Introduction {#Sec1}
============

A function $\documentclass[12pt]{minimal}
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                \begin{document}$f:I\to\mathbb{R}$\end{document}$ is said to be convex if $$\documentclass[12pt]{minimal}
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                \begin{document}$$f\bigl((1-t)x+ty\bigr)\leq(1-t)f(x)+tf(y),\quad\forall x,y\in I,t\in[0,1]. $$\end{document}$$ Convexity plays a pivotal role in different fields of pure and applied sciences. Another fact that makes it more attractive is its close relationship with theory of inequalities. In particular, integral inequalities have been obtained via convex functions. Inspired by the research work in this field, many authors introduced new extensions of classical convex functions; see, for example, \[[@CR1]--[@CR4]\] and the references therein. Recently, Işcan \[[@CR3]\] introduced and investigated the notion of harmonically convex functions. These days the class of harmonically convex functions is receiving much attention by many researchers. For more details, see \[[@CR3]--[@CR9]\]. Hermite and Hadamard independently obtained an integral inequality that provides us a necessary and sufficient condition for a function to be convex. This famous result reads as follows.

Let $\documentclass[12pt]{minimal}
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                \begin{document}$f:I\supseteq[a,b]\rightarrow\mathbb{R}$\end{document}$ be a convex function, then $$\documentclass[12pt]{minimal}
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                \begin{document}$$ f \biggl(\frac{a+b}{2} \biggr)\leq\frac{1}{b-a} \int _{a}^{b} f(x)\,\mathrm{d}x\leq\frac{f(a)+f(b)}{2}. $$\end{document}$$ For more details on Hermite-Hadamard-type inequalities, see \[[@CR1], [@CR2], [@CR4], [@CR10], [@CR11]\]. Sarikaya et al. \[[@CR12]\] developed a new generalization of Hermite-Hadamard-type inequality using fractional calculus approach. This opened a new venue of research in this field. Utilizing the concepts of fractional calculus, Işcan et al. \[[@CR6]\] derived new refinements of fractional Hermite-Hadamard's inequality via harmonically convex functions. The main motivation of this paper is to obtain new refinements of fractional Hermite-Hadamard-type inequalities via harmonically convex functions in connection with the generalized Mittag-Leffler function, which even generalizes the classical Riemann-Liouville fractional integral operators. We also discuss some particular cases.

Preliminaries {#Sec2}
=============

In this section, we discuss some preliminary concepts and facts. Recently, Işcan \[[@CR3]\] obtained several inequalities of Hermite-Hadamard type via harmonic convex functions.

The class of harmonic convex functions is defined as follows.

Definition 2.1 {#FPar1}
--------------

(\[[@CR3]\])

Let $\documentclass[12pt]{minimal}
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                \begin{document}$f:I\subseteq\mathbb{R}\setminus \{ 0 \} \rightarrow \mathbb{R}$\end{document}$, where *I* a real interval. The function *f* is said to be harmonic convex if $$\documentclass[12pt]{minimal}
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                \begin{document}$$ f \biggl( \frac{xy}{tx+(1-t)y} \biggr) \leq tf(y)+(1-t)f(x) $$\end{document}$$ for all $\documentclass[12pt]{minimal}
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                \begin{document}$x,y\in I$\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$t\in{[}0,1]$\end{document}$. If ([2.1](#Equ2){ref-type=""}) holds in the reversed sense, then *f* is said to be harmonic concave.

Işcan et al. \[[@CR6]\] established new fractional estimates of Hermite-Hadamard-type inequalities via harmonic convex functions. For more details on Hermite-Hadamard inequalities involving fractional integrals; see Mihai \[[@CR13], [@CR14]\], Mihai et al. \[[@CR15]\], Awan et al. \[[@CR16]\], Kunt et al. \[[@CR7]\], Sarikaya et al. \[[@CR17]\], Nisan et al. \[[@CR18]\], and references therein. Latif et al. \[[@CR8]\] gave the following definition.

Definition 2.2 {#FPar2}
--------------

A function $\documentclass[12pt]{minimal}
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                \begin{document}$g:[a,b]\subseteq\mathbb{R}\setminus \lbrace0 \rbrace\to\mathbb{R}$\end{document}$ is said to be harmonically symmetric with respect to $\documentclass[12pt]{minimal}
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                \begin{document}$\frac{2ab}{a+b}$\end{document}$ if $$\documentclass[12pt]{minimal}
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                \begin{document}$$ g(x)=g \biggl( \frac{1}{\frac{1}{a}+\frac{1}{b}-\frac{1}{x}} \biggr) $$\end{document}$$ for all $\documentclass[12pt]{minimal}
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                \begin{document}$x\in[a,b]$\end{document}$.

For details and the definition of Riemann-Liouville fractional integrals, see \[[@CR19], [@CR20]\].

Salim et al. \[[@CR21]\] have defined the generalized fractional integral operators containing Mittag-Leffler function:

Definition 2.3 {#FPar3}
--------------

Let $\documentclass[12pt]{minimal}
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                \begin{document}$\mu, \nu, k, l, \gamma>0$\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$\omega\in\mathbb{R} $\end{document}$. Then the generalized fractional integral operators containing the Mittag-Leffler functions $\documentclass[12pt]{minimal}
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                \begin{document}$\varepsilon_{\mu, \nu, l, \omega, a^{+}}^{\gamma, \delta, k}$\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$\varepsilon_{\mu, \nu, l, \omega, b^{-}}^{\gamma, \delta, k}$\end{document}$ for a real-valued continuous function *f* are defined by $$\documentclass[12pt]{minimal}
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                \begin{document}$$ \bigl( \varepsilon_{\mu, \nu, l, \omega, a^{+}}^{\gamma, \delta, k}f \bigr) (x)= \int_{a}^{x}(x-t)^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta , k} \bigl( \omega(x-t)^{\mu} \bigr) f(t)\,\mathrm{d}t $$\end{document}$$ and $$\documentclass[12pt]{minimal}
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                \begin{document}$$ \bigl( \varepsilon_{\mu, \nu, l, \omega, b^{-}}^{\gamma, \delta, k}f \bigr) (x)= \int_{x}^{b}(t-x)^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta , k} \bigl( \omega(t-x)^{\mu} \bigr) f(t)\,\mathrm{d}t, $$\end{document}$$ respectively, where $\documentclass[12pt]{minimal}
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                \begin{document}$E_{\mu, \nu, l}^{\gamma, \delta, k}$\end{document}$ is the generalized Mittag-Leffler function defined as $$\documentclass[12pt]{minimal}
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                \begin{document}$$ E_{\mu, \nu, l}^{\gamma, \delta, k}(t)=\sum_{n=0}^{\infty} \frac{(\gamma )_{kn}}{\Gamma(\mu n+\nu)}\frac{t^{n}}{(\delta)_{ln}}, $$\end{document}$$ and $\documentclass[12pt]{minimal}
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                \begin{document}$(a)_{n}$\end{document}$ is the Pochhammer symbol defined as $$\documentclass[12pt]{minimal}
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                \begin{document}$$(a)_{n}=a(a+1)\cdots(a+n-1),\quad(a)_{0}=1. $$\end{document}$$

Remark {#FPar4}
------

If $\documentclass[12pt]{minimal}
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                \begin{document}$k=l=1$\end{document}$ in ([2.2](#Equ3){ref-type=""}), then the integral operator $\documentclass[12pt]{minimal}
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                \begin{document}$( \varepsilon_{\mu, \nu, 1, \omega, a^{+}}^{\gamma, \delta, k}f )$\end{document}$ reduces to the integral operator $\documentclass[12pt]{minimal}
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                \begin{document}$( \varepsilon_{\mu, \nu, l, \omega, a^{+}}^{\gamma, \delta, 1}f )$\end{document}$ containing the generalized Mittag-Leffler function $\documentclass[12pt]{minimal}
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                \begin{document}$E_{\mu, \nu, 1}^{\gamma, \delta, 1}$\end{document}$ introduced by Srivastava and Tomovski \[[@CR22]\]. If $\documentclass[12pt]{minimal}
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                \begin{document}$k=l=1$\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$\delta=1$\end{document}$, then ([2.2](#Equ3){ref-type=""}) reduces to the integral operator defined by Prabhaker \[[@CR23]\] and containing the Mittag-Leffler function $\documentclass[12pt]{minimal}
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                \begin{document}$E_{\mu, \nu}^{\gamma}$\end{document}$. For $\documentclass[12pt]{minimal}
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                \begin{document}$\omega=0$\end{document}$ in ([2.2](#Equ3){ref-type=""}), the integral operator $\documentclass[12pt]{minimal}
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                \begin{document}$( \varepsilon_{\mu, \nu, l, \omega, a^{+}}^{\gamma, \delta, k}f )$\end{document}$ reduces to the Riemann-Liouville fractional integral operator \[[@CR21]\].

In \[[@CR21]\] the properties of the generalized integral operator and the generalized Mittag-Leffler function are studied. It is proved that $\documentclass[12pt]{minimal}
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                \begin{document}$E_{\mu, \nu, l}^{\gamma, \delta, k} ( t )$\end{document}$ is absolutely convergent for all $\documentclass[12pt]{minimal}
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                \begin{document}$t\in\mathbb{R}$\end{document}$ if $\documentclass[12pt]{minimal}
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                \begin{document}$k< l+\mu$\end{document}$. Since $$\documentclass[12pt]{minimal}
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                \begin{document}$$\bigl\vert E_{\mu, \nu, l}^{\gamma, \delta, k} ( t ) \bigr\vert \leq\sum _{n=0}^{\infty} \biggl\vert \frac{(\gamma)_{kn}}{\Gamma(\mu n+\nu )} \frac{t^{n}}{(\delta)_{ln}} \biggr\vert $$\end{document}$$ with $$\documentclass[12pt]{minimal}
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                \begin{document}$$\sum_{n=0}^{\infty} \biggl\vert \frac{(\gamma)_{kn}}{\Gamma(\mu n+\nu )}\frac{t^{n}}{(\delta)_{ln}} \biggr\vert =S, $$\end{document}$$ we have $$\documentclass[12pt]{minimal}
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                \begin{document}$$\bigl\vert E_{\mu, \nu, l}^{\gamma, \delta, k} ( t ) \bigr\vert \leq S. $$\end{document}$$

Results and discussions {#Sec3}
=======================

In this section, we discuss our main results. We write $\documentclass[12pt]{minimal}
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                \begin{document}$\Vert g \Vert _{\infty}=\sup_{t\in[a,b]} \vert g(t) \vert $\end{document}$ for a continuous function $\documentclass[12pt]{minimal}
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                \begin{document}$g:[a,b]\to\mathbb{R}$\end{document}$.

Lemma 3.1 {#FPar5}
---------

*If* $\documentclass[12pt]{minimal}
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                \begin{document}$g:[a,b]\subseteq\mathbb{R}\setminus \lbrace0 \rbrace\to \mathbb{R}$\end{document}$ *is integrable and harmonically symmetric with respect to* $\documentclass[12pt]{minimal}
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                \begin{document}$\frac{2ab}{a+b}$\end{document}$, *then* $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} \bigl( \varepsilon_{\mu, \nu, l, \omega, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr) &= \bigl( \varepsilon_{\mu, \nu, l, \omega, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr) \\ &=\frac{1}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega, (\frac {a+b}{2ab} )^{+}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac {1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega, (\frac {a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac {1}{b} \biggr) \biggr] , \end{aligned}$$ \end{document}$$ *where* $\documentclass[12pt]{minimal}
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                \begin{document}$h(x)=\frac{1}{x}$\end{document}$, $\documentclass[12pt]{minimal}
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                \begin{document}$x\in [\frac{1}{b}, \frac{1}{a} ] $\end{document}$.

Proof {#FPar6}
-----

Since *g* is harmonically symmetric with respect to $\documentclass[12pt]{minimal}
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                \begin{document}$\frac{2ab}{b-a}$\end{document}$, using Definition [2.2](#FPar2){ref-type="sec"}, we have $\documentclass[12pt]{minimal}
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                \begin{document}$g (\frac{1}{t} ) =g ( \frac{1}{\frac{1}{a}+\frac{1}{b}-t} )$\end{document}$ for all $\documentclass[12pt]{minimal}
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                \begin{document}$t\in [\frac {1}{b}, \frac{1}{a} ] $\end{document}$. Hence, in the following integral, setting $\documentclass[12pt]{minimal}
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                \begin{document}$u=\frac{1}{a}+\frac{1}{b}-t$\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$\mathrm{d}u=-\mathrm{d}t$\end{document}$ gives $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} & \bigl( \varepsilon_{\mu, \nu, l, \omega, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr) \\ &\quad = \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-t \biggr)^{\nu -1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega \biggl( \frac {1}{a}-t \biggr) ^{\mu} \biggr)g \biggl(\frac{1}{t} \biggr)\,\mathrm{d}t \\ &\quad = \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-t \biggr)^{\nu -1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega \biggl( \frac {1}{a}-t \biggr) ^{\mu} \biggr)g \biggl( \frac{1}{\frac{1}{a}+\frac {1}{b}-t} \biggr)\,\mathrm{d}t \\ &\quad = \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(u- \frac{1}{b} \biggr)^{\nu -1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega \biggl(u- \frac {1}{b} \biggr) ^{\mu} \biggr)g \biggl( \frac{1}{u} \biggr)\,\mathrm{d}u= \bigl( \varepsilon_{\mu, \nu, l, \omega, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr) . \end{aligned}$$ \end{document}$$ This completes the proof. □

Lemma 3.2 {#FPar7}
---------

*Let* $\documentclass[12pt]{minimal}
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                \begin{document}$f:I\subset(0, +\infty)\to\mathbb{R}$\end{document}$ *be a differentiable function on* $\documentclass[12pt]{minimal}
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                \begin{document}$I^{\circ}$\end{document}$, *the interior of* *I*, *such that* $\documentclass[12pt]{minimal}
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                \begin{document}$f^{\prime}\in L[a,b]$\end{document}$, *where* $\documentclass[12pt]{minimal}
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                \begin{document}$a,b \in I$\end{document}$. *If* $\documentclass[12pt]{minimal}
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                \begin{document}$g:[a,b]\to\mathbb{R}$\end{document}$ *is integrable and harmonically symmetric with respect to* $\documentclass[12pt]{minimal}
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                \begin{document}$\frac{2ab}{a+b}$\end{document}$, *then the following equality holds*: $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} &f \biggl(\frac{2ab}{a+b} \biggr) \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu , l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\qquad{}- \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad = \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl( \int_{\frac{1}{b}}^{t} \biggl(s-\frac{1}{b} \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(s- \frac{1}{b} \biggr) ^{\mu} \biggr) (g\circ h ) (s) \,\mathrm{d}s \biggr) (f\circ h ) ^{\prime }(t)\,\mathrm{d}t \\ &\qquad{}- \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl( \int_{t}^{\frac {1}{a}} \biggl(\frac{1}{a}-s \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-s \biggr) ^{\mu} \biggr) (g\circ h ) (s) \,\mathrm{d}s \biggr) (f\circ h ) ^{\prime}(t)\,\mathrm{d}t, \end{aligned}$$ \end{document}$$ *where* $\documentclass[12pt]{minimal}
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                \begin{document}$h(x)=\frac{1}{x}$\end{document}$, $\documentclass[12pt]{minimal}
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                \begin{document}$x\in [\frac{1}{b}, \frac{1}{a} ]$\end{document}$, *and* $\documentclass[12pt]{minimal}
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                \begin{document}$\omega^{\prime}= ( \frac{2ab}{b-a} ) ^{\mu}\omega$\end{document}$.

Proof {#FPar8}
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It suffices to show that $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} I&= \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl( \int_{\frac{1}{b}}^{t} \biggl(s-\frac{1}{b} \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(s- \frac{1}{b} \biggr) ^{\mu} \biggr) (g\circ h ) (s) \,\mathrm{d}s \biggr) (f\circ h ) ^{\prime }(t)\,\mathrm{d}t \\ &\quad{}- \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl( \int_{t}^{\frac {1}{a}} \biggl(\frac{1}{a}-s \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-s \biggr) ^{\mu} \biggr) (g\circ h ) (s) \,\mathrm{d}s \biggr) (f\circ h ) ^{\prime}(t)\,\mathrm{d}t \\ &=I_{1}-I_{2}. \end{aligned}$$ \end{document}$$ By Lemma [3.1](#FPar5){ref-type="sec"}, integrating by parts, we have $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} I_{1}&= \int_{\frac{1}{b}}^{t} \biggl(s-\frac{1}{b} \biggr) ^{\nu -1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(s- \frac{1}{b} \biggr) ^{\mu} \biggr) (g\circ h ) (s) \,\mathrm{d}s \cdot (f\circ h ) (t) \bigg\vert _{\frac{1}{b}}^{\frac {a+b}{2ab}} \\ &\quad{}- \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(t-\frac{1}{b} \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(t- \frac{1}{b} \biggr) ^{\mu} \biggr) (g\circ h ) (t) (f \circ h ) (t)\,\mathrm{d}t \\ &=f \biggl(\frac{2ab}{a+b} \biggr) \int_{\frac{1}{b}}^{\frac {2ab}{a+b}} \biggl(s-\frac{1}{b} \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(s- \frac{1}{b} \biggr) ^{\mu} \biggr) (g\circ h ) (s) \,\mathrm{d}s \\ &\quad{}- \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(t-\frac{1}{b} \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(t- \frac{1}{b} \biggr) ^{\mu} \biggr) (fg\circ h ) (t) \,\mathrm{d}t. \end{aligned}$$ \end{document}$$ Thus $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} I_{1}&=f \biggl(\frac{2ab}{a+b} \biggr) \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr)- \bigl( \varepsilon_{\mu, \nu , l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg\circ h \bigr) \biggl( \frac{1}{b} \biggr) \\ &=\frac{1}{2}f \biggl(\frac{2ab}{a+b} \biggr) \biggl[ \bigl( \varepsilon_{\mu , \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon _{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad{}- \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac {a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg\circ h \bigr) \biggl( \frac {1}{b} \biggr). \end{aligned}$$ \end{document}$$ Analogously, $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} I_{2}&= \int_{t}^{\frac{1}{a}} \biggl(\frac{1}{a}-s \biggr) ^{\nu -1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac {1}{a}-s \biggr) ^{\mu} \biggr) (g\circ h ) (s) \,\mathrm{d}s \cdot (f\circ h ) (t) \bigg\vert _{\frac{a+b}{2ab}}^{\frac {1}{a}} \\ &\quad{}- \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-t \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-t \biggr) ^{\mu} \biggr) (g\circ h ) (t) (f \circ h ) (t)\,\mathrm{d}t \\ &=-f \biggl(\frac{2ab}{a+b} \biggr) \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-s \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-s \biggr) ^{\mu} \biggr) (g\circ h ) (s) \,\mathrm{d}s \\ &\quad{}+ \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-t \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-t \biggr) ^{\mu} \biggr) (fg\circ h ) (t) \,\mathrm{d}t. \end{aligned}$$ \end{document}$$ Also, $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} I_{2}&=-f \biggl(\frac{2ab}{a+b} \biggr) \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu , l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg\circ h \bigr) \biggl( \frac{1}{a} \biggr) \\ &=-\frac{1}{2}f \biggl(\frac{2ab}{a+b} \biggr) \biggl[ \bigl( \varepsilon _{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon _{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad{}+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac {a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg\circ h \bigr) \biggl( \frac {1}{a} \biggr). \end{aligned}$$ \end{document}$$ Inserting ([3.4](#Equ8){ref-type=""}) and ([3.6](#Equ10){ref-type=""}) into ([3.2](#Equ6){ref-type=""}), we get ([3.1](#Equ5){ref-type=""}), and the proof is complete. □

Remark {#FPar9}
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Taking $\documentclass[12pt]{minimal}
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                \begin{document}$\omega=0$\end{document}$ in Lemmas [3.1](#FPar5){ref-type="sec"} and [3.2](#FPar7){ref-type="sec"}, we have Lemmas 2 and 3 from \[[@CR7]\].

The next result is an Hermite-Hadamard-type inequality for a generalized fractional integral operator containing the generalized Mittag-Leffler function.

Theorem 3.3 {#FPar10}
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*Let* $\documentclass[12pt]{minimal}
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                \begin{document}$f:I\subset(0, +\infty)\to\mathbb{R}$\end{document}$ *be a function such that* $\documentclass[12pt]{minimal}
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                \begin{document}$f\in L[a,b]$\end{document}$, *where* $\documentclass[12pt]{minimal}
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                \begin{document}$a,b \in I$\end{document}$. *If* *f* *is a harmonically convex function on* $\documentclass[12pt]{minimal}
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                \begin{document}$[a,b]$\end{document}$, *then the following inequality for fractional integrals holds*: $$\documentclass[12pt]{minimal}
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                \begin{document}$$ \begin{aligned}[b]&f \biggl(\frac{2ab}{a+b} \biggr) \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}1 \bigr) \biggl(\frac{1}{b} \biggr) \\ &\quad \leq\frac{1}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}f \circ h \bigr) \biggl(\frac{1}{b} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}f \circ h \bigr) \biggl(\frac{1}{a} \biggr) \biggr] \\ &\quad \leq\frac{f(a)+f(b)}{2} \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime }, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}1 \bigr) \biggl(\frac {1}{a} \biggr), \end{aligned} $$\end{document}$$ *where* $\documentclass[12pt]{minimal}
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                \begin{document}$h(x)=\frac{1}{x}$\end{document}$, $\documentclass[12pt]{minimal}
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                \begin{document}$x\in [\frac{1}{b}, \frac{1}{a} ]$\end{document}$, *and* $\documentclass[12pt]{minimal}
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                \begin{document}$\omega^{\prime}= ( \frac{2ab}{b-a} ) ^{\mu}\omega$\end{document}$.

Proof {#FPar11}
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Since *f* is a harmonically convex function on $\documentclass[12pt]{minimal}
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                \begin{document}$[a,b]$\end{document}$, we have $$\documentclass[12pt]{minimal}
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                \begin{document}$$ f \biggl(\frac{2xy}{x+y} \biggr) \leq\frac{f(x)+f(y)}{2} $$\end{document}$$ for $\documentclass[12pt]{minimal}
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                \begin{document}$x,y\in[a,b]$\end{document}$. Substituting $\documentclass[12pt]{minimal}
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                \begin{document}$x=\frac{ab}{\frac{t}{2}a+\frac{2-t}{2}b}$\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$y=\frac {ab}{\frac{t}{2}b+\frac{2-t}{2}a} $\end{document}$ into inequality ([3.8](#Equ12){ref-type=""}), we get $$\documentclass[12pt]{minimal}
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                \begin{document}$$ f \biggl(\frac{2ab}{a+b} \biggr) \leq\frac{f (\frac{ab}{\frac {t}{2}a+\frac{2-t}{2}b} ) +f (\frac{ab}{\frac{t}{2}b+\frac {2-t}{2}a} ) }{2}. $$\end{document}$$ Multiplying both sides of ([3.9](#Equ13){ref-type=""}) by $\documentclass[12pt]{minimal}
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                \begin{document}$t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} ( \omega t^{\mu} )$\end{document}$ and integrating over $\documentclass[12pt]{minimal}
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                \begin{document}$[0,1]$\end{document}$, we get $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} &2f \biggl(\frac{2ab}{a+b} \biggr) \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)\,\mathrm{d}t \\ &\quad \leq \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)f \biggl(\frac{ab}{\frac{t}{2}a+\frac {2-t}{2}b} \biggr) \,\mathrm{d}t+ \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)f \biggl(\frac {ab}{\frac{t}{2}b+\frac{2-t}{2}a} \biggr) \,\mathrm{d}t . \end{aligned}$$ \end{document}$$ Setting $$\documentclass[12pt]{minimal}
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                \begin{document}$$ u=\frac{1}{ab} \biggl(\frac{t}{2}a+ \frac{2-t}{2}b \biggr),\qquad v=\frac {1}{ab} \biggl(\frac{t}{2}b+ \frac{2-t}{2}a \biggr), $$\end{document}$$ we have $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} &2f \biggl(\frac{2ab}{a+b} \biggr) \int_{\frac{1}{b}}^{\frac {a+b}{2ab}} \biggl(v-\frac{1}{b} \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(v-\frac{1}{b} \biggr) ^{\mu} \biggr)\,\mathrm{d}v \\ &\quad \leq \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-u \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-u \biggr) ^{\mu} \biggr) (f\circ h ) (u)\,\mathrm {d}u \\ &\qquad{}+ \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(v-\frac{1}{b} \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(v-\frac{1}{b} \biggr) ^{\mu} \biggr) (f\circ h ) (v)\,\mathrm {d}v. \end{aligned}$$ \end{document}$$ This implies $$\documentclass[12pt]{minimal}
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                \begin{document}$$ \begin{aligned}[b] &f \biggl(\frac{2ab}{a+b} \biggr) \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}1 \bigr) \biggl(\frac{1}{b} \biggr) \\ &\quad \leq\frac{1}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}f \circ h \bigr) \biggl(\frac{1}{b} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}f \circ h \bigr) \biggl(\frac{1}{a} \biggr) \biggr]. \end{aligned} $$\end{document}$$ On the other hand, the harmonic convexity of *f* yields $$\documentclass[12pt]{minimal}
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                \begin{document}$$ f \biggl(\frac{ab}{\frac{t}{2}a+\frac{2-t}{2}b} \biggr) +f \biggl(\frac {ab}{\frac{t}{2}b+\frac{2-t}{2}a} \biggr)\leq f(a)+f(b) $$\end{document}$$ for all $\documentclass[12pt]{minimal}
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                \begin{document}$t\in[0,1]$\end{document}$. Multiplying both sides of ([3.12](#Equ16){ref-type=""}) by $\documentclass[12pt]{minimal}
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                \begin{document}$t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} ( \omega t^{\mu} )$\end{document}$ and integrating over $\documentclass[12pt]{minimal}
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                \begin{document}$[0,1]$\end{document}$, we have $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} & \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)f \biggl(\frac{ab}{\frac{t}{2}a+\frac{2-t}{2}b} \biggr) \,\mathrm{d}t+ \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)f \biggl(\frac{ab}{\frac{t}{2}b+\frac {2-t}{2}a} \biggr) \,\mathrm{d}t \\ &\quad \leq \bigl(f(a)+f(b) \bigr) \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)\,\mathrm{d}t . \end{aligned}$$ \end{document}$$ This implies, using substitutions ([3.10](#Equ14){ref-type=""}), $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} & \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-u \biggr) ^{\nu -1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac {1}{a}-u \biggr) ^{\mu} \biggr) (f\circ h ) (u) \,\mathrm{d}u \\ &\qquad{}+ \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(v-\frac{1}{b} \biggr) ^{\nu -1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(v-\frac{1}{b} \biggr) ^{\mu} \biggr) (f\circ h ) (v)\,\mathrm {d}v. \\ &\quad \leq \bigl(f(a)+f(b) \bigr) \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-u \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-u \biggr) ^{\mu} \biggr) (f\circ h ) (u) \,\mathrm{d}u . \end{aligned}$$ \end{document}$$ So $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} &\frac{1}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}f \circ h \bigr) \biggl(\frac {1}{b} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}f \circ h \bigr) \biggl(\frac {1}{a} \biggr) \biggr] \\ &\quad \leq\frac{f(a)+f(b)}{2} \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime }, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}1 \bigr) \biggl(\frac {1}{a} \biggr). \end{aligned}$$ \end{document}$$ Combining ([3.11](#Equ15){ref-type=""}) and ([3.13](#Equ17){ref-type=""}), we get ([3.7](#Equ11){ref-type=""}). □

Remark {#FPar12}
------

In Theorem [3.3](#FPar10){ref-type="sec"}, if we take $\documentclass[12pt]{minimal}
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                \begin{document}$\omega=0$\end{document}$, then we get the known inequality of Işan et al. \[[@CR7]\] $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} &f \biggl(\frac{2ab}{a+b} \biggr)\leq\frac{\Gamma(\nu+1)}{2^{1-\nu }} \biggl[J_{\frac{a+b}{2ab}+}(f \circ h) \biggl(\frac{1}{a} \biggr)+J_{\frac {a+b}{2ab}-}(f\circ h) \biggl( \frac{1}{b} \biggr) \biggr] \leq\frac {f(a)+f(b)}{2}. \end{aligned}$$ \end{document}$$

Theorem 3.4 {#FPar13}
-----------

*Let* $\documentclass[12pt]{minimal}
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                \begin{document}$f:I\subset(0, +\infty)\to\mathbb{R}$\end{document}$ *be a harmonically convex functsuch that* $\documentclass[12pt]{minimal}
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                \begin{document}$f\in L[a,b]$\end{document}$. *If* $\documentclass[12pt]{minimal}
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                \begin{document}$g:[a,b]\to\mathbb{R}$\end{document}$ *is nonnegative*, *integrable*, *and harmonically symmetric with respect to* $\documentclass[12pt]{minimal}
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                \begin{document}$\frac {2ab}{a+b}$\end{document}$, *then the following inequality holds*: $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} &f \biggl(\frac{2ab}{a+b} \biggr) \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu , l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad \leq\frac{1}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime }, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad \leq\frac{f(a)+f(b)}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr], \end{aligned}$$ \end{document}$$ *where* $\documentclass[12pt]{minimal}
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                \begin{document}$h(x)=\frac{1}{x}$\end{document}$, $\documentclass[12pt]{minimal}
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                \begin{document}$x\in [\frac{1}{b}, \frac{1}{a} ] $\end{document}$, *and* $\documentclass[12pt]{minimal}
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                \begin{document}$\omega^{\prime}= ( \frac{2ab}{b-a} ) ^{\mu}\omega$\end{document}$.

Proof {#FPar14}
-----

Since *f* ia a harmonically convex function on $\documentclass[12pt]{minimal}
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                \begin{document}$[a,b]$\end{document}$, multiplying both sides of ([3.9](#Equ13){ref-type=""}) by $\documentclass[12pt]{minimal}
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                \begin{document}$2t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} ( \omega t^{\mu} )g (\frac {ab}{\frac{t}{2}a+\frac{2-t}{2}b} )$\end{document}$ and then integrating the resulting inequality over $\documentclass[12pt]{minimal}
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                \begin{document}$[0,1]$\end{document}$, we obtain $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} &2f \biggl(\frac{2ab}{a+b} \biggr) \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)g \biggl(\frac {ab}{\frac{t}{2}a+\frac{2-t}{2}b} \biggr) \,\mathrm{d}t \\ &\quad \leq \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)f \biggl(\frac{ab}{\frac{t}{2}a+\frac {2-t}{2}b} \biggr)g \biggl( \frac{ab}{\frac{t}{2}a+\frac{2-t}{2}b} \biggr)\,\mathrm{d}t \\ &\qquad{}+ \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)f \biggl(\frac{ab}{\frac{t}{2}b+\frac {2-t}{2}a} \biggr)g \biggl( \frac{ab}{\frac{t}{2}a+\frac{2-t}{2}b} \biggr)\,\mathrm{d}t . \end{aligned}$$ \end{document}$$ Since *g* is harmonically symmetric with respect to $\documentclass[12pt]{minimal}
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                \begin{document}$\frac{2ab}{a+b}$\end{document}$, using Definition [2.2](#FPar2){ref-type="sec"}, we have $\documentclass[12pt]{minimal}
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                \begin{document}$g ( \frac{1}{x} )=g (\frac{1}{\frac{1}{a}+\frac{1}{b}-x} ) $\end{document}$ for all $\documentclass[12pt]{minimal}
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                \begin{document}$x\in [\frac{1}{b}, \frac{1}{a} ] $\end{document}$. Setting $\documentclass[12pt]{minimal}
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                \begin{document}$x=\frac{1}{ab} (\frac{t}{2}a+\frac{2-t}{2}b ) $\end{document}$ gives $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} &2f \biggl(\frac{2ab}{a+b} \biggr) \int_{\frac{a+b}{2ab}}^{\frac {1}{a}} \biggl(\frac{1}{a}-x \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-x \biggr)^{\mu} \biggr)g \biggl( \frac{1}{x} \biggr)\,\mathrm{d}x \\ &\quad \leq\frac{1}{2} \biggl[ \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac {1}{a}-x \biggr)^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega ^{\prime} \biggl(\frac{1}{a}-x \biggr)^{\mu} \biggr)f \biggl( \frac{1}{x} \biggr)g \biggl(\frac{1}{x} \biggr)\,\mathrm{d}x \\ &\qquad{}+ \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac {1}{a}-x \biggr)^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega ^{\prime} \biggl(\frac{1}{a}-x \biggr)^{\mu} \biggr)f \biggl( \frac{1}{\frac {1}{a}+\frac{1}{b}-x} \biggr)g \biggl(\frac{1}{x} \biggr)\,\mathrm {d}x \biggr] . \end{aligned}$$ \end{document}$$ Using substitution $\documentclass[12pt]{minimal}
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                \begin{document}$u=\frac{1}{a}+\frac{1}{b}-x$\end{document}$, we have $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} &2f \biggl(\frac{2ab}{a+b} \biggr) \int_{\frac{a+b}{2ab}}^{\frac {1}{a}} \biggl(\frac{1}{a}-x \biggr) ^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-x \biggr)^{\mu} \biggr)g \biggl( \frac{1}{x} \biggr)\,\mathrm{d}x \\ &\quad \leq\frac{1}{2} \biggl[ \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac {1}{a}-x \biggr)^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega ^{\prime} \biggl(\frac{1}{a}-x \biggr)^{\mu} \biggr)f \biggl( \frac{1}{x} \biggr)g \biggl(\frac{1}{x} \biggr)\,\mathrm{d}x \\ &\qquad{}+ \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(u-\frac {1}{b} \biggr)^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega ^{\prime} \biggl(u-\frac{1}{b} \biggr)^{\mu} \biggr)f \biggl( \frac{1}{u} \biggr)g \biggl(\frac{1}{u} \biggr)\,\mathrm{d}x \biggr]. \end{aligned}$$ \end{document}$$ Hence, using Lemma [3.1](#FPar5){ref-type="sec"}, we obtain $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} &f \biggl(\frac{2ab}{a+b} \biggr) \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad \leq\frac{1}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime }, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr]. \end{aligned}$$ \end{document}$$ For the proof of the second inequality in ([3.14](#Equ18){ref-type=""}), we first note that *f* is a harmonically convex function. Then, multiplying both sides of ([3.12](#Equ16){ref-type=""}) by $\documentclass[12pt]{minimal}
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                \begin{document}$2t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} ( \omega t^{\mu} )g (\frac{ab}{\frac{t}{2}a+\frac {2-t}{2}b} )$\end{document}$ and integrating the resulting inequality over $\documentclass[12pt]{minimal}
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                \begin{document}$[0,1]$\end{document}$, we obtain $$\documentclass[12pt]{minimal}
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                \begin{document}$$\begin{aligned} & \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)f \biggl(\frac{ab}{\frac{t}{2}a+\frac{2-t}{2}b} \biggr)g \biggl( \frac{ab}{\frac{t}{2}a+\frac{2-t}{2}b} \biggr)\,\mathrm{d}t \\ &\qquad{}+ \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma, \delta, k} \bigl( \omega t^{\mu} \bigr)f \biggl(\frac{ab}{\frac{t}{2}b+\frac{2-t}{2}a} \biggr)g \biggl( \frac{ab}{\frac{t}{2}a+\frac{2-t}{2}b} \biggr)\,\mathrm{d}t \\ &\quad \leq \bigl(f(a)+f(b) \bigr) \int_{0}^{1}t^{\nu-1}E_{\mu, \nu, l}^{\gamma , \delta, k} \bigl( \omega t^{\mu} \bigr)g \biggl(\frac{ab}{\frac {t}{2}a+\frac{2-t}{2}b} \biggr) \,\mathrm{d}t. \end{aligned} $$\end{document}$$ From this, using Lemma [3.1](#FPar5){ref-type="sec"}, we get $$\documentclass[12pt]{minimal}
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                \begin{document}$$ \begin{aligned} &\frac{1}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad \leq\frac{f(a)+f(b)}{2} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr]. \end{aligned} $$\end{document}$$ From ([3.15](#Equ19){ref-type=""}) and ([3.16](#Equ20){ref-type=""}) we obtain ([3.14](#Equ18){ref-type=""}). The proof is completed. □

Theorem 3.5 {#FPar15}
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*Let* $\documentclass[12pt]{minimal}
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                \begin{document}$f:I\subset(0, +\infty)\to\mathbb{R}$\end{document}$ *be a differentiable function such that* $\documentclass[12pt]{minimal}
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                \begin{document}$f^{\prime}\in L[a,b]$\end{document}$, *where* $\documentclass[12pt]{minimal}
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                \begin{document}$a,b\in I$\end{document}$, $\documentclass[12pt]{minimal}
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                \begin{document}$a< b$\end{document}$. *If* $\documentclass[12pt]{minimal}
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                \begin{document}$\vert f^{\prime} \vert $\end{document}$ *is a harmonically convex function and* $\documentclass[12pt]{minimal}
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                \begin{document}$g:[a,b]\to\mathbb{R}$\end{document}$ *is continuous and harmonically symmetric with respect to* $\documentclass[12pt]{minimal}
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                \begin{document}$\frac{2ab}{a+b}$\end{document}$, *then the following inequality holds*: $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} & \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta , k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\qquad{}- \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime }, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \biggr\vert \\ &\quad \leq \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}1 \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}1 \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\qquad{}+S \Vert g \Vert _{\infty} \biggl( \frac{b-a}{2ab} \biggr) ^{\nu}\cdot\frac{\nu+4}{4\nu (\nu+1 ) } \bigl( \bigl\vert f(a) \bigr\vert + \bigl\vert f(b) \bigr\vert \bigr), \end{aligned}$$ \end{document}$$ *where* $\documentclass[12pt]{minimal}
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                \begin{document}$h(x)=\frac{1}{x}$\end{document}$, $\documentclass[12pt]{minimal}
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                \begin{document}$x\in [\frac{1}{b}, \frac{1}{a} ] $\end{document}$, $\documentclass[12pt]{minimal}
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                \begin{document}$\omega^{\prime}= ( \frac{2ab}{b-a} ) ^{\mu}\omega$\end{document}$, *and* $\documentclass[12pt]{minimal}
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                \begin{document}$\vert E_{\mu, \nu, l}^{\gamma, \delta, k} ( t ) \vert \leq S$\end{document}$.

Proof {#FPar16}
-----

From Lemma [3.2](#FPar7){ref-type="sec"}, relationships ([3.3](#Equ7){ref-type=""}) and ([3.5](#Equ9){ref-type=""}), and the property of the modulus we have $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} J&= \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta , k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad{}- \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime }, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \biggr\vert \\ &\leq \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \int_{\frac {1}{b}}^{\frac{2ab}{a+b}} \biggl(s-\frac{1}{b} \biggr) ^{\nu-1} \biggl\vert E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(s- \frac {1}{b} \biggr) ^{\mu} \biggr) \biggr\vert \bigl\vert (g\circ h ) (s) \bigr\vert \,\mathrm{d}s \\ &\quad{}+ \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(t-\frac{1}{b} \biggr) ^{\nu -1} \biggl\vert E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime } \biggl(t- \frac{1}{b} \biggr) ^{\mu} \biggr) \biggr\vert \bigl\vert (fg\circ h ) (t) \bigr\vert \,\mathrm{d}t \\ &\begin{aligned} &\quad{}+ \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \int_{\frac {a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-s \biggr) ^{\nu-1} \biggl\vert E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-s \biggr) ^{\mu} \biggr) \biggr\vert \bigl\vert (g\circ h ) (s) \bigr\vert \,\mathrm{d}s \\ &\quad{}+ \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-t \biggr) ^{\nu -1} \biggl\vert E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime } \biggl(\frac{1}{a}-t \biggr) ^{\mu} \biggr) \biggr\vert \bigl\vert (fg\circ h ) (t) \bigr\vert \,\mathrm{d}t. \end{aligned} \end{aligned}$$ \end{document}$$ Since $\documentclass[12pt]{minimal}
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                \begin{document}$\Vert g \Vert _{\infty}=\sup_{t\in[a,b]} \vert g(t) \vert $\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$\vert E_{\mu, \nu, l}^{\gamma, \delta, k} ( t ) \vert \leq S$\end{document}$, we have $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} J&\leq \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \int_{\frac{1}{b}}^{\frac{2ab}{a+b}} \biggl(s-\frac {1}{b} \biggr) ^{\nu-1} E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega ^{\prime} \biggl(s- \frac{1}{b} \biggr) ^{\mu} \biggr)\,\mathrm{d}s \\ &\quad{}+S \Vert g \Vert _{\infty} \int_{\frac{1}{b}}^{\frac {a+b}{2ab}} \biggl(t-\frac{1}{b} \biggr) ^{\nu-1} \bigl\vert (f\circ h ) (t) \bigr\vert \,\mathrm{d}t \\ &\quad{}+ \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac {1}{a}-s \biggr) ^{\nu-1} E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-s \biggr) ^{\mu} \biggr)\,\mathrm{d}s \\ &\quad{}+S \Vert g \Vert _{\infty} \int_{\frac{a+b}{2ab}}^{\frac {1}{a}} \biggl(\frac{1}{a}-t \biggr) ^{\nu-1} \bigl\vert (f\circ h ) (t) \bigr\vert \,\mathrm{d}t \\ &= \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad{}+S \Vert g \Vert _{\infty} (J_{1}+J_{2} ). \end{aligned}$$ \end{document}$$ Setting $\documentclass[12pt]{minimal}
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                \begin{document}$t=\frac{1}{ab} (\frac{u}{2}a+\frac{2-u}{2}b ) $\end{document}$ and using the harmonic convexity of $\documentclass[12pt]{minimal}
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                \begin{document}$\vert f \vert $\end{document}$, we have $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} J_{1}&= \int_{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(t-\frac{1}{b} \biggr) ^{\nu-1} \bigl\vert (f\circ h ) (t) \bigr\vert \,\mathrm{d}t= \int _{\frac{1}{b}}^{\frac{a+b}{2ab}} \biggl(t-\frac{1}{b} \biggr) ^{\nu -1} \biggl\vert f \biggl(\frac{1}{t} \biggr) \biggr\vert \,\mathrm{d}t \\ &=\frac{1}{2} \biggl( \frac{b-a}{ab} \biggr) ^{\nu} \int_{1}^{2} \biggl(1-\frac{1}{2}u \biggr) ^{\nu-1} \biggl\vert f \biggl(\frac{ab}{\frac {u}{2}a+\frac{2-u}{2}b} \biggr) \biggr\vert \,\mathrm{d}u \\ &\leq\frac{1}{2} \biggl( \frac{b-a}{ab} \biggr) ^{\nu} \int_{1}^{2} \biggl(1-\frac{1}{2}u \biggr) ^{\nu-1} \biggl( \frac{u}{2} \bigl\vert f(b) \bigr\vert + \frac{2-u}{2} \bigl\vert f(a) \bigr\vert \biggr)\,\mathrm{d}u. \end{aligned}$$ \end{document}$$ This implies $$\documentclass[12pt]{minimal}
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                \begin{document}$$ J_{1}\leq\frac{1}{4} \biggl( \frac{b-a}{ab} \biggr) ^{\nu} \biggl(\frac {2\nu+4}{\nu(\nu+1)} \bigl\vert f(b) \bigr\vert -\frac{1}{\nu+1} \bigl\vert f(a) \bigr\vert \biggr). $$\end{document}$$ Similarly, using the substitution $\documentclass[12pt]{minimal}
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                \begin{document}$t=\frac{1}{ab} (\frac {v}{2}b+\frac{2-v}{2}a ) $\end{document}$, we get $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} J_{2}&= \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-t \biggr) ^{\nu-1} \bigl\vert (f\circ h ) (t) \bigr\vert \,\mathrm{d}t= \int _{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac{1}{a}-t \biggr) ^{\nu -1} \biggl\vert f \biggl(\frac{1}{t} \biggr) \biggr\vert \,\mathrm{d}t \\ &\leq\frac{1}{4} \biggl( \frac{b-a}{ab} \biggr) ^{\nu} \biggl(\frac{2\nu +4}{\nu(\nu+1)} \bigl\vert f(a) \bigr\vert -\frac{1}{\nu+1} \bigl\vert f(b) \bigr\vert \biggr). \end{aligned}$$ \end{document}$$ Substituting ([3.19](#Equ23){ref-type=""}) and ([3.20](#Equ24){ref-type=""}) into ([3.18](#Equ22){ref-type=""}), we obtain ([3.17](#Equ21){ref-type=""}). The proof is completed. □

Theorem 3.6 {#FPar17}
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*Let* $\documentclass[12pt]{minimal}
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                \begin{document}$f:I\subset(0, +\infty)\to\mathbb{R}$\end{document}$ *be a differentiable function on* $\documentclass[12pt]{minimal}
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                \begin{document}$I^{\circ}$\end{document}$, *the interior of* *I*, *such that* $\documentclass[12pt]{minimal}
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                \begin{document}$f^{\vert}\in L[a,b]$\end{document}$, *where* $\documentclass[12pt]{minimal}
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                \begin{document}$a,b\in I$\end{document}$. *If* $\documentclass[12pt]{minimal}
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                \begin{document}$\vert f \vert ^{q}$\end{document}$, $\documentclass[12pt]{minimal}
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                \begin{document}$q\geq1$\end{document}$, *is a harmonically convex function on* $\documentclass[12pt]{minimal}
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                \begin{document}$[a,b]$\end{document}$, $\documentclass[12pt]{minimal}
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                \begin{document}$g:[a,b]\to\mathbb{R}$\end{document}$ *is continuous and harmonically symmetric with respect to* $\documentclass[12pt]{minimal}
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                \begin{document}$\frac {2ab}{a+b}$\end{document}$, *then the following inequality holds*: $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} & \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta , k}g\circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\qquad{}- \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime }, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}fg\circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}fg \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \biggr\vert \\ &\quad \leq \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}1 \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}1 \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\qquad{}+S \Vert g \Vert _{\infty} \biggl( \frac{1}{\nu} \biggr) ^{1-1/q} \biggl( \frac{1}{\nu+1 } \biggr) ^{1/q} \biggl( \frac{1}{2} \biggr) ^{\nu+1/q} \\ &\qquad{}\times \biggl[ \biggl( \frac{1}{\nu} \bigl\vert f(b) \bigr\vert ^{q} + \bigl\vert f(a) \bigr\vert ^{q} \biggr)^{1/q}+ \biggl(\frac{1}{\nu} \bigl\vert f(a) \bigr\vert ^{q} + \bigl\vert f(b) \bigr\vert ^{q} \biggr)^{1/q} \biggr] , \end{aligned}$$ \end{document}$$ *where* $\documentclass[12pt]{minimal}
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                \begin{document}$h(x)=\frac{1}{x}$\end{document}$, $\documentclass[12pt]{minimal}
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                \begin{document}$x\in [\frac{1}{b}, \frac{1}{a} ] $\end{document}$, $\documentclass[12pt]{minimal}
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                \begin{document}$\omega^{\prime}= ( \frac{2ab}{b-a} ) ^{\mu}\omega$\end{document}$, *and* $\documentclass[12pt]{minimal}
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                \begin{document}$\vert E_{\mu, \nu, l}^{\gamma, \delta, k} ( t ) \vert \leq S$\end{document}$.

Proof {#FPar18}
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From ([3.18](#Equ22){ref-type=""}) we have $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} J&\leq \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \int_{\frac{1}{b}}^{\frac{2ab}{a+b}} \biggl(s-\frac {1}{b} \biggr) ^{\nu-1} E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega ^{\prime} \biggl(s- \frac{1}{b} \biggr) ^{\mu} \biggr)\,\mathrm{d}s \\ &\quad{}+S \Vert g \Vert _{\infty} \int_{\frac{1}{b}}^{\frac {a+b}{2ab}} \biggl(t-\frac{1}{b} \biggr) ^{\nu-1} \bigl\vert (f\circ h ) (t) \bigr\vert \,\mathrm{d}t \\ &\quad{}+ \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \int_{\frac{a+b}{2ab}}^{\frac{1}{a}} \biggl(\frac {1}{a}-s \biggr) ^{\nu-1} E_{\mu, \nu, l}^{\gamma, \delta, k} \biggl( \omega^{\prime} \biggl(\frac{1}{a}-s \biggr) ^{\mu} \biggr)\,\mathrm{d}s \\ &\quad{}+S \Vert g \Vert _{\infty} \int_{\frac{a+b}{2ab}}^{\frac {1}{a}} \biggl(\frac{1}{a}-t \biggr) ^{\nu-1} \bigl\vert (f\circ h ) (t) \bigr\vert \,\mathrm{d}t \\ &= \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad{}+S \Vert g \Vert _{\infty} \biggl[ \int_{\frac{1}{b}}^{\frac {a+b}{2ab}} \biggl(t-\frac{1}{b} \biggr) ^{\nu-1} \biggl\vert f \biggl(\frac {1}{t} \biggr) \biggr\vert \,\mathrm{d}t+ \int_{\frac{a+b}{2ab}}^{\frac {1}{a}} \biggl(\frac{1}{a}-t \biggr) ^{\nu-1} \biggl\vert f \biggl(\frac {1}{t} \biggr) \biggr\vert \,\mathrm{d}t \biggr] . \end{aligned}$$ \end{document}$$ Using the substitutions $\documentclass[12pt]{minimal}
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                \begin{document}$t=\frac{1}{ab} (\frac{u}{2}a+\frac {2-u}{2}b )$\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$t=\frac{1}{ab} (\frac{v}{2}b+\frac {2-v}{2}a ) $\end{document}$, the power mean inequality, and the harmonicity of $\documentclass[12pt]{minimal}
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                \begin{document}$\vert f \vert ^{q}$\end{document}$, it follows that $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} J&\leq \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad{}+\frac{S \Vert g \Vert _{\infty}}{2} \biggl(\frac {b-a}{ab} \biggr) ^{\nu} \\ &\quad{}\times \biggl[ \int_{1}^{2} \biggl(1-\frac{1}{2}u \biggr) ^{\nu -1} \biggl\vert f \biggl(\frac{ab}{\frac{u}{2}a+\frac{2-u}{2}b } \biggr) \biggr\vert \,\mathrm{d}u+ \int_{1}^{2} \biggl(1-\frac{1}{2}v \biggr) ^{\nu -1} \biggl\vert f \biggl(\frac{ab}{\frac{v}{2}b+\frac{2-v}{2}a} \biggr) \biggr\vert \,\mathrm{d}v \biggr] \\ &\leq \biggl\vert f \biggl(\frac{2ab}{a+b} \biggr) \biggr\vert \Vert g \Vert _{\infty} \biggl[ \bigl( \varepsilon_{\mu, \nu, l, \omega ^{\prime}, (\frac{a+b}{2ab} )^{+}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{a} \biggr)+ \bigl( \varepsilon_{\mu, \nu, l, \omega^{\prime}, (\frac{a+b}{2ab} )^{-}}^{\gamma, \delta, k}g \circ h \bigr) \biggl( \frac{1}{b} \biggr) \biggr] \\ &\quad{}+\frac{S \Vert g \Vert _{\infty}}{2} \biggl(\frac {b-a}{ab} \biggr) ^{\nu} \biggl[ \biggl( \int_{1}^{2} \biggl(1-\frac{1}{2}u \biggr) ^{\nu -1}\,\mathrm{d}u \biggr) ^{1-1/q} \\ &\quad{}\times \biggl( \int_{1}^{2} \biggl(1-\frac{1}{2}u \biggr) ^{\nu-1} \biggl( \frac{u}{2} \bigl\vert f(b) \bigr\vert ^{q}+\frac {2-u}{2} \bigl\vert f(a) \bigr\vert ^{q} \biggr)\,\mathrm{d}u \biggr) ^{1/q} \\ &\quad{}+ \biggl( \int_{1}^{2} \biggl(1-\frac{1}{2}v \biggr) ^{\nu -1}\,\mathrm{d}v \biggr) ^{1-1/q} \biggl( \int_{1}^{2} \biggl(1-\frac{1}{2}v \biggr) ^{\nu-1} \biggl( \frac{v}{2} \bigl\vert f(a) \bigr\vert ^{q}+\frac {2-v}{2} \bigl\vert f(b) \bigr\vert ^{q} \biggr)\,\mathrm{d}v \biggr) ^{1/q} \biggr]. \end{aligned}$$ \end{document}$$ Making the substitutions $\documentclass[12pt]{minimal}
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                \begin{document}$x=1-\frac{1}{2}u$\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$y=1-\frac{1}{2}v $\end{document}$ and simple calculations, we get inequality ([3.21](#Equ25){ref-type=""}). □

Conclusions {#Sec4}
===========

We have obtained several new refinements of Hermite-Hadamard-type inequalities via harmonic convex functions. These results involve particularly the generalized Mittag-Leffler function. We also discussed several particular cases. We expect that the results obtained in this paper may stimulate further research in the field. We would like to specify here that the generalized fractional integral operators containing the Mittag-Leffler function generalize the integrals of Riemann-Liouville type, and the results obtained in this paper can be developed via other different types of convexities.

**Publisher's Note**

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The authors are thankful to the editor and anonymous referees for their helpful comments and suggestions.

MVM, MUA, MAN, and KIN worked jointly. All the authors read and approved the final manuscript.

Competing interests {#FPar19}
===================

The authors declare that they have no competing interests.
