01Δ1σ−Δτp11/p1.
Since the functionϒ′r1 is harmonic convex on k1,k2, with the help of (7), we get(43)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1Δ1∫01Δτ−Δ1ρp1dτ1/p1×ϒ′k1r12∫011−τ1+τk1+1−τk22r1dτ+ϒ′k2r12∫011+τ1+τk1+1−τk22r1dτ1/r1+∫01Δ1σ−Δτp1dτ1/p1×ϒ′k1r12∫011+τ1−τk1+1+τk22dτ+Υ′k2r12∫011−τ1−τk1+1+τk22dτ1/r1=2k1k2k2−k1Δ1ϕ1/p1ρϒ′k1r1μ1+ϒ′k2r1μ221/r1+ϕ1/p1σϒ′k1r1μ3+ϒ′k2r1μ421/r1.
This completes the proof.


### 3.2. Special Cases

We now discuss some special cases of results discussed in the main results section.First, we consider some particular values ofρ and σ in Theorem 2.(1)
Forρ=σ=2/3, we have(44)16ϒk1+4ϒ2k1k2k1+k2+ϒk2−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤k1k2k2−k1Δ1ϒ′k1π1φ23+π2φ23+ϒ′k2π3φ23+π4φ23,where(45)π1φ23≔∫011−τΔτ/2−Δ1/31+τk1+1−τk22dτ,π2φ23≔∫011+τΔ1/3−Δτ/21−τk1+1+τk22dτ,π3φ23≔∫011+τΔτ/2−Δ1/3−1+τk1+1−τk22dτ,(46)π4φ23≔∫011−τΔ1/3−Δτ/21−τk1+1+τk22dτ.(2)
Forσ=ρ=1, we get(47)ϒ2k1k2k1+k2−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤k1k2k2−k12Δ1ϒ′k1π1φ1+π2φ1+ϒ′k2π3φ1+π4φ1,where(48)π1φ1≔∫011−τΔτ−Δ11+τk1+1−τk22dτ,π2φ1≔∫011+τΔ1−Δτ1−τk1+1+τk22dτ,π3φ1≔∫011+τΔτ−Δ11+τk1+1−τk22dτ,(49)π4φ1≔∫011−τΔ1−Δτ1−τk1+1+τk22dτ.(3)
Forρ=σ=0, we obtain(50)ϒk1+ϒk22−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1+ϒ∘Λ2k1k2k1+k2≤k1k2k2−k12Δ1ϒ′k1π1φ0+π2φ0+ϒ′k2π3φ0+π4φ0,where(51)π1φ0≔∫011−τΔτ1+τk1+1−τk22dτ,π2φ0≔∫011+τΔτ1−τk1+1+τk22dτ,π3φ0≔∫011+τΔτ1+τk1+1−τk22dτ,(52)π4φ0≔∫011−τΔτ1−τk1+1+τk22dτ.We now discuss some particular cases of Theorem2 when choosing special functions φ.(I
) If we takeφτ=τ, then we have(53)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2+2k1k2k2−k1∫1/k11/k2ϒ∘Λ1xdx,(54)≤k1k2k2−k1ϒ′k1π1ρ+π2σ+ϒ′k2π3ρ+π4σ,where(55)π1ρ≔k2−k12k2k1∫011−ττ−ρ1+τk1+1−τk22dτ=k2−k12k2k1k1+k22162ℱ12,2,4,k2−k1k1+k2−ρ2ℱ12,1,3,k2−k1k1+k2+2ρ22ℱ12,1,2,ρk2−k1k1+k2−ρ2ρ+12ℱ12,2,3,ρk2−k1k1+k2+2ρ332ℱ12,3,4,ρk2−k1k1+k2,(56)π2σ≔k2−k12k2k1∫011+τσ−τ1−τk1+1+τk22dτ=k2−k12k2k1k1+k22−σℱ122,1,2,k1−k2k1+k2+1−σ2ℱ122,2,3,k1−k2k1+k2+13ℱ122,3,4,k1−k2k1+k2+2σ2ℱ122,1,2,σk1−k2k1+k2−σ21−σ2ℱ12,2,3,σk1−k2k1+k2−2σ332ℱ12,3,4,σk1−k2k1+k2,(57)π3ρ≔k2−k12k2k1∫011+ττ−ρ1+τk1+1−τk22dτ=k2−k12k2k1k1+k22−ρℱ122,1,2,k2−k1k1+k2+1−ρ2ℱ122,2,3,k2−k1k1+k2+13ℱ122,3,4,k2−k1k1+k2+2ρ2ℱ122,1,2,ρk2−k1k1+k2−ρ21−ρℱ122,2,3,ρk2−k1k1+k2−2ρ33π3ρ≔k2−k12k2k1∫011+ττ−ρ1+τk1+1−τk22dτ=k2−k12k2k1k1+k22−ρℱ122,1,2,k2−k1k1+k2+1−ρ2ℱ122,2,3,k2−k1k1+k2+13ℱ122,3,4,k2−k1k1+k2+2ρ2ℱ122,1,2,ρk2−k1k1+k2−ρ21−ρℱ122,2,3,ρk2−k1k1+k2−2ρ33ℱ122,3,4,ρk2−k1k1+k2.2,3,4,ρk2−k1k1+k2.(58)π4σ≔k2−k12k2k1∫011−τσ−τ1−τk1+1+τk22dτ=k2−k12k2k1k1+k2216ℱ122,2,4,k1−k2k1+k2−σ2ℱ122,1,3,k1−k2k1+k2+2σ2ℱ122,1,2,σk1−k2k1+k2−σ2σ+1ℱ122,2,3,σk1−k2k1+k2+2σ33π4σ≔k2−k12k2k1∫011−τσ−τ1−τk1+1+τk22dτ=k2−k12k2k1k1+k22+16ℱ122,2,4,k1−k2k1+k2−σ2ℱ122,1,3,k1−k2k1+k2+2σ2ℱ122,1,2,σk1−k2k1+k2+σ2σ+1ℱ122,2,3,σk1−k2k1+k2+2σ332ℱ12,3,4,σk1−k2k1+k2.2,3,4,σk1−k2k1+k2.(II)
If we chooseφτ=τα/Γα, then we get(59)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−2αk1k2αΓα+1k2−k1αJ1/k2+αϒ∘Λ2k1k2k1+k2+J1/k1−αϒ∘Λ2k1k2k1+k2≤k1k2k2−k1ϒ′k1π1αρ+π2ασ+ϒ′k2π3αρ+π4ασ,where(60)π1αρ≔k2−k1αΓα+12k2k1α∫011−ττα−ρ1+τk1+1−τk22dτ=k2−k1αΓα+12k2k1αk2+k121α+1α+22,α+1,α+3,k2−k1k1+k2−ρ22,1,3,k2−k1k1+k2+2ρ1/α+12,1,2,ρ1/αk2−k1k1+k2−2ρ1/α+1α+1ℱ12,α+1,α+2,ρ1/αk2−k1k1+k2−ρ2/α+1ℱ12,2,3,ρ1/αk2−k1k1+k2+2ρ2/α+1α+2ℱ12,α+2,α+3,ρ1/αk2−k1k1+k2,(61)π2ασ≔k2−k1αΓα+12k2k1α∫011+τσ−τα1−τk1+1+τk22dτ=k2−k1αΓα+12k2k1αk1+k22−σF12,1,2,k1−k2k1+k2+1α+2F12,α+2,α+3,k1−k2k1+k2+σ2F12,2,3,k1−k2k1+k2−1α+1π2ασ≔k2−k1αΓα+12k2k1α∫011+τσ−τα1−τk1+1+τk22dτ=k2−k1αΓα+12k2k1αk1+k22−σF12,1,2,k1−k2k1+k2+1α+2F12,α+2,α+3,k1−k2k1+k2+σ2F12,2,3,k1−k2k1+k2−1α+12F12,α+1,α+2,k1−k2k1+k2+2σ1/α+1F12,1,2,σ1/αk1−k2k1+k2−2σ1/α+1α+1F12,α+1,α+2,σ1/αk1−k2k1+k2+σ2/α+1F12,2,3,σ1/αk1−k2k1+k2−2σ2/α+1α+2F12,α+2,α+3,σ1/αk1−k2k1+k2,2,α+1,α+2,k1−k2k1+k2+2σ1/α+1F12,1,2,σ1/αk1−k2k1+k2−2σ1/α+1α+1F12,α+1,α+2,σ1/αk1−k2k1+k2+σ2/α+1F12,2,3,σ1/αk1−k2k1+k2−2σ2/α+1α+2F12,α+2,α+3,σ1/αk1−k2k1+k2,(62)π3αρ≔k2−k1αΓα+12k2k1α∫011+ττα−ρ1+τk1+1−τk22dτ=k2−k1αΓα+12k2k1αk1+k22−ρ2ℱ12,1,2,k2−k1k1+k2+1α+22ℱ12,α+2,α+3,k2−k1k1+k2+ρ22ℱ12,2,3,k2−k1k1+k2−1α+12ℱ12,α+1,α+2,k2−k1k1+k2+2ρ1/α+12ℱ12,1,2,ρ1/αk2−k1k1+k2−2ρ1/α+1α+12ℱ12,α+1,α+2,ρ1/αk2−k1k1+k2+ρ2/α+12ℱ12,2,3,ρ1/αk2−k1k1+k2−2ρ2/α+1α+22ℱ12,α+2,α+3,ρ1/αk2−k1k1+k2.(63)π4ασ≔k2−k1αΓα+12k2k1α∫011−τσ−τα1−τk1+1+τk22dτ=k2−k1αΓα+12k2k1αk1+k221α+1α+22ℱ12,α+1,α+3,k1−k2k1+k2−σ22ℱ12,1,3,k1−k2k1+k2+2σ1/α+1ℱ12,1,2,σ1/αk1−k2k1+k2−2σ1/α+1α+1ℱ12,α+1,α+2,σ1/αk1−k2k1+k2−σ2/α+1ℱ12,2,3,σ1/αk1−k2k1+k2+2ρ2/α+1α+2ℱ12,α+2,α+3,σ1/αk1−k2k1+k2.(III)
Withφτ=τα/k/kΓkα, we obtain(64)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−2α/kk1k2α/kΓkα+kk2−k1α/kJ1/k2+,kαϒ∘Λ2k1k2k1+k2+J1/k1−,kαϒ∘Λ2k1k2k1+k2≤k1k2k2−k1ϒ′k1π1α/kρ+π2α/kσ+ϒ′k2π3α/kρ+π4α/kσ,where(65)π1α/kρ≔k2−k1α/kΓkα+k2k2k1α/k∫011−τ|τα/k−ρ|1+τk1+1−τk22dτ=k2−k1α/kΓkα+k2k2k1α/kk2+k12k2α+kα+2k2,α+kk,α+3kk,k2−k1k1+k2−ρ22ℱ12,1,3,k2−k1k1+k2+2ρk/α+1ℱ122,1,2,ρk/αk2−k1k1+k2−2kρk/α+1α+k2ℱ12,α+kk,α+2kk,ρk/αk2−k1k1+k2−ρk/α+2ℱ122,2,3,ρk/αk2−k1k1+k2+2kρk/α+2α+2k2ℱ12,α+2kk,α+3kk,ρk/αk2−k1k1+k2,(66)π2α/kσ≔k2−k1α/kΓkα+k2k2k1α/k∫011+τ|σ−τα/k|1−τk1+1+τk22dτ=k2−k1α/kΓkα+k2k1k2α/kk2+k12−σ2,1,2,k1−k2k1+k2+kα+2k2,α+2kk,α+3kk,k1−k2k1+k2+σ2ℱ12,2,3,k1−k2k1+k2−kα+kℱ12,α+kk,α+2kk,k1−k2k1+k2+2σk/α+1ℱ12,1,2,σk/αk1−k2k1+k2−2kσk/α+1α+kℱ12,α+kk,α+2kk,σk/αk1−k2k1+k2+σ2k/α+1ℱ12,2,3,σk/αk1−k2k1+k2−2kσ2k/α+1α+2kℱ12,α+2kk,α+3kk,σk/αk1−k2k1+k2,(67)π3α/kρ≔k2−k1α/kΓkα+k2k2k1α/k∫011+ττα−ρ1+τk1+1−τk22dτ=k2−k1α/kΓkα+k2k2k1α/kk1+k22−ρ2ℱ12,1,2,k2−k1k1+k2+kα+2k2ℱ12,α+2kk,α+3kk,k2−k1k1+k2+ρ22ℱ12,2,3,k2−k1k1+k2−kα+k2ℱ12,α+kk,α+2kk,k2−k1k1+k2+2ρk/α+1ℱ12,1,2,ρk/αk2−k1k1+k2−2kρk/α+1α+k2ℱ12,α+kk,α+2kk,ρk/αk2−k1k1+k2+ρ2k/α+1ℱ12,2,3,ρk/αk2−k1k1+k2−2kρ2k/α+1α+2k2ℱ12,α+2kk,α+3kk,ρk/αk2−k1k1+k2,(68)π4α/kσ≔k2−k1α/kΓkα+k2k2k1α/k∫011−τσ−τα1−τk1+1+τk22dτ=k2−k1α/kΓkα+k2k2k1α/kk1+k22k2α+kα+2kℱ122,α+kk,α+3kk,k1−k2k1+k2−σ2ℱ122,1,3,k1−k2k1+k2+2σk/α+1ℱ122,1,2,σk/αk1−k2k1+k2−2kσk/α+1α+kℱ122,α+kk,α+2kk,σk/αk1−k2k1+k2−σ2k/α+1ℱ122,2,3,σ1/αk1−k2k1+k2+2kρ2k/α+1α+2kℱ122,α+2kk,α+3kk,σk/αk1−k2k1+k2.We now discuss some special cases of Theorem3First, we consider some particular values ofρ and σ in Theorem 3.(1)
Forρ=σ=2/3, we have(69)16ϒk1+4ϒ2k1k2k1+k2+ϒk2−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤k1k2k2−k1Δ1ψ11−1/p123ϒ′k1p1π1φ2/3+Υ′k2p1π3φ2/321/p1+ψ21−1/p123ϒ′k1p1π2φ2/3+ϒ′k2p1π4φ2/321/p1.where(70)ψ1φ23≔∫01Δτ−2/3Δ11+τk1+1−τk22dτ,ψ2φ23≔∫012/3Δ1−Δτ1−τk1+1+τk22dτ,π1φ23≔∫011−τΔτ−2/3Δ11+τk1+1−τk22dτ,π2φ23≔∫011+τ2/3Δ1−Δτ1−τk1+1+τk22dτ,π3φ23≔∫011+τΔτ−2/3Δ11+τk1+1−τk22dτ,(71)π4φ23≔∫011−τ2/3Δ1−Δτ1−τk1+1+τk22dτ.(2)
Forσ=ρ=1, we get(72)ϒ2k1k2k1+k2−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤k1k2k2−k1Δ1ψ11−1/p11ϒ′k1p1π1φ1+ϒ′k2p1π3φ121/p1+ψ21−1/p11ϒ′k1p1π2φ1+ϒ′k2p1π4φ121/p1,where(73)ψ1φ1≔∫01Δτ−Δ11+τk1+1−τk22dτ,ψ2φ1≔∫01Δ1−Δτ1−τk1+1+τk22dτ,π1φ1≔∫011−τΔτ−Δ11+τk1+1−τk22dτ,π2φ1≔∫011+τΔ1−Δτ1−τk1+1+τk22dτ,π3φ1≔∫011+τΔτ−Δ11+τk1+1−τk22dτ,(74)π4φ1≔∫011−τΔ1−Δτ1−τk1+1+τk22dτ.(3)
Forσ=ρ=0, we obtain(75)ϒk1+ϒk22−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤k1k2k2−k1Δ1ψ11−1/p1ϒ′k1p1π1φ0+ϒ′k2p1π3φ021/p1+ψ21−1/p1ϒ′k1p1π2φ0+ϒ′k2p1π4φ021/p1.where(76)ψ1φ≔∫01Δτ1+τk1+1−τk22dτ,ψ2φ≔∫01Δτ1−τk1+1+τk22dτ,π1φ0≔∫011−τΔτ1+τk1+1−τk22dτ,π2φ0≔∫011−τΔτ1+τk1+1−τk22dτ,π3φ0≔∫011−τΔτ1+τk1+1−τk22dτ,(77)π4φ0≔∫011−τΔτ1−τk1+1+τk22dτ.We now discuss some particular cases of Theorem3 for special functions φ.(I)
If we takeφτ=τ, then we have(78)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2+2k1k2k2−k1∫1/k11/k2ϒ∘Λ1xdx(79)≤2k1k2k2−k1ψ11−1/p1ρϒ′k1p1π1ρ+ϒ′k2p1π3ρ1/p1,+ψ21−1/p1σϒ′k1p1π2σ+ϒ′k2p1π4σ1/p1,whereπ1ρ,π2σ,π3ρ,π4σ are given by (55), (56), (57), and (58), respectively,(80)ψ1ρ≔k2−k12k2k1∫01τ−ρ1+τk1+1−τk22dτ=k2−k12k2k1k2+k12ρ2ℱ12,1,3,ρk2−k1k1+k2+122ℱ12,2,3,k2−k1k1+k2−2ℱ12,1,2,k2−k1k1+k2.(81)ψ2σ≔2k2−k12k2k1∫01σ−τ1−τk1+1+τk22dτ=k2−k1k2k1k2+k12σ22ℱ12,1,3,σk1−k2k1+k2+122ℱ12,2,3,k1−k2k1+k2−2ℱ12,1,2,k1−k2k1+k2.(II)
If we chooseφτ=τα/Γα, then we get(82)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−2αk1k2αΓα+1k2−k1αJ1/k2+αϒ∘Λ2k1k2k1+k2+J1/k1−αϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1ψ1αρ1−1/p1ϒ′k1p1π1αρ+ϒ′k2p1π3αρ21/p1+ψ2ασ1−1/p1ϒ′k1p1π2ασ+ϒ′k2p1π4ασ21/p1,whereπ1αρ,π2ασ,π3αρ,π4α0 are given by (60), (61), (62), and (63), respectively, and(83)ψ1αρ≔k2−k1α2k1k2αΓα+1∫01τα−ρ1+τk1+1−τk22dτ=k2−k1α2k1k2αk2+k12Γα+12ρ1/α+12ℱ12,1,2,ρ1/αk2−k1k1+k2−2ρ1/α+1α+1ℱ12,α+1,α+2,ρ1/αk2−k1k1+k2−ρ2ℱ12,1,2,k2−k1k1+k2+1α+1ℱ12,α+1,α+2,k2−k1k1+k2.ψ2ασ≔k2−k1α2k1k2αΓα+1∫01σ−τα1−τk1+1+τk22dτ=−k2−k1α−22k1k2αk2+k12Γα+12σ1/α+12ℱ12,1,2,σ1/αk1−k2k1+k2−2σ1/α+1α+12ℱ12,α+1,α+2,σ1/αk1−k2k1+k2−σ2ℱ12,1,2,k1−k2k1+k2+1α+12ℱ12,α+1,α+2,k1−k2k1+k2.(III)
Withφτ=τα/k/kΓkα, we obtain(84)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−2α/kk1k2α/kΓkα+kk2−k1α/kJ1/k2+,kαϒ∘Λ2k1k2k1+k2+J1/k1−,kαϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1ψ1α/kρ1−1/p1ϒ′k1p1π1α/kρ+ϒ′k2p1π3α/kρ21/p1+ψ2α/kσ1−1/p1ϒ′k1p1π2α/kσ+ϒ′k2p1π4α/kσ21/p1,.where π1α/kρ,π2α/kσ,π3α/kρ,π4α/kσ are given by (65), (66), (67), and (68), respectively, and(85)ψ1α/kρ≔k2−k1α/kΓkα+k2k2k1α/k∫01τα/k−ρ1+τk1+1−τk22dτ=k2−k1α/k−2Γkα+k2k2k1α/kk2+k122ρk/α+1ℱ12,1,2,ρk/αk2−k1k1+k2−2kρk/α+1α+k2ℱ12,α+kk,α+2kk,ρk/αk2−k1k1+k2−ρ2ℱ12,1,2,k2−k1k1+k2+kα+k2ℱ12,α+kk,α+2kk,k2−k1k1+k2,ψ2α/kσ≔k2−k1α/kΓkα+k2k2k1α/k∫01σ−τα/k1−τk1+1+τk22dτ=k2−k1α/k−2Γkα+k2k2k1α/kk2+k122σk/α+1ℱ12,1,2,σk/αk1−k2k1+k2−2kσk/α+1α+k2ℱ12,α+kk,α+2kk,σk/αk1−k2k1+k2−σ2ℱ12,1,2,k1−k2k1+k2+kα+k2ℱ12,α+kk,α+2kk,k1−k2k1+k2.We now consider some particular values ofρ and σ in Theorem 4.(1)
Forρ=σ=2/3, we have(86)16ϒk1+4Υ2k1k2k1+k2+ϒk2−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1Δ1ϕ1/p123ϒ′k1r1μ1+ϒ′k2r1μ221/r1+ϒ′k1r1μ3+ϒ′k2r1μ421/r1,whereμ1,μ2,μ3,μ4 are given by (38), (39), (40), and (41), respectively, and(87)ϕ23≔∫01Δτ−2Δ13p1dτ.(2)
Forρ=σ=1, we get(88)ϒ2k1k2k1+k2−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1Δ1ϕ1/p11ϒ′k1r1μ1+ϒ′k2r1μ221/r1+ϒ′k1r1μ3+ϒ′k2r1μ421/r1,whereμ1,μ2,μ3,μ4 are given by (38), (39), (40), and (41), respectively, and(89)ϕ1≔∫01Δτ−Δ1p1dτ.(3)
Forρ=σ=0, we obtain(90)ϒk1+ϒk22−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1Δ1ϕ1/p10ϒ′k1r1μ1+ϒ′k2r1μ221/r1+ϒ′k1r1μ3+ϒ′k2r1μ421/r1,where μ1,μ2,μ3,μ4 are given by (38), (39), (40), and (41), respectively, and(91)ϕ0≔∫01Δτp1dτ.(I)
Under the assumptions of Theorem4, if we take φτ=τ, then we have(92)1−σϒk1+σ+ρΥ2k1k2k1+k2+1−ρϒk2+2k1k2k2−k1∫1/k11/k2ϒ∘Λ1xdx,(93)≤2k1k2k2−k1ϕ1/p1ρϒ′k1r1μ1+ϒ′k2r1μ221/r1,+ϕ1/p1σϒ′k1r1μ3+ϒ′k2r1μ421/r1,whereϕ is defined, for t≥0, by(94)ϕt=k2−k12k2k1∫01τ−tp1dτ=k2−k12k2k1tp1+1+1−tp1+1p1+1,if0≤t≤1,k2−k12k2k1tp1+1−t−1p1+1p1+1,ift≥1.andμ1,μ2,μ3,μ4 are given by (38), (39), (40), and (41), respectively.(II)
Under the assumptions of Theorem4, if we choose φτ=τα/Γα, then we get(95)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−2αk1k2αΓα+1k2−k1αJ1/k2+αϒ∘Λ2k1k2k1+k2+J1/k1−αϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1ϕαρ1/p1ϒ′k1r1μ1+ϒ′k2r1μ221/r1+ϕασ1/p1ϒ′k1r1μ3+ϒ′k2r1μ421/r1,whereϕα is defined, for t≥0, by(96)ϕαt≔k2−k1α2k1k2αΓα+1∫01τα−tp1dτ.andμ1,μ2,μ3,μ4 are given by (38), (39), (40), and (41), respectively.(III)
Under the assumptions of Theorem4, if we substitute φτ=τα/k/kΓkα, then we obtain(97)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−2α/kk1k2α/kΓkα+kk2−k1α/kJ1/k2+,kαϒ∘Λ2k1k2k1+k2+J1/k1−kαϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1ϕα/kρ1/p1ϒ′k1r1μ1+ϒ′k2r1μ221/r1+ϕα/kσ1/p1ϒ′k1r1μ3+ϒ′k2r1μ421/r1,where, for t≥0 we set(98)ϕα/kt≔k2−k1α/k2k1k2α/kΓkα+k∫01τα/k−tp1dτ,and μ1,μ2,μ3,μ4 are given by (38), (39), (40), and (41), respectively.Remark 3.
. If we chooseφτ=τx−τα−1 or φτ=τ/αexp−1−α/ατ for α∈0,1, in Theorems 2, 3, and Theorem 4, we can establish new inequalities regarding conformable fractional integrals and fractional integrals with the exponential kernel. We omit their proofs here and the details are left to the interested reader.


## 3.1. Main Results

Our first main result in this section is as follows.Theorem 2.
Under the assumptions of Lemma1, if the function ϒ′ is harmonic convex on k1,k2, then the following inequality holds for Sarikaya fractional integrals(28)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤k1k2k2−k1Δ1ϒ′k1π1φρ+π2φσ+ϒ′k2π3φρ+π4φσ,where(29)π1φρ≔∫011−τΔτ−Δ1ρ1+τk1+1−τk22dτ,π2φσ≔∫011+τΔ1σ−Δτ1−τk1+1+τk22dτ,π3φρ≔∫011+τΔτ−Δ1ρ1+τk1+1−τk22dτ,(30)π4φσ≔∫011−τΔ1σ−Δτ1−τk1+1+τk22dτ.Proof.
Using Lemma1 and the properties of the modulus, we have(31)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1Δ1∫01ϒ′2k1k21+τk1+1−τk2dτ+∫01Δ1σ−Δτ1−τk1+1+τk22ϒ′2k1k21−τk1+1+τk2dτ
Since the functionϒ′ is harmonic convex on k1,k2, with the help of (7), we get(32)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤k1k2k2−k1Δ1×ϒ′k1∫011−τΔτ−Δ1ρ1+τk1+1−τk22dτ+∫011+τΔ1σ−Δτ1−τk1+1+τk22dτ+ϒ′k2∫011+τΔτ−Δ1ρ1−τk1+1+τk22dτ+∫011−τΔ1σ−Δτ1+τk1+1−τk22dτ=k1k2k2−k1Δ1ϒ′k1π1φρ+π2φσ+Υ′k2π3φρ+π4φσ.
This completes the proof.
We now state our second main result as recited in the following.Theorem 3.
Under the assumptions of Lemma1, if the function ϒ′p1 is harmonic convex on k1,k2 for some p1≥1, then(33)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1Δ1ψ11−1/p1ρϒ′k1p1π1φρ+ϒ′k2p1π3φρ21/p1+ψ21−1/p1σϒ′k1p1π2φσ+ϒ′k2p1π4φσ21/p1,where(34)ψ1ρ≔ψ1φρ≔∫01Δτ−Δ1ρ1+τk1+1−τk22dτ,ψ2σ≔ψ2φσ≔∫01Δ1σ−Δτ1−τk1+1+τk22dτ,and π1φρ,π2φσ,π3φρ,π4φσ are defined as in Theorem 2.Proof.
Using Lemma1, the properties of the modulus and Lemma 2, respectively, we have(35)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1Δ1∫01Δτ−Δ1ρdτ1+τk1+1−τk221−1/p1×∫01Δτ−Δ1ρ1+τk1+1−τk22ϒ′2k1k21+τk1+1−τk2p1dτ1/p1+∫01Δ1σ−Δτ1−τk1+1+τk221−1/p1×∫01Δ1σ−Δτ1−τk1+1+τk22ϒ′2k1k21−τk1+1+τk2p1dτ1/p1.
Since the functionϒ′p1 is harmonic convex on k1,k2, with the help of (7), we get(36)|1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/1/k1−ϒ∘Λ2k1k2k1+k2|≤k1k2k2−k1Δ1×ϒ′k1∫011−τΔτ−Δ1ρ1+τk1+1−τk22dτ+∫011+τΔ1σ−Δτ1−τk1+1+τk22dτ+|ϒ′k2|∫011+τΔτ−Δ1ρ1−τk1+1+τk22dτ+∫011−τΔ1σ−Δτ1+τk1+1−τk22dτ=k1k2k2−k1Δ1|ϒ′k1|π1φρ+π2φσ+|ϒ′k2|π3φρ+π4φσ.
This completes the proof.
We end this section by stating our third main result, which reads as follows:Theorem 4.
Under the assumptions of Lemma1, let us assume that the function |ϒ′|r1 is harmonic convex on k1,k2 for some r1>1. Let p1>1 be such that 1/p1+1/r1=1. Then we have(37)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ°Λ2k1k2k1+k2+Iφ1/k1−ϒ°Λ2k1k2k1+k2≤2k1k2k2−k1Δ1ϕ1/p1ρϒ′k1r1μ1+ϒ′k2r1μ221/r1+ϕ1/p1σϒ′k1r1μ3+ϒ′k2r1μ421/r1,where ϕ is defined, for t≥0, by(38)ϕt≔∫01Δτ−Δ1tp1dτ,μ1≔∫011−τ1+τk1+1−τk22r1dτ=k2−k1−2r122ℱ12r1,1,3,k2−k1k1+k2,(39)μ2≔∫011+τ1+τk1+1−τk22r1dτk2−k1−2r1ℱ122r1,1,2,k2−k1k1+k2+12−ℱ122r1,2,3,k2−k1k1+k2,(40)μ3≔∫011+τ1−τk1+1+τk22r1dτ=k1−k2−2r1ℱ122r1,1,2,k1−k2k1+k2+122ℱ12r1,2,3,k1−k2k1+k2,(41)μ4≔∫011−τ1−τk1+1+τk22r1dτ=k1−k2−2r12ℱ122r1,1,3,k1−k2k1+k2.Proof.
Using Lemma1, the properties of the modulus and Hölder’s inequality, we have(42)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1Δ1∫01Δτ−Δ1ρp1dτ1/p1×∫0111+τk1+1−τk22r1ϒ′2k1k21+τk1+1−τk2r1dτ1/r1+∫01Δ1σ−Δτp11/p1+∫01Δ1σ−Δτp11/p1.
Since the functionϒ′r1 is harmonic convex on k1,k2, with the help of (7), we get(43)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−1Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1Δ1∫01Δτ−Δ1ρp1dτ1/p1×ϒ′k1r12∫011−τ1+τk1+1−τk22r1dτ+ϒ′k2r12∫011+τ1+τk1+1−τk22r1dτ1/r1+∫01Δ1σ−Δτp1dτ1/p1×ϒ′k1r12∫011+τ1−τk1+1+τk22dτ+Υ′k2r12∫011−τ1−τk1+1+τk22dτ1/r1=2k1k2k2−k1Δ1ϕ1/p1ρϒ′k1r1μ1+ϒ′k2r1μ221/r1+ϕ1/p1σϒ′k1r1μ3+ϒ′k2r1μ421/r1.
This completes the proof.


## 3.2. Special Cases

We now discuss some special cases of results discussed in the main results section.First, we consider some particular values ofρ and σ in Theorem 2.(1)
Forρ=σ=2/3, we have(44)16ϒk1+4ϒ2k1k2k1+k2+ϒk2−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤k1k2k2−k1Δ1ϒ′k1π1φ23+π2φ23+ϒ′k2π3φ23+π4φ23,where(45)π1φ23≔∫011−τΔτ/2−Δ1/31+τk1+1−τk22dτ,π2φ23≔∫011+τΔ1/3−Δτ/21−τk1+1+τk22dτ,π3φ23≔∫011+τΔτ/2−Δ1/3−1+τk1+1−τk22dτ,(46)π4φ23≔∫011−τΔ1/3−Δτ/21−τk1+1+τk22dτ.(2)
Forσ=ρ=1, we get(47)ϒ2k1k2k1+k2−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤k1k2k2−k12Δ1ϒ′k1π1φ1+π2φ1+ϒ′k2π3φ1+π4φ1,where(48)π1φ1≔∫011−τΔτ−Δ11+τk1+1−τk22dτ,π2φ1≔∫011+τΔ1−Δτ1−τk1+1+τk22dτ,π3φ1≔∫011+τΔτ−Δ11+τk1+1−τk22dτ,(49)π4φ1≔∫011−τΔ1−Δτ1−τk1+1+τk22dτ.(3)
Forρ=σ=0, we obtain(50)ϒk1+ϒk22−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1+ϒ∘Λ2k1k2k1+k2≤k1k2k2−k12Δ1ϒ′k1π1φ0+π2φ0+ϒ′k2π3φ0+π4φ0,where(51)π1φ0≔∫011−τΔτ1+τk1+1−τk22dτ,π2φ0≔∫011+τΔτ1−τk1+1+τk22dτ,π3φ0≔∫011+τΔτ1+τk1+1−τk22dτ,(52)π4φ0≔∫011−τΔτ1−τk1+1+τk22dτ.We now discuss some particular cases of Theorem2 when choosing special functions φ.(I
) If we takeφτ=τ, then we have(53)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2+2k1k2k2−k1∫1/k11/k2ϒ∘Λ1xdx,(54)≤k1k2k2−k1ϒ′k1π1ρ+π2σ+ϒ′k2π3ρ+π4σ,where(55)π1ρ≔k2−k12k2k1∫011−ττ−ρ1+τk1+1−τk22dτ=k2−k12k2k1k1+k22162ℱ12,2,4,k2−k1k1+k2−ρ2ℱ12,1,3,k2−k1k1+k2+2ρ22ℱ12,1,2,ρk2−k1k1+k2−ρ2ρ+12ℱ12,2,3,ρk2−k1k1+k2+2ρ332ℱ12,3,4,ρk2−k1k1+k2,(56)π2σ≔k2−k12k2k1∫011+τσ−τ1−τk1+1+τk22dτ=k2−k12k2k1k1+k22−σℱ122,1,2,k1−k2k1+k2+1−σ2ℱ122,2,3,k1−k2k1+k2+13ℱ122,3,4,k1−k2k1+k2+2σ2ℱ122,1,2,σk1−k2k1+k2−σ21−σ2ℱ12,2,3,σk1−k2k1+k2−2σ332ℱ12,3,4,σk1−k2k1+k2,(57)π3ρ≔k2−k12k2k1∫011+ττ−ρ1+τk1+1−τk22dτ=k2−k12k2k1k1+k22−ρℱ122,1,2,k2−k1k1+k2+1−ρ2ℱ122,2,3,k2−k1k1+k2+13ℱ122,3,4,k2−k1k1+k2+2ρ2ℱ122,1,2,ρk2−k1k1+k2−ρ21−ρℱ122,2,3,ρk2−k1k1+k2−2ρ33π3ρ≔k2−k12k2k1∫011+ττ−ρ1+τk1+1−τk22dτ=k2−k12k2k1k1+k22−ρℱ122,1,2,k2−k1k1+k2+1−ρ2ℱ122,2,3,k2−k1k1+k2+13ℱ122,3,4,k2−k1k1+k2+2ρ2ℱ122,1,2,ρk2−k1k1+k2−ρ21−ρℱ122,2,3,ρk2−k1k1+k2−2ρ33ℱ122,3,4,ρk2−k1k1+k2.2,3,4,ρk2−k1k1+k2.(58)π4σ≔k2−k12k2k1∫011−τσ−τ1−τk1+1+τk22dτ=k2−k12k2k1k1+k2216ℱ122,2,4,k1−k2k1+k2−σ2ℱ122,1,3,k1−k2k1+k2+2σ2ℱ122,1,2,σk1−k2k1+k2−σ2σ+1ℱ122,2,3,σk1−k2k1+k2+2σ33π4σ≔k2−k12k2k1∫011−τσ−τ1−τk1+1+τk22dτ=k2−k12k2k1k1+k22+16ℱ122,2,4,k1−k2k1+k2−σ2ℱ122,1,3,k1−k2k1+k2+2σ2ℱ122,1,2,σk1−k2k1+k2+σ2σ+1ℱ122,2,3,σk1−k2k1+k2+2σ332ℱ12,3,4,σk1−k2k1+k2.2,3,4,σk1−k2k1+k2.(II)
If we chooseφτ=τα/Γα, then we get(59)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−2αk1k2αΓα+1k2−k1αJ1/k2+αϒ∘Λ2k1k2k1+k2+J1/k1−αϒ∘Λ2k1k2k1+k2≤k1k2k2−k1ϒ′k1π1αρ+π2ασ+ϒ′k2π3αρ+π4ασ,where(60)π1αρ≔k2−k1αΓα+12k2k1α∫011−ττα−ρ1+τk1+1−τk22dτ=k2−k1αΓα+12k2k1αk2+k121α+1α+22,α+1,α+3,k2−k1k1+k2−ρ22,1,3,k2−k1k1+k2+2ρ1/α+12,1,2,ρ1/αk2−k1k1+k2−2ρ1/α+1α+1ℱ12,α+1,α+2,ρ1/αk2−k1k1+k2−ρ2/α+1ℱ12,2,3,ρ1/αk2−k1k1+k2+2ρ2/α+1α+2ℱ12,α+2,α+3,ρ1/αk2−k1k1+k2,(61)π2ασ≔k2−k1αΓα+12k2k1α∫011+τσ−τα1−τk1+1+τk22dτ=k2−k1αΓα+12k2k1αk1+k22−σF12,1,2,k1−k2k1+k2+1α+2F12,α+2,α+3,k1−k2k1+k2+σ2F12,2,3,k1−k2k1+k2−1α+1π2ασ≔k2−k1αΓα+12k2k1α∫011+τσ−τα1−τk1+1+τk22dτ=k2−k1αΓα+12k2k1αk1+k22−σF12,1,2,k1−k2k1+k2+1α+2F12,α+2,α+3,k1−k2k1+k2+σ2F12,2,3,k1−k2k1+k2−1α+12F12,α+1,α+2,k1−k2k1+k2+2σ1/α+1F12,1,2,σ1/αk1−k2k1+k2−2σ1/α+1α+1F12,α+1,α+2,σ1/αk1−k2k1+k2+σ2/α+1F12,2,3,σ1/αk1−k2k1+k2−2σ2/α+1α+2F12,α+2,α+3,σ1/αk1−k2k1+k2,2,α+1,α+2,k1−k2k1+k2+2σ1/α+1F12,1,2,σ1/αk1−k2k1+k2−2σ1/α+1α+1F12,α+1,α+2,σ1/αk1−k2k1+k2+σ2/α+1F12,2,3,σ1/αk1−k2k1+k2−2σ2/α+1α+2F12,α+2,α+3,σ1/αk1−k2k1+k2,(62)π3αρ≔k2−k1αΓα+12k2k1α∫011+ττα−ρ1+τk1+1−τk22dτ=k2−k1αΓα+12k2k1αk1+k22−ρ2ℱ12,1,2,k2−k1k1+k2+1α+22ℱ12,α+2,α+3,k2−k1k1+k2+ρ22ℱ12,2,3,k2−k1k1+k2−1α+12ℱ12,α+1,α+2,k2−k1k1+k2+2ρ1/α+12ℱ12,1,2,ρ1/αk2−k1k1+k2−2ρ1/α+1α+12ℱ12,α+1,α+2,ρ1/αk2−k1k1+k2+ρ2/α+12ℱ12,2,3,ρ1/αk2−k1k1+k2−2ρ2/α+1α+22ℱ12,α+2,α+3,ρ1/αk2−k1k1+k2.(63)π4ασ≔k2−k1αΓα+12k2k1α∫011−τσ−τα1−τk1+1+τk22dτ=k2−k1αΓα+12k2k1αk1+k221α+1α+22ℱ12,α+1,α+3,k1−k2k1+k2−σ22ℱ12,1,3,k1−k2k1+k2+2σ1/α+1ℱ12,1,2,σ1/αk1−k2k1+k2−2σ1/α+1α+1ℱ12,α+1,α+2,σ1/αk1−k2k1+k2−σ2/α+1ℱ12,2,3,σ1/αk1−k2k1+k2+2ρ2/α+1α+2ℱ12,α+2,α+3,σ1/αk1−k2k1+k2.(III)
Withφτ=τα/k/kΓkα, we obtain(64)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−2α/kk1k2α/kΓkα+kk2−k1α/kJ1/k2+,kαϒ∘Λ2k1k2k1+k2+J1/k1−,kαϒ∘Λ2k1k2k1+k2≤k1k2k2−k1ϒ′k1π1α/kρ+π2α/kσ+ϒ′k2π3α/kρ+π4α/kσ,where(65)π1α/kρ≔k2−k1α/kΓkα+k2k2k1α/k∫011−τ|τα/k−ρ|1+τk1+1−τk22dτ=k2−k1α/kΓkα+k2k2k1α/kk2+k12k2α+kα+2k2,α+kk,α+3kk,k2−k1k1+k2−ρ22ℱ12,1,3,k2−k1k1+k2+2ρk/α+1ℱ122,1,2,ρk/αk2−k1k1+k2−2kρk/α+1α+k2ℱ12,α+kk,α+2kk,ρk/αk2−k1k1+k2−ρk/α+2ℱ122,2,3,ρk/αk2−k1k1+k2+2kρk/α+2α+2k2ℱ12,α+2kk,α+3kk,ρk/αk2−k1k1+k2,(66)π2α/kσ≔k2−k1α/kΓkα+k2k2k1α/k∫011+τ|σ−τα/k|1−τk1+1+τk22dτ=k2−k1α/kΓkα+k2k1k2α/kk2+k12−σ2,1,2,k1−k2k1+k2+kα+2k2,α+2kk,α+3kk,k1−k2k1+k2+σ2ℱ12,2,3,k1−k2k1+k2−kα+kℱ12,α+kk,α+2kk,k1−k2k1+k2+2σk/α+1ℱ12,1,2,σk/αk1−k2k1+k2−2kσk/α+1α+kℱ12,α+kk,α+2kk,σk/αk1−k2k1+k2+σ2k/α+1ℱ12,2,3,σk/αk1−k2k1+k2−2kσ2k/α+1α+2kℱ12,α+2kk,α+3kk,σk/αk1−k2k1+k2,(67)π3α/kρ≔k2−k1α/kΓkα+k2k2k1α/k∫011+ττα−ρ1+τk1+1−τk22dτ=k2−k1α/kΓkα+k2k2k1α/kk1+k22−ρ2ℱ12,1,2,k2−k1k1+k2+kα+2k2ℱ12,α+2kk,α+3kk,k2−k1k1+k2+ρ22ℱ12,2,3,k2−k1k1+k2−kα+k2ℱ12,α+kk,α+2kk,k2−k1k1+k2+2ρk/α+1ℱ12,1,2,ρk/αk2−k1k1+k2−2kρk/α+1α+k2ℱ12,α+kk,α+2kk,ρk/αk2−k1k1+k2+ρ2k/α+1ℱ12,2,3,ρk/αk2−k1k1+k2−2kρ2k/α+1α+2k2ℱ12,α+2kk,α+3kk,ρk/αk2−k1k1+k2,(68)π4α/kσ≔k2−k1α/kΓkα+k2k2k1α/k∫011−τσ−τα1−τk1+1+τk22dτ=k2−k1α/kΓkα+k2k2k1α/kk1+k22k2α+kα+2kℱ122,α+kk,α+3kk,k1−k2k1+k2−σ2ℱ122,1,3,k1−k2k1+k2+2σk/α+1ℱ122,1,2,σk/αk1−k2k1+k2−2kσk/α+1α+kℱ122,α+kk,α+2kk,σk/αk1−k2k1+k2−σ2k/α+1ℱ122,2,3,σ1/αk1−k2k1+k2+2kρ2k/α+1α+2kℱ122,α+2kk,α+3kk,σk/αk1−k2k1+k2.We now discuss some special cases of Theorem3First, we consider some particular values ofρ and σ in Theorem 3.(1)
Forρ=σ=2/3, we have(69)16ϒk1+4ϒ2k1k2k1+k2+ϒk2−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤k1k2k2−k1Δ1ψ11−1/p123ϒ′k1p1π1φ2/3+Υ′k2p1π3φ2/321/p1+ψ21−1/p123ϒ′k1p1π2φ2/3+ϒ′k2p1π4φ2/321/p1.where(70)ψ1φ23≔∫01Δτ−2/3Δ11+τk1+1−τk22dτ,ψ2φ23≔∫012/3Δ1−Δτ1−τk1+1+τk22dτ,π1φ23≔∫011−τΔτ−2/3Δ11+τk1+1−τk22dτ,π2φ23≔∫011+τ2/3Δ1−Δτ1−τk1+1+τk22dτ,π3φ23≔∫011+τΔτ−2/3Δ11+τk1+1−τk22dτ,(71)π4φ23≔∫011−τ2/3Δ1−Δτ1−τk1+1+τk22dτ.(2)
Forσ=ρ=1, we get(72)ϒ2k1k2k1+k2−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤k1k2k2−k1Δ1ψ11−1/p11ϒ′k1p1π1φ1+ϒ′k2p1π3φ121/p1+ψ21−1/p11ϒ′k1p1π2φ1+ϒ′k2p1π4φ121/p1,where(73)ψ1φ1≔∫01Δτ−Δ11+τk1+1−τk22dτ,ψ2φ1≔∫01Δ1−Δτ1−τk1+1+τk22dτ,π1φ1≔∫011−τΔτ−Δ11+τk1+1−τk22dτ,π2φ1≔∫011+τΔ1−Δτ1−τk1+1+τk22dτ,π3φ1≔∫011+τΔτ−Δ11+τk1+1−τk22dτ,(74)π4φ1≔∫011−τΔ1−Δτ1−τk1+1+τk22dτ.(3)
Forσ=ρ=0, we obtain(75)ϒk1+ϒk22−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤k1k2k2−k1Δ1ψ11−1/p1ϒ′k1p1π1φ0+ϒ′k2p1π3φ021/p1+ψ21−1/p1ϒ′k1p1π2φ0+ϒ′k2p1π4φ021/p1.where(76)ψ1φ≔∫01Δτ1+τk1+1−τk22dτ,ψ2φ≔∫01Δτ1−τk1+1+τk22dτ,π1φ0≔∫011−τΔτ1+τk1+1−τk22dτ,π2φ0≔∫011−τΔτ1+τk1+1−τk22dτ,π3φ0≔∫011−τΔτ1+τk1+1−τk22dτ,(77)π4φ0≔∫011−τΔτ1−τk1+1+τk22dτ.We now discuss some particular cases of Theorem3 for special functions φ.(I)
If we takeφτ=τ, then we have(78)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2+2k1k2k2−k1∫1/k11/k2ϒ∘Λ1xdx(79)≤2k1k2k2−k1ψ11−1/p1ρϒ′k1p1π1ρ+ϒ′k2p1π3ρ1/p1,+ψ21−1/p1σϒ′k1p1π2σ+ϒ′k2p1π4σ1/p1,whereπ1ρ,π2σ,π3ρ,π4σ are given by (55), (56), (57), and (58), respectively,(80)ψ1ρ≔k2−k12k2k1∫01τ−ρ1+τk1+1−τk22dτ=k2−k12k2k1k2+k12ρ2ℱ12,1,3,ρk2−k1k1+k2+122ℱ12,2,3,k2−k1k1+k2−2ℱ12,1,2,k2−k1k1+k2.(81)ψ2σ≔2k2−k12k2k1∫01σ−τ1−τk1+1+τk22dτ=k2−k1k2k1k2+k12σ22ℱ12,1,3,σk1−k2k1+k2+122ℱ12,2,3,k1−k2k1+k2−2ℱ12,1,2,k1−k2k1+k2.(II)
If we chooseφτ=τα/Γα, then we get(82)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−2αk1k2αΓα+1k2−k1αJ1/k2+αϒ∘Λ2k1k2k1+k2+J1/k1−αϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1ψ1αρ1−1/p1ϒ′k1p1π1αρ+ϒ′k2p1π3αρ21/p1+ψ2ασ1−1/p1ϒ′k1p1π2ασ+ϒ′k2p1π4ασ21/p1,whereπ1αρ,π2ασ,π3αρ,π4α0 are given by (60), (61), (62), and (63), respectively, and(83)ψ1αρ≔k2−k1α2k1k2αΓα+1∫01τα−ρ1+τk1+1−τk22dτ=k2−k1α2k1k2αk2+k12Γα+12ρ1/α+12ℱ12,1,2,ρ1/αk2−k1k1+k2−2ρ1/α+1α+1ℱ12,α+1,α+2,ρ1/αk2−k1k1+k2−ρ2ℱ12,1,2,k2−k1k1+k2+1α+1ℱ12,α+1,α+2,k2−k1k1+k2.ψ2ασ≔k2−k1α2k1k2αΓα+1∫01σ−τα1−τk1+1+τk22dτ=−k2−k1α−22k1k2αk2+k12Γα+12σ1/α+12ℱ12,1,2,σ1/αk1−k2k1+k2−2σ1/α+1α+12ℱ12,α+1,α+2,σ1/αk1−k2k1+k2−σ2ℱ12,1,2,k1−k2k1+k2+1α+12ℱ12,α+1,α+2,k1−k2k1+k2.(III)
Withφτ=τα/k/kΓkα, we obtain(84)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−2α/kk1k2α/kΓkα+kk2−k1α/kJ1/k2+,kαϒ∘Λ2k1k2k1+k2+J1/k1−,kαϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1ψ1α/kρ1−1/p1ϒ′k1p1π1α/kρ+ϒ′k2p1π3α/kρ21/p1+ψ2α/kσ1−1/p1ϒ′k1p1π2α/kσ+ϒ′k2p1π4α/kσ21/p1,.where π1α/kρ,π2α/kσ,π3α/kρ,π4α/kσ are given by (65), (66), (67), and (68), respectively, and(85)ψ1α/kρ≔k2−k1α/kΓkα+k2k2k1α/k∫01τα/k−ρ1+τk1+1−τk22dτ=k2−k1α/k−2Γkα+k2k2k1α/kk2+k122ρk/α+1ℱ12,1,2,ρk/αk2−k1k1+k2−2kρk/α+1α+k2ℱ12,α+kk,α+2kk,ρk/αk2−k1k1+k2−ρ2ℱ12,1,2,k2−k1k1+k2+kα+k2ℱ12,α+kk,α+2kk,k2−k1k1+k2,ψ2α/kσ≔k2−k1α/kΓkα+k2k2k1α/k∫01σ−τα/k1−τk1+1+τk22dτ=k2−k1α/k−2Γkα+k2k2k1α/kk2+k122σk/α+1ℱ12,1,2,σk/αk1−k2k1+k2−2kσk/α+1α+k2ℱ12,α+kk,α+2kk,σk/αk1−k2k1+k2−σ2ℱ12,1,2,k1−k2k1+k2+kα+k2ℱ12,α+kk,α+2kk,k1−k2k1+k2.We now consider some particular values ofρ and σ in Theorem 4.(1)
Forρ=σ=2/3, we have(86)16ϒk1+4Υ2k1k2k1+k2+ϒk2−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1Δ1ϕ1/p123ϒ′k1r1μ1+ϒ′k2r1μ221/r1+ϒ′k1r1μ3+ϒ′k2r1μ421/r1,whereμ1,μ2,μ3,μ4 are given by (38), (39), (40), and (41), respectively, and(87)ϕ23≔∫01Δτ−2Δ13p1dτ.(2)
Forρ=σ=1, we get(88)ϒ2k1k2k1+k2−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1Δ1ϕ1/p11ϒ′k1r1μ1+ϒ′k2r1μ221/r1+ϒ′k1r1μ3+ϒ′k2r1μ421/r1,whereμ1,μ2,μ3,μ4 are given by (38), (39), (40), and (41), respectively, and(89)ϕ1≔∫01Δτ−Δ1p1dτ.(3)
Forρ=σ=0, we obtain(90)ϒk1+ϒk22−12Δ1Iφ1/k2+ϒ∘Λ2k1k2k1+k2+Iφ1/k1−ϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1Δ1ϕ1/p10ϒ′k1r1μ1+ϒ′k2r1μ221/r1+ϒ′k1r1μ3+ϒ′k2r1μ421/r1,where μ1,μ2,μ3,μ4 are given by (38), (39), (40), and (41), respectively, and(91)ϕ0≔∫01Δτp1dτ.(I)
Under the assumptions of Theorem4, if we take φτ=τ, then we have(92)1−σϒk1+σ+ρΥ2k1k2k1+k2+1−ρϒk2+2k1k2k2−k1∫1/k11/k2ϒ∘Λ1xdx,(93)≤2k1k2k2−k1ϕ1/p1ρϒ′k1r1μ1+ϒ′k2r1μ221/r1,+ϕ1/p1σϒ′k1r1μ3+ϒ′k2r1μ421/r1,whereϕ is defined, for t≥0, by(94)ϕt=k2−k12k2k1∫01τ−tp1dτ=k2−k12k2k1tp1+1+1−tp1+1p1+1,if0≤t≤1,k2−k12k2k1tp1+1−t−1p1+1p1+1,ift≥1.andμ1,μ2,μ3,μ4 are given by (38), (39), (40), and (41), respectively.(II)
Under the assumptions of Theorem4, if we choose φτ=τα/Γα, then we get(95)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−2αk1k2αΓα+1k2−k1αJ1/k2+αϒ∘Λ2k1k2k1+k2+J1/k1−αϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1ϕαρ1/p1ϒ′k1r1μ1+ϒ′k2r1μ221/r1+ϕασ1/p1ϒ′k1r1μ3+ϒ′k2r1μ421/r1,whereϕα is defined, for t≥0, by(96)ϕαt≔k2−k1α2k1k2αΓα+1∫01τα−tp1dτ.andμ1,μ2,μ3,μ4 are given by (38), (39), (40), and (41), respectively.(III)
Under the assumptions of Theorem4, if we substitute φτ=τα/k/kΓkα, then we obtain(97)1−σϒk1+σ+ρϒ2k1k2k1+k2+1−ρϒk2−2α/kk1k2α/kΓkα+kk2−k1α/kJ1/k2+,kαϒ∘Λ2k1k2k1+k2+J1/k1−kαϒ∘Λ2k1k2k1+k2≤2k1k2k2−k1ϕα/kρ1/p1ϒ′k1r1μ1+ϒ′k2r1μ221/r1+ϕα/kσ1/p1ϒ′k1r1μ3+ϒ′k2r1μ421/r1,where, for t≥0 we set(98)ϕα/kt≔k2−k1α/k2k1k2α/kΓkα+k∫01τα/k−tp1dτ,and μ1,μ2,μ3,μ4 are given by (38), (39), (40), and (41), respectively.Remark 3.
. If we chooseφτ=τx−τα−1 or φτ=τ/αexp−1−α/ατ for α∈0,1, in Theorems 2, 3, and Theorem 4, we can establish new inequalities regarding conformable fractional integrals and fractional integrals with the exponential kernel. We omit their proofs here and the details are left to the interested reader.


## 4. Simpson Quadrature Formula

In this section we will present some applications of the integral inequalities obtained above, to find new error bounds for the Simpson quadrature formula. We preserve the same notations as in the previous sections.First, we fix two parametersρ,σ≥0.Fork2>k1>0, let U:1/k2=χ0<χ1<⋯∈<χn−1<χn=1/k1 be a partition of 1/k2,1/k1.We set(99)SU,Υ≔∑i=0n−11−σϒχi+σ+ρϒ2χiχi+1χi+χi+1+1−ρϒχi+1χi+1−χi2χiχi+1.Our aim here is to approximate the following integral by demanding(100)∫k1k2ϒ∘Λ1τdτ≔SU,ϒ+ℛU,ϒ,where ℛU,ϒ is the remainder term for i=0,1,2,…,n−1.Using the above notations, we are in the position to prove the following error estimations.Proposition 1.
Under the assumptions of Theorem2, the following inequality holds:(101)ℛU,ϒ≤∑i=0n−1χi+1−χi22ϒ′χiπi,1ρ+πi,2σ+ϒ′χi+1πi,3ρ+πi,4σ,where(102)πi,1ρ≔χi+1−χi2χi+1χi∫011−ττ−ρ1+τχi+1−τχi+12dτ=χi+1−χi2χi+1χiχi+χi+12162ℱ12,2,4,χi+1−χiχi+χi+1−ρ22ℱ12,1,3,χi+1−χiχi+χi+1+2ρ22ℱ12,1,2,ρχi+1−χiχi+χi+1−ρ2ρ+12ℱ12,2,3,ρχi+1−χiχi+χi+1+2ρ332ℱ12,3,4,ρχi+1−χiχi+χi+1,πi,2σ≔χi+1−χi2χi+1χi∫011+τσ−τ1−τχi+1+τχi+12dτ=χi+1−χi2χi+1χiχi+χi+12−σ2ℱ12,1,2,χi−χi+1χi+χi+1+1−σ22ℱ12,2,3,χi−χi+1χi+χi+1+132ℱ12,3,4,χi−χi+1χi+χi+1+2σ22ℱ12,1,2,σχi−χi+1χi+χi+1−σ21−σ2ℱ12,2,3,σχi−χi+1χi+χi+1−2σ332ℱ12,3,4,σχi−χi+1χi+χi+1,πi,3ρ≔χi+1−χi2χi+1χi∫011+ττ−ρ1+τχi+1−τχi+12dτ=χi+1−χi2χi+1χiχi+χi+12−ρ2ℱ12,1,2,χi+1−χiχi+χi+1+1−ρ22ℱ12,2,3,χi+1−χiχi+χi+1+132ℱ12,3,4,χi+1−χiχi+χi+1+2ρ22ℱ12,1,2,ρχi+1−χiχi+χi+1−ρ21−ρ2ℱ12,2,3,ρχi+1−χiχi+χi+1−2ρ332ℱ12,3,4,ρχi+1−χiχi+χi+1.and(103)πi,4σ≔χi+1−χi2χi+1χi∫011−τσ−τ1−τχi+1+τχi+12dτ=χi+1−χi2χi+1χiχi+χi+12162ℱ12,2,4,χi−χi+1χi+χi+1−σ22ℱ12,1,3,χi−χi+1χi+χi+1+2σ22ℱ12,1,2,σχi−χi+1χi+χi+1−σ2σ+12ℱ12,2,3,σχi−χi+1χi+χi+1+2σ332ℱ12,3,4,σχi−χi+1χi+χi+1.Proof.
We use Theorem2 on the subinterval χi,χi+1 of the closed interval 1/k2,1/k1 when φτ=τ. Following (53) we have, for all i=0,1,2,…,n−1,(104)1−σϒχi+1+σ+ρϒ2χiχi+1χi+χi+1+1−ρϒχiχi+1−χi2χiχi+1−∫1/χi+11/χiϒ∘Λ1τdτ(105)≤χi+1−χi22ϒ′χiπi,1ρ+πi,2σ+ϒ′χi+1πi,3ρ+πi,4σ.
Summing inequality (104) over i from 0 to n−1 and using the properties of the modulus, we obtain the desired inequality.Proposition 2.
. Under the assumptions of Theorem3, the following inequality holds:(106)ℛU,ϒ≤∑i=0n−1χi+1−χi2ψi,11−1/p1ρϒ′χip1πi,1ρ+ϒ′χi+1p1πi,3ρ1/p1+ψi,21−1/p1σϒ′χip1πi,2σ+ϒ′χi+1p1πi,4σ1/p1.where πi,1ρ,πi,2σπi,3ρ,πi,4σ are as in Proposition 1,(107)ψi,1ρ≔χi+1−χi2χi+1χi∫01τ−ρ1+τχi+1−τχi+12dτ=χi+1−χi2χi+1χiχi+1+χi2ρ22ℱ12,1,3,ρχi+1−χiχi+χi+1+122ℱ12,2,3,χi+1−χiχi+χi+1−2ℱ12,1,2,χi+1−χiχi+χi+1.and(108)ψi,2σ≔χi+1−χi2χi+1χi∫01σ−τ1−τχi+1+τχi+12dτ=χi+1−χi2χi+1χiχi+1+χi2σ22ℱ12,1,3,σχi−χi+1χi+χi+1+122ℱ12,2,3,χi−χi+1χi+χi+1−2ℱ12,1,2,χi−χi+1χi+χi+1.Proof.
We apply the same technique as in the proof of Proposition1, by using Theorem 3 with φτ=τ and with the help of (78).Proposition 3.
. Under the assumptions of Theorem4, the following inequality holds:(109)ℛU,Υ≤∑i=0n−1χi+1−χi2ϕi1/p1ρϒ′χir1μi,1+ϒ′χi+1r1μi,221/r1+ϕi1/p1σϒ′χir1μi,3+ϒ′χi+1r1μi,421/r1.where ϕi is defined, for t≥0, by(110)ϕit≔χi+1−χi2χi+1χi∫01πτ−tp1dτ=χi+1−χi2χi+1χitp1+1+1−tp1+1p1+1,if0≤t≤1,χi+1−χi2χi+1χitp1+1−t−1p1+1p1+1,ift≥1,(111)μi,1≔∫011−τ1+τχi+1−τχi+12r1dτ=χi+1−χi−2r122r1,1,3,χi+1−χiχi+χi+1,μi,2≔∫011+τ1+τχi+1−τχi+12r1dτ=χi+1−χi−2r1ℱ122r1,1,2,χi+1−χiχi+χi+1+12ℱ122r1,2,3,χi+1−χiχi+χi+1,μi,3≔∫011+τ1−τχi+1+τχi+12r1dτ=χi−χi+1−2r1ℱ122r1,1,2,χi−χi+1χi+χi+1+12ℱ122r1,2,3,χi−χi+1χi+χi+1,μi,4≔∫011−τ1−τχi+1+τχi+12r1dτ=χi−χi+1−2r12ℱ122r1,1,3,χi−χi+1χi+χi+1.Proof.
Applying the same technique as in the proof of Proposition1 but via Theorem 4 with φτ=τ and using (92).


## 5. Conclusion

We have derived a new generalized fractional integral identity using the Sarikaya fractional integral. Using this, we have established some new associated fractional integral inequalities of the Simpson, midpoint, and trapezoid types using the class of harmonic convex functions. To show that our results are quite unifying, we have discussed several new special cases. In order to illustrate the significance of our main results, some applications regarding error estimations for the Simpson quadrature formula have been discussed.



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