# Fractional Versions of Hadamard-Type Inequalities for Strongly Exponentiallyα,h−m-Convex Functions

Authors: Shasha Li; Ghulam Farid; Atiq Ur Rehman; Hafsa Yasmeen
Journal: Journal of Mathematics
 (2021)

---

## Abstract

In this article, we prove some fractional versions of Hadamard-type inequalities for strongly exponentiallyα,h−m-convex functions via generalized Riemann–Liouville fractional integrals. The outcomes of this paper provide inequalities of strongly convex, strongly m-convex, strongly s-convex, strongly α,m-convex, strongly s,m-convex, strongly h−m-convex, strongly α,h−m-convex, strongly exponentially convex, strongly exponentially m-convex, strongly exponentially s-convex, strongly exponentially s,m-convex, strongly exponentially h−m-convex, and exponentially α,h−m-convex functions. The error estimations are also studied by applying two fractional integral identities.


---

## Body

## 1. Introduction

Fractional calculus is the study of derivatives and integrals of any arbitrary real or complex order. It is the generalization of ordinary calculus in which operations are mainly focused on integers. Its history starts when Leibniz and l’Hospital discussed the meaning of fractional order in 1695. This is the first discussion of fractional calculus. Many mathematicians devoted their efforts to make the foundation of fractional calculus. At that time, it was considered only in mathematics but now it has several applications in Science and Engineering, signal processing, mathematical biology, and rheology. In mathematics, many fractional integral operators have been introduced by researchers, see [1, 2]. Using these fractional operators, extensive inequalities are established for different types of convexity, see [3–5] and reference therein. The convex function is defined as follows:A functionf:I⟶ℝ, where I is an interval in ℝ, is called the convex function if the following inequality holds:(1)fxt+1−ty≤tfx+1−tfy,t∈0,1andx,y∈I.Hadamard inequality is geometrical interpretation of the convex function, and it is stated as follows:(2)fa+b2≤1b−a∫abfνdν≤fa+fb2.This inequality gives estimates of the mean value of a convex function. Recently, many mathematicians investigated different versions of Hadamard inequality and discussed its basic properties with corresponding fractional integral operators (see [6–9] and reference therein). Our aim is to establish Hadamard inequalities for the strongly exponentially α,h−m-convex function via generalized Riemann–Liouville fractional integrals. Also, we have obtained error estimations for this convexity by using two fractional integral identities. Now, we recall the definition of the strongly exponentially α,h−m-convex function.Definition 1.
(see [10]). Let J⊆ℝ be an interval containing 0,1, and let h:J⟶ℝ be a nonnegative function. A function f:0,b⟶ℝ is called a strongly exponentially α,h−m-convex function if f is nonnegative and for all xandy∈0,b, t∈0,1, η∈ℝ, and α,m∈0,12, with modulus λ≥0, one has(3)ftx+m1−ty≤htαfxeηx+mh1−tαfyeηy−mλex+yηhtαh1−tαy−x2.
The above definition provides some kinds of exponential convexities as follows:Remark 1.
(i)
If we substituteα=1,λ=0 and ht=ts, then the exponentially s,m-convex function in the second sense introduced by Qiang et al. in [11] can be obtained(ii)
If we substituteα=1,λ=0, ht=ts, and m=1, then the exponentially s-convex function introduced by Mehreen et al. in [12] can be obtained(iii)
If we substituteα=1,λ=0, ht=t, and m=1, then the exponentially convex function introduced by Awan et al. in [13] can be obtained
The classical Riemann–Liouville fractional integrals are given as follows:Definition 2.
(see [14]). Let f∈L1a,b. Then, left-sided and right-sided Riemann–Liouville fractional integrals of a function f of the order ξ∈ℂ and ℜξ>0 are given by(4)Ia+ξfx=1Γξ∫axx−tξ−1ftdt,x>a,Ib−ξfx=1Γξ∫xbt−xξ−1ftdt,x<b,where ℜξ denotes the real part of ξ and Γξ=∫0∞e−zzμ−1dz.
Following two theorems are the fractional versions of Hadamard inequalities via Riemann–Liouville fractional integrals.Theorem 1.
(see [15]). Let f:a,b⟶ℝ be a positive function with 0≤a<b and f∈L1a,b. If f is a convex function on a,b, then the following fractional integral inequality holds:(5)fa+b2≤Γξ+12b−aξIa+ξfb+Ib−ξfa≤fa+fb2,with ξ>0.Theorem 2.
(see [16]). Under the assumptions of Theorem 1, the following fractional integral inequality holds:(6)fa+b2≤2ξ−1Γξ+1b−aξIa+b/2+ξfb+Ia+b/2−ξfa≤fa+fb2,with ξ>0.Following theorem is the error estimation of inequality (5).Theorem 3.
(see [15]). Let f:a,b⟶ℝ be a differentiable mapping on a,b with a<b. If f′ is convex on a,b, then the following fractional integral inequality holds:(7)fa+fb2−Γξ+12b−aξIa+ξfb+Ib−ξfa≤b−a2ξ+11−12ξf′a+f′b.Thek-analogue of the Riemann–Liouville fractional integral is defined as follows:Definition 3.
(see [17]). Let f∈L1a,b. Then, k-fractional Riemann–Liouville integrals of order ξ, where ℜξ>0 and k>0, are defined as(8)Ika+ξfx=1kΓkξ∫axx−tξ/k−1ftdt,x>a,Ikb−ξfx=1kΓkξ∫xbt−xξ/k−1ftdt,x<b,where Γk. is defined by [18]:(9)Γkξ=∫0∞tξ−1e−tk/kdt,ℜξ>0.
Twok-fractional versions of Hadamard inequality are given in next two theorems.Theorem 4.
(see [19]). Let f:a,b⟶ℝ be a positive function with 0≤a<b. If f is a convex function on a,b, then the following inequality for k-fractional integral holds:(10)fa+b2≤Γkξ+k2b−aξ/kIka+ξfb+Ikb−ξfa≤fa+fb2.Theorem 5 (see [20]).
Under the assumptions of Theorem4, the following inequality for k-fractional integral holds:(11)fa+b2≤2ξ/k−1Γkξ+kb−aξ/kIka+b/2+ξfb+kIa+b/2−ξfa≤fa+fb2.The error estimation of inequality (10) is given in the following theorem.Theorem 6.
(see [19]). Let f:a,b⟶ℝ be a differentiable mapping on a,b with 0≤a<b. If f′ is convex on a,b, then the following inequality for k-fractional integral holds:(12)fa+fb2−Γkξ+k2b−aξ/kIka+ξfb+Ikb−ξfa≤b−a2ξ/k+11−12ξ/kf′a+f′b.
Now, we recall generalized Riemann–Liouville fractional integrals by a monotonically increasing function.Definition 4.
(see [21]). Let f∈L1a,b. Also let ψ be an increasing and positive monotone function on a,b; further, ψ has a continuous derivative ψ′ on a,b. Therefore, left as well as right fractional integral operators of order ξ where ℜξ>0 of f with respect to ψ on a,b are defined by(13)Ia+ξ,ψfx=1Γξ∫axψ′tψx−ψtξ−1ftdt,x>a,Ib−ξ,ψfx=1Γξ∫xbψ′tψt−ψxξ−1ftdt,x<b.
Thek-analogue of generalized Riemann–Liouville fractional integrals is defined as follows.Definition 5.
(see [22]). Let f∈L1a,b. Also let ψ be an increasing and positive monotone function on a,b; further, ψ has a continuous derivative ψ′ on a,b. Therefore, left as well as right k-fractional integral operators of order ξ where ℜξ>0 of f with respect to ψ on a,b are defined by(14)Ika+ξ,ψfx=1kΓkξ∫axψ′tψx−ψtξ/k−1ftdt,x>a,(15)Ikb−ξ,ψfx=1kΓkξ∫xbψ′tψt−ψxξ/k−1ftdt,x<b.
For more details of fractional integrals, see [14, 23, 24]. We will utilize the following well-known hypergeometric, Beta, and incomplete Beta functions in our results [25]:(16)F21a,b;c;z=1Bb,c−b∫01tb−11−tc−b−11−zt−adt,c>b>0andz<1,Bx,y=∫01tx−11−ty−1dt=ΓxΓyΓx+y,Bx,y;z=∫0ztx−11−ty−1dt.
In Section2, we established Hadamard inequality for strongly exponentially α,h−m-convex functions via generalized Riemann–Liouville fractional integrals. The special cases of these inequalities are associated with previously published papers. In Section 3, error estimations of fractional Hadamard inequality for strongly exponentially α,h−m are obtained with the help of two fractional integral identities. The outcomes of this article are connected with already established results given in [15, 16, 19, 20, 26–37].


## 2. Main Results

This section is concerned with two fractional versions of Hadamard inequalities for strongly exponentiallyα,h−m-convex functions. One of them is given in the following theorem.Theorem 7.
Letf:a,b⟶ℝ be a positive function with 0≤a<mb and f∈L1a,b. Also, suppose that f is the strongly exponentially α,h−m-convex function on a,b with modulus λ≥0. Then, for k>0 and α,m∈0,12, the following k-fractional integral inequality holds for operators given in (14) and (15):(17)fa+mb2+mg1ηλh1/2αh2α−1/2αξ+kξ+2kξξ+kb−a2+2k2am−mb2+2kξb−aam−mb≤Γkξ+kmb−aξ/kg2ηh12αIkψ−1a+ξ,ψf∘ψψ−1mb+g3ηh2α−12αmξ/k+1Ikψ−1b−ξ,ψf∘ψψ−1am≤ξkh1/2αg2ηfaeηa+mh2α−1/2αg3ηfbeηb∫01htαtξ/k−1dt+mξkh1/2αg2ηfbeηb+mh2α−1/2αg3ηfa/m2eηa/m2∫01tξ/k−1h1−tαdt−mλξkg2ηh1/2αb−a2eηa+b+mg3ηh2α−1/2αb−a/m22eηa/m2+b∫01htαh1−tαtξ/k−1dt,where ξ>0 and(18)g1η=e−ηa+b,ifη>0,e−ηmb+a/m,ifη<0,g2η=e−ηmb,ifη<0,e−ηa,ifη>0,g3η=e−ηb,ifη<0,e−ηa/m,ifη>0.Proof.
From strongly exponentiallyα,h−m-convexity of f, we have(19)fx+my2≤h12αfxeηx+mh2α−12αfyeηy−mλeηx+yh12αh2α−12αy−x2.
By settingx=at+m1−tb and y=a/m1−t+bt, t∈0,1, in (19), multiplying the resulting inequality with tξ/k−1, and then integrating with respect to t, we get(20)kξfa+mb2≤h12α∫01fat+m1−tbeηat+m1−tbtξ/k−1dt+mh2α−12α×∫01fa/m1−t+bteηa/m1−t+bttξ/k−1dt−mλh12αh2α−12α×∫01tb−a+1−ta/m−mb2eηta+b+1−tmb+a/mtξ/k−1dt.
Let(21)g1t=e−ηta+b+1−tmb+a/m,g1′t=η1−mam−be−ηta+b+1−tmb+a/m.
Now one can see thatg1 will be increasing if η<0 and decreasing if η>0. Therefore, from inequality (20), we can have(22)kξfa+mb2≤h12α∫01fat+m1−tbeηat+m1−tbtξ/k−1dt+mh2α−12α×∫01fa/m1−t+bteηa/m1−t+bttξ/k−1dt−mkg1ηλξξ+kξ+2kh12αh2α−12αξξ+kb−a2+2k2am−mb2+2kξb−aam−mb.
By settingψu=at+m1−tb and ψv=a/m1−t+bt in (22), we get the following inequality:(23)kξfa+mb2≤1mb−aξ/kh12α∫ψ−1aψ−1mbfψueηψumb−ψuξ/k−1ψ′udu+mh2α−12α∫ψ−1a/mψ−1bfψveηψvψv−amξ/k−1ψ′vdv−mkg1ηλξξ+kξ+2k×h12αh2α−12αξξ+kb−a2+2k2am−mb2+2kξb−aam−mb.
Further, multiplying byξ/k and using Definition 5, we get(24)fa+mb2≤Γkξ+kmb−aξ/kg2ηh12αIkψ−1a+ξ,ψf∘ψψ−1mb+g3ηmξ/k+1h2α−12αIkψ−1b−ξ,ψf∘ψψ−1am−mg1ηλξ+kξ+2k×h12αh2α−12αξξ+kb−a2+2k2am−mb2+2kξb−aam−mb.
The above inequality leads to the first inequality of (17). Again, using strongly exponentially α,h−m-convexity of f, for t∈0,1, we have(25)g2ηh12αfat+m1−tb+mg3ηh2α−12αfam1−t+bt≤htαh1/2αg2ηfaeηa+mh2α−1/2αg3ηfbeηb+mh1−tαh1/2αg2ηfbeηb+mh2α−1/2αg3ηfa/m2eηa/m2−mλhtαh1−tαg2ηh1/2αb−a2eηa+b+mg3ηh2α−1/2αb−a/m22eηa/m2+b.
By integrating (25) over the interval 0,1 after multiplying with tξ/k−1, we get(26)g2ηh12α∫01fta+m1−tbtξ/k−1dt+mg3ηh2α−12α∫01fam1−t+tbtξ/k−1dt≤h1/2αg2ηfaeηa+mh2α−1/2αg3ηfbeηb∫01htαtξ/k−1dt+mh1/2αg2ηfbeηb+mh2α−1/2αg3ηfa/m2eηa/m2∫01tξ/k−1h1−tαdt−mλg2ηh1/2αb−a2eηa+b+mg3ηh2α−1/2αb−a/m22eηa/m2+b∫01htαh1−tαtξ/k−1dt.
Again using substitutions as considered in (22), the above inequality leads to the second inequality of (17).
In the following remark, we give the connection of inequality (17) with already established results.Remark 2.
(i)
If we takeη=0, λ=0, α=1, and ψ as the identity function in (17), then the inequality stated in Theorem 2.1 in [29] is obtained(ii)
If we takeη=0, k=1, ht=t, m=1, λ=0, α=1, and ψ as the identity function in (17), then Theorem 1 is obtained(iii)
If we takeη=0, ht=t,m=1, α=1, λ=0, and ψ as the identity function in (17), then refinement of Theorem 1 is obtained(iv)
If we takeα=1, ξ=1, k=1, ht=t, m=1, η=0, λ=0, and ψ as the identity function in (17), then Hadamard inequality is obtained(v)
If we takeη=0, m=1, α=1, λ=0, and ht=t in (17), then the inequality stated in Theorem 1 in [26] is obtained(vi)
If we takeη=0, m=1, α=1, and ht=t in (17), then the inequality stated in Theorem 10 in [32] is obtained(vii)
If we takeη=0, k=1, m=1, α=1, λ=0, and ht=t in (17), then the inequality stated in Theorem 2.1 in [34] is obtained(viii)
If we takeα=1, k=1, ht=t, η=0, λ=0, and ψ as the identity function in (17), then the inequality stated in Theorem 2.1 in [31] is obtained(ix)
If we takeα=1, λ=0, ht=ts, and ψ as the identity function in (17), then the inequality stated in Theorem 2 in [36] is obtained(x)
If we takeα=1, η=0, λ=0, and ht=ts in (17), then the inequality stated in Corollary 1 in [35] is obtained(xi)
If we takeη=0 and k=1 in (17), then the inequality stated in Theorem 4 in [37] is obtained(xii)
If we takeη=0, k=1, and α=1 in (17), then the inequality stated in Corollary 1 in [37] is obtained(xiii)
If we takeλ=0 in (17), then the inequality stated in Theorem 7 in [38] is obtained
Now, we give inequality (17) for strongly exponentially h−m-convex, strongly exponentially s,m-convex, strongly exponentially m-convex, and strongly exponentially convex functions.Corollary 1.
If we takeα=1 in (17), then the following inequality holds for strongly exponentially h−m-convex functions:(27)fa+mb2+mg1ηλh21/2ξ+kξ+2kξξ+kb−a2+2k2am−mb2+2kξb−aam−mb≤h1/2Γkξ+kmb−aξ/kg2ηIkψ−1a+ξ,ψf∘ψψ−1mb+g3ηmξ/k+1Ikψ−1b−ξ,ψf∘ψψ−1am≤h1/2ξkg2ηfaeηa+mg3ηfbeηb×∫01httξ/k−1dt+mh1/2ξkg2ηfbeηb+mg3ηfa/m2eηa/m2∫01tξ/k−1h1−tdt−mλh1/2ξkg2ηb−a2eηa+b+mg3ηb−a/m22eηa/m2+b∫01hth1−ttξ/k−1dt.Corollary 2.
If we takeα=1 and ht=ts in (17), then the following inequality holds for strongly exponentially s,m-convex functions:(28)fa+mb2+mg1ηλ22sξ+kξ+2kξξ+kb−a2+2k2am−mb2+2kξb−aam−mb≤Γkξ+k2smb−aξ/kg2ηIkψ−1a+ξ,ψf∘ψψ−1mb+g3ηmξ/k+1Ikψ−1b−ξ,ψf∘ψψ−1am≤ξ2sξ+skg2ηfaeηa+mg3ηfbeηb+mξB1+s,ξ/k2skg2ηfbeηb+mg3ηfa/m2eηa/m2−mλξB1+s,s+ξ/k2sk×g2ηb−a2eηa+b+mg3ηb−a/m22eηa/m2+b.Corollary 3.
If we takeα=1 and ht=t in (17), then the following inequality holds for strongly exponentially m-convex functions:(29)fa+mb2+mg1ηλ4ξ+kξ+2kξξ+kb−a2+2k2am−mb2+2kξb−aam−mb≤Γkξ+k2mb−aξ/kg2ηIkψ−1a+ξ,ψf∘ψψ−1mb+g3ηmξ/k+1Ikψ−1b−ξ,ψf∘ψψ−1am≤ξ2ξ+kg2ηfaeηa+mg3ηfbeηb+mk2ξ+kg2ηfbeηb+mg3ηfa/m2eηa/m2−mλξk2ξ+kξ+2kg2ηb−a2eηa+b+mg3ηb−a/m22eηa/m2+b.Corollary 4.
If we takeα=1, m=1, and ht=t in (17), then the following inequality holds for strongly exponentially convex functions:(30)fa+b2+g1ηk2λ2ξ+kξ+2k≤Γkξ+k2b−aξ/kg2ηIkψ−1a+ξ,ψf∘ψψ−1b+g3ηIkψ−1b−ξ,ψf∘ψψ−1a,≤ξ2ξ+kg2ηfaeηa+mg3ηfbeηb+k2ξ+kg2ηfbeηa+mg3ηfaeηb−λξkb−a2g2η+g3η2ξ+kξ+2keηa+b.The next theorem is another version of Hadamard inequality for strongly exponentiallyα,h−m-convex functions.Theorem 8.
Under the assumptions of Theorem7, the following k-fractional integral inequality holds:(31)fa+mb2+h12αh2α−12αmλg1η4ξ+kξ+2kξξ+kb−a2+ξ2+5kξ+8k2am−mb2+2ξξ+3kb−aam−mb≤2ξ/kΓkξ+kmb−aξ/kg2ηh12αIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+g3ηh2α−12αmξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am≤ξkh1/2αg2ηfaeηa+mh2α−1/2αg3ηfbeηb∫0hhtα2αtξ/k−1dt+ξmkh1/2αg2ηfbeηb+mh2α−1/2αg3ηfa/m2eηa/m2∫01h2α−tα2αtξ/k−1dt−mλξkh1/2αg2ηb−a2eηa+b+mh2α−1/2αg3ηb−a/m22eηa/m2+b∫01htα2αh2α−tα2αtξ/k−1dt,where ξ>0 and(32)g1η=e−η/2a1+1/m+b1+m,ifη>0,e−ηmb+a/m,ifη<0,g2η=e−ηmb,ifη<0,e−ηa+mb/2,ifη>0,g3η=e−ηa+mb/2m,ifη<0,e−ηa/m,ifη>0.Proof.
By settingx=at/2+m2−t/2b and y=a/m2−t/2+bt/2, t∈0,1, in (19), multiplying the resulting inequality with tξ/k−1, and then integrating with respect to t, we get(33)kξfa+mb2≤h12α∫01fat/2+m1−t/2beηat/2+m1−t/2btξ/k−1dt+mh2α−12α∫01fa/m2−t/2+bt/2eηa/m2−t/2+bt/2tξ/k−1dt−mλg1ηh12αh2α−12α×∫01tb−a+1−ta/m−mb2eηt/2a+b+1−t/2mb+a/mtξ/k−1dt.
Let(34)g1t=e−ηt/2a+b+1−t/2mb+a/m,g1′t=η21−mam−be−ηt/2a+b+1−t/2mb+a/m.
Now one can see thatg1 will be increasing if η<0 and decreasing if η>0. Therefore, from inequality (33), we can have(35)kξfa+mb2≤h12α∫01fat/2+m1−t/2beηat/2+m1−t/2btξ/k−1dt+mh2α−12α∫01fa/m2−t/2+bt/2eηa/m2−t/2+bt/2tξ/k−1dt−mλg1ηh12αh2α−12α×k4ξξ+kξ+2kξξ+kb−a2+ξ2+5kξ+8k2am−mb2+2ξξ+3kb−aam−mb.
By settingψu=at/2+m2−t/2b and ψv=a/m2−t/2+bt/2 in (35), we get the following inequality:(36)kξfa+mb2≤2ξ/kmb−aξ/kh12α∫ψ−1a+mb/2ψ−1mbfψueηψumb−ψuξ/k−1ψ′udu+mξ/k+1h2α−12α∫ψ−1a/mψ−1a+mb/2mfψveηψvψv−amξ/k−1ψ′vdv−mλg1ηh12αh2α−12αk4ξξ+kξ+2kξξ+kb−a2+ξ2+5kξ+8k2am−mb2+2ξξ+3kb−aam−mb.
Further, multiplying above inequality byξ/k and using Definition 5, we get(37)fa+mb2≤2ξ/kΓkξ+kmb−aξ/kg2ηh12αIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+g3ηh2α−12αmξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am−h12αh2α−12αmλg1η4ξ+kξ+2kξξ+kb−a2+ξ2+5kξ+8k2am−mb2+2ξξ+3kb−aam−mb.
The above inequality leads to the first inequality of (31). Again using strongly exponentially α,h−m-convexity of f, for t∈0,1, we have(38)g2ηh12αfat2+m2−t2b+mg3ηh2α−12αfam2−t2+bt2≤htα2αh1/2αg2ηfaeηa+mh2α−1/2αfbeηb+mh2α−tα2αh1/2αg2ηfbeηb+mh2α−1/2αg3ηfa/m2eηa/m2−mλhtα2αh2α−tα2αh1/2αg2ηb−a2eηa+b+h2α−1/2αg3ηb−a/m22eηa/m2+b.
By integrating (38) over 0,1 after multiplying with tξ/k−1, the following inequality holds:(39)g2ηh12α∫01fat2+m2−t2btξ/k−1dt+mg3ηh2α−12α∫01fam2−t2+bt2tξ/k−1dt≤h1/2αg2ηfaeηa+mh2α−1/2αg3ηfbeηb∫01htα2αtξ/k−1dt+mh1/2αg2ηfb/eηb+mh2α−1/2αg2ηfa/m2/eηam2∫01h2α−tα2αtξ/k−1dt−mλh1/2αg2ηb−a2eηa+b+h2α−1/2αg3ηb−a/m22eηa/m2+b∫01htα2αh2α−tα2αtξ/k−1dt.
Again, using substitutions as considered in (35), the above inequality leads to the second inequality of (31).
In the following remark, we give the connection of inequality (31) with already established results.Remark 3.
(i)
If we takeη=0 and k=1 in (31), then the inequality stated in Theorem 5 in [37] is obtained(ii)
If we takeη=0 and α=1 in (31), then the inequality stated in Corollary 3 in [37] is obtained(iii)
If we takeα=1, ht=t, m=1, λ=0, k=1, η=0, and ψ as the identity function in (31), then Theorem 2 is obtained(iv)
If we takeα=1, m=1, ht=t, λ=0, η=0, and ψ as the identity function in inequality (31), then refinement of Theorem 2 is obtained(v)
If we takeht=t, m=1, λ=0, k=1, ξ=1,η=0, α=1, and ψ as the identity function in (31), then the Hadamard inequality is obtained(vi)
If we takeht=t,m=1, α=1, and η=0 in (31), then the inequality stated in Theorem 11 in [32] is obtained(vii)
If we takeα=1, ht=t, k=1, λ=0, η=0, and ψ as the identity function in (31), then the inequality stated in Theorem 2.1 in [30] is obtained(viii)
If we takeλ=0, α=1, ht=ts, and η=0 in (31), then the inequality stated in Corollary 3 in [35] is obtained(ix)
If we takeλ=0 in (31), then the inequality stated in Theorem 8 in [38] is obtained
Now, we give inequality (31) for strongly exponentially h−m-convex, strongly exponentially s,m-convex, strongly exponentially m-convex, and strongly exponentially convex functions.Corollary 5.
If we takeα=1 in (31), then the following inequality holds for strongly exponentially h−m-convex functions:(40)fa+mb2+mh21/2λg1η4ξ+kξ+2kξξ+kb−a2+ξ2+5kξ+8k2am−mb2+2ξξ+3kb−aam−mb≤h1/22ξ/kΓkξ+kmb−aξ/kg2ηIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+g3ηmξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am≤h1/2ξkg2ηfaeηa+mg3ηfbeηb×∫01ht2tξ/k−1dtr+h1/2ξmkg2ηfbeηb+mg3ηfa/m2eηa/m2∫01h2−t2tξ/k−1dt−mλh12g2ηb−a2eηa+b+g3ηb−a/m22eηa/m2+b∫01ht2h2−t2tξ/k−1dt.Corollary 6.
If we takeα=1 and ht=ts in (31), then the following inequality holds for strongly exponentially s,m-convex functions:(41)fa+mb2+2−2smλg1η4ξ+2kξ+kξξ+kb−a2+ξ2+5kξ+8k2am−mb2+2ξξ+3kb−aam−mb≤2ξ/k−sΓkξ+kmb−aξ/kg2ηIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+g3ηmξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am≤ξ22sξ+kg2ηfaeηa+g3ηfbeηb+m2F1−s,ξk,ξk+1;12g2ηfbeηb+mg3ηfa/m2eηa/m2−2ξ/k−smλξB1/2,s+ξ/k,1+sk×g2ηb−a2eηa+b+g3ηb−a/m22eηa/m2+b.Corollary 7.
If we takeα=1 and ht=t in (31), then the following inequality holds for strongly exponentially m-convex functions:(42)fa+mb2+mλg1η16ξ+2kξ+kξξ+kb−a2+ξ2+5kξ+8k2am−mb2+2ξξ+3kb−aam−mb≤2ξ/k−1Γkξ+kmb−aξ/kg2ηIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+g3ηmξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am≤ξ4ξ+kg2ηfaeηa+mg3ηfbeηb+mξ+2k4ξ+kg2ηfbeηb+mg3ηfa/m2eηa/m2−mλξξ+3k8ξ+kξ+2kg2ηb−a2eηa+b+g3ηb−a/m22eηa/m2+b.Corollary 8.
If we takeα=1, m=1, and ht=t in (31), then the following inequality holds for strongly exponentially convex functions:(43)fa+b2+λk2g1η2ξ+2kξ+k≤2ξ/k−1Γkξ+kb−aξ/kg2ηIkψ−1a+b/2+ξ,ψf∘ψψ−1b+g3ηIkψ−1a+b/2−ξ,ψf∘ψψ−1a,≤ξ4ξ+kg2ηfaeηa+g3ηfbeηb+ξ+2k4ξ+kg2ηfbeηb+g3ηfaeηa−λb−a2ξξ+3kg2η+g3η8ξ+kξ+2keηa+b.


## 3. Error Estimations of Hadamard Inequalities for Strongly Exponentiallyα,h−m-Convex Functions

This section deals with error bounds of Hadamard inequalities for strongly exponentiallyα,h−m-convex functions using generalized Riemann–Liouville fractional integrals. Estimations obtained here provide refinements of several well-known inequalities for different types of convexity. The following identity is used to prove the next theorem.Lemma 1.
(see [26]). Let a<b and f:a,b⟶ℝ be a differentiable mapping on a,b. Also, suppose that f′∈L1a,b. Then, for k>0, the following identity holds for the operators given in (14) and (15):(44)fa+fb2−Γkξ+k2b−aξ/kIkψ−1a+ξ,ψf∘ψψ−1b+Ikψ−1b−ξ,ψf∘ψψ−1a=b−a2∫011−tξ/k−tξ/kf′ta+1−tbdt.Theorem 9.
Letf:a,b⟶ℝ be a differentiable mapping on a,b such that f′∈L1a,b. Also, suppose that f′ is strongly exponentially α,h−m-convex on a,b. Then, for k>0 and α,m∈0,12, the following k-fractional integral inequality holds for the operators given in (14) and (15):(45)fa+fb2−Γkξ+k2b−aξ/kIkψ−1a+ξ,ψf∘ψψ−1b+Ikψ−1b−ξ,ψf∘ψψ−1a≤b−a2f′aeηa∫01/2htα1−tξ/k−tξ/kdt+∫1/21htαtξ/k−1−tξ/kdt+mf′b/meηb/m∫01/2h1−tα1−tξ/k−tξ/kdt+∫1/21h1−tαtξ/k−1−tξ/kdt−mλb−a2eηa+b∫01/2htαh1−tα1−tξ/k−tξ/kdt−∫121htαh1−tαtξ/k−1−tξ/kdt,where ξ>0.Proof.
From Lemma1, it follows that(46)fa+fb2−Γkξ+k2b−aξ/kIkψ−1a+ξ,ψf∘ψψ−1b+Ikψ−1b−ξ,ψf∘ψψ−1a≤b−a2∫011−tξ/k−tξ/kf′ta+1−tbdt.
By using strongly exponentiallyα,h−m-convexity of f′ and for t∈0,1, we have(47)f′ta+1−tb≤htαf′aeηa+mh1−tαf′b/meηb/m−mλhtαh1−tαeηa+b/mbm−a2.
Using (47) in (46), we get(48)fa+fb2−Γkξ+k2b−aξ/kIkψ−1a+ξ,ψf∘ψψ−1b+Ikψ−1b−ξ,ψf∘ψψ−1a≤b−a2∫011−tξ/k−tξ/khtαf′aeηa+mh1−tαf′b/meηb/m−mλhtαh1−tαeηa+b/mbm−a2=b−a2f′aeηa∫01/2htα1−tξ/k−tξ/kdt+∫1/21htαtξ/k−1−tξ/kdt+mf′b/meηb/m∫01/2h1−tα1−tξ/k−tξ/kdt+∫1/21h1−tαtξ/k−1−tξ/kdt−mλb−a2eηa+b/m∫01/2htαh1−tα1−tξ/k−tξ/kdt+∫1/21htαh1−tαtξ/k−1−tξ/kdt.
In the following remark, we give the connection of inequality (45) with already established results.Remark 4.
(i)
If we takeη=0 and k=1 in (45), then the inequality stated in Theorem 6 in [37] is obtained(ii)
If we takeη=0 and α=1 in (45), then the inequality stated in Corollary 7 in [37] is obtained(iii)
If we takeht=t, m=1, α=1, and η=0 in (45), then the inequality stated in Theorem 12 in [32] is obtained(iv)
If we takem=1, α=1, η=0, λ=0, and ht=ts in (45), then the inequality stated in Theorem 2 in [26] is obtained(v)
If we takem=1, α=1, λ=0, ht=t, η=0, and ψ as the identity function in (45), then Theorem 6 is obtained(vi)
If we takek=1, m=1, α=1, λ=0, ht=t, η=0, and ψ as the identity function in (45), then Theorem 3 is obtained(vii)
If we takek=1, α=1, ξ=1, λ=0, ht=t, m=1, η=0, and ψ as the identity function in (45), then the inequality stated in Theorem 2.2 in [27] is obtained(viii)
If we takeη=0, α=1, λ=0, and ht=ts in (45), then the inequality stated in Corollary 5 in [35] is obtained(ix)
If we takeα=1, ht=t, m=1, and η=0 in (45), then the inequality stated in Theorem 12 in [32] is obtained(x)
If we takeλ=0 in (45), then the inequality stated in Theorem 9 in [38] is obtained
Now, we give inequality (45) for strongly exponentially h−m-convex, strongly exponentially m-convex, and strongly exponentially convex functions.Corollary 9.
If we takeα=1 in (45), then the following inequality holds for strongly exponentially h−m-convex functions:(49)fa+fb2−Γkξ+k2b−aξ/kIkψ−1a+ξ,ψf∘ψψ−1b+Ikψ−1b−ξ,ψf∘ψψ−1a≤b−a2f′aeηa∫01/2ht1−tξ/k−tξ/kdt+∫1/21httξ/k−1−tξ/kdt+mf′b/meηb/m∫01/2h1−t1−tξ/k−tξ/kdt+∫1/21h1−ttξ/k−1−tξ/kdt−mλb−a2eηa+b∫01/2hth1−t1−tξ/k−tξ/kdt−∫1/21hth1−ttξ/k−1−tξ/kdt.Corollary 10.
If we takeα=1 and ht=t in (45), then the following inequality holds for strongly exponentially m-convex functions:(50)fa+fb2−Γkξ+k2b−aξ/kIkψ−1a+ξ,ψf∘ψψ−1b+Ikψ−1b−ξ,ψf∘ψψ−1a≤b−a2ξ/k+11−12ξ/kf′aeηa+mf′b/meηb/m−λb/m−a31−ξ/k+4/2ξ/k+2eηb/m+aξ/k+2ξ/k+3.Corollary 11.
If we takeα=1, m=1, and ht=t in (45), then the following inequality holds for strongly exponentially convex functions:(51)fa+fb2−Γkξ+k2b−aξ/kIkψ−1a+ξ,ψf∘ψψ−1b+Ikψ−1b−ξ,ψf∘ψψ−1a≤b−a2ξ/k+11−12ξ/kf′aeηa+f′beηb−λb−a31−ξ/k+4/2ξ/k+2eηb+aξ/k+2ξ/k+3.
The following integral identity is useful to get our next theorem.Lemma 2.
(see [35]). Let f:a,b⟶ℝ be a differentiable mapping on a,b such that f′∈L1a,b. Then, for k>0 and m∈0,1, the following integral identity holds for operators given in (14) and (15):(52)2ξ/k−1Γkξ+kmb−aξ/kIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+mξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am−12fa+mb2+mfa+mb2m=mb−a4∫01tξ/kf′at2+m2−t2bdt−∫01tξ/kf′am2−t2+bt2dt.Theorem 10.
Letf:a,b⟶ℝ, a,b⊂0,b, be a differentiable mapping on a,b such that f′∈L1a,b. Also suppose that f′q is a strongly exponentially α,h−m-convex function on a,b for q≥1. Then, for k>0 and α,m∈0,12, the following fractional integral inequality holds for operators given in (14) and (15):(53)|2ξ/k−1Γkξ+kmb−aξ/kIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+mξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am−12fa+mb2+mfa+mb2m|≤mb−a4ξ/k+11−1/qf′aqeηa∫01htα2αtξ/kdt+mf′bqeηb∫01h2α−tα2αtξ/kdt−mλb−a2eηa+b∫01htα2αh2α−tα2αtξ/kdt1/q+mf′a/m2qeηa/m2∫01h2α−tα2αtξ/kdt+mf′bqeηb∫01htα2αtξ/kdt−mλb−a/m22eηa/m2+b∫01htα2αh2α−tα2αtξ/kdt1/q.Proof.
We divide the proof in two cases:Case 1 (forq=1). Applying Lemma 2 and using strongly exponentially α,h−m-convexity of f′, we have(54)|2ξ/k−1Γkξ+kmb−aξ/kIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+mξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am−12fa+mb2+mfa+mb2m≤mb−a4∫01tξ/kf′at2+m2−t2bdt+∫01tξ/kf′am2−t2+bt2dt≤mb−a4f′aeηa+f′beηb∫01htα2αtξ/kdt+mf′beηb+f′a/m2eηa/m2∫01h2α−tα2αtξ/k−mλb−a2eηa+b+b−a/m22eηa/m2+b×∫01htα2αh2α−tα2αdt.Case 2 (forq>1). From Lemma 2 and using power mean inequality, we get(55)|2ξ/k−1Γkξ+kmb−aξ/kIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+mξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am−12fa+mb2+mfa+mb2m|≤mb−a4∫01tξ/kdt1−1/q∫01tξ/kf′at2+m2−t2bqdt1/q+∫01tξ/kf′am2−t2+bt2qdt1/q≤mb−a4ξ/k+11−1/qf′aqeηa∫01htα2αtξ/kdt+mf′bqeηb∫01h2α−tα2αtξ/kdt−mλb−a2eηa+b∫01htα2αh2α−tα2αtξ/kdt1/q+mf′a/m2qeηa/m2∫01h2α−tα2αtξ/kdt+f′bqeηb∫01htα2αtξ/kdt−mλb−a/m22eηa/m2+b∫01htα2αh2α−tα2αtξ/kdt1/q.
Hence, equation (53) is obtained.
In the following remark, we give the connection of inequality (53) with already established results.Remark 5.
(i)
If we takeλ=0 in (53), then the inequality stated in Theorem 10 in [38] is obtained(ii)
If we takeη=0, α=1, λ=0, and ht=ts in (53), then the inequality stated in Corollary 7 in [35] is obtained(iii)
If we takeht=t, k=1, η=0, α=1, λ=0, and ψ as the identity function in (53), then the inequality stated in Theorem 2.4 in [30] is obtained(iv)
If we takeη=0, ht=t,m=1, α=1, and ψ as the identity function in (53), then the inequality stated in Theorem 3.1 in [20] is obtained(v)
If we takeht=t, m=1, η=0, k=1, α=1, λ=0, and ψ as the identity function in (53), then the inequality stated in Theorem 5 in [16] is obtained(vi)
If we takeq=1, ht=t, m=1,η=0, k=1, ξ=1, α=1, λ=0, and ψ as the identity function in (53), then the inequality stated in Theorem 2.2 in [33] is obtained(vii)
If we takeη=0, α=1, m=1, and ht=t in (53), then the inequality stated in Theorem 13 in [32] is obtained(viii)
If we takeη=0 and k=1 in (53), then the inequality stated in Theorem 7 in [37] is obtained(ix)
If we takeη=0, k=1, and α=1 in (53), then the inequality stated in Corollary 10 in [37] is obtained
Now, we give inequality (53) for strongly exponentially h−m-convex, strongly exponentially s,m-convex, strongly exponentially m-convex, and strongly exponentially convex functions.Corollary 12.
If we takeα=1 in (53), then the following inequality holds for strongly exponentially h−m-convex functions:(56)|2ξ/k−1Γkξ+kmb−aξ/kIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+mξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am−12fa+mb2+mfa+mb2m|≤mb−a4ξ/k+11−1/qf′aqeηa∫01ht2tξ/kdt+mf′bqeηb∫01h2−t2tξ/kdt−mλb−a2eηa+b∫01ht2h2−t2tξ/kdt1/q+mf′a/m2qeηa/m2∫01h2−t2tξ/kdt+f′bqeηb∫01ht2tξ/kdt−mλb−a/m22eηa/m2+b×∫01ht2h2−t2tξ/kdt1/q.Corollary 13.
If we takeα=1 and ht=ts in (53), then the following inequality holds for strongly exponentially s,m-convex functions:(57)|2ξ/k−1Γkξ+kmb−aξ/kIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+mξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am−12fa+mb2+mfa+mb2m|≤mb−a4ξ/k+12sξ/k+s+11/qξ/k+1f′aqeηa+2smf′bqξ/k+s+1F21−s,1+ξ/k,2+ξ/k;1/2eηb−21+s+μ/kmλb−a2B1/2,1+s+μ/k,1+sξ/k+1ξ/k+s+1eηa+b1/q+2sm2F1−s,1+ξ/k,2+ξ/k;1/2ξ/k+s+1f′a/m2qeηa/m2+ξ/k+1f′bqeηb−21+s+μ/kmλb−a2B1/2,1+s+μ/k,1+sξ/k+1ξ/k+s+1eηa+b1/q.Corollary 14.
If we takeα=1, m=1, and ht=t in (53), then the following inequality holds for strongly exponentially m-convex functions:(58)2ξ/k−1Γkξ+kb−aξ/kIkψ−1a+b/2+ξ,ψf∘ψψ−1b+Ikψ−1a+b/2−ξ,ψf∘ψψ−1a−fa+b2≤b−a4ξ/k+12ξ/k+21/qξ/k+1f′aqeηa+f′bqξ/k+3eηb−λb−a2ξ/k+1ξ/k+42eηa+bξ/k+31/q+ξ/k+3f′a/m2qeηa/m2+ξ/k+1f′bqeηb−λb−a2ξ/k+1ξ/k+4eηa+bξ/k+31/q.Corollary 15.
If we takeα=1, k=1, m=1, q=1, ξ=1, ht=t, and ψ as the identity function in (53), then the following inequality is obtained:(59)1b−a∫01fξdξ−fa+b2≤b−a8f′aeηa+f′beηb−5λb−a212eηa+b.Theorem 11.
Letf:a,b⟶ℝ, a,b⊂0,b, be a differentiable mapping on a,b with a<b. Also, suppose that f′q is the strongly exponentially α,h−m-convex function on a,b for q>1. Then, for k>0 and α,m∈0,12, the following fractional integral inequality holds for the operators given in (14) and (15):(60)|2ξ/k−1Γkξ+kmb−aξ/kIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+mξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am−12fa+mb2+mfa+mb2m|≤mb−a4ξp/k+11/pf′aeηaq∫01htα2αdt+mf′bqeηb∫01h2α−tα2αdt−mλb−a2eηa+b∫01htα2αh2α−tα2αdt1/q+mf′a/m2qeηa/m2∫01h2α−tα2αdt+f′bqeηb∫01htα2αdt−mλb−a/m22eηa/m2+b∫01htα2αh2α−tα2αdt1/q,with 1/p+1/q=1.Proof.
Applying Lemma2 and using the property of modulus, we get(61)|2ξ/k−1Γkξ+kmb−aξ/kIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+mξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am−12fa+mb2+mfa+mb2m|≤mb−a4∫01tξ/kf′at2+m1−t2bdt+∫01tξ/kf′am1−t2+bt2dt.
Now applying Hölder’s inequality for integrals, we get(62)|2ξ/k−1Γkξ+kmb−aξ/kIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+mξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am−12fa+mb2+mfa+mb2m|≤mb−a4ξp/k+11/p∫01f′at2+m2−t2bqdt1/q+∫01f′am2−t2+bt2qdt1/q.
Using strongly exponentiallyα,h−m-convexity of f′q, we get(63)|2ξ/k−1Γkξ+kmb−aξ/kIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+mξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am−12fa+mb2+mfa+mb2m|≤mb−a4ξp/k+11/pf′aeηaq∫01htα2αdt+mf′bqeηb∫01h1−tα2αdt−mλb−a2eηa+b∫01htα2αh2α−tα2αdt1/q+mf′a/m2qeηa/m2∫01h1−tα2αdt+f′bqeηb∫01htα2αdt−mλb−a/m22eηa/m2+b∫01htα2αh2α−tα2αdt1/q.
In the following remark, we give the connection of inequality (60) with already established results.Remark 6.
(i)
If we takeλ=0 in (60), then the inequality stated in Theorem 11 in [38] is obtained(ii)
If we takeη=0 and k=1 in (60), then the inequality stated in Theorem 8 in [37] is obtained(iii)
If we takeη=0 and α=1 in (60), then the inequality stated in Corollary 12 in [37] is obtained(iv)
If we takeht=t, k=1, η=0, α=1, λ=0, and ψ as the identity function in (60), then the inequality stated in Theorem 2.7 in [30] is obtained(v)
If we takeη=0, λ=0, ht=t, α=1, m=1, and ψ as the identity function in (60), then the inequality stated in Theorem 3.2 in [20] is obtained(vi)
If we takek=1, ht=t, m=1, η=0, ξ=1, α=1, λ=0, and ψ as the identity function in (60), then the inequality stated in Theorem 2.4 in [33] is obtained(vii)
If we takeη=0, α=1, and ht=ts in (60), then the inequality stated in Corollary 9 in [35] is obtained(viii)
If we takeα=1, m=1, ht=t, and η=0 in (60), then the inequality stated in Theorem 14 in [32] is obtained
Now, we give inequality (60) for strongly exponentially h−m-convex, strongly exponentially s,m-convex, strongly exponentially m-convex, and strongly exponentially convex functions.Corollary 16.
If we takeα=1 in (60), then the following inequality holds for strongly exponentially h−m-convex functions:(64)|2ξ/k−1Γkξ+kmb−aξ/kIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+mξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am−12fa+mb2+mfa+mb2m|≤mb−a4ξp/k+11/pf′aeηaq∫01ht2dt+mf′bqeηb∫01h2−t2dt−mλb−a2eηa+b∫01ht2h2−t2dt1/q+mf′a/m2qeηa/m2∫01h2−t2dt+f′bqeηb∫01ht2dt−mλb−a/m22eηa/m2+b∫01ht2h2−t2dt1/q.Corollary 17.
If we takeα=1 and ht=ts in (60), then the following inequality holds for strongly exponentially s,m-convex functions:(65)|2ξ/k−1Γkξ+kmb−aξ/kIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+mξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am−12fa+mb2+mfa+mb2m|≤mb−a41+s1/qξp/k+11/pf′aqeηa+2smf′bq−1+21+s2seηb−2s+11+smB1/2,1+s,1+sλb−a2eηa+b1/q+mf′a/m2qeηa/m2+f′bq−1+21+s2seηb−2s+1m1+sB1/2,1+s,1+sλb−a/m22eηa+b1/q.Corollary 18.
If we takeα=1 and ht=t in (60), then the following inequality holds for strongly exponentially m-convex functions:(66)|2ξ/k−1Γkξ+kmb−aξ/kIkψ−1a+mb/2+ξ,ψf∘ψψ−1mb+mξ/k+1Ikψ−1a+mb/2−ξ,ψf∘ψψ−1am−12fa+mb2+mfa+mb2m|≤mb−a164ξp/k+11/pf′aeηaq+3mf′bqeηb−2mλb−a23eηa+b1/q+3mf′a/m2qeηa/m2+f′bqeηb−2mλb−a/m223eηa+b1/q.Corollary 19.
If we takeα=1, m=1, and ht=t in (60), then the following inequality holds for strongly exponentially convex functions:(67)2ξ/k−1Γkξ+kmb−aξ/kIkψ−1a+b/2+ξ,ψf∘ψψ−1b+Ikψ−1a+b/2−ξ,ψf∘ψψ−1a−fa+b2≤b−a164ξp/k+11/pf′aeηaq+3f′bqeηb−2λb−a23eηa+b1/q+3f′aqeηa+f′beηb−2λb−a/m223eηa+b1/q.


## 4. Conclusion

In this paper, we have proved fractional versions of the Hadamard inequality and their estimations for strongly exponentiallyα,h−m-convex functions via generalized Riemann–Liouville fractional integrals. The outcomes of this article give refinements and generalizations of fractional integral inequalities for different types of convex functions deducible from the definition of the exponentially α,h−m-convex function. [39–41].



---

*Source: 2555974-2021-08-11.xml*