## 1 Introduction

Let a real function f be defined on some nonempty interval I of a real line ℝ. The function f is said to be convex on I if inequality

$f\left(tx+\left(1-t\right)y\right)\le tf\left(x\right)+\left(1-t\right)f\left(y\right)$
(1)

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

We recall that the notion of quasi-convex function generalizes the notion of convex function. More exactly, a function $f:\left[a,b\right]\subset \mathbb{R}\to \mathbb{R}$ is said to be quasi-convex on $\left[a,b\right]$ if

$f\left(tx+\left(1-t\right)y\right)\le max\left\{f\left(x\right),f\left(y\right)\right\}$

for all $x,y\in \left[a,b\right]$ and $t\in \left[0,1\right]$. Clearly, any convex function is a quasi-convex function. Furthermore, there exist quasi-convex functions which are not convex (see [1]).

The following inequalities are well known in the literature as the Hermite-Hadamard inequality, the Ostrowski inequality and the Simpson inequality, respectively.

Theorem 1.1 Let $f:I\subseteq \mathbb{R}\to \mathbb{R}$ be a convex function defined on the interval I of real numbers, and let $a,b\in I$ with $a. The following double inequality holds:

$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\le \frac{f\left(a\right)+f\left(b\right)}{2}.$
(2)

Theorem 1.2 Let $f:I\subseteq \mathbb{R}\to \mathbb{R}$ be a mapping differentiable in ${I}^{\circ }$, the interior of I, and let $a,b\in {I}^{\circ }$ with $a. If $|{f}^{\prime }\left(x\right)|\le M$, $x\in \left[a,b\right]$, then the following inequality holds:

$|f\left(x\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(t\right)\phantom{\rule{0.2em}{0ex}}dt|\le \frac{M}{b-a}\left[\frac{{\left(x-a\right)}^{2}+{\left(b-x\right)}^{2}}{2}\right]$

for all $x\in \left[a,b\right]$.

Theorem 1.3 Let $f:\left[a,b\right]\to \mathbb{R}$ be a four times continuously differentiable mapping on $\left(a,b\right)$ and ${\parallel {f}^{\left(4\right)}\parallel }_{\mathrm{\infty }}={sup}_{x\in \left(a,b\right)}|{f}^{\left(4\right)}\left(x\right)|<\mathrm{\infty }$. Then the following inequality holds:

$|\frac{1}{3}\left[\frac{f\left(a\right)+f\left(b\right)}{2}+2f\left(\frac{a+b}{2}\right)\right]-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\le \frac{1}{2\text{,}880}{\parallel {f}^{\left(4\right)}\parallel }_{\mathrm{\infty }}{\left(b-a\right)}^{4}.$

The following definitions are well known in the literature.

Definition 1.1 ([2, 3])

A function $f:I\subseteq \left(0,\mathrm{\infty }\right)\to \mathbb{R}$ is said to be GA-convex (geometric-arithmetically convex) if

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

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

Definition 1.2 ([2, 3])

A function $f:I\subseteq \left(0,\mathrm{\infty }\right)\to \left(0,\mathrm{\infty }\right)$ is said to be GG-convex (called in [4] a geometrically convex function) if

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

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

We will now give definitions of the right-hand side and left-hand side Hadamard fractional integrals which are used throughout this paper.

Definition 1.3 Let $f\in L\left[a,b\right]$. The right-hand side and left-hand side Hadamard fractional integrals ${J}_{{a}^{+}}^{\alpha }f$ and ${J}_{{b}^{-}}^{\alpha }f$ of order $\alpha >0$ with $b>a\ge 0$ are defined by

${J}_{a+}^{\alpha }f\left(x\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{a}^{x}{\left(ln\frac{x}{t}\right)}^{\alpha -1}f\left(t\right)\frac{dt}{t},\phantom{\rule{1em}{0ex}}a

and

${J}_{b-}^{\alpha }f\left(x\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{x}^{b}{\left(ln\frac{t}{x}\right)}^{\alpha -1}f\left(t\right)\frac{dt}{t},\phantom{\rule{1em}{0ex}}a

respectively, where $\mathrm{\Gamma }\left(\alpha \right)$ is the Gamma function defined by $\mathrm{\Gamma }\left(\alpha \right)={\int }_{0}^{\mathrm{\infty }}{e}^{-t}{t}^{\alpha -1}\phantom{\rule{0.2em}{0ex}}dt$ (see [5]).

In recent years, many authors have studied error estimations for Hermite-Hadamard, Ostrowski and Simpson inequalities; for refinements, counterparts, generalization see [4, 620].

In this paper, the concept of the quasi-geometrically convex function is introduced, Hermite-Hadamard’s inequalities for GA-convex functions in fractional integral forms are established, and a new identity for Hadamard fractional integrals is defined. By using this identity, author obtains a generalization of Hadamard, Ostrowski and Simpson type inequalities for quasi-geometrically convex functions via Hadamard fractional integrals.

## 2 Main results

Let $f:I\subseteq \left(0,\mathrm{\infty }\right)\to \mathbb{R}$ be a differentiable function on ${I}^{\circ }$, the interior of I, throughout this section we will take

$\begin{array}{r}{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)\\ \phantom{\rule{1em}{0ex}}=\left(1-\lambda \right)\left[{ln}^{\alpha }\frac{x}{a}+{ln}^{\alpha }\frac{b}{x}\right]f\left(x\right)+\lambda \left[f\left(a\right){ln}^{\alpha }\frac{x}{a}+f\left(b\right){ln}^{\alpha }\frac{b}{x}\right]\\ \phantom{\rule{2em}{0ex}}-\mathrm{\Gamma }\left(\alpha +1\right)\left[{J}_{x-}^{\alpha }f\left(a\right)+{J}_{x+}^{\alpha }f\left(b\right)\right],\end{array}$

where $a,b\in I$ with $a, $x\in \left[a,b\right]$, $\lambda \in \left[0,1\right]$, $\alpha >0$ and Γ is the Euler Gamma function.

Definition 2.1 A function $f:I\subseteq \left(0,\mathrm{\infty }\right)\to \mathbb{R}$ is said to be quasi-geometrically convex on I if

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

for any $x,y\in I$ and $t\in \left[0,1\right]$.

Remark 2.1 Clearly, any GA-convex and geometrically convex functions are quasi-geometrically convex functions. Furthermore, there exist quasi-geometrically convex functions which are neither GA-convex nor geometrically convex. In that context, we point out an elementary example. The function $f:\left(0,4\right]\to \mathbb{R}$,

$f\left(x\right)=\left\{\begin{array}{ll}1,& x\in \left(0,1\right],\\ {\left(x-2\right)}^{2},& x\in \left[1,4\right]\end{array}$

is neither GA-convex nor geometrically convex on $\left(0,4\right]$, but it is a quasi-geometrically convex function on $\left(0,4\right]$.

Proposition 2.1 If $f:I\subseteq \left(0,\mathrm{\infty }\right)\to \mathbb{R}$ is convex and nondecreasing, then it is quasi-geometrically convex on I.

Proof This follows from

$\begin{array}{rcl}f\left({x}^{t}{y}^{1-t}\right)& \le & f\left(tx+\left(1-t\right)y\right)\\ \le & tf\left(x\right)+\left(1-t\right)f\left(y\right)\le sup\left\{f\left(x\right),f\left(y\right)\right\},\end{array}$

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

Proposition 2.2 If $f:I\subseteq \left(0,\mathrm{\infty }\right)\to \mathbb{R}$ is quasi-convex and nondecreasing, then it is quasi-geometrically convex on I. If $f:I\subseteq \left(0,\mathrm{\infty }\right)\to \mathbb{R}$ is quasi-geometrically convex and nonincreasing, then it is quasi-convex on I.

Proof These conclusions follows from

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

and

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

for all $x,y\in I$ and $t\in \left[0,1\right]$, respectively. □

Hermite-Hadamard’s inequalities can be represented for GA-convex functions in fractional integral forms as follows.

Theorem 2.1 Let $f:I\subseteq \left(0,\mathrm{\infty }\right)\to \mathbb{R}$ be a function such that $f\in L\left[a,b\right]$, where $a,b\in I$ with $a. If f is a GA-convex function on $\left[a,b\right]$, then the following inequalities for fractional integrals hold:

$f\left(\sqrt{ab}\right)\le \frac{\mathrm{\Gamma }\left(\alpha +1\right)}{2{\left(ln\frac{b}{a}\right)}^{\alpha }}\left\{{J}_{a+}^{\alpha }f\left(b\right)+{J}_{b-}^{\alpha }f\left(a\right)\right\}\le \frac{f\left(a\right)+f\left(b\right)}{2}$
(3)

with $\alpha >0$.

Proof Since f is a GA-convex function on $\left[a,b\right]$, we have for all $x,y\in \left[a,b\right]$ (with $t=1/2$ in inequality (1)),

$f\left(\sqrt{xy}\right)\le \frac{f\left(x\right)+f\left(y\right)}{2}.$

Choosing $x={a}^{t}{b}^{1-t}$, $y={b}^{t}{a}^{1-t}$, we get

$f\left(\sqrt{ab}\right)\le \frac{f\left({a}^{t}{b}^{1-t}\right)+f\left({b}^{t}{a}^{1-t}\right)}{2}.$
(4)

Multiplying both sides of (4) by ${t}^{\alpha -1}$, then integrating the resulting inequality with respect to t over $\left[0,1\right]$, we obtain

$\begin{array}{rcl}f\left(\sqrt{ab}\right)& \le & \frac{\alpha }{2}\left\{{\int }_{0}^{1}f\left({a}^{t}{b}^{1-t}\right)\phantom{\rule{0.2em}{0ex}}dt+{\int }_{0}^{1}f\left({b}^{t}{a}^{1-t}\right)\phantom{\rule{0.2em}{0ex}}dt\right\}\\ =& \frac{\alpha }{2}\left\{{\int }_{a}^{b}{\left(\frac{lnb-lnu}{lnb-lna}\right)}^{\alpha -1}f\left(u\right)\frac{du}{uln\frac{b}{a}}+{\int }_{a}^{b}{\left(\frac{lnu-lna}{lnb-lna}\right)}^{\alpha -1}f\left(u\right)\frac{du}{uln\frac{b}{a}}\right\}\\ =& \frac{\alpha \mathrm{\Gamma }\left(\alpha \right)}{2{\left(ln\frac{b}{a}\right)}^{\alpha }}\left\{{J}_{a+}^{\alpha }f\left(b\right)+{J}_{b-}^{\alpha }f\left(a\right)\right\}\\ =& \frac{\mathrm{\Gamma }\left(\alpha +1\right)}{2{\left(ln\frac{b}{a}\right)}^{\alpha }}\left\{{J}_{a+}^{\alpha }f\left(b\right)+{J}_{b-}^{\alpha }f\left(a\right)\right\},\end{array}$

and the first inequality is proved.

For the proof of the second inequality in (3), we first note that if f is a convex function, then for $t\in \left[0,1\right]$, it yields

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

and

$f\left({b}^{t}{a}^{1-t}\right)\le tf\left(b\right)+\left(1-t\right)f\left(a\right).$

By adding these inequalities, we have

$f\left({a}^{t}{b}^{1-t}\right)+f\left({b}^{t}{a}^{1-t}\right)\le f\left(a\right)+f\left(b\right).$
(5)

Then multiplying both sides of (5) by ${t}^{\alpha -1}$, and integrating the resulting inequality with respect to t over $\left[0,1\right]$, we obtain

${\int }_{0}^{1}f\left({a}^{t}{b}^{1-t}\right){t}^{\alpha -1}\phantom{\rule{0.2em}{0ex}}dt+{\int }_{0}^{1}f\left({b}^{t}{a}^{1-t}\right){t}^{\alpha -1}\phantom{\rule{0.2em}{0ex}}dt\le \left[f\left(a\right)+f\left(b\right)\right]{\int }_{0}^{1}{t}^{\alpha -1}\phantom{\rule{0.2em}{0ex}}dt,$

i.e.,

$\frac{\mathrm{\Gamma }\left(\alpha +1\right)}{{\left(ln\frac{b}{a}\right)}^{\alpha }}\left\{{J}_{a+}^{\alpha }f\left(b\right)+{J}_{b-}^{\alpha }f\left(a\right)\right\}\le f\left(a\right)+f\left(b\right).$

The proof is completed. □

In order to prove our main results, we need the following identity.

Lemma 2.1 Let $f:I\subseteq \left(0,\mathrm{\infty }\right)\to \mathbb{R}$ be a differentiable function on ${I}^{\circ }$ such that ${f}^{\prime }\in L\left[a,b\right]$, where $a,b\in I$ with $a. Then for all $x\in \left[a,b\right]$, $\lambda \in \left[0,1\right]$ and $\alpha >0$, we have:

$\begin{array}{rcl}{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)& =& a{\left(ln\frac{x}{a}\right)}^{\alpha +1}{\int }_{0}^{1}\left({t}^{\alpha }-\lambda \right){\left(\frac{x}{a}\right)}^{t}{f}^{\prime }\left({x}^{t}{a}^{1-t}\right)\phantom{\rule{0.2em}{0ex}}dt\\ -b{\left(ln\frac{b}{x}\right)}^{\alpha +1}{\int }_{0}^{1}\left({t}^{\alpha }-\lambda \right){\left(\frac{x}{b}\right)}^{t}{f}^{\prime }\left({x}^{t}{b}^{1-t}\right)\phantom{\rule{0.2em}{0ex}}dt.\end{array}$
(6)

Proof By integration by parts and twice changing the variable, for $x\ne a$, we can state that

$\begin{array}{r}aln\frac{x}{a}{\int }_{0}^{1}\left({t}^{\alpha }-\lambda \right){\left(\frac{x}{a}\right)}^{t}{f}^{\prime }\left({x}^{t}{a}^{1-t}\right)\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{1em}{0ex}}={\int }_{0}^{1}\left({t}^{\alpha }-\lambda \right)\phantom{\rule{0.2em}{0ex}}df\left({x}^{t}{a}^{1-t}\right)\\ \phantom{\rule{1em}{0ex}}=\left({t}^{\alpha }-\lambda \right)f\left({x}^{t}{a}^{1-t}\right){|}_{0}^{1}-\frac{\alpha }{{\left(ln\frac{x}{a}\right)}^{\alpha }}{\int }_{a}^{x}{\left(ln\frac{u}{a}\right)}^{\alpha -1}\frac{f\left(u\right)}{u}\phantom{\rule{0.2em}{0ex}}du\\ \phantom{\rule{1em}{0ex}}=\left(1-\lambda \right)f\left(x\right)+\lambda f\left(a\right)-\frac{\mathrm{\Gamma }\left(\alpha +1\right)}{{\left(ln\frac{x}{a}\right)}^{\alpha }}{J}_{x-}^{\alpha }f\left(a\right),\end{array}$
(7)

and for $x\ne b$, similarly, we get

$\begin{array}{r}-bln\frac{b}{x}{\int }_{0}^{1}\left({t}^{\alpha }-\lambda \right){\left(\frac{x}{b}\right)}^{t}{f}^{\prime }\left({x}^{t}{b}^{1-t}\right)\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{1em}{0ex}}={\int }_{0}^{1}\left({t}^{\alpha }-\lambda \right)\phantom{\rule{0.2em}{0ex}}df\left({x}^{t}{b}^{1-t}\right)\\ \phantom{\rule{1em}{0ex}}=\left({t}^{\alpha }-\lambda \right)f\left({x}^{t}{b}^{1-t}\right){|}_{0}^{1}-\frac{\alpha }{{\left(ln\frac{b}{x}\right)}^{\alpha }}{\int }_{x}^{b}{\left(ln\frac{b}{u}\right)}^{\alpha -1}\frac{f\left(u\right)}{u}\phantom{\rule{0.2em}{0ex}}du\\ \phantom{\rule{1em}{0ex}}=\left(1-\lambda \right)f\left(x\right)+\lambda f\left(b\right)-\frac{\mathrm{\Gamma }\left(\alpha +1\right)}{{\left(ln\frac{b}{x}\right)}^{\alpha }}{J}_{x+}^{\alpha }f\left(b\right).\end{array}$
(8)

Multiplying both sides of (7) and (8) by ${\left(ln\frac{x}{a}\right)}^{\alpha }$ and ${\left(ln\frac{b}{x}\right)}^{\alpha }$, respectively, and adding the resulting identities, we obtain the desired result. For $x=a$ and $x=b$, the identities

${I}_{f}\left(a,\lambda ,\alpha ;a,b\right)=b{\left(ln\frac{b}{a}\right)}^{\alpha +1}{\int }_{0}^{1}\left({t}^{\alpha }-\lambda \right){\left(\frac{a}{b}\right)}^{t}{f}^{\prime }\left({a}^{t}{b}^{1-t}\right)\phantom{\rule{0.2em}{0ex}}dt,$

and

${I}_{f}\left(b,\lambda ,\alpha ;a,b\right)=a{\left(ln\frac{b}{a}\right)}^{\alpha +1}{\int }_{0}^{1}\left({t}^{\alpha }-\lambda \right){\left(\frac{b}{a}\right)}^{t}{f}^{\prime }\left({b}^{t}{a}^{1-t}\right),$

can be proved respectively easily by performing an integration by parts in the integrals from the right-hand side and changing the variable. □

Theorem 2.2 Let $f:I\subset \left(0,\mathrm{\infty }\right)\to \mathbb{R}$ be a differentiable function on ${I}^{\circ }$ such that ${f}^{\prime }\in L\left[a,b\right]$, where $a,b\in {I}^{\circ }$ with $a. If ${|{f}^{\prime }|}^{q}$ is quasi-geometrically convex on $\left[a,b\right]$ for some fixed $q\ge 1$, $x\in \left[a,b\right]$, $\lambda \in \left[0,1\right]$ and $\alpha >0$, then the following inequality for fractional integrals holds:

$\begin{array}{r}|{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)|\\ \phantom{\rule{1em}{0ex}}\le {A}_{1}^{1-\frac{1}{q}}\left(\alpha ,\lambda \right)\left\{a{\left(ln\frac{x}{a}\right)}^{\alpha +1}{\left(sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right)}^{\frac{1}{q}}{B}_{1}^{\frac{1}{q}}\left(x,\alpha ,\lambda ,q\right)\\ \phantom{\rule{2em}{0ex}}+b{\left(ln\frac{b}{x}\right)}^{\alpha +1}{\left(sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right)}^{\frac{1}{q}}{B}_{2}^{\frac{1}{q}}\left(x,\alpha ,\lambda ,q\right)\right\},\end{array}$
(9)

where

$\begin{array}{c}{A}_{1}\left(\alpha ,\lambda \right)=\frac{2\alpha {\lambda }^{1+\frac{1}{\alpha }}+1}{\alpha +1}-\lambda ,\hfill \\ {B}_{1}\left(x,\alpha ,\lambda ,q\right)={\int }_{0}^{1}|{t}^{\alpha }-\lambda |{\left(\frac{x}{a}\right)}^{qt}\phantom{\rule{0.2em}{0ex}}dt,\hfill \\ {B}_{2}\left(x,\alpha ,\lambda ,q\right)={\int }_{0}^{1}|{t}^{\alpha }-\lambda |{\left(\frac{x}{b}\right)}^{qt}\phantom{\rule{0.2em}{0ex}}dt.\hfill \end{array}$

Proof Since $|{f}^{\prime }{|}^{q}$ is quasi-geometrically convex on $\left[a,b\right]$, for all $t\in \left[0,1\right]$,

$|{f}^{\prime }\left({x}^{t}{a}^{1-t}\right){|}^{q}\le sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}$

and

$|{f}^{\prime }\left({x}^{t}{b}^{1-t}\right){|}^{q}\le sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}.$

Hence, using Lemma 2.1 and power mean inequality, we get

$\begin{array}{r}|{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)|\\ \phantom{\rule{1em}{0ex}}\le a{\left(ln\frac{x}{a}\right)}^{\alpha +1}{\left({\int }_{0}^{1}|{t}^{\alpha }-\lambda |\phantom{\rule{0.2em}{0ex}}dt\right)}^{1-\frac{1}{q}}{\left({\int }_{0}^{1}|{t}^{\alpha }-\lambda |{\left(\frac{x}{a}\right)}^{qt}sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+b{\left(ln\frac{b}{x}\right)}^{\alpha +1}{\left({\int }_{0}^{1}|{t}^{\alpha }-\lambda |\phantom{\rule{0.2em}{0ex}}dt\right)}^{1-\frac{1}{q}}{\left({\int }_{0}^{1}|{t}^{\alpha }-\lambda |{\left(\frac{x}{b}\right)}^{qt}sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}},\\ |{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)|\le {\left({\int }_{0}^{1}|{t}^{\alpha }-\lambda |\phantom{\rule{0.2em}{0ex}}dt\right)}^{1-\frac{1}{q}}\\ \phantom{|{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)|\le }×\left\{a{\left(ln\frac{x}{a}\right)}^{\alpha +1}{\left(sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right)}^{\frac{1}{q}}{\left({\int }_{0}^{1}|{t}^{\alpha }-\lambda |{\left(\frac{x}{a}\right)}^{qt}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}}\\ \phantom{|{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)|\le }+b{\left(ln\frac{b}{x}\right)}^{\alpha +1}{\left(sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right)}^{\frac{1}{q}}{\left({\int }_{0}^{1}|{t}^{\alpha }-\lambda |{\left(\frac{x}{b}\right)}^{qt}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}}\right\}\\ \phantom{|{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)|}\le {A}_{1}^{1-\frac{1}{q}}\left(\alpha ,\lambda \right)\left\{a{\left(ln\frac{x}{a}\right)}^{\alpha +1}{\left(sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right)}^{\frac{1}{q}}{B}_{1}^{\frac{1}{q}}\left(x,\alpha ,\lambda ,q\right)\\ \phantom{|{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)|\le }+b{\left(ln\frac{b}{x}\right)}^{\alpha +1}{\left(sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right)}^{\frac{1}{q}}{B}_{2}^{\frac{1}{q}}\left(x,\alpha ,\lambda ,q\right)\right\},\end{array}$

which completes the proof. □

Corollary 2.1 Under the assumptions of Theorem  2.2 with $q=1$, inequality (9) reduces to the following inequality:

$\begin{array}{rcl}|{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)|& \le & \left\{a{\left(ln\frac{x}{a}\right)}^{\alpha +1}{B}_{1}\left(x,\alpha ,\lambda ,1\right)sup\left\{|{f}^{\prime }\left(x\right)|,|{f}^{\prime }\left(a\right)|\right\}\\ +b{\left(ln\frac{b}{x}\right)}^{\alpha +1}{B}_{2}\left(x,\alpha ,\lambda ,1\right)sup\left\{|{f}^{\prime }\left(x\right)|,|{f}^{\prime }\left(b\right)|\right\}\right\}.\end{array}$

Corollary 2.2 Under the assumptions of Theorem  2.2 with $\alpha =1$, inequality (9) reduces to the following inequality:

$\begin{array}{r}{\left(ln\frac{b}{a}\right)}^{-1}|{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)|\\ \phantom{\rule{1em}{0ex}}\le |\left(1-\lambda \right)f\left(x\right)+\lambda \left[\frac{f\left(a\right)ln\frac{x}{a}+f\left(b\right)ln\frac{b}{x}}{ln\frac{b}{a}}\right]-\frac{1}{ln\frac{b}{a}}{\int }_{a}^{b}\frac{f\left(u\right)}{u}\phantom{\rule{0.2em}{0ex}}du|\\ \phantom{\rule{1em}{0ex}}\le {\left(ln\frac{b}{a}\right)}^{-1}{\left(\frac{2{\lambda }^{2}-2\lambda +1}{2}\right)}^{1-\frac{1}{q}}\left\{a{\left(ln\frac{x}{a}\right)}^{2}{B}_{1}^{\frac{1}{q}}\left(x,1,\lambda ,q\right){\left(sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+b{\left(ln\frac{b}{x}\right)}^{2}{B}_{2}^{\frac{1}{q}}\left(x,1,\lambda ,q\right){\left(sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right)}^{\frac{1}{q}}\right\},\end{array}$

where

$\begin{array}{c}{B}_{1}\left(x,1,\lambda ,q\right)={h}_{\lambda }\left({\left(\frac{x}{a}\right)}^{q}\right),\phantom{\rule{2em}{0ex}}{B}_{2}\left(x,1,\lambda ,q\right)={h}_{\lambda }\left({\left(\frac{x}{b}\right)}^{q}\right),\hfill \\ h\left(u,\lambda \right)=\frac{2{u}^{\lambda }-u-1}{{\left(lnu\right)}^{2}}+\frac{\left(1-\lambda \right)u-\lambda }{lnu},\phantom{\rule{1em}{0ex}}u\in \left(0,\mathrm{\infty }\right)\mathrm{\setminus }\left\{1\right\},\hfill \end{array}$
(10)

specially for $x=\sqrt{ab}$, we get

$\begin{array}{r}|\left(1-\lambda \right)f\left(\sqrt{ab}\right)+\lambda \left(\frac{f\left(a\right)+f\left(b\right)}{2}\right)-\frac{1}{ln\frac{b}{a}}{\int }_{a}^{b}\frac{f\left(u\right)}{u}\phantom{\rule{0.2em}{0ex}}du|\\ \phantom{\rule{1em}{0ex}}\le \frac{ln\frac{b}{a}}{4}{\left(\frac{2{\lambda }^{2}-2\lambda +1}{2}\right)}^{1-\frac{1}{q}}\left\{a{h}^{\frac{1}{q}}\left({\left(\frac{b}{a}\right)}^{\frac{q}{2}},\lambda \right){\left(sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+b{h}^{\frac{1}{q}}\left({\left(\frac{a}{b}\right)}^{\frac{q}{2}},\lambda \right){\left(sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right)}^{\frac{1}{q}}\right\}.\end{array}$
(11)

Corollary 2.3 In Theorem  2.2,

1. 1.

If we take $x=\sqrt{ab}$, $\lambda =\frac{1}{3}$, then we get the following Simpson-type inequality for fractional integrals:

$\begin{array}{r}|\frac{1}{6}\left[f\left(a\right)+4f\left(\sqrt{ab}\right)+f\left(b\right)\right]-\frac{{2}^{\alpha -1}\mathrm{\Gamma }\left(\alpha +1\right)}{{\left(ln\frac{b}{a}\right)}^{\alpha }}\left[{J}_{\sqrt{ab}-}^{\alpha }f\left(a\right)+{J}_{\sqrt{ab}+}^{\alpha }f\left(b\right)\right]|\\ \phantom{\rule{1em}{0ex}}\le \frac{ln\frac{b}{a}}{4}{A}_{1}^{1-\frac{1}{q}}\left(\alpha ,\frac{1}{3}\right)\left\{a{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{B}_{1}^{\frac{1}{q}}\left(\sqrt{ab},\alpha ,\frac{1}{3},q\right)\\ \phantom{\rule{2em}{0ex}}+b{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{B}_{2}^{\frac{1}{q}}\left(\sqrt{ab},\alpha ,\frac{1}{3},q\right)\right\},\end{array}$

specially for $\alpha =1$, we get

$\begin{array}{r}|\frac{1}{6}\left[f\left(a\right)+4f\left(\sqrt{ab}\right)+f\left(b\right)\right]-\frac{1}{ln\frac{b}{a}}{\int }_{a}^{b}\frac{f\left(u\right)}{u}\phantom{\rule{0.2em}{0ex}}du|\\ \phantom{\rule{1em}{0ex}}\le \frac{ln\frac{b}{a}}{4}{\left(\frac{5}{18}\right)}^{1-\frac{1}{q}}\left\{a{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right)|,|{f}^{\prime }\left(a\right)|\right\}\right]}^{\frac{1}{q}}{h}^{\frac{1}{q}}\left({\left(\frac{b}{a}\right)}^{\frac{q}{2}},\frac{1}{3}\right)\\ \phantom{\rule{2em}{0ex}}+b{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{h}^{\frac{1}{q}}\left({\left(\frac{a}{b}\right)}^{\frac{q}{2}},\frac{1}{3}\right)\right\},\end{array}$

where h is defined as in (10).

Remark 2.2

1. 1.

If we take $x=\sqrt{ab}$, $\lambda =0$, then we get the following midpoint- type inequality for fractional integrals:

$\begin{array}{r}|f\left(\sqrt{ab}\right)-\frac{{2}^{\alpha -1}\mathrm{\Gamma }\left(\alpha +1\right)}{{\left(ln\frac{b}{a}\right)}^{\alpha }}\left[{J}_{\sqrt{ab}-}^{\alpha }f\left(a\right)+{J}_{\sqrt{ab}+}^{\alpha }f\left(b\right)\right]|\\ \phantom{\rule{1em}{0ex}}\le \frac{ln\frac{b}{a}}{4}{\left(\frac{1}{\alpha +1}\right)}^{1-\frac{1}{q}}\left\{a{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{B}_{1}^{\frac{1}{q}}\left(\sqrt{ab},1,0,q\right)\\ \phantom{\rule{2em}{0ex}}+b{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{B}_{2}^{\frac{1}{q}}\left(\sqrt{ab},1,0,q\right)\right\},\end{array}$

specially for $\alpha =1$, we get

$\begin{array}{r}|f\left(\sqrt{ab}\right)-\frac{1}{ln\frac{b}{a}}{\int }_{a}^{b}\frac{f\left(u\right)}{u}\phantom{\rule{0.2em}{0ex}}du|\\ \phantom{\rule{1em}{0ex}}\le \frac{{2}^{\frac{1}{q}}ln\frac{b}{a}}{8}\left\{a{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{h}^{\frac{1}{q}}\left({\left(\frac{b}{a}\right)}^{\frac{q}{2}},0\right)\\ \phantom{\rule{2em}{0ex}}+b{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{h}^{\frac{1}{q}}\left({\left(\frac{a}{b}\right)}^{\frac{q}{2}},0\right)\right\},\end{array}$

where h is defined as in (10).

1. 2.

If we take $x=\sqrt{ab}$, $\lambda =1$, then we get the following trapezoid-type inequality for fractional integrals:

$\begin{array}{r}|\frac{f\left(a\right)+f\left(b\right)}{2}-\frac{{2}^{\alpha -1}\mathrm{\Gamma }\left(\alpha +1\right)}{{\left(ln\frac{b}{a}\right)}^{\alpha }}\left[{J}_{\sqrt{ab}-}^{\alpha }f\left(a\right)+{J}_{\sqrt{ab}+}^{\alpha }f\left(b\right)\right]|\\ \phantom{\rule{1em}{0ex}}\le \frac{ln\frac{b}{a}}{4}{\left(\frac{1}{\alpha +1}\right)}^{1-\frac{1}{q}}\left\{a{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{B}_{1}^{\frac{1}{q}}\left(\sqrt{ab},\alpha ,1,q\right)\\ \phantom{\rule{2em}{0ex}}+b{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{B}_{2}^{\frac{1}{q}}\left(\sqrt{ab},\alpha ,1,q\right)\right\},\end{array}$

specially for $\alpha =1$, we get

$\begin{array}{r}|\frac{f\left(a\right)+f\left(b\right)}{2}-\frac{1}{ln\frac{b}{a}}{\int }_{a}^{b}\frac{f\left(u\right)}{u}\phantom{\rule{0.2em}{0ex}}du|\\ \phantom{\rule{1em}{0ex}}\le \frac{{2}^{\frac{1}{q}}ln\frac{b}{a}}{8}\left\{a{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{h}^{\frac{1}{q}}\left({\left(\frac{b}{a}\right)}^{\frac{q}{2}},1\right)\\ \phantom{\rule{2em}{0ex}}+b{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{h}^{\frac{1}{q}}\left({\left(\frac{a}{b}\right)}^{\frac{q}{2}},1\right)\right\},\end{array}$

where h is defined as in (10).

Corollary 2.4 Let the assumptions of Theorem  2.2 hold. If $|{f}^{\prime }\left(x\right)|\le M$ for all $x\in \left[a,b\right]$ and $\lambda =0$, then we get the following Ostrowski-type inequality for fractional integrals from inequality (9):

$\begin{array}{r}|\left[{\left(ln\frac{x}{a}\right)}^{\alpha }+{\left(ln\frac{b}{x}\right)}^{\alpha }\right]f\left(x\right)-\mathrm{\Gamma }\left(\alpha +1\right)\left[{J}_{\sqrt{ab}-}^{\alpha }f\left(a\right)+{J}_{\sqrt{ab}+}^{\alpha }f\left(b\right)\right]|\\ \phantom{\rule{1em}{0ex}}\le \frac{M}{{\left(\alpha +1\right)}^{1-\frac{1}{q}}}\left[a{\left(ln\frac{x}{a}\right)}^{\alpha +1}{B}_{1}^{\frac{1}{q}}\left(x,\alpha ,0,q\right)+b{\left(ln\frac{b}{x}\right)}^{\alpha +1}{B}_{2}^{\frac{1}{q}}\left(x,\alpha ,0,q\right)\right].\end{array}$

Theorem 2.3 Let $f:I\subset \left(0,\mathrm{\infty }\right)\to \mathbb{R}$ be a differentiable function on ${I}^{\circ }$ such that ${f}^{\prime }\in L\left[a,b\right]$, where $a,b\in {I}^{\circ }$ with $a. If ${|{f}^{\prime }|}^{q}$ is quasi-geometrically convex on $\left[a,b\right]$ for some fixed $q>1$, $x\in \left[a,b\right]$, $\lambda \in \left[0,1\right]$ and $\alpha >0$, then the following inequality for fractional integrals holds:

$\begin{array}{r}|{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)|\\ \phantom{\rule{1em}{0ex}}\le {A}_{2}^{\frac{1}{p}}\left(\alpha ,\lambda ,p\right)\left\{a{\left(ln\frac{x}{a}\right)}^{\alpha +\frac{1}{p}}{\left(sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right)}^{\frac{1}{q}}{\left(\frac{{x}^{q}-{a}^{q}}{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+b{\left(ln\frac{b}{x}\right)}^{\alpha +\frac{1}{p}}{\left(sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right)}^{\frac{1}{q}}{\left(\frac{{b}^{q}-{x}^{q}}{q}\right)}^{\frac{1}{q}}\right\},\end{array}$
(12)

where

$\begin{array}{r}{A}_{2}\left(\alpha ,\lambda ,p\right)\\ \phantom{\rule{1em}{0ex}}=\left\{\begin{array}{ll}\frac{1}{\alpha p+1},& \lambda =0,\\ \frac{{\lambda }^{\frac{\alpha p+1}{\alpha }}}{\alpha }\left\{\beta \left(\frac{1}{\alpha },p+1\right)+\frac{{\left(1-\lambda \right)}^{p+1}}{p+1}\\ \phantom{\rule{1em}{0ex}}{×}_{2}{F}_{1}\left(\frac{1}{\alpha }+p+1,p+1,p+2;1-\lambda \right)\right\},& 0<\lambda <1,\\ \frac{1}{\alpha }\beta \left(p+1,\frac{1}{\alpha }\right),& \lambda =1,\end{array}\end{array}$

${}_{2}F_{1}$ is hypergeometric function defined by

and $\frac{1}{p}+\frac{1}{q}=1$.

Proof Using Lemma 2.1, the Hölder inequality and quasi-geometrical convexity of ${|{f}^{\prime }|}^{q}$, we get

$\begin{array}{r}|{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)|\\ \phantom{\rule{1em}{0ex}}\le a{\left(ln\frac{x}{a}\right)}^{\alpha +1}{\left({\int }_{0}^{1}|{t}^{\alpha }-\lambda {|}^{p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{p}}{\left({\int }_{0}^{1}{\left(\frac{x}{a}\right)}^{qt}sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+b{\left(ln\frac{b}{x}\right)}^{\alpha +1}{\left({\int }_{0}^{1}|{t}^{\alpha }-\lambda {|}^{p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{p}}{\left({\int }_{0}^{1}{\left(\frac{x}{b}\right)}^{qt}sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}},\\ |{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)|\le {\left({\int }_{0}^{1}|{t}^{\alpha }-\lambda {|}^{p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{p}}\\ \phantom{|{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)|\le }×\left\{a{\left(ln\frac{x}{a}\right)}^{\alpha +1}{\left(sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right)}^{\frac{1}{q}}{\left({\int }_{0}^{1}{\left(\frac{x}{a}\right)}^{qt}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}}\\ \phantom{|{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)|\le }+b{\left(ln\frac{b}{x}\right)}^{\alpha +1}{\left(sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right)}^{\frac{1}{q}}{\left({\int }_{0}^{1}{\left(\frac{x}{b}\right)}^{qt}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}}\right\}\\ \phantom{|{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)|}\le {A}_{2}^{\frac{1}{p}}\left(\alpha ,\lambda ,p\right)\left\{a{\left(ln\frac{x}{a}\right)}^{\alpha +1-\frac{1}{q}}{\left(sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right)}^{\frac{1}{q}}{\left(\frac{{x}^{q}-{a}^{q}}{q}\right)}^{\frac{1}{q}}\\ \phantom{|{I}_{f}\left(x,\lambda ,\alpha ,a,b\right)|\le }+b{\left(ln\frac{b}{x}\right)}^{\alpha +1-\frac{1}{q}}{\left(sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right)}^{\frac{1}{q}}{\left(\frac{{b}^{q}-{x}^{q}}{q}\right)}^{\frac{1}{q}}\right\},\end{array}$

here, it is seen by a simple computation that

$\begin{array}{rcl}{A}_{2}\left(\alpha ,\lambda ,p\right)& =& {\int }_{0}^{1}|{t}^{\alpha }-\lambda {|}^{p}\phantom{\rule{0.2em}{0ex}}dt\\ =& \left\{\begin{array}{ll}\frac{1}{\alpha p+1},& \lambda =0,\\ \frac{{\lambda }^{\frac{\alpha p+1}{\alpha }}}{\alpha }\left\{\beta \left(\frac{1}{\alpha },p+1\right)+\frac{{\left(1-\lambda \right)}^{p+1}}{p+1}\\ \phantom{\rule{1em}{0ex}}{×}_{2}{F}_{1}\left(\frac{1}{\alpha }+p+1,p+1,2+p;1-\lambda \right)\right\},& 0<\lambda <1,\\ \frac{1}{\alpha }\beta \left(p+1,\frac{1}{\alpha }\right),& \lambda =1.\end{array}\end{array}$

Hence, the proof is completed. □

Corollary 2.5 Under the assumptions of Theorem  2.3 with $\alpha =1$, inequality (12) reduces to the following inequality:

$\begin{array}{r}|\left(1-\lambda \right)f\left(x\right)+\lambda \left[\frac{f\left(a\right)ln\frac{x}{a}+f\left(b\right)ln\frac{b}{x}}{ln\frac{b}{a}}\right]-\frac{1}{ln\frac{b}{a}}{\int }_{a}^{b}\frac{f\left(u\right)}{u}\phantom{\rule{0.2em}{0ex}}du|\\ \phantom{\rule{1em}{0ex}}\le {\left(ln\frac{b}{a}\right)}^{-1}{\left(\frac{{\lambda }^{p+1}+{\left(1-\lambda \right)}^{p+1}}{p+1}\right)}^{\frac{1}{p}}\\ \phantom{\rule{2em}{0ex}}×\left\{a{\left(ln\frac{x}{a}\right)}^{1+\frac{1}{p}}{\left(sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right)}^{\frac{1}{q}}{\left(\frac{{x}^{q}-{a}^{q}}{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+b{\left(ln\frac{b}{x}\right)}^{1+\frac{1}{p}}{\left(sup\left\{|{f}^{\prime }\left(x\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right)}^{\frac{1}{q}}{\left(\frac{{b}^{q}-{x}^{q}}{q}\right)}^{\frac{1}{q}}\right\},\end{array}$

specially for $x=\sqrt{ab}$, we get

$\begin{array}{r}|\left(1-\lambda \right)f\left(\sqrt{ab}\right)+\lambda \left(\frac{f\left(a\right)+f\left(b\right)}{2}\right)-\frac{1}{ln\frac{b}{a}}{\int }_{a}^{b}\frac{f\left(u\right)}{u}\phantom{\rule{0.2em}{0ex}}du|\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{2}{\left(\frac{ln\frac{b}{a}\left({\lambda }^{p+1}+{\left(1-\lambda \right)}^{p+1}\right)}{2\left(p+1\right)}\right)}^{\frac{1}{p}}\left\{a{\left(sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right)}^{\frac{1}{q}}{\left(\frac{{\sqrt{ab}}^{q}-{a}^{q}}{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+b{\left(sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right)}^{\frac{1}{q}}{\left(\frac{{b}^{q}-{\sqrt{ab}}^{q}}{q}\right)}^{\frac{1}{q}}\right\}.\end{array}$
(13)

Corollary 2.6 In Theorem  2.3,

1. 1.

If we take $x=\sqrt{ab}$, $\lambda =\frac{1}{3}$, then we get the following Simpson-type inequality for fractional integrals:

$\begin{array}{r}|\frac{1}{6}\left[f\left(a\right)+4f\left(\sqrt{ab}\right)+f\left(b\right)\right]-\frac{{2}^{\alpha -1}\mathrm{\Gamma }\left(\alpha +1\right)}{{\left(ln\frac{b}{a}\right)}^{\alpha }}\left[{J}_{\sqrt{ab}-}^{\alpha }f\left(a\right)+{J}_{\sqrt{ab}+}^{\alpha }f\left(b\right)\right]|\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{2}{\left(\frac{ln\frac{b}{a}\left(1+{2}^{p+1}\right)}{{3}^{p+1}\left(p+1\right)2}\right)}^{\frac{1}{p}}\left\{a{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{\left(\frac{{\sqrt{ab}}^{q}-{a}^{q}}{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+b{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{\left(\frac{{b}^{q}-{\sqrt{ab}}^{q}}{q}\right)}^{\frac{1}{q}}\right\},\end{array}$

specially for $\alpha =1$, we get

$\begin{array}{r}|\frac{1}{6}\left[f\left(a\right)+4f\left(\sqrt{ab}\right)+f\left(b\right)\right]-\frac{1}{ln\frac{b}{a}}{\int }_{a}^{b}\frac{f\left(u\right)}{u}\phantom{\rule{0.2em}{0ex}}du|\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{2}{\left(\frac{ln\frac{b}{a}\left(1+{2}^{p+1}\right)}{{3}^{p+1}\left(p+1\right)2}\right)}^{\frac{1}{p}}\\ \phantom{\rule{2em}{0ex}}×\left\{a{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{\left(\frac{{\sqrt{ab}}^{q}-{a}^{q}}{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+b{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{\left(\frac{{b}^{q}-{\sqrt{ab}}^{q}}{q}\right)}^{\frac{1}{q}}\right\}.\end{array}$

Remark 2.3

1. 1.

If we take $x=\sqrt{ab}$, $\lambda =0$, then we get the following midpoint- type inequality for fractional integrals:

$\begin{array}{r}|f\left(\sqrt{ab}\right)-\frac{{2}^{\alpha -1}\mathrm{\Gamma }\left(\alpha +1\right)}{{\left(ln\frac{b}{a}\right)}^{\alpha }}\left[{J}_{\sqrt{ab}-}^{\alpha }f\left(a\right)+{J}_{\sqrt{ab}+}^{\alpha }f\left(b\right)\right]|\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{2}{\left(\frac{ln\frac{b}{a}}{2\left(\alpha p+1\right)}\right)}^{\frac{1}{p}}\left\{a{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{\left(\frac{{\sqrt{ab}}^{q}-{a}^{q}}{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+b{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{\left(\frac{{b}^{q}-{\sqrt{ab}}^{q}}{q}\right)}^{\frac{1}{q}}\right\},\end{array}$

specially for $\alpha =1$, we get

$\begin{array}{r}|f\left(\sqrt{ab}\right)-\frac{1}{ln\frac{b}{a}}{\int }_{a}^{b}\frac{f\left(u\right)}{u}\phantom{\rule{0.2em}{0ex}}du|\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{2}{\left(\frac{ln\frac{b}{a}}{2\left(p+1\right)}\right)}^{\frac{1}{p}}\left\{a{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{\left(\frac{{\sqrt{ab}}^{q}-{a}^{q}}{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+b{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{\left(\frac{{b}^{q}-{\sqrt{ab}}^{q}}{q}\right)}^{\frac{1}{q}}\right\}.\end{array}$
1. 2.

If we take $x=\sqrt{ab}$, $\lambda =1$, then we get the following trapezoid-type inequality for fractional integrals $\frac{1}{\alpha }\beta \left(p+1,\frac{1}{\alpha }\right)$:

$\begin{array}{r}|\frac{f\left(a\right)+f\left(b\right)}{2}-\frac{{2}^{\alpha -1}\mathrm{\Gamma }\left(\alpha +1\right)}{{\left(ln\frac{b}{a}\right)}^{\alpha }}\left[{J}_{\sqrt{ab}-}^{\alpha }f\left(a\right)+{J}_{\sqrt{ab}+}^{\alpha }f\left(b\right)\right]|\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{2}{\left(\frac{ln\frac{b}{a}\beta \left(p+1,\frac{1}{\alpha }\right)}{2\alpha }\right)}^{\frac{1}{p}}\\ \phantom{\rule{2em}{0ex}}×\left\{a{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{\left(\frac{{\sqrt{ab}}^{q}-{a}^{q}}{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+b{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{\left(\frac{{b}^{q}-{\sqrt{ab}}^{q}}{q}\right)}^{\frac{1}{q}}\right\},\end{array}$

specially for $\alpha =1$, we get

$\begin{array}{r}|\frac{f\left(a\right)+f\left(b\right)}{2}-\frac{1}{ln\frac{b}{a}}{\int }_{a}^{b}\frac{f\left(u\right)}{u}\phantom{\rule{0.2em}{0ex}}du|\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{2}{\left(\frac{ln\frac{b}{a}}{2\left(p+1\right)}\right)}^{\frac{1}{p}}\left\{a{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(a\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{\left(\frac{{\sqrt{ab}}^{q}-{a}^{q}}{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+b{\left[sup\left\{|{f}^{\prime }\left(\sqrt{ab}\right){|}^{q},|{f}^{\prime }\left(b\right){|}^{q}\right\}\right]}^{\frac{1}{q}}{\left(\frac{{b}^{q}-{\sqrt{ab}}^{q}}{q}\right)}^{\frac{1}{q}}\right\}.\end{array}$

Corollary 2.7 Let the assumptions of Theorem  2.3 hold. If $|{f}^{\prime }\left(x\right)|\le M$ for all $x\in \left[a,b\right]$ and $\lambda =0$, then we get the following Ostrowski-type inequality for fractional integrals from inequality (12):

$\begin{array}{r}|\left[{\left(ln\frac{x}{a}\right)}^{\alpha }+{\left(ln\frac{b}{x}\right)}^{\alpha }\right]f\left(x\right)-\mathrm{\Gamma }\left(\alpha +1\right)\left[{J}_{\sqrt{ab}-}^{\alpha }f\left(a\right)+{J}_{\sqrt{ab}+}^{\alpha }f\left(b\right)\right]|\\ \phantom{\rule{1em}{0ex}}\le \frac{M}{{\left(\alpha p+1\right)}^{\frac{1}{p}}}\left[a{\left(ln\frac{x}{a}\right)}^{\alpha +\frac{1}{p}}{\left(\frac{{x}^{q}-{a}^{q}}{q}\right)}^{\frac{1}{q}}+b{\left(ln\frac{b}{x}\right)}^{\alpha +\frac{1}{p}}{\left(\frac{{b}^{q}-{x}^{q}}{q}\right)}^{\frac{1}{q}}\right].\end{array}$

## 3 Application to special means

Let us recall the following special means of two nonnegative numbers a, b with $b>a$:

1. 1.

The arithmetic mean

$A=A\left(a,b\right):=\frac{a+b}{2}.$
2. 2.

The geometric mean

$G=G\left(a,b\right):=\sqrt{ab}.$
3. 3.

The logarithmic mean

$L=L\left(a,b\right):=\frac{b-a}{lnb-lna}.$
4. 4.

The p-logarithmic mean

${L}_{p}={L}_{p}\left(a,b\right):={\left(\frac{{b}^{p+1}-{a}^{p+1}}{\left(p+1\right)\left(b-a\right)}\right)}^{\frac{1}{p}},\phantom{\rule{1em}{0ex}}p\in \mathbb{R}\mathrm{\setminus }\left\{-1,0\right\}.$

Proposition 3.1 For $b>a>0$, $n>0$ and $q\ge 1$, we have

$\begin{array}{r}|\left(1-\lambda \right){G}^{n+1}\left(a,b\right)+\lambda A\left({a}^{n+1},{b}^{n+1}\right)-\left(n+1\right)L\left(a,b\right){L}_{n}^{n}\left(a,b\right)|\\ \phantom{\rule{1em}{0ex}}\le \frac{\left(n+1\right)ln\frac{b}{a}}{4}{\left(\frac{2{\lambda }^{2}-2\lambda +1}{2}\right)}^{1-\frac{1}{q}}\left\{a{G}^{n}\left(a,b\right){h}^{\frac{1}{q}}\left({\left(\frac{b}{a}\right)}^{\frac{q}{2}},\lambda \right)\\ \phantom{\rule{1em}{0ex}}+{b}^{n+1}{h}^{\frac{1}{q}}\left({\left(\frac{a}{b}\right)}^{\frac{q}{2}},\lambda \right)\right\},\end{array}$

where h is defined as in (10).

Proof Let $f\left(x\right)=\frac{{x}^{n+1}}{n+1}$, $x>0$, $n>0$ and $q\ge 1$. Then the function $|{f}^{\prime }\left(x\right){|}^{q}={x}^{nq}$ is quasi-geometrically convex on $\left(0,\mathrm{\infty }\right)$. Thus, by inequality (11), Proposition 3.1 is proved. □

Proposition 3.2 For $b>a>0$, $n>0$ and $q>1$, we have

$\begin{array}{r}|\left(1-\lambda \right){G}^{n+1}\left(a,b\right)+\lambda A\left({a}^{n+1},{b}^{n+1}\right)-\left(n+1\right)L\left(a,b\right){L}_{n}^{n}\left(a,b\right)|\\ \phantom{\rule{1em}{0ex}}\le \frac{n+1}{2}{\left(\frac{ln\frac{b}{a}\left({\lambda }^{p+1}+{\left(1-\lambda \right)}^{p+1}\right)}{2\left(p+1\right)}\right)}^{\frac{1}{p}}\left\{a{G}^{n}\left(a,b\right){\left(\frac{{G}^{q}\left(a,b\right)-{a}^{q}}{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+{b}^{n+1}{\left(\frac{{b}^{q}-{G}^{q}\left(a,b\right)}{q}\right)}^{\frac{1}{q}}\right\}.\end{array}$

Proof Let $f\left(x\right)=\frac{{x}^{n+1}}{n+1}$, $x>0$, $n>0$ and $q>1$. Then the function $|{f}^{\prime }\left(x\right){|}^{q}={x}^{nq}$ is quasi-geometrically convex on $\left(0,\mathrm{\infty }\right)$. Thus, by inequality (13), Proposition 3.2 is proved. □