Maclaurin-Type Integral Inequalities for GA-Convex Functions Involving Confluent Hypergeometric Function via Hadamard Fractional Integrals

Abstract

In this manuscript, by using a new identity, we establish some new Maclaurin-type inequalities for functions whose modulus of the first derivatives are $GA$-convex functions via Hadamard fractional integrals.

1. Introduction

It is well known that convexity plays an important and central role in many fields, such as economics, finance, optimization, and game theory. Due to its various applications, this concept has been extended and generalized in several directions.

This concept is closely related to integral inequalities. The literature in this context is rich. One can easily find papers that deal with different types of inequalities via different kinds of convexity.

Over the past few years, numerous scholars have investigated the error estimates associated with specific quadrature formulas. Their aim has been to develop new refinements, generalizations, and variants. For additional details, readers are encouraged to consult references [1,2,3,4,5,6,7,8,9,10] for classical inequalities, and [11,12,13,14] for fractional inequalities.

In [15], İşcan gave the analogue fractional of Hermite–Hadamard inequality for $GA$-convex functions as follows:

$$f\left(\sqrt{ab}\right)\le \frac{\mathsf{\Gamma }\left(\alpha +1\right)}{2{\left(lnb-lna\right)}^{\alpha }}\left\{{}_H{J}_{a^{+}}^{\alpha }f\left(b\right){+}_H{J}_{b^{-}}^{\alpha }f\left(a\right)\right\}\le \frac{f\left(a\right)+f\left(b\right)}{2},$$

where $\alpha >0$ and $0<a<b$ and f is an integrable and $GA$-convex function on $\left[a,b\right]$.

Qi and Xi [16] have derived specific Simpson-type inequalities for $GA$-$\epsilon$-convex functions. Within the outcomes obtained for differentiable function f: $\left[a,b\right]\to \mathbb{R}$, with $0<a<b$ and $f^{\prime }\in L\left[a,b\right]$ and ${\left|f^{\prime }\right|}^q$ is $GA$-$\epsilon$-convex, we have

$$\begin{array}{cc}\hfill & \left|\frac{f\left(a\right)+4f\left(\sqrt{ab}\right)+f\left(b\right)}{6}-\frac{1}{lnb-lna}\underset{a}{\stackrel{b}{\int }}f\left(u\right)\frac{du}{u}\right|\hfill \\ \hfill \le & \frac{lnb-lna}{4}\left\{M_1^{1-\frac{1}{q}}\left(a,b\right)\left(\left(M_1\left(a,b\right)-M_2\left(a,b\right)\right){\left|f^{\prime }\left(a\right)\right|}^q\right.\right.\hfill \\ \hfill & +{\left.M_2\left(a,b\right){\left|f^{\prime }\left(b\right)\right|}^q+\epsilon M_1\left(a,b\right)\right)}^{\frac{1}{q}}\hfill \\ \hfill & +M_1^{1-\frac{1}{q}}\left(b,a\right)\left(M_2\left(b,a\right){\left|f^{\prime }\left(a\right)\right|}^q\right.\hfill \\ \hfill & +\left.{\left.\left(M_1\left(b,a\right)-M_2\left(b,a\right)\right){\left|f^{\prime }\left(b\right)\right|}^q+\epsilon M_1\left(b,a\right)\right)}^{\frac{1}{q}}\right\},\hfill \end{array}$$

where $q\ge 1,$

$$M_1\left(x,y\right)=\frac{2\left(x^{\frac{5}{6}}L\left(x^{\frac{1}{6}},y^{\frac{1}{6}}\right)+x^{\frac{1}{2}}\left(2y^{\frac{1}{2}}-x^{\frac{1}{2}}\right)-2x^{\frac{1}{2}}{y}^{\frac{1}{6}}L\left(x^{\frac{1}{3}},y^{\frac{1}{3}}\right)\right)}{3\left(lny-lnx\right)}$$

and

$$M_1\left(x,y\right)=\frac{2x^{\frac{1}{2}}\left(4y^{\frac{1}{6}}L\left(x^{\frac{1}{3}},y^{\frac{1}{3}}\right)+y^{\frac{1}{2}}\left(lny-lnx\right)-2x^{\frac{1}{3}}L\left(x^{\frac{1}{6}},y^{\frac{1}{6}}\right)-5y^{\frac{1}{2}}+2x^{\frac{1}{3}}{y}^{\frac{1}{6}}+x^{\frac{1}{2}}\right)}{3{\left(lny-lnx\right)}^2},$$

with

$$L\left(x,y\right)=\left\{\begin{array}{c}\frac{x-y}{lnx-lny}\mathrm{if}x\ne y\\ x\mathrm{if}x=y.\end{array}\right.$$

Motivated by the above results, we propose in this work to study one of the open three-point Newton–Cotes formulas called Maclaurin inequality, which can be declared as follows:

$$\left|\frac{1}{8}\left(3f\left(\frac{5a+b}{6}\right)+2f\left(\frac{a+b}{2}\right)+3f\left(\frac{a+5b}{6}\right)\right)-\frac{1}{b-a}\stackrel{b}{\underset{a}{\int }}f\left(u\right)du\right|\le \frac{7{\left(\tau -\upsilon \right)}^4}{51840}{∥f^{\left(4\right)}∥}_{\infty }{\left(b-a\right)}^4,$$

where f is four times continuously differentiable function on $\left(a,b\right)$, and ${∥f^{\left(4\right)}∥}_{\infty }$ = ${\displaystyle \underset{x\in \left(a,b\right)}{sup}}\left|f^{\left(4\right)}\left(x\right)\right|$, (see [17]).

For this, we first prove a new identity involving Hadamard fractional integrals. On the basis of this identity, we establish some new Maclaurin-type inequalities for functions whose modulus of the first derivatives are $GA$-convex.

2. Preliminaries

This section recalls some known definitions. We denote by $\mathbb{R}$ the set of real numbers, and by ${\mathbb{R}}^{{}^{+}}$ the set of non-negative real numbers.

Definition 1

([18]). Let I be a subintervals of $\left(0,+\infty \right)$. A function f: $I\to {\mathbb{R}}^{{}^{+}}$ is said to be $GA$-convex on I, if

$$f\left(x^t{y}^{1-t}\right)\le tf\left(x\right)+\left(1-t\right)f\left(y\right)$$

holds for all $x,y\in I$ and $t\in [0,1]$.

Definition 2

([19]). The integral representation of the confluent hypergeometric function is given by

$${}_1{F}_1\left(a;b;z\right)=\frac{1}{B\left(a,b-a\right)}\underset{0}{\stackrel{1}{\int }}{t}^{a-1}{\left(1-t\right)}^{b-a-1}{e}^{zt}dt,$$

where $\mathfrak{R}b>\mathfrak{R}a>0$ and B is the beta function.

Definition 3

([20]). The integral representation of the incomplete confluent hypergeometric function is given by

$$\begin{array}{cc}\hfill {}_1{F}_1\left(\left[a,b;y\right],z\right)=& \frac{1}{B\left(a,b-a\right)}\underset{0}{\stackrel{y}{\int }}{u}^{a-1}{\left(1-u\right)}^{b-a-1}{e}^{zu}du\hfill \\ \hfill =& \frac{y^a}{B\left(a,b-a\right)}\underset{0}{\stackrel{1}{\int }}{u}^{a-1}{\left(1-uy\right)}^{b-a-1}{e}^{zuy}du,\hfill \end{array}$$

where $\mathfrak{R}b>\mathfrak{R}a>0$ and B is the Beta function.

Definition 4

([21]). The left-sided and right-sided Hadamard fractional integrals of order $\alpha \in {\mathbb{R}}^{+}$ of function $f\left(x\right)$ are defined by

$${}_H{J}_{a^{+}}^{\alpha }f\left(x\right)=\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}\underset{a}{\stackrel{x}{\int }}{\left(lnx-lnu\right)}^{\alpha -1}f\left(u\right)\frac{du}{u},\left(0<a<x\le b\right),$$

and

$${}_H{J}_{b^{-}}^{\alpha }f\left(x\right)=\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}\underset{x}{\stackrel{b}{\int }}{\left(lnu-lnx\right)}^{\alpha -1}f\left(u\right)\frac{du}{u},\left(0<a\le x<b\right).$$

Lemma 1

([22]). For any $0\le a<b$ in $\mathbb{R}$, and a fixed $p\ge 1$, we have

$${\left(b-a\right)}^p\le b^p-a^p.$$

3. Auxiliary Results

We provide certain lemmas in this section that help with the computations and are utilized in the following section. The following lemma is crucial to establish our main results

Lemma 2.

Let f: $[a,b]\to \mathbb{R}$ be a differentiable mapping on $\left[a,b\right]$ with $0<a<b$. Assume that $f^{\prime }\in L\left[a,b\right]$. Then, the following equality for fractional integrals holds:

$$\begin{array}{cc}\hfill & \frac{1}{8}\left(3f\left(a^{\frac{5}{6}}{b}^{\frac{1}{6}}\right)+2f\left(a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)+3f\left(a^{\frac{1}{6}}{b}^{\frac{5}{6}}\right)\right)-\frac{6^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(lnb-lna\right)}^{\alpha }}\mathrm{{\rm Y}}\hfill \\ \hfill =& \frac{lnb-lna}{9}\left(\underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\varkappa }^{\alpha }{a}^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}{f}^{\prime }\left(a^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}\right)d\varkappa \right.\hfill \\ \hfill & -\underset{0}{\stackrel{1}{\int }}\left({\left(1-\varkappa \right)}^{\alpha }-\frac{3}{8}\right)a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}{f}^{\prime }\left(a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}\right)d\varkappa \hfill \\ \hfill & +\underset{0}{\stackrel{1}{\int }}\left({\varkappa }^{\alpha }-\frac{3}{8}\right)a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}{f}^{\prime }\left(a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}\right)d\varkappa \hfill \\ \hfill & -\left.\underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\left(1-\varkappa \right)}^{\alpha }{a}^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}{f}^{\prime }\left(a^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}\right)d\varkappa \right)\hfill \end{array}$$

where $\alpha >0$ and

$$\begin{array}{cc}\hfill \mathrm{{\rm Y}}=& {}_H{J}_{{\left(a^{\frac{5}{6}}{b}^{\frac{1}{6}}\right)}^{-}}^{\alpha }f\left(a\right){+}_H{J}_{{\left(a^{\frac{1}{6}}{b}^{\frac{5}{6}}\right)}^{+}}^{\alpha }f\left(b\right)\hfill \\ \hfill & +\frac{1}{2^{\alpha -1}}\left({}_H{J}_{{\left(a^{\frac{5}{6}}{b}^{\frac{1}{6}}\right)}^{+}}^{\alpha }f\left(a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right){+}_H{J}_{{\left(a^{\frac{1}{6}}{b}^{\frac{5}{6}}\right)}^{-}}^{\alpha }f\left(a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)\right).\hfill \end{array}$$

(1)

Proof. 

Let

$$I=I_1-I_2+I_3-I_4,$$

(2)

where

$$\begin{array}{cc}\hfill I_1=& \underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\varkappa }^{\alpha }{a}^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}{f}^{\prime }\left(a^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}\right)d\varkappa ,\hfill \\ \hfill I_2=& \underset{0}{\stackrel{1}{\int }}\left({\left(1-\varkappa \right)}^{\alpha }-\frac{3}{8}\right)a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}{f}^{\prime }\left(a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}\right)d\varkappa ,\hfill \\ \hfill I_3=& \underset{0}{\stackrel{1}{\int }}\left({\varkappa }^{\alpha }-\frac{3}{8}\right)a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}{f}^{\prime }\left(a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}\right)d\varkappa \hfill \end{array}$$

and

$$I_4=\underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\left(1-\varkappa \right)}^{\alpha }{a}^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}{f}^{\prime }\left(a^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}\right)d\varkappa ,$$

Integrating by parts $I_1$, we have

$$\begin{array}{cc}\hfill I_1=& \underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\varkappa }^{\alpha }{a}^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}{f}^{\prime }\left(a^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}\right)d\varkappa \hfill \\ \hfill =& {\left.\frac{6{\varkappa }^{\alpha }}{4\left(lnb-lna\right)}f\left(a^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}\right)\right|}_{t=0}^{t=1}-\frac{6\alpha }{4\left(lnb-lna\right)}\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha -1}f\left(a^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}\right)d\varkappa \hfill \\ \hfill =& \frac{3}{2\left(lnb-lna\right)}f\left(a^{\frac{5}{6}}{b}^{\frac{1}{6}}\right)-\frac{6^{\alpha +1}\alpha }{4{\left(lnb-lna\right)}^{\alpha +1}}\underset{a}{\stackrel{a^{\frac{5}{6}}{b}^{\frac{1}{6}}}{\int }}{\left(lnu-lna\right)}^{\alpha -1}f\left(u\right)\frac{du}{u}\hfill \\ \hfill =& \frac{3}{2\left(lnb-lna\right)}f\left(a^{\frac{5}{6}}{b}^{\frac{1}{6}}\right)-{\frac{6^{\alpha +1}\mathsf{\Gamma }\left(\alpha +1\right)}{4{\left(lnb-lna\right)}^{\alpha +1}}}_H{J}_{{\left(a^{\frac{5}{6}}{b}^{\frac{1}{6}}\right)}^{-}}^{\alpha }f\left(a\right).\hfill \end{array}$$

(3)

Similarly, we have

$$\begin{array}{cc}\hfill I_2=& \underset{0}{\stackrel{1}{\int }}\left({\left(1-\varkappa \right)}^{\alpha }-\frac{3}{8}\right)a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}{f}^{\prime }\left(a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}\right)d\varkappa \hfill \\ \hfill =& {\left.\frac{3}{lnb-lna}\left({\left(1-\varkappa \right)}^{\alpha }+\frac{3}{8}\right)f\left(a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}\right)\right|}_{t=0}^{t=1}\hfill \\ \hfill & +\frac{3\alpha }{lnb-lna}\underset{0}{\stackrel{1}{\int }}{\left(1-\varkappa \right)}^{\alpha -1}f\left(a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}\right)d\varkappa \hfill \\ \hfill =& -\frac{9}{8\left(lnb-lna\right)}f\left(a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)-\frac{15}{8\left(lnb-lna\right)}f\left(a^{\frac{5}{6}}{b}^{\frac{1}{6}}\right)\hfill \\ \hfill & +\frac{3^{\alpha +1}\alpha }{{\left(lnb-lna\right)}^{\alpha +1}}\underset{a^{\frac{5}{6}}{b}^{\frac{1}{6}}}{\stackrel{a^{\frac{1}{2}}{b}^{\frac{1}{2}}}{\int }}{\left(ln{a}^{\frac{1}{2}}{b}^{\frac{1}{2}}-lnu\right)}^{\alpha -1}f\left(u\right)\frac{du}{u}\hfill \\ \hfill =& -\frac{9}{8\left(lnb-lna\right)}f\left(a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)-\frac{15}{8\left(lnb-lna\right)}f\left(a^{\frac{5}{6}}{b}^{\frac{1}{6}}\right)\hfill \\ \hfill & +{\frac{3^{\alpha +1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(lnb-lna\right)}^{\alpha +1}}}_H{J}_{{\left(a^{\frac{5}{6}}{b}^{\frac{1}{6}}\right)}^{+}}^{\alpha }f\left(a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right),\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill I_3=& \underset{0}{\stackrel{1}{\int }}\left({\varkappa }^{\alpha }-\frac{3}{8}\right)a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}{f}^{\prime }\left(a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}\right)d\varkappa \hfill \\ \hfill =& {\left.\frac{3}{lnb-lna}\left({\varkappa }^{\alpha }-\frac{3}{8}\right)f\left(a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}\right)\right|}_{t=0}^{t=1}\hfill \\ \hfill & -\frac{3\alpha }{lnb-lna}\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha -1}f\left(a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}\right)d\varkappa \hfill \\ \hfill =& \frac{15}{8\left(lnb-lna\right)}f\left(a^{\frac{1}{6}}{b}^{\frac{5}{6}}\right)+\frac{9}{8\left(lnb-lna\right)}f\left(a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)\hfill \\ \hfill & -\frac{3^{\alpha +1}\alpha }{{\left(lnb-lna\right)}^{\alpha +1}}\underset{a^{\frac{1}{2}}{b}^{\frac{1}{2}}}{\stackrel{a^{\frac{1}{6}}{b}^{\frac{5}{2}}}{\int }}{\left(lnu-ln{a}^{\frac{5}{6}}{b}^{\frac{1}{6}}\right)}^{\alpha -1}f\left(u\right)\frac{du}{u}\hfill \\ \hfill =& \frac{15}{8\left(lnb-lna\right)}f\left(a^{\frac{1}{6}}{b}^{\frac{5}{6}}\right)+\frac{9}{8\left(lnb-lna\right)}f\left(a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)\hfill \\ \hfill & -{\frac{3^{\alpha +1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(lnb-lna\right)}^{\alpha +1}}}_H{J}_{{\left(a^{\frac{1}{6}}{b}^{\frac{5}{6}}\right)}^{-}}^{\alpha }f\left(a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill I_4=& \underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\left(1-\varkappa \right)}^{\alpha }{a}^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}{f}^{\prime }\left(a^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}\right)d\varkappa \hfill \\ \hfill =& {\left.\frac{3}{2\left(lnb-lna\right)}{\left(1-\varkappa \right)}^{\alpha }f\left(a^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}\right)\right|}_{t=0}^{t=1}\hfill \\ \hfill & +\frac{3\alpha }{2\left(lnb-lna\right)}\underset{0}{\stackrel{1}{\int }}{\left(1-\varkappa \right)}^{\alpha -1}f\left(a^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}\right)d\varkappa \hfill \\ \hfill =& -\frac{6}{4\left(lnb-lna\right)}f\left(a^{\frac{1}{6}}{b}^{\frac{5}{6}}\right)+\frac{6^{\alpha +1}\alpha }{4{\left(lnb-lna\right)}^{\alpha +1}}\underset{a^{\frac{1}{6}}{b}^{\frac{5}{6}}}{\stackrel{b}{\int }}{\left(lnb-lnu\right)}^{\alpha -1}f\left(u\right)\frac{du}{u}\hfill \\ \hfill =& -\frac{6}{4\left(lnb-lna\right)}f\left(a^{\frac{1}{6}}{b}^{\frac{5}{6}}\right)+{\frac{6^{\alpha +1}\mathsf{\Gamma }\left(\alpha +1\right)}{4{\left(lnb-lna\right)}^{\alpha +1}}}_H{J}_{{\left(a^{\frac{1}{6}}{b}^{\frac{5}{6}}\right)}^{+}}^{\alpha }f\left(b\right).\hfill \end{array}$$

(6)

Using (3)–(6) in (2), and then multiplying the resulting equality by $\frac{lnb-lna}{9}$, we obtain the desired result. □

In order for the paper to be well organized, we calculated the resulting integrals separately, so that there is no confusion.

Lemma 3.

Let λ and η be two positive numbers. Then, the following equality holds:

$$I_1\left(\lambda ,\eta \right)=\underset{0}{\stackrel{\lambda }{\int }}{e}^{\eta \varkappa }d\varkappa =\frac{1}{\eta }\left(e^{\eta \lambda }-1\right)$$

(7)

and

$$I_2\left(\lambda ,\eta \right)=\underset{0}{\stackrel{\lambda }{\int }}\varkappa e^{\eta \varkappa }d\varkappa =\frac{1}{\eta }\lambda e^{\eta \lambda }-\frac{1}{{\eta }^2}\left(e^{\eta \lambda }-1\right).$$

(8)

Proof. 

By computing directly, we have

$$\underset{0}{\stackrel{\lambda }{\int }}{e}^{\eta \varkappa }d\varkappa ={\left.\frac{1}{\eta }{e}^{\eta \varkappa }\right|}_{\varkappa =0}^{\varkappa =\lambda }=\frac{1}{\eta }\left(e^{\eta \lambda }-1\right).$$

By using the integration by parts, we have

$$\begin{array}{cc}\hfill \underset{0}{\stackrel{\lambda }{\int }}\varkappa e^{\eta \varkappa }d\varkappa =& {\left.\frac{1}{\eta }\varkappa e^{\eta \varkappa }\right|}_{\varkappa =0}^{\varkappa =\lambda }-\underset{0}{\stackrel{\lambda }{\int }}\frac{1}{\eta }{e}^{\eta \varkappa }d\varkappa \hfill \\ \hfill =& \frac{1}{\eta }\lambda e^{\eta \lambda }-{\left.\frac{1}{{\eta }^2}{e}^{\eta \varkappa }\right|}_{\varkappa =0}^{\varkappa =\lambda }=\frac{1}{\eta }\lambda e^{\eta \lambda }-\frac{1}{{\eta }^2}\left(e^{\eta \lambda }-1\right).\hfill \end{array}$$

The proof is completed. □

Lemma 4.

Let α and θ be two positive numbers. Then, the following equality holds:

$${\vartheta }_1\left(\alpha ,\theta \right)=\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha }\left(1-\frac{1}{6}\varkappa \right){\theta }^{\frac{\varkappa }{6}}d\varkappa =\frac{6^{\alpha +1}{.}_1{F}_1\left(\left[\alpha +1,\alpha +3;\frac{1}{6}\right],ln\theta \right)}{\left(\alpha +1\right)\left(\alpha +2\right)}.$$

(9)

Proof. 

By computing directly, we have

$$\begin{array}{cc}\hfill & \underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha }\left(1-\frac{1}{6}\varkappa \right){\theta }^{\frac{\varkappa }{6}}d\varkappa =\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha }\left(1-\frac{1}{6}\varkappa \right)e^{\frac{1}{6}\varkappa ln\theta }d\varkappa \hfill \\ \hfill =& \frac{6^{\alpha +1}{.}_1{F}_1\left(\left[\alpha +1,\alpha +3;\frac{1}{6}\right],ln\theta \right)}{\left(\alpha +1\right)\left(\alpha +2\right)}.\hfill \end{array}$$

The proof is completed. □

Lemma 5.

Let α and θ be two positive numbers. Then, the following equality holds:

$$\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|\left(3+2\varkappa \right){\theta }^{\frac{\varkappa }{3}}d\varkappa ={\mu }_1\left(\alpha ,\theta \right)+{\mu }_2\left(\alpha ,\theta \right).$$

(10)

Proof. 

Clearly, we have

$$\begin{array}{cc}\hfill & \underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|\left(3+2\varkappa \right){\theta }^{\frac{\varkappa }{3}}d\varkappa \hfill \\ \hfill =& \underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\left(\frac{3}{8}-{\varkappa }^{\alpha }\right)\left(3+2\varkappa \right){\theta }^{\frac{\varkappa }{3}}d\varkappa +\underset{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\stackrel{1}{\int }}\left({\varkappa }^{\alpha }-\frac{3}{8}\right)\left(3+2\varkappa \right){\theta }^{\frac{\varkappa }{3}}d\varkappa .\hfill \end{array}$$

(11)

By computing directly, we obtain

$$\begin{array}{cc}\hfill {\mu }_1\left(\alpha ,\theta \right)=& \underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\left(\frac{3}{8}-{\varkappa }^{\alpha }\right)\left(3+2\varkappa \right){\theta }^{\frac{\varkappa }{3}}d\varkappa \hfill \\ \hfill =& \frac{1}{8}\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\left(9+6\varkappa -24{\varkappa }^{\alpha }-16{\varkappa }^{\alpha +1}\right)e^{\frac{\varkappa }{3}ln\theta }d\varkappa \hfill \\ \hfill =& \frac{9}{8}\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}{e}^{\frac{\varkappa }{3}ln\theta }d\varkappa +\frac{6}{8}\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\varkappa e^{\frac{\varkappa }{3}ln\theta }d\varkappa \hfill \\ \hfill & -3\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}{\varkappa }^{\alpha }{e}^{\frac{\varkappa }{3}ln\theta }d\varkappa -2\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}{\varkappa }^{\alpha +1}{e}^{\frac{\varkappa }{3}ln\theta }d\varkappa \hfill \\ \hfill =& \frac{9}{8}{I}_1\left({\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }},\frac{1}{3}ln\theta \right)+\frac{6}{8}{I}_2\left({\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }},\frac{1}{3}ln\theta \right)\hfill \\ \hfill & -\frac{3{.}_1{F}_1\left(\left[\alpha +1,\alpha +2;{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{1}{3}ln\theta \right)}{\alpha +1}-\frac{2{.}_1{F}_1\left(\left[\alpha +2,\alpha +3;{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{1}{3}ln\theta \right)}{\alpha +2}.\hfill \end{array}$$

(12)

On the other hand, we have

$$\begin{array}{cc}\hfill {\mu }_2\left(\alpha ,\theta \right)=& \underset{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\stackrel{1}{\int }}\left({\varkappa }^{\alpha }-\frac{3}{8}\right)\left(3+2\varkappa \right){\theta }^{\frac{\varkappa }{3}}d\varkappa \hfill \\ \hfill =& \frac{{\theta }^{\frac{1}{3}}}{8}\underset{0}{\stackrel{1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\left(40{\left(1-\varkappa \right)}^{\alpha }-16\varkappa {\left(1-\varkappa \right)}^{\alpha }-15+6\varkappa \right)e^{-\frac{\varkappa }{3}ln\theta }d\varkappa \hfill \\ \hfill =& 5{\theta }^{\frac{1}{3}}\underset{0}{\stackrel{1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}{\left(1-\varkappa \right)}^{\alpha }{e}^{-\frac{\varkappa }{3}ln\theta }d\varkappa -2{\theta }^{\frac{1}{3}}\underset{0}{\stackrel{1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\varkappa {\left(1-\varkappa \right)}^{\alpha }{e}^{-\frac{\varkappa }{3}ln\theta }d\varkappa \hfill \\ \hfill & -\frac{15{\theta }^{\frac{1}{3}}}{8}\underset{0}{\stackrel{1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}{e}^{-\frac{\varkappa }{3}ln\theta }d\varkappa +\frac{3{\theta }^{\frac{1}{3}}}{4}\underset{0}{\stackrel{1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\varkappa e^{-\frac{\varkappa }{3}ln\theta }d\varkappa \hfill \\ \hfill =& 5{\theta }^{\frac{1}{3}}\frac{{}_1{F}_1\left(\left[1,\alpha +2;1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],-\frac{1}{3}ln\theta \right)}{\alpha +1}-2{\theta }^{\frac{1}{3}}\frac{{}_1{F}_1\left(\left[2,\alpha +3;1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],-\frac{1}{3}ln\theta \right)}{\left(\alpha +1\right)\left(\alpha +2\right)}\hfill \\ \hfill & -\frac{15{\theta }^{\frac{1}{3}}}{8}{I}_1\left(1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }},-\frac{1}{3}ln\theta \right)+\frac{3{\theta }^{\frac{1}{3}}}{4}{I}_2\left(1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }},-\frac{1}{3}ln\theta \right).\hfill \end{array}$$

(13)

Using (12) and (13) in (11), we obtain the desired result. The proof is completed. □

Lemma 6.

Let α and θ be two positive numbers. Then, the following equality holds:

$$\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|\left(3-2\varkappa \right){\theta }^{\frac{\varkappa }{3}}d\varkappa ={\mu }_3\left(\alpha ,\theta \right)+{\mu }_4\left(\alpha ,\theta \right).$$

(14)

Proof. 

Clearly, we have

$$\begin{array}{cc}\hfill & \underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|\left(3-2\varkappa \right){\theta }^{\frac{\varkappa }{3}}d\varkappa \hfill \\ \hfill =& \underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\left(\frac{3}{8}-{\varkappa }^{\alpha }\right)\left(3-2\varkappa \right){\theta }^{\frac{\varkappa }{3}}d\varkappa +\underset{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\stackrel{1}{\int }}\left({\varkappa }^{\alpha }-\frac{3}{8}\right)\left(3-2\varkappa \right){\theta }^{\frac{\varkappa }{3}}d\varkappa .\hfill \end{array}$$

(15)

By computing directly, we obtain

$$\begin{array}{cc}\hfill {\mu }_3\left(\alpha ,\theta \right)=& \underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\left(\frac{3}{8}-{\varkappa }^{\alpha }\right)\left(3-2\varkappa \right){\theta }^{\frac{\varkappa }{3}}d\varkappa \hfill \\ \hfill =& \frac{1}{8}\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\left(9-6\varkappa -24{\varkappa }^{\alpha }+16{\varkappa }^{\alpha +1}\right)e^{\frac{\varkappa }{3}ln\theta }d\varkappa \hfill \\ \hfill =& \frac{9}{8}\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}{e}^{\frac{\varkappa }{3}ln\theta }d\varkappa -\frac{3}{4}\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\varkappa e^{\frac{\varkappa }{3}ln\theta }d\varkappa \hfill \\ \hfill & -3\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}{\varkappa }^{\alpha }{e}^{\frac{\varkappa }{3}ln\theta }d\varkappa +2\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}{\varkappa }^{\alpha +1}{e}^{\frac{\varkappa }{3}ln\theta }d\varkappa \hfill \\ \hfill =& \frac{9}{8}{I}_1\left({\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }},\frac{1}{3}ln\theta \right)-\frac{3}{4}{I}_2\left({\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }},\frac{1}{3}ln\theta \right)\hfill \\ \hfill & -3\frac{{}_1{F}_1\left(\left[\alpha +1,\alpha +2;{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{1}{3}ln\theta \right)}{\alpha +1}+2\frac{{}_1{F}_1\left(\left[\alpha +2,\alpha +3;{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{1}{3}ln\theta \right)}{\alpha +2}.\hfill \end{array}$$

(16)

And on the other hand, we have

$$\begin{array}{cc}\hfill {\mu }_4\left(\alpha ,\theta \right)=& \underset{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\stackrel{1}{\int }}\left({\varkappa }^{\alpha }-\frac{3}{8}\right)\left(3-2\varkappa \right){\theta }^{\frac{\varkappa }{3}}d\varkappa \hfill \\ \hfill =& \frac{{\theta }^{\frac{1}{3}}}{8}\underset{0}{\stackrel{1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\left(8{\left(1-\varkappa \right)}^{\alpha }+16\varkappa {\left(1-\varkappa \right)}^{\alpha }-3-6\varkappa \right)e^{-\frac{\varkappa }{3}ln\theta }d\varkappa \hfill \\ \hfill =& {\theta }^{\frac{1}{3}}\underset{0}{\stackrel{1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}{\left(1-\varkappa \right)}^{\alpha }{e}^{-\frac{\varkappa }{3}ln\theta }d\varkappa +2{\theta }^{\frac{1}{3}}\underset{0}{\stackrel{1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\varkappa {\left(1-\varkappa \right)}^{\alpha }{e}^{-\frac{\varkappa }{3}ln\theta }d\varkappa \hfill \\ \hfill & -\frac{3{\theta }^{\frac{1}{3}}}{8}\underset{0}{\stackrel{1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}{e}^{-\frac{\varkappa }{3}ln\theta }d\varkappa -\frac{3{\theta }^{\frac{1}{3}}}{4}\underset{0}{\stackrel{1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\varkappa e^{-\frac{\varkappa }{3}ln\theta }d\varkappa \hfill \\ \hfill =& {\theta }^{\frac{1}{3}}\frac{{}_1{F}_1\left(\left[1,\alpha +2;1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],-\frac{1}{3}ln\theta \right)}{\alpha +1}+2{\theta }^{\frac{1}{3}}\frac{{}_1{F}_1\left(\left[2,\alpha +3;1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],-\frac{1}{3}ln\theta \right)}{\left(\alpha +1\right)\left(\alpha +2\right)}\hfill \\ \hfill & -\frac{3{\theta }^{\frac{1}{3}}}{8}{I}_1\left(1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }},-\frac{1}{3}ln\theta \right)-\frac{3{\theta }^{\frac{1}{3}}}{4}{I}_2\left(1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }},-\frac{1}{3}ln\theta \right).\hfill \end{array}$$

(17)

Using (15) and (16) in (14), we obtain the desired result. The proof is completed. □

Lemma 7.

Let α and θ be two positive numbers. Then, the following equality holds:

$${\vartheta }_2\left(\alpha ,\theta \right)=\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha +1}{\theta }^{\frac{\varkappa }{6}}d\varkappa =\frac{{}_1{F}_1\left(\alpha +2,\alpha +3,\frac{1}{6}ln\theta \right)}{\alpha +2}.$$

(18)

Proof. 

By computing directly, we obtain

$$\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha +1}{\theta }^{\frac{\varkappa }{6}}d\varkappa =\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha +1}{e}^{\frac{\varkappa }{6}ln\theta }d\varkappa =\frac{{}_1{F}_1\left(\alpha +2,\alpha +3,\frac{1}{6}ln\theta \right)}{\alpha +2}.$$

The proof is completed. □

Lemma 8.

Let $\lambda ,\beta$ and η be a positive numbers. Then, the following equality holds

$$J_1\left(\lambda ,\eta \right)=\underset{0}{\stackrel{\lambda }{\int }}{\varkappa }^{\beta }{\theta }^{\eta \varkappa }d\varkappa =\frac{{}_1{F}_1\left(\left[\beta +1,\beta +2;\lambda \right],\eta ln\theta \right)}{\beta +1}$$

(19)

and

$$J_2\left(\lambda ,\eta \right)=\underset{0}{\stackrel{\lambda }{\int }}{\left(1-\varkappa \right)}^{\beta }{\theta }^{\eta \varkappa }d\varkappa =\frac{{}_1{F}_1\left(\left[1,\beta +2;\lambda \right],\eta ln\theta \right)}{\beta +1}.$$

(20)

Proof. 

By computing directly, we have

$$\begin{array}{cc}\hfill J_1\left(\lambda ,\eta \right)=& \underset{0}{\stackrel{\lambda }{\int }}{\varkappa }^{\beta }{\theta }^{\eta \varkappa }d\varkappa =J_1\left(\lambda ,\eta \right)=\underset{0}{\stackrel{\lambda }{\int }}{\varkappa }^{\beta }{e}^{\eta \varkappa ln\theta }d\varkappa \hfill \\ \hfill =& \frac{{}_1{F}_1\left(\left[\beta +1,\beta +2;\lambda \right],\eta ln\theta \right)}{\beta +1}.\hfill \end{array}$$

For $J_2\left(\lambda ,\eta \right)$, we have

$$\begin{array}{cc}\hfill J_2\left(\lambda ,\eta \right)=& \underset{0}{\stackrel{\lambda }{\int }}{\left(1-\varkappa \right)}^{\beta }{\theta }^{\eta \varkappa }d\varkappa =\underset{0}{\stackrel{\lambda }{\int }}{\left(1-\varkappa \right)}^{\beta }{e}^{\eta \varkappa ln\theta }d\varkappa \hfill \\ \hfill =& \frac{{}_1{F}_1\left(\left[1,\beta +2;\lambda \right],\eta ln\theta \right)}{\beta +1}.\hfill \end{array}$$

The proof is completed. □

Lemma 9.

Let α and θ be two positive numbers. Then, the following equality holds:

$$\begin{array}{cc}\hfill \mathsf{\Phi }\left(\alpha ,\theta \right)=& \underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|{\theta }^{\frac{\varkappa }{3}}d\varkappa \hfill \\ \hfill =& \frac{9}{8ln\theta }\left(e^{\frac{1}{3}{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}ln\theta }+{\theta }^{\frac{1}{3}}{e}^{\frac{1}{3}\left(1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right)ln{\theta }^{-1}}-{\theta }^{\frac{1}{3}}-1\right)\hfill \\ \hfill & +\frac{{\theta }^{\frac{1}{3}}{.}_1{F}_1\left(\left[1,\alpha +2;1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{1}{3}ln{\theta }^{-1}\right){-}_1{F}_1\left(\left[\alpha +1,\alpha +2;{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{1}{3}ln\theta \right)}{\alpha +1}.\hfill \end{array}$$

(21)

Proof. 

By computing directly, we have

$$\begin{array}{cc}\hfill & \underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|{\theta }^{\frac{\varkappa }{3}}d\varkappa =\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\left(\frac{3}{8}-{\varkappa }^{\alpha }\right){\theta }^{\frac{\varkappa }{3}}d\varkappa +\underset{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\stackrel{1}{\int }}\left({\varkappa }^{\alpha }-\frac{3}{8}\right){\theta }^{\frac{\varkappa }{3}}d\varkappa \hfill \\ \hfill =& \underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\left(\frac{3}{8}-{\varkappa }^{\alpha }\right){\theta }^{\frac{\varkappa }{3}}d\varkappa +{\theta }^{\frac{1}{3}}\underset{0}{\stackrel{1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\left({\left(1-\varkappa \right)}^{\alpha }-\frac{3}{8}\right){\theta }^{-\frac{\varkappa }{3}}d\varkappa \hfill \\ \hfill =& \frac{3}{8}\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}{e}^{\frac{\varkappa }{3}ln\theta }d\varkappa -\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}{\varkappa }^{\alpha }{\theta }^{\frac{\varkappa }{3}}d\varkappa +{\theta }^{\frac{1}{3}}\underset{0}{\stackrel{1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}{\left(1-\varkappa \right)}^{\alpha }{\theta }^{-\frac{\varkappa }{3}}d\varkappa \hfill \\ \hfill & -\frac{3{\theta }^{\frac{1}{3}}}{8}\underset{0}{\stackrel{1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}{e}^{-\frac{\varkappa }{3}ln\theta }d\varkappa \hfill \\ \hfill =& \frac{9}{8ln\theta }\left(e^{\frac{1}{3}{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}ln\theta }-1\right)+\frac{9{\theta }^{\frac{1}{3}}}{8ln\theta }\left(e^{\frac{1}{3}\left(1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right)ln{\theta }^{-1}}-1\right)\hfill \\ \hfill & +\frac{{\theta }^{\frac{1}{3}}{.}_1{F}_1\left(\left[1,\alpha +2;1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{1}{3}ln{\theta }^{-1}\right){-}_1{F}_1\left(\left[\alpha +1,\alpha +2;{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{1}{3}ln\theta \right)}{\alpha +1}.\hfill \end{array}$$

(22)

The proof is completed. □

4. Main Results

Our first result concerns functions whose absolute values of the first derivatives are $GA$-convex functions.

Theorem 1.

Let f: $[a,b]\to \mathbb{R}$ be a differentiable mapping on $\left[a,b\right]$ with $0<a<b$, and $f^{\prime }\in L^1\left[a,b\right]$. If $\left|f^{\prime }\right|$ is $GA$-convex function, then the following inequality for fractional integrals holds:

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left(3f\left(a^{\frac{5}{6}}{b}^{\frac{1}{6}}\right)+2f\left(a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)+3f\left(a^{\frac{1}{6}}{b}^{\frac{5}{6}}\right)\right)-\frac{6^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(lnb-lna\right)}^{\alpha }}\mathrm{{\rm Y}}\right|\hfill \\ \hfill \le & \frac{lnb-lna}{9}\left(\left|f^{\prime }\left(a\right)\right|\left(\frac{a\times 6^{\alpha +1}{.}_1{F}_1\left(\left[\alpha +1,\alpha +3;\frac{1}{6}\right],ln\frac{b}{a}\right)}{4\left(\alpha +1\right)\left(\alpha +2\right)}+\frac{b{\times }_1{F}_1\left(\alpha +2,\alpha +3,\frac{1}{6}ln\frac{a}{b}\right)}{24\left(\alpha +2\right)}\right.\right.\hfill \\ \hfill & +\left.\frac{a^{\frac{1}{2}}{b}^{\frac{1}{2}}}{6}\left({\mu }_1\left(\alpha ,\frac{a}{b}\right)+{\mu }_2\left(\alpha ,\frac{a}{b}\right)+{\mu }_3\left(\alpha ,\frac{b}{a}\right)+{\mu }_4\left(\alpha ,\frac{b}{a}\right)\right)\right)\hfill \\ \hfill & +\left|f^{\prime }\left(b\right)\right|\left(\frac{a{\times }_1{F}_1\left(\alpha +2,\alpha +3,\frac{1}{6}ln\frac{b}{a}\right)}{24\left(\alpha +2\right)}+\frac{b\times 6^{\alpha +1}{.}_1{F}_1\left(\left[\alpha +1,\alpha +3;\frac{1}{6}\right],ln\frac{a}{b}\right)}{4\left(\alpha +1\right)\left(\alpha +2\right)}\right.\hfill \\ \hfill & +\left.\left.\frac{a^{\frac{1}{2}}{b}^{\frac{1}{2}}}{6}\left({\mu }_1\left(\alpha ,\frac{b}{a}\right)+{\mu }_2\left(\alpha ,\frac{b}{a}\right)+{\mu }_3\left(\alpha ,\frac{a}{b}\right)+{\mu }_4\left(\alpha ,\frac{a}{b}\right)\right)\right)\right),\hfill \end{array}$$

where $\alpha >0,\mathrm{{\rm Y}},{\mu }_1,{\mu }_2,{\mu }_3,{\mu }_4$ are defined by (1), (12), (13), (16), and (17), respectively, where ${}_1{F}_1\left(.;.;.\right)$ and ${}_1{F}_1\left(\left[.,.;.\right],.\right)$ are the confluent and the incomplete confluent hypergeometric functions, respectively.

Proof. 

From Lemma 2, and the properties of modulus and $GA$-convexity of $\left|f^{\prime }\right|$, we obtain

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left(3f\left(a^{\frac{5}{6}}{b}^{\frac{1}{6}}\right)+2f\left(a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)+3f\left(a^{\frac{1}{6}}{b}^{\frac{5}{6}}\right)\right)-\frac{6^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(lnb-lna\right)}^{\alpha }}\mathrm{{\rm Y}}\right|\hfill \\ \hfill \le & \frac{lnb-lna}{9}\left(\underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\varkappa }^{\alpha }{a}^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}\left|f^{\prime }\left(a^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}\right)\right|d\varkappa \right.\hfill \\ \hfill & +\underset{0}{\stackrel{1}{\int }}\left|{\left(1-\varkappa \right)}^{\alpha }-\frac{3}{8}\right|a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}\left|f^{\prime }\left(a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}\right)\right|d\varkappa \hfill \\ \hfill & +\underset{0}{\stackrel{1}{\int }}\left(\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|\right)a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}\left|f^{\prime }\left(a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}\right)\right|d\varkappa \hfill \\ \hfill & +\left.\underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\left(1-\varkappa \right)}^{\alpha }{a}^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}\left|f^{\prime }\left(a^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}\right)\right|d\varkappa \right)\hfill \\ \hfill \le & \frac{lnb-lna}{9}\left(\underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\varkappa }^{\alpha }{a}^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}\left(\frac{6-\varkappa }{6}\left|f^{\prime }\left(a\right)\right|+\frac{\varkappa }{6}\left|f^{\prime }\left(b\right)\right|\right)d\varkappa \right.\hfill \\ \hfill & +\underset{0}{\stackrel{1}{\int }}\left|{\left(1-\varkappa \right)}^{\alpha }-\frac{3}{8}\right|a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}\left(\frac{5-2\varkappa }{6}\left|f^{\prime }\left(a\right)\right|+\frac{1+2\varkappa }{6}\left|f^{\prime }\left(b\right)\right|\right)d\varkappa \hfill \\ \hfill & +\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}\left(\frac{3-2\varkappa }{6}\left|f^{\prime }\left(a\right)\right|+\frac{3+2\varkappa }{6}\left|f^{\prime }\left(b\right)\right|\right)d\varkappa \hfill \\ \hfill & +\left.\underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\left(1-\varkappa \right)}^{\alpha }{a}^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}\left(\frac{1-\varkappa }{6}\left|f^{\prime }\left(a\right)\right|+\frac{5+\varkappa }{6}\left|f^{\prime }\left(b\right)\right|\right)d\varkappa \right)\hfill \\ \hfill =& \frac{lnb-lna}{9}\left(\left|f^{\prime }\left(a\right)\right|\underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\varkappa }^{\alpha }\left(1-\frac{1}{6}\varkappa \right)a^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}d\varkappa +\left|f^{\prime }\left(b\right)\right|\underset{0}{\stackrel{1}{\int }}\frac{1}{24}{\varkappa }^{\alpha +1}{a}^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}d\varkappa \right.\hfill \\ \hfill & +\left|f^{\prime }\left(a\right)\right|\underset{0}{\stackrel{1}{\int }}\left|{\left(1-\varkappa \right)}^{\alpha }-\frac{3}{8}\right|a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}\left(\frac{5-2\varkappa }{6}\right)d\varkappa \hfill \\ \hfill & +\left|f^{\prime }\left(b\right)\right|\underset{0}{\stackrel{1}{\int }}\left|{\left(1-\varkappa \right)}^{\alpha }-\frac{3}{8}\right|a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}\left(\frac{1+2\varkappa }{6}\right)d\varkappa \hfill \\ \hfill & +\left|f^{\prime }\left(a\right)\right|\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}\left(\frac{3-2\varkappa }{6}\right)d\varkappa \hfill \\ \hfill & +\left|f^{\prime }\left(b\right)\right|\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}\left(\frac{3+2\varkappa }{6}\right)d\varkappa \hfill \\ \hfill & +\left|f^{\prime }\left(a\right)\right|\underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\left(1-\varkappa \right)}^{\alpha }{a}^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}\left(\frac{1-\varkappa }{6}\right)d\varkappa \hfill \\ \hfill & +\left.\left|f^{\prime }\left(b\right)\right|\underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\left(1-\varkappa \right)}^{\alpha }{a}^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}\left(\frac{5+\varkappa }{6}\right)d\varkappa \right)\hfill \\ \hfill =& \frac{lnb-lna}{9}\left(\left|f^{\prime }\left(a\right)\right|\left(\frac{a}{4}\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha }\left(1-\frac{1}{6}\varkappa \right){\left(\frac{b}{a}\right)}^{\frac{\varkappa }{6}}d\varkappa \right.\right.\hfill \\ \hfill & +\frac{a^{\frac{1}{2}}{b}^{\frac{1}{2}}}{6}\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|\left(3+2\varkappa \right){\left(\frac{a}{b}\right)}^{\frac{\varkappa }{3}}d\varkappa \hfill \\ \hfill & +\left.\frac{a^{\frac{1}{2}}{b}^{\frac{1}{2}}}{6}\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|\left(3-2\varkappa \right){\left(\frac{b}{a}\right)}^{\frac{\varkappa }{3}}d\varkappa +\frac{b}{24}\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha +1}{\left(\frac{a}{b}\right)}^{\frac{\varkappa }{6}}d\varkappa \right)\hfill \\ \hfill & +\left|f^{\prime }\left(b\right)\right|\left(\frac{a}{24}\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha +1}{\left(\frac{b}{a}\right)}^{\frac{\varkappa }{6}}d\varkappa +\frac{a^{\frac{1}{2}}{b}^{\frac{1}{2}}}{6}\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|\left(3-2\varkappa \right){\left(\frac{a}{b}\right)}^{\frac{\varkappa }{3}}d\varkappa \right.\hfill \\ \hfill & +\left.\left.\frac{a^{\frac{1}{2}}{b}^{\frac{1}{2}}}{6}\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|\left(3+2\varkappa \right){\left(\frac{b}{a}\right)}^{\frac{\varkappa }{3}}d\varkappa +\frac{b}{4}\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha }\left(1-\frac{1}{6}\varkappa \right){\left(\frac{a}{b}\right)}^{\frac{\varkappa }{6}}d\varkappa \right)\right)\hfill \\ \hfill =& \frac{lnb-lna}{9}\left(\left|f^{\prime }\left(a\right)\right|\left(\frac{a\times 6^{\alpha +1}{.}_1{F}_1\left(\left[\alpha +1,\alpha +3;\frac{1}{6}\right],ln\frac{b}{a}\right)}{4\left(\alpha +1\right)\left(\alpha +2\right)}+\frac{b{\times }_1{F}_1\left(\alpha +2,\alpha +3,\frac{1}{6}ln\frac{a}{b}\right)}{24\left(\alpha +2\right)}\right.\right.\hfill \\ \hfill & +\left.\frac{a^{\frac{1}{2}}{b}^{\frac{1}{2}}}{6}\left({\mu }_1\left(\alpha ,\frac{a}{b}\right)+{\mu }_2\left(\alpha ,\frac{a}{b}\right)+{\mu }_3\left(\alpha ,\frac{b}{a}\right)+{\mu }_4\left(\alpha ,\frac{b}{a}\right)\right)\right)\hfill \\ \hfill & +\left|f^{\prime }\left(b\right)\right|\left(\frac{a{\times }_1{F}_1\left(\alpha +2,\alpha +3,\frac{1}{6}ln\frac{b}{a}\right)}{24\left(\alpha +2\right)}+\frac{b\times 6^{\alpha +1}{.}_1{F}_1\left(\left[\alpha +1,\alpha +3;\frac{1}{6}\right],ln\frac{a}{b}\right)}{4\left(\alpha +1\right)\left(\alpha +2\right)}\right.\hfill \\ \hfill & +\left.\left.\frac{a^{\frac{1}{2}}{b}^{\frac{1}{2}}}{6}\left({\mu }_1\left(\alpha ,\frac{b}{a}\right)+{\mu }_2\left(\alpha ,\frac{b}{a}\right)+{\mu }_3\left(\alpha ,\frac{a}{b}\right)+{\mu }_4\left(\alpha ,\frac{a}{b}\right)\right)\right)\right),\hfill \end{array}$$

which we have used. The proof is completed. □

The following result deals with the case where the absolute values of the first derivatives at a certain power q are $GA$-convex functions.

Theorem 2.

Let f: $[a,b]\to \mathbb{R}$ be a differentiable mapping on $\left[a,b\right]$ with $0<a<b$, and $f^{\prime }\in L^1\left[a,b\right]$. If ${\left|f^{\prime }\right|}^q$ is $GA$-convex function and $q>1$ with $\frac{1}{p}+\frac{1}{q}=1$, then the following inequality for fractional integrals holds:

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left(3f\left(a^{\frac{5}{6}}{b}^{\frac{1}{6}}\right)+2f\left(a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)+3f\left(a^{\frac{1}{6}}{b}^{\frac{5}{6}}\right)\right)-\frac{6^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(lnb-lna\right)}^{\alpha }}\mathrm{{\rm Y}}\right|\hfill \\ \hfill \le & \frac{lnb-lna}{9}\left(\frac{a}{4}{\left(\frac{{}_1{F}_1\left(p\alpha +1,p\alpha +2,\frac{p}{6}ln\left(\frac{b}{a}\right)\right)}{p\alpha +1}\right)}^{\frac{1}{p}}{\left(\frac{11{\left|f^{\prime }\left(a\right)\right|}^q+{\left|f^{\prime }\left(b\right)\right|}^q}{12}\right)}^{\frac{1}{q}}\right.\hfill \\ \hfill & +a^{\frac{1}{2}}{b}^{\frac{1}{2}}{\left(\frac{2{\left|f^{\prime }\left(a\right)\right|}^q+{\left|f^{\prime }\left(b\right)\right|}^q}{3}\right)}^{\frac{1}{q}}\left(\frac{3^{p+1}\left(e^{\frac{p}{3}{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}ln\left(\frac{a}{b}\right)}-1\right)}{8^{p}pln\left(\frac{a}{b}\right)}-\frac{{}_1{F}_1\left(\left[p\alpha +1,p\alpha +2;{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{p}{3}ln\frac{a}{b}\right)}{p\alpha +1}\right.\hfill \\ \hfill & +{\left.{\left(\frac{a}{b}\right)}^{\frac{p}{3}}\left(\frac{{}_1{F}_1\left(\left[1,p\alpha +2;1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{p}{3}ln\frac{b}{a}\right)}{p\alpha +1}-\frac{3^{p+1}}{8^{p}pln\left(\frac{a}{b}\right)}\left(e^{\frac{p}{3}\left(1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right)ln\left(\frac{a}{b}\right)}-1\right)\right)\right)}^{\frac{1}{p}}\hfill \\ \hfill & +a^{\frac{1}{2}}{b}^{\frac{1}{2}}{\left(\frac{{\left|f^{\prime }\left(a\right)\right|}^q+2{\left|f^{\prime }\left(b\right)\right|}^q}{3}\right)}^{\frac{1}{q}}\left(\frac{3^{p+1}\left(e^{\frac{p}{3}{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}ln\left(\frac{b}{a}\right)}-1\right)}{8^{p}pln\left(\frac{a}{b}\right)}-\frac{{}_1{F}_1\left(\left[p\alpha +1,p\alpha +2;{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{p}{3}ln\frac{b}{a}\right)}{p\alpha +1}\right.\hfill \\ \hfill & +{\left.{\left(\frac{b}{a}\right)}^{\frac{p}{3}}\left(\frac{{}_1{F}_1\left(\left[1,p\alpha +2;1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{p}{3}ln\frac{a}{b}\right)}{p\alpha +1}-{\left(\frac{3}{8}\right)}^p\frac{{}_1{F}_1\left(\left[p\alpha +1,p\alpha +2;1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{p}{3}ln\frac{a}{b}\right)}{p\alpha +1}\right)\right)}^{\frac{1}{p}}\hfill \\ \hfill & +\left.\frac{b}{4}{\left(\frac{{}_1{F}_1\left(p\alpha +1,p\alpha +2,\frac{p}{6}ln\frac{a}{b}\right)}{p\alpha +1}\right)}^{\frac{1}{p}}{\left(\frac{{\left|f^{\prime }\left(a\right)\right|}^q+11{\left|f^{\prime }\left(b\right)\right|}^q}{12}\right)}^{\frac{1}{q}}\right),\hfill \end{array}$$

where $\alpha >0,$$\mathrm{{\rm Y}}$ is given by (1), and ${}_1{F}_1\left(.;.;.\right)$ and ${}_1{F}_1\left(\left[.,.;.\right],.\right)$ are the confluent and the incomplete confluent hypergeometric functions, respectively.

Proof. 

From Lemma 2, the modulus, Hölder’s inequality, $GA$-convexity of ${\left|f^{\prime }\right|}^q$, and Lemma 1, we have

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left(3f\left(a^{\frac{5}{6}}{b}^{\frac{1}{6}}\right)+2f\left(a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)+3f\left(a^{\frac{1}{6}}{b}^{\frac{5}{6}}\right)\right)-\frac{6^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(lnb-lna\right)}^{\alpha }}\mathrm{{\rm Y}}\right|\hfill \\ \hfill \le & \frac{lnb-lna}{9}\left({\left(\underset{0}{\stackrel{1}{\int }}{\left(\frac{1}{4}{\varkappa }^{\alpha }{a}^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}\right)}^{p}d\varkappa \right)}^{\frac{1}{p}}{\left(\underset{0}{\stackrel{1}{\int }}{\left|f^{\prime }\left(a^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}\right)\right|}^{q}d\varkappa \right)}^{\frac{1}{q}}\right.\hfill \\ \hfill & +{\left(\underset{0}{\stackrel{1}{\int }}{\left(\left|{\left(1-\varkappa \right)}^{\alpha }-\frac{3}{8}\right|a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}\right)}^{p}d\varkappa \right)}^{\frac{1}{p}}{\left(\underset{0}{\stackrel{1}{\int }}{\left|f^{\prime }\left(a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}\right)\right|}^{q}d\varkappa \right)}^{\frac{1}{q}}\hfill \\ \hfill & +{\left(\underset{0}{\stackrel{1}{\int }}{\left(\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}\right)}^{p}d\varkappa \right)}^{\frac{1}{p}}{\left(\underset{0}{\stackrel{1}{\int }}{\left|f^{\prime }\left(a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}\right)\right|}^{q}d\varkappa \right)}^{\frac{1}{q}}\hfill \\ \hfill & +\left.{\left(\underset{0}{\stackrel{1}{\int }}{\left(\frac{1}{4}{\left(1-\varkappa \right)}^{\alpha }{a}^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}\right)}^{p}d\varkappa \right)}^{\frac{1}{p}}{\left(\underset{0}{\stackrel{1}{\int }}{\left|f^{\prime }\left(a^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}\right)\right|}^{q}d\varkappa \right)}^{\frac{1}{q}}\right)\hfill \\ \hfill \le & \frac{lnb-lna}{9}\left(\frac{a}{4}{\left(\underset{0}{\stackrel{1}{\int }}{\varkappa }^{p\alpha }{\left(\frac{b}{a}\right)}^{\frac{p}{6}\varkappa }d\varkappa \right)}^{\frac{1}{p}}{\left(\underset{0}{\stackrel{1}{\int }}\left(\frac{6-\varkappa }{6}{\left|f^{\prime }\left(a\right)\right|}^q+\frac{\varkappa }{6}{\left|f^{\prime }\left(b\right)\right|}^q\right)d\varkappa \right)}^{\frac{1}{q}}\right.\hfill \\ \hfill & +a^{\frac{1}{2}}{b}^{\frac{1}{2}}{\left(\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}{\left(\frac{3}{8}-{\varkappa }^{\alpha }\right)}^p{\left(\frac{a}{b}\right)}^{\frac{p}{3}\varkappa }d\varkappa +\underset{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\stackrel{1}{\int }}{\left({\varkappa }^{\alpha }-\frac{3}{8}\right)}^p{\left(\frac{a}{b}\right)}^{\frac{p}{3}\varkappa }d\varkappa \right)}^{\frac{1}{p}}\hfill \\ \hfill & \times {\left(\underset{0}{\stackrel{1}{\int }}\left(\frac{5-2\varkappa }{6}{\left|f^{\prime }\left(a\right)\right|}^q+\frac{1+2\varkappa }{6}{\left|f^{\prime }\left(b\right)\right|}^q\right)d\varkappa \right)}^{\frac{1}{q}}\hfill \\ \hfill & +a^{\frac{1}{2}}{b}^{\frac{1}{2}}{\left(\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}{\left(\frac{3}{8}-{\varkappa }^{\alpha }\right)}^p{\left(\frac{b}{a}\right)}^{\frac{p}{3}\varkappa }d\varkappa +\underset{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\stackrel{1}{\int }}{\left({\varkappa }^{\alpha }-\frac{3}{8}\right)}^p{\left(\frac{b}{a}\right)}^{\frac{p}{3}\varkappa }d\varkappa \right)}^{\frac{1}{p}}\hfill \\ \hfill & \times {\left(\underset{0}{\stackrel{1}{\int }}\left(\frac{3-2\varkappa }{6}{\left|f^{\prime }\left(a\right)\right|}^q+\frac{3+2\varkappa }{6}{\left|f^{\prime }\left(b\right)\right|}^q\right)d\varkappa \right)}^{\frac{1}{q}}\hfill \\ \hfill & +\left.\frac{b}{4}{\left(\underset{0}{\stackrel{1}{\int }}{\varkappa }^{p\alpha }{\left(\frac{a}{b}\right)}^{\frac{p}{6}\varkappa }d\varkappa \right)}^{\frac{1}{p}}{\left(\underset{0}{\stackrel{1}{\int }}\left(\frac{1-\varkappa }{6}{\left|f^{\prime }\left(a\right)\right|}^q+\frac{5+\varkappa }{6}{\left|f^{\prime }\left(b\right)\right|}^q\right)d\varkappa \right)}^{\frac{1}{q}}\right)\hfill \\ \hfill \le & \frac{lnb-lna}{9}\left(\frac{a}{4}{\left(\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\left(p\alpha -1\right)+1}{\left({\left(\frac{b}{a}\right)}^p\right)}^{\frac{1}{6}\varkappa }d\varkappa \right)}^{\frac{1}{p}}{\left({\left|f^{\prime }\left(a\right)\right|}^q\underset{0}{\stackrel{1}{\int }}\frac{6-\varkappa }{6}d\varkappa +{\left|f^{\prime }\left(b\right)\right|}^q\underset{0}{\stackrel{1}{\int }}\frac{\varkappa }{6}d\varkappa \right)}^{\frac{1}{q}}\right.\hfill \\ \hfill & +a^{\frac{1}{2}}{b}^{\frac{1}{2}}\left(\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\left({\left(\frac{3}{8}\right)}^p{e}^{\frac{p}{3}\varkappa ln\left(\frac{a}{b}\right)}-{\varkappa }^{p\alpha }{\left(\frac{a}{b}\right)}^{\frac{p}{3}\varkappa }\right)d\varkappa \right.\hfill \\ \hfill & +{\left.{\left(\frac{a}{b}\right)}^{\frac{p}{3}}\underset{0}{\stackrel{1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\left({\left(1-\varkappa \right)}^{p\alpha }{\left(\frac{b}{a}\right)}^{\frac{p}{3}\varkappa }-{\left(\frac{3}{8}\right)}^p{e}^{\frac{p}{3}\varkappa ln\left(\frac{b}{a}\right)}\right)d\varkappa \right)}^{\frac{1}{p}}\hfill \\ \hfill & \times {\left({\left|f^{\prime }\left(a\right)\right|}^q\underset{0}{\stackrel{1}{\int }}\frac{5-2\varkappa }{6}d\varkappa +{\left|f^{\prime }\left(b\right)\right|}^q\underset{0}{\stackrel{1}{\int }}\frac{1+2\varkappa }{6}d\varkappa \right)}^{\frac{1}{q}}\hfill \\ \hfill & +a^{\frac{1}{2}}{b}^{\frac{1}{2}}\left(\underset{0}{\stackrel{{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\left({\left(\frac{3}{8}\right)}^p{e}^{\frac{p}{3}\varkappa ln\left(\frac{b}{a}\right)}-{\varkappa }^{p\alpha }{\left(\frac{b}{a}\right)}^{\frac{p}{3}\varkappa }\right)d\varkappa \right.\hfill \\ \hfill & +{\left.{\left(\frac{b}{a}\right)}^{\frac{p}{3}}\underset{0}{\stackrel{1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}}{\int }}\left({\left(1-\varkappa \right)}^{p\alpha }{\left(\frac{a}{b}\right)}^{\frac{p}{3}\varkappa }-{\left(\frac{3}{8}\right)}^p{\varkappa }^{p\alpha }{\left(\frac{a}{b}\right)}^{\frac{p}{3}\varkappa }\right)d\varkappa \right)}^{\frac{1}{p}}\hfill \\ \hfill & \times {\left({\left|f^{\prime }\left(a\right)\right|}^q\underset{0}{\stackrel{1}{\int }}\frac{3-2\varkappa }{6}d\varkappa +{\left|f^{\prime }\left(b\right)\right|}^q\underset{0}{\stackrel{1}{\int }}\frac{3+2\varkappa }{6}d\varkappa \right)}^{\frac{1}{q}}\hfill \\ \hfill & +\left.\frac{b}{4}{\left(\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\left(p\alpha -1\right)+1}{\left({\left(\frac{a}{b}\right)}^p\right)}^{\frac{1}{6}\varkappa }d\varkappa \right)}^{\frac{1}{p}}{\left({\left|f^{\prime }\left(a\right)\right|}^q\underset{0}{\stackrel{1}{\int }}\frac{1-\varkappa }{6}d\varkappa +{\left|f^{\prime }\left(b\right)\right|}^q\underset{0}{\stackrel{1}{\int }}\frac{5+\varkappa }{6}d\varkappa \right)}^{\frac{1}{q}}\right)\hfill \\ \hfill =& \frac{lnb-lna}{9}\left(\frac{a}{4}{\left(\frac{{}_1{F}_1\left(p\alpha +1,p\alpha +2,\frac{p}{6}ln\left(\frac{b}{a}\right)\right)}{p\alpha +1}\right)}^{\frac{1}{p}}{\left(\frac{11{\left|f^{\prime }\left(a\right)\right|}^q+{\left|f^{\prime }\left(b\right)\right|}^q}{12}\right)}^{\frac{1}{q}}\right.\hfill \\ \hfill & +a^{\frac{1}{2}}{b}^{\frac{1}{2}}{\left(\frac{2{\left|f^{\prime }\left(a\right)\right|}^q+{\left|f^{\prime }\left(b\right)\right|}^q}{3}\right)}^{\frac{1}{q}}\left(\frac{3^{p+1}\left(e^{\frac{p}{3}{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}ln\left(\frac{a}{b}\right)}-1\right)}{8^{p}pln\left(\frac{a}{b}\right)}-\frac{{}_1{F}_1\left(\left[p\alpha +1,p\alpha +2;{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{p}{3}ln\frac{a}{b}\right)}{p\alpha +1}\right.\hfill \\ \hfill & +{\left.{\left(\frac{a}{b}\right)}^{\frac{p}{3}}\left(\frac{{}_1{F}_1\left(\left[1,p\alpha +2;1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{p}{3}ln\frac{b}{a}\right)}{p\alpha +1}-\frac{3^{p+1}}{8^{p}pln\left(\frac{a}{b}\right)}\left(e^{\frac{p}{3}\left(1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right)ln\left(\frac{a}{b}\right)}-1\right)\right)\right)}^{\frac{1}{p}}\hfill \\ \hfill & +a^{\frac{1}{2}}{b}^{\frac{1}{2}}{\left(\frac{{\left|f^{\prime }\left(a\right)\right|}^q+2{\left|f^{\prime }\left(b\right)\right|}^q}{3}\right)}^{\frac{1}{q}}\left(\frac{3^{p+1}\left(e^{\frac{p}{3}{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}ln\left(\frac{b}{a}\right)}-1\right)}{8^{p}pln\left(\frac{a}{b}\right)}-\frac{{}_1{F}_1\left(\left[p\alpha +1,p\alpha +2;{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{p}{3}ln\frac{b}{a}\right)}{p\alpha +1}\right.\hfill \\ \hfill & +{\left.{\left(\frac{b}{a}\right)}^{\frac{p}{3}}\left(\frac{{}_1{F}_1\left(\left[1,p\alpha +2;1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{p}{3}ln\frac{a}{b}\right)}{p\alpha +1}-{\left(\frac{3}{8}\right)}^p\frac{{}_1{F}_1\left(\left[p\alpha +1,p\alpha +2;1-{\left(\frac{3}{8}\right)}^{\frac{1}{\alpha }}\right],\frac{p}{3}ln\frac{a}{b}\right)}{p\alpha +1}\right)\right)}^{\frac{1}{p}}\hfill \\ \hfill & +\left.\frac{b}{4}{\left(\frac{{}_1{F}_1\left(p\alpha +1,p\alpha +2,\frac{p}{6}ln\frac{a}{b}\right)}{p\alpha +1}\right)}^{\frac{1}{p}}{\left(\frac{{\left|f^{\prime }\left(a\right)\right|}^q+11{\left|f^{\prime }\left(b\right)\right|}^q}{12}\right)}^{\frac{1}{q}}\right),\hfill \end{array}$$

The proof is completed. □

The following theorem represents a variation of Theorem 2.

Theorem 3.

Let f: $[a,b]\to \mathbb{R}$ be a differentiable mapping on $\left[a,b\right]$ with $0<a<b$, and $f^{\prime }\in L^1\left[a,b\right]$. If ${\left|f^{\prime }\right|}^q$ is $GA$-convex function and $q\ge 1$, then the following inequality for fractional integrals holds:

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left(3f\left(a^{\frac{5}{6}}{b}^{\frac{1}{6}}\right)+2f\left(a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)+3f\left(a^{\frac{1}{6}}{b}^{\frac{5}{6}}\right)\right)-\frac{6^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(lnb-lna\right)}^{\alpha }}\mathrm{{\rm Y}}\right|\hfill \\ \hfill \le & \frac{lnb-lna}{9}\left(\frac{a}{4}{\left(\frac{{}_1{F}_1\left(\alpha +1,\alpha +2,\frac{1}{6}ln\frac{b}{a}\right)}{\alpha +1}\right)}^{1-\frac{1}{q}}\right.\hfill \\ \hfill & \times {\left(\frac{6^{\alpha +2}{.}_1{F}_1\left(\left[\alpha +1,\alpha +3;\frac{1}{6}\right],ln\frac{b}{a}\right){\left|f^{\prime }\left(a\right)\right|}^q+{\left(\alpha +1\right)}_1{F}_1\left(\alpha +2,\alpha +3,\frac{1}{6}ln\frac{b}{a}\right){\left|f^{\prime }\left(b\right)\right|}^q}{6\left(\alpha +1\right)\left(\alpha +2\right)}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +a^{\frac{1}{2}}{b}^{\frac{1}{2}}{\left(\mathsf{\Phi }\left(\alpha ,\frac{a}{b}\right)\right)}^{1-\frac{1}{q}}{\left(\frac{\left({\mu }_1\left(\alpha ,\frac{a}{b}\right)+{\mu }_2\left(\alpha ,\frac{a}{b}\right)\right){\left|f^{\prime }\left(a\right)\right|}^q+\left({\mu }_3\left(\alpha ,\frac{a}{b}\right)+{\mu }_4\left(\alpha ,\frac{a}{b}\right)\right){\left|f^{\prime }\left(b\right)\right|}^q}{6}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +a^{\frac{1}{2}}{b}^{\frac{1}{2}}{\left(\mathsf{\Phi }\left(\alpha ,\frac{b}{a}\right)\right)}^{1-\frac{1}{q}}{\left(\frac{\left({\mu }_3\left(\alpha ,\frac{b}{a}\right)+{\mu }_4\left(\alpha ,\frac{b}{a}\right)\right){\left|f^{\prime }\left(a\right)\right|}^q+\left({\mu }_1\left(\alpha ,\frac{b}{a}\right)+{\mu }_2\left(\alpha ,\frac{b}{a}\right)\right){\left|f^{\prime }\left(b\right)\right|}^q}{6}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{b}{4}{\left(\frac{{}_1{F}_1\left(\alpha +1,\alpha +2,\frac{1}{6}ln\frac{a}{b}\right)}{\alpha +1}\right)}^{1-\frac{1}{q}}\hfill \\ \hfill & \times \left.{\left(\frac{{\left(\alpha +1\right)}_1{F}_1\left(\alpha +2,\alpha +3,\frac{1}{6}ln\frac{a}{b}\right){\left|f^{\prime }\left(a\right)\right|}^q+6^{\alpha +2}{.}_1{F}_1\left(\left[\alpha +1,\alpha +3;\frac{1}{6}\right],ln\frac{a}{b}\right){\left|f^{\prime }\left(b\right)\right|}^q}{6\left(\alpha +1\right)\left(\alpha +2\right)}\right)}^{\frac{1}{q}}\right),\hfill \end{array}$$

where $\alpha >0,\mathrm{{\rm Y}},{\mu }_1,{\mu }_2,{\mu }_3,{\mu }_4,\mathsf{\Phi }$ are defined by (1), (12), (13), (16), (17) and (20), respectively, where ${}_1{F}_1\left(.;.;.\right)$ and ${}_1{F}_1\left(\left[.,.;.\right],.\right)$ are the confluent and the incomplete confluent hypergeometric functions, respectively.

Proof. 

From Lemma 2, the modulus, the power mean inequality, and the $GA$-convexity of ${\left|f^{\prime }\right|}^q$, we have

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left(3f\left(a^{\frac{5}{6}}{b}^{\frac{1}{6}}\right)+2f\left(a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)+3f\left(a^{\frac{1}{6}}{b}^{\frac{5}{6}}\right)\right)-\frac{6^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(lnb-lna\right)}^{\alpha }}\mathrm{{\rm Y}}\right|\hfill \\ \hfill \le & \frac{lnb-lna}{9}\left({\left(\underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\varkappa }^{\alpha }{a}^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}d\varkappa \right)}^{1-\frac{1}{q}}{\left(\underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\varkappa }^{\alpha }{a}^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}{\left|f^{\prime }\left(a^{\frac{6-\varkappa }{6}}{b}^{\frac{\varkappa }{6}}\right)\right|}^{q}d\varkappa \right)}^{\frac{1}{q}}\right.\hfill \\ \hfill & +{\left(\underset{0}{\stackrel{1}{\int }}\left|{\left(1-\varkappa \right)}^{\alpha }-\frac{3}{8}\right|a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}d\varkappa \right)}^{1-\frac{1}{q}}\hfill \\ \hfill & \times {\left(\underset{0}{\stackrel{1}{\int }}\left|{\left(1-\varkappa \right)}^{\alpha }-\frac{3}{8}\right|a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}{\left|f^{\prime }\left(a^{\frac{5-2\varkappa }{6}}{b}^{\frac{1+2\varkappa }{6}}\right)\right|}^{q}d\varkappa \right)}^{\frac{1}{q}}\hfill \\ \hfill & +{\left(\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}d\varkappa \right)}^{1-\frac{1}{q}}\hfill \\ \hfill & \times {\left(\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}{\left|f^{\prime }\left(a^{\frac{3-2\varkappa }{6}}{b}^{\frac{3+2\varkappa }{6}}\right)\right|}^{q}d\varkappa \right)}^{\frac{1}{q}}\hfill \\ \hfill & +{\left(\underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\left(1-\varkappa \right)}^{\alpha }{a}^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}d\varkappa \right)}^{1-\frac{1}{q}}\hfill \\ \hfill & \times \left.{\left(\underset{0}{\stackrel{1}{\int }}\frac{1}{4}{\left(1-\varkappa \right)}^{\alpha }{a}^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}{\left|f^{\prime }\left(a^{\frac{1-\varkappa }{6}}{b}^{\frac{5+\varkappa }{6}}\right)\right|}^{q}d\varkappa \right)}^{\frac{1}{q}}\right)\hfill \\ \hfill & \le \frac{lnb-lna}{9}\left(\frac{a}{4}{\left(\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha }{\left(\frac{b}{a}\right)}^{\frac{\varkappa }{6}}d\varkappa \right)}^{1-\frac{1}{q}}\right.\hfill \\ \hfill & \times {\left({\left|f^{\prime }\left(a\right)\right|}^q\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha }\left(1-\frac{\varkappa }{6}\right){\left(\frac{b}{a}\right)}^{\frac{\varkappa }{6}}d\varkappa +\frac{1}{6}{\left|f^{\prime }\left(b\right)\right|}^q\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha +1}{\left(\frac{b}{a}\right)}^{\frac{\varkappa }{6}}d\varkappa \right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{a^{\frac{1}{2}}{b}^{\frac{1}{2}}}{6}{\left(6\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|{\left(\frac{a}{b}\right)}^{\frac{\varkappa }{3}}d\varkappa \right)}^{1-\frac{1}{q}}\hfill \\ \hfill & \times {\left({\left|f^{\prime }\left(a\right)\right|}^q\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|\left(3+2\varkappa \right){\left(\frac{a}{b}\right)}^{\frac{\varkappa }{3}}d\varkappa +{\left|f^{\prime }\left(b\right)\right|}^q\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|\left(3-2\varkappa \right){\left(\frac{a}{b}\right)}^{\frac{\varkappa }{3}}d\varkappa \right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{a^{\frac{1}{2}}{b}^{\frac{1}{2}}}{6}{\left(6\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|{\left(\frac{b}{a}\right)}^{\frac{\varkappa }{3}}d\varkappa \right)}^{1-\frac{1}{q}}\hfill \\ \hfill & \times {\left({\left|f^{\prime }\left(a\right)\right|}^q\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|\left(3-2\varkappa \right){\left(\frac{b}{a}\right)}^{\frac{\varkappa }{3}}d\varkappa +{\left|f^{\prime }\left(b\right)\right|}^q\underset{0}{\stackrel{1}{\int }}\left|{\varkappa }^{\alpha }-\frac{3}{8}\right|\left(3+2\varkappa \right){\left(\frac{b}{a}\right)}^{\frac{\varkappa }{3}}d\varkappa \right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{b}{4}{\left(\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha }{\left(\frac{a}{b}\right)}^{\frac{\varkappa }{6}}d\varkappa \right)}^{1-\frac{1}{q}}\hfill \\ \hfill & \times \left.{\left(\frac{1}{6}{\left|f^{\prime }\left(a\right)\right|}^q\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha +1}{\left(\frac{a}{b}\right)}^{\frac{\varkappa }{6}}d\varkappa +{\left|f^{\prime }\left(b\right)\right|}^q\underset{0}{\stackrel{1}{\int }}{\varkappa }^{\alpha }\left(1-\frac{1}{6}\varkappa \right){\left(\frac{a}{b}\right)}^{\frac{\varkappa }{6}}d\varkappa \right)}^{\frac{1}{q}}\right)\hfill \\ \hfill =& \frac{lnb-lna}{9}\left(\frac{a}{4}{\left(\frac{{}_1{F}_1\left(\alpha +1,\alpha +2,\frac{1}{6}ln\frac{b}{a}\right)}{\alpha +1}\right)}^{1-\frac{1}{q}}\right.\hfill \\ \hfill & \times {\left(\frac{6^{\alpha +2}{.}_1{F}_1\left(\left[\alpha +1,\alpha +3;\frac{1}{6}\right],ln\frac{b}{a}\right){\left|f^{\prime }\left(a\right)\right|}^q+{\left(\alpha +1\right)}_1{F}_1\left(\alpha +2,\alpha +3,\frac{1}{6}ln\frac{b}{a}\right){\left|f^{\prime }\left(b\right)\right|}^q}{6\left(\alpha +1\right)\left(\alpha +2\right)}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +a^{\frac{1}{2}}{b}^{\frac{1}{2}}{\left(\mathsf{\Phi }\left(\alpha ,\frac{a}{b}\right)\right)}^{1-\frac{1}{q}}{\left(\frac{\left({\mu }_1\left(\alpha ,\frac{a}{b}\right)+{\mu }_2\left(\alpha ,\frac{a}{b}\right)\right){\left|f^{\prime }\left(a\right)\right|}^q+\left({\mu }_3\left(\alpha ,\frac{a}{b}\right)+{\mu }_4\left(\alpha ,\frac{a}{b}\right)\right){\left|f^{\prime }\left(b\right)\right|}^q}{6}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +a^{\frac{1}{2}}{b}^{\frac{1}{2}}{\left(\mathsf{\Phi }\left(\alpha ,\frac{b}{a}\right)\right)}^{1-\frac{1}{q}}{\left(\frac{\left({\mu }_3\left(\alpha ,\frac{b}{a}\right)+{\mu }_4\left(\alpha ,\frac{b}{a}\right)\right){\left|f^{\prime }\left(a\right)\right|}^q+\left({\mu }_1\left(\alpha ,\frac{b}{a}\right)+{\mu }_2\left(\alpha ,\frac{b}{a}\right)\right){\left|f^{\prime }\left(b\right)\right|}^q}{6}\right)}^{\frac{1}{q}}\hfill \\ \hfill & +\frac{b}{4}{\left(\frac{{}_1{F}_1\left(\alpha +1,\alpha +2,\frac{1}{6}ln\frac{a}{b}\right)}{\alpha +1}\right)}^{1-\frac{1}{q}}\hfill \\ \hfill & \times \left.{\left(\frac{{\left(\alpha +1\right)}_1{F}_1\left(\alpha +2,\alpha +3,\frac{1}{6}ln\frac{a}{b}\right){\left|f^{\prime }\left(a\right)\right|}^q+6^{\alpha +2}{.}_1{F}_1\left(\left[\alpha +1,\alpha +3;\frac{1}{6}\right],ln\frac{a}{b}\right){\left|f^{\prime }\left(b\right)\right|}^q}{6\left(\alpha +1\right)\left(\alpha +2\right)}\right)}^{\frac{1}{q}}\right),\hfill \end{array}$$

The proof is completed. □

5. Conclusions

This study deals with the fractional Newton–Cotes-type inequalities involving three points by applying one of a novel generalizations of convexity, called geometrically arithmetically convexity. To study this, we have firstly proved a new integral identity. Based on this identity, we have establish some new Maclaurin-type inequalities for functions whose modulus of the first derivatives are geometrically arithmetically convex via Hadamard fractional integral operators, which are very useful and important fractional integral operators in fractional calculus. We hope that the obtained results could be motivation researchers working in the of fractional calculus, and serve as inspiration for academics to prove novel results using more generalized forms of convexity together with other fractional integral operators.