General fractional integral inequalities for convex and m-convex functions via an extended generalized Mittag-Leffler function

Abstract

In this paper some new general fractional integral inequalities for convex and m-convex functions by involving an extended Mittag-Leffler function are presented. These results produce inequalities for several kinds of fractional integral operators. Some interesting special cases of our main results are also pointed out.

Introduction, definitions, and preliminaries

Convex functions are very important in the field of integral inequalities. A lot of fractional integral inequalities and novel results have been established due to convex functions (for more details, see [1, 8, 13, 14]).

Definition 1

A function $f:I\to \mathbb{R}$, where I is an interval in $\mathbb{R}$, is said to be a convex function if

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

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

A convex function $f:I\to \mathbb{R}$ is also equivalently defined by the Hadamard inequality

$$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f(x)dx\le \frac{f(a)+f(b)}{2},$$

where $a,b\in I$, $a<b$.

The concept of m-convexity was introduced in [17] and since then many properties, especially inequalities, have been obtained for this class of functions (see [3, 6, 7, 18]).

Definition 2

A function $f:[0,b]\to \mathbb{R}$, $b>0$ is called m-convex, where $m\in [0,1]$, if for every $x,y\in [0,b]$ and $t\in [0,1]$, we have

$$f(tx+m(1-t)y)\le tf(x)+m(1-t)f(y).$$

For $m=1$, we recapture the definition of convex functions, and for $m=0$, the definition of star-shaped functions defined on $[0,b]$. We recall that a function $f:[0,b]\to \mathbb{R}$ is called star-shaped if

$$f(tx)\le tf(x)\text{for all }t\in [0,1]\text{ and }x\in [0,b].$$

If we denote by $K_m(b)$ the set of m-convex functions defined on $[0,b]$ for which $f(0)<0$, then

$$K_1(b)\subset K_m(b)\subset K_0(b),$$

whenever $m\in (0,1)$. Note that in the class $K_1(b)$ there are only convex functions $f:[0,b]\to \mathbb{R}$ for which $f(0)\le 0$ (see [4]), while $k_0(b)$ contains star-shaped functions.

Example 1.1

([6])

The function $f:[0,\mathrm{\infty })\to \mathbb{R}$, given by

$$f(x)=\frac{1}{12}(x^4-5x^3+9x^2-5x),$$

is a $\frac{16}{17}$-convex function but it is not m-convex for any $m\in (\frac{16}{17},1]$.

For more results and inequalities related to m-convex functions, one can consult, for example, [3, 6, 7] along with the references therein.

Recently in [2] Andrić et al. defined an extended generalized Mittag-Leffler function $E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\cdot ;p)$ as follows.

Definition 3

([2])

Let $\mu ,\alpha ,l,\gamma ,c\in \mathbb{C}$, $\mathrm{\Re }(\mu ),\mathrm{\Re }(\alpha ),\mathrm{\Re }(l)>0$, $\mathrm{\Re }(c)>\mathrm{\Re }(\gamma )>0$ with $p\ge 0$, $\delta >0$, and $0<k\le \delta +\mathrm{\Re }(\mu )$. Then the extended generalized Mittag-Leffler function $E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(t;p)$ is defined by

$$E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(t;p)=\sum _{n=0}^{\mathrm{\infty }}\frac{{\beta }_p(\gamma +nk,c-\gamma )}{\beta (\gamma ,c-\gamma )}\frac{{(c)}_{nk}}{\mathrm{\Gamma }(\mu n+\alpha )}\frac{t^n}{{(l)}_{n\delta }},$$ 2

where ${\beta }_p$ is the generalized beta function defined by

$${\beta }_p(x,y)={\int }_0^1{t}^{x-1}{(1-t)}^{y-1}{e}^{-\frac{p}{t(1-t)}}dt$$

and ${(c)}_{nk}$ is the Pochhammer symbol defined as ${(c)}_{nk}=\frac{\mathrm{\Gamma }(c+nk)}{\mathrm{\Gamma }(c)}$.

In [2] properties of the generalized Mittag-Leffler function are discussed, and it is given that $E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(t;p)$ is absolutely convergent for $k<\delta +\mathrm{\Re }(\mu )$. Let S be the sum of series of absolute terms of the Mittag-Leffler function $E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(t;p)$, then we have $|E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(t;p)|\le S$. We use this property of Mittag-Leffler function in our results where we need.

The corresponding left and right sided extended generalized fractional integral operators are defined as follows.

Definition 4

([2])

Let $\omega ,\mu ,\alpha ,l,\gamma ,c\in \mathbb{C}$, $\mathrm{\Re }(\mu ),\mathrm{\Re }(\alpha ),\mathrm{\Re }(l)>0$, $\mathrm{\Re }(c)>\mathrm{\Re }(\gamma )>0$ with $p\ge 0$, $\delta >0$ and $0<k\le \delta +\mathrm{\Re }(\mu )$. Let $f\in L_1[a,b]$ and $x\in [a,b]$. Then the extended generalized fractional integral operators ${ϵ}_{\mu ,\alpha ,l,\omega ,a^{+}}^{\gamma ,\delta ,k,c}f$ and ${ϵ}_{\mu ,\alpha ,l,\omega ,b^{-}}^{\gamma ,\delta ,k,c}f$ are defined by

$$\left({ϵ}_{\mu ,\alpha ,l,\omega ,a^{+}}^{\gamma ,\delta ,k,c}f\right)(x;p)={\int }_a^x{(x-t)}^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega {(x-t)}^{\mu };p)f(t)dt$$ 3

and

$$\left({ϵ}_{\mu ,\alpha ,l,\omega ,b^{-}}^{\gamma ,\delta ,k,c}f\right)(x;p)={\int }_x^b{(t-x)}^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega {(t-x)}^{\mu };p)f(t)dt.$$ 4

From extended generalized fractional integral operators, we have

$$\begin{array}{rl}& ({ϵ}_{\mu ,\alpha ,l,\omega ,a^{+}}^{\gamma ,\delta ,k,c}1)(x;p)\\ & ={\int }_a^x{(x-t)}^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(w{(x-t)}^{\mu };p)dt\\ & ={\int }_a^x{(x-t)}^{\alpha -1}\sum _{n=0}^{\mathrm{\infty }}\frac{{\mathrm{B}}_p(\gamma +nk,c-\gamma )}{\mathrm{B}(\gamma ,c-\gamma )}\frac{{(c)}_{nk}}{\mathrm{\Gamma }(\mu n+\alpha )}\frac{w^n{(x-t)}^{\mu n}}{{(l)}_{n\delta }}dt\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{{\mathrm{B}}_p(\gamma +nk,c-\gamma )}{\mathrm{B}(\gamma ,c-\gamma )}\frac{{(c)}_{nk}}{\mathrm{\Gamma }(\mu n+\alpha )}\frac{w^n}{{(l)}_{n\delta }}{\int }_a^x{(x-t)}^{\mu n+\alpha -1}dt\\ & ={(x-a)}^{\alpha }\sum _{n=0}^{\mathrm{\infty }}\frac{{\mathrm{B}}_p(\gamma +nk,c-\gamma )}{\mathrm{B}(\gamma ,c-\gamma )}\frac{{(c)}_{nk}}{\mathrm{\Gamma }(\mu n+\alpha )}\frac{w^n}{{(l)}_{n\delta }}{(x-a)}^{\mu n}\frac{1}{\mu n+\alpha }.\end{array}$$

Hence

$$\left({ϵ}_{\mu ,\alpha ,l,\omega ,a^{+}}^{\gamma ,\delta ,k,c}1\right)(x;p)={(x-a)}^{\alpha }{E}_{\mu ,\alpha +1,l}^{\gamma ,\delta ,k,c}(w{(x-a)}^{\mu };p),$$

and similarly

$$\left({ϵ}_{\mu ,\alpha ,l,\omega ,b^{-}}^{\gamma ,\delta ,k,c}1\right)(x;p)={(b-x)}^{\alpha }{E}_{\mu ,\alpha +1,l}^{\gamma ,\delta ,k,c}(w{(b-x)}^{\mu };p).$$

We use the following notations in our results:

$$C_{\alpha ,a^{+}}(x;p)=\left({ϵ}_{\mu ,\alpha ,l,\omega ,a^{+}}^{\gamma ,\delta ,k,c}1\right)(x;p)$$ 5

and

$$C_{\alpha ,b^{-}}(x;p)=\left({ϵ}_{\mu ,\alpha ,l,\omega ,b^{-}}^{\gamma ,\delta ,k,c}1\right)(x;p).$$ 6

For more information related to Mittag-Leffler functions and corresponding fractional integral operators, the readers are referred to [9–12, 15, 16, 19].

In this paper we give general fractional integral inequalities for convex and m-convex functions by involving an extended Mittag-Leffler function and deduce some results already published in [1, 5, 6, 8, 13]. Also we give a Hadamard type inequality for convex and m-convex functions by involving an extended Mittag-Leffler function.

Main results

Here we give some fractional integral inequalities for convex and m-convex functions via an extended generalized Mittag-Leffler function and corresponding fractional integral operators given in (3) and (4). The following lemma is useful to establish the results.

Lemma 2.1

Let $f:[a,mb]\to \mathbb{R}$ be a differentiable function such that $f^{\prime }\in L_1[a,mb]$ with $0\le a<mb$. Also let $g:[a,mb]\to \mathbb{R}$ be a continuous function on $[a,mb]$, then the following identity for extended generalized fractional integral operators holds:

$$\begin{array}{rl}& {({\int }_a^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }[f(a)+f(mb)]\\ & -\alpha {\int }_a^{mb}{({\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt\\ & -\alpha {\int }_a^{mb}{({\int }_t^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt\\ & ={\int }_a^{mb}{({\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }{f}^{\prime }(t)dt\\ & -{\int }_a^{mb}{({\int }_t^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }{f}^{\prime }(t)dt.\end{array}$$ 7

Proof

On integrating by parts one can have

$$\begin{array}{rl}& {\int }_a^{mb}{({\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }{f}^{\prime }(t)dt\\ & ={({\int }_a^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }f(mb)\\ & -\alpha {\int }_a^{mb}{({\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt\end{array}$$ 8

and

$$\begin{array}{rl}& {\int }_a^{mb}{({\int }_t^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }{f}^{\prime }(t)dt\\ & =-{({\int }_a^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }f(a)\\ & +\alpha {\int }_a^{mb}{({\int }_t^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt.\end{array}$$ 9

Subtracting (9) from (8), we get (7) which is the required identity. □

If we take $m=1$ in (7), then we get the following identity for a convex function.

Corollary 2.2

Let $f:[a,b]\subseteq [0,\mathrm{\infty })\to \mathbb{R}$ be a differentiable function such that $f^{\prime }\in L_1[a,b]$ with $a<b$. Also let $g:[a,b]\to \mathbb{R}$ be continuous on $[a,b]$, then the following identity for extended generalized fractional integral operators holds:

$$\begin{array}{rl}& {({\int }_a^{b}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }[f(a)+f(b)]\\ & -\alpha {\int }_a^b{({\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt\\ & -\alpha {\int }_a^b{({\int }_t^{b}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt\\ & ={\int }_a^b{({\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }{f}^{\prime }(t)dt\\ & -{\int }_a^b{({\int }_t^{b}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }{f}^{\prime }(t)dt.\end{array}$$ 10

We use identity (7) to establish the following fractional integral inequality.

Theorem 2.3

Let $f:[a,mb]\to \mathbb{R}$ be a differentiable function such that $f^{\prime }\in L_1[a,mb]$ with $0\le a<mb$. Also let $g:[a,mb]\to \mathbb{R}$ be a continuous function on $[a,mb]$. If $|f^{\prime }|$ is an m-convex function on $[a,mb]$, then the following inequality for extended generalized fractional integral operators holds:

$$\begin{array}{rl}& |{({\int }_a^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }(f(a)+f(mb))\\ & -\alpha {\int }_a^{mb}{({\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt\\ & -\alpha {\int }_a^{mb}{({\int }_t^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt|\\ & \le \frac{{(mb-a)}^{\alpha +1}{\parallel g\parallel }_{\mathrm{\infty }}^{\alpha }{S}^{\alpha }}{(\alpha +1)}\left(\right|f^{\prime }(a)|+m|f^{\prime }(b)\left|\right)\end{array}$$ 11

for $k<\delta +\mathrm{\Re }(\mu )$ and ${\parallel g\parallel }_{\mathrm{\infty }}={sup}_{t\in [a,mb]}|g(t)|$.

Proof

From Lemma 2.1, we have

$$\begin{array}{rl}& |{({\int }_a^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }(f(a)+f(mb))\\ & -\alpha {\int }_a^{mb}{({\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt\\ & -\alpha {\int }_a^{mb}{({\int }_t^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt|\\ & \le {\int }_a^{mb}|{\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds{|}^{\alpha }|f^{\prime }(t)|dt\\ & +{\int }_a^{mb}|{\int }_t^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds{|}^{\alpha }|f^{\prime }(t)|dt.\end{array}$$ 12

Using absolute convergence of the Mittag-Leffler function and ${\parallel g\parallel }_{\mathrm{\infty }}={sup}_{t\in [a,b]}|g(t)|$, we have

$$\begin{array}{rl}& |{({\int }_a^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }(f(a)+f(mb))\\ & -\alpha {\int }_a^{mb}{({\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt\\ & -\alpha {\int }_a^{mb}{({\int }_t^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt|\\ & \le {\parallel g\parallel }_{\mathrm{\infty }}^{\alpha }{S}^{\alpha }\left({\int }_a^{mb}{(t-a)}^{\alpha }\right|f^{\prime }(t)|dt+{\int }_a^{mb}{(mb-t)}^{\alpha }|f^{\prime }(t)\left|dt\right).\end{array}$$ 13

Since $|f^{\prime }|$ is an m-convex function, we have

$$|f^{\prime }(t)|\le \frac{mb-t}{mb-a}|f^{\prime }(a)|+m\frac{t-a}{mb-a}|f^{\prime }(b)|$$ 14

for $t\in [a,mb]$.

Using (14) in (13), we have

$$\begin{array}{rl}& |{({\int }_a^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }(f(a)+f(mb))\\ & -\alpha {\int }_a^{mb}{({\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt\\ & -\alpha {\int }_a^{mb}{({\int }_t^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt|\\ & \le {\parallel g\parallel }_{\mathrm{\infty }}^{\alpha }{S}^{\alpha }({\int }_a^{mb}{(t-a)}^{\alpha }\left(\frac{mb-t}{mb-a}\right|f^{\prime }(a)|+m\frac{t-a}{mb-a}|f^{\prime }(b)\left|\right)dt\\ & +{\int }_a^{mb}{(mb-t)}^{\alpha }\left(\frac{mb-t}{mb-a}\right|f^{\prime }(a)|+m\frac{t-a}{mb-a}|f^{\prime }(b)\left|\right)dt).\end{array}$$ 15

After simple calculation of the above inequality, we get (11) which is required. □

If we take $m=1$ in (11), then we get the following result for a convex function.

Corollary 2.4

Let $f:[a,b]\subseteq [0,\mathrm{\infty })\to \mathbb{R}$ be a differentiable function such that $f^{\prime }\in L_1[a,b]$ with $a<b$. Also let $g:[a,b]\to \mathbb{R}$ be a continuous function on $[a,b]$. If $|f^{\prime }|$ is a convex function on $[a,b]$, then the following inequality for extended generalized fractional integral operators holds:

$$\begin{array}{rl}& |{({\int }_a^{b}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }[f(a)+f(b)]\\ & -\alpha {\int }_a^b{({\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt\\ & -\alpha {\int }_a^b{({\int }_t^{b}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt|\\ & \le \frac{{(b-a)}^{\alpha +1}{\parallel g\parallel }_{\mathrm{\infty }}^{\alpha }{S}^{\alpha }}{(\alpha +1)}\left[\right|f^{\prime }(a)|+|f^{\prime }(b)\left|\right]\end{array}$$ 16

for $k<\delta +\mathrm{\Re }(\mu )$ and ${\parallel g\parallel }_{\mathrm{\infty }}={sup}_{t\in [a,b]}|g(t)|$.

Remark 2.5

In Theorem 2.3.

• (i)If we put $p=0$, then we get [6, Theorem 3.2].
• (ii)If we put $\omega =p=0$ and $m=1$, then we get [13, Theorem 6].
• (iii)If we take $\omega =p=0$, $m=1$, $\alpha =\frac{\mu }{k}$, and $g(s)=1$, then we get [8, Corollary 2.3].
• (iv)For $g(s)=1$ along with $\omega =p=0$, $m=1$, and $\alpha =\mu$, we get [13, Corollary 2].

Remark 2.6

In Corollary 2.4.

• (i)If we put $p=0$, then we get [1, Theorem 3.2].
• (ii)If we put $\omega =p=0$, then we get [13, Theorem 6].
• (iii)For $\omega =p=0$, $\alpha =\frac{\mu }{k}$, and $g(s)=1$, we get [8, Corollary 2.3].
• (iv)For $g(s)=1$ along with $\omega =p=0$, we get [13, Corollary 2].

Next we give the following fractional integral inequality.

Theorem 2.7

Let $f:[a,mb]\to \mathbb{R}$ be a differentiable function such that $f\in L_1[a,mb]$ with $0\le a<mb$. Also let $g:[a,mb]\to \mathbb{R}$ be a continuous function on $[a,mb]$. If ${|f^{\prime }|}^q$ is a convex function on $[a,mb]$, then for $q>0$ the following inequality for extended generalized fractional integral operators holds:

$$\begin{array}{rl}& |{({\int }_a^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }(f(a)+f(mb))\\ & -\alpha {\int }_a^{mb}{({\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt\\ & -\alpha {\int }_a^{mb}{({\int }_t^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt|\\ & \le \frac{2{(mb-a)}^{\alpha +1}{\parallel g\parallel }_{\mathrm{\infty }}^{\alpha }{S}^{\alpha }}{{(\alpha p+1)}^{\frac{1}{q}}}{\left(\frac{|f^{\prime }(a){|}^q+m|f^{\prime }(b){|}^q}{2}\right)}^{\frac{1}{q}}\end{array}$$ 17

for $k<\delta +\mathrm{\Re }(\mu )$ and ${\parallel g\parallel }_{\mathrm{\infty }}={sup}_{t\in [a,mb]}|g(t)|$ and $\frac{1}{p}+\frac{1}{q}=1$.

Proof

From Lemma 2.1 and by using Hölder’s inequality, we have

$$\begin{array}{rl}& |{({\int }_a^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }(f(a)+f(mb))\\ & -\alpha {\int }_a^{mb}{({\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt\\ & -\alpha {\int }_a^{mb}{({\int }_t^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt|\\ & \le {\left({\int }_a^{mb}\right|{\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds{|}^{\alpha p}dt)}^{\frac{1}{p}}{\left({\int }_a^{mb}\right|f^{\prime }(t){|}^{q}dt)}^{\frac{1}{q}}\\ & +{\left({\int }_a^{mb}\right|{\int }_t^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds{|}^{\alpha p}dt)}^{\frac{1}{p}}{\left({\int }_a^{mb}\right|f^{\prime }(t){|}^{q}dt)}^{\frac{1}{q}}.\end{array}$$ 18

Using absolute convergence of the Mittag-Leffler function and ${\parallel g\parallel }_{\mathrm{\infty }}={sup}_{t\in [a,b]}|g(t)|$, we have

$$\begin{array}{rl}& |{({\int }_a^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }(f(a)+f(mb))\\ & -\alpha {\int }_a^{mb}{({\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt\\ & -\alpha {\int }_a^{mb}{({\int }_t^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt|\\ & \le {\parallel g\parallel }_{\mathrm{\infty }}^{\alpha }{S}^{\alpha }({({\int }_a^{mb}|t-a{|}^{\alpha p}dt)}^{\frac{1}{p}}\\ & +{({\int }_a^{mb}|mb-t{|}^{\alpha p}dt)}^{\frac{1}{p}}){\left({\int }_a^{mb}\right|f^{\prime }(t){|}^{q}dt)}^{\frac{1}{q}}.\end{array}$$ 19

Since ${|f^{\prime }(t)|}^q$ is an m-convex function, we have

$$|f^{\prime }(t){|}^q\le \frac{mb-t}{mb-a}|f^{\prime }(a){|}^q+m\frac{t-a}{mb-a}|f^{\prime }(b){|}^q.$$ 20

Using (20) in (19), we have

$$\begin{array}{rl}& |{({\int }_a^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }(f(a)+f(mb))\\ & -\alpha {\int }_a^{mb}{({\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt\\ & -\alpha {\int }_a^{mb}{({\int }_t^{mb}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt|\\ & \le {\parallel g\parallel }_{\mathrm{\infty }}^{\alpha }{S}^{\alpha }({({\int }_a^{mb}|t-a{|}^{\alpha p}dt)}^{\frac{1}{p}}+{({\int }_a^{mb}|mb-t{|}^{\alpha p}dt)}^{\frac{1}{p}})\\ & \times {\left({\int }_a^{mb}\frac{mb-t}{mb-a}\right|f^{\prime }(a){|}^q+m\frac{t-a}{mb-a}|f^{\prime }(b){|}^q)}^{\frac{1}{q}}.\end{array}$$ 21

After simple calculation of the above inequality, we get (17) which is required. □

If we take $m=1$ in (17), then we get the following result for a convex function.

Corollary 2.8

Let $f:[a,b]\subseteq [0,\mathrm{\infty })\to \mathbb{R}$ be a differentiable function such that $f^{\prime }\in L_1[a,b]$ with $a<b$. Also let $g:[a,b]\to \mathbb{R}$ be a continuous function on $[a,b]$. If ${|f^{\prime }|}^q$ is a convex function on $[a,b]$, then for $q>0$ the following inequality for extended generalized fractional integral operators holds:

$$\begin{array}{rl}& |{({\int }_a^{b}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha }[f(a)+f(b)]\\ & -\alpha {\int }_a^b{({\int }_a^{t}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt\\ & -\alpha {\int }_a^b{({\int }_t^{b}g(s)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega s^{\mu };p)ds)}^{\alpha -1}g(t)E_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(t)dt|\\ & \le \frac{2{(b-a)}^{\alpha +1}{\parallel g\parallel }_{\mathrm{\infty }}^{\alpha }{S}^{\alpha }}{{(\alpha p+1)}^{\frac{1}{q}}}{\left[\frac{|f^{\prime }(a){|}^q+|f^{\prime }(b){|}^q}{2}\right]}^{\frac{1}{q}}\end{array}$$ 22

for $k<\delta +\mathrm{\Re }(\mu )$ and ${\parallel g\parallel }_{\mathrm{\infty }}={sup}_{t\in [a,b]}|g(t)|$ and $\frac{1}{p}+\frac{1}{q}=1$.

Remark 2.9

In Theorem 2.7.

• (i)If we put $p=0$, then we get [6, Theorem 3.6].
• (ii)If we put $\omega =p=0$ and $m=1$, then we get [13, Theorem 7].
• (iii)If we take $\omega =p=0$, $m=1$ along with $\alpha =\frac{\mu }{k}$, then we get [8, Theorem 2.5].
• (iv)If we take $g(s)=1$, $m=1$, and $\omega =p=0$, then we get [5, Theorem 2.3].
• (v)If we put $\omega =p=0$, $m=1$, and $\alpha =1$, then we get [5, Corollary 3].

Remark 2.10

In Corollary 2.8.

• (i)If we put $p=0$, then we get [1, Theorem 3.5].
• (ii)If we put $\omega =p=0$, then we get [13, Theorem 7].
• (iii)If we put $\omega =p=0$, $\alpha =1$, then we get [13, Corollary 3].
• (iv)If we take $\omega =p=0$ along with $\alpha =\frac{\mu }{k}$, then we get [8, Theorem 2.5].
• (v)If we take $g(s)=1$ and $\omega =p=0$, then we get [5, Theorem 2.3].

In the next result we give Hadamard type inequalities for m-convex functions via an extended Mittag-Leffler function.

Theorem 2.11

Let $f:[a,mb]\to \mathbb{R}$ be a function such that $f\in L_1[a,mb]$ with $0\le a<mb$. If f is m-convex on $[a,mb]$, then the following inequalities for extended generalized fractional integral operators hold:

$$\begin{array}{rl}& 2f\left(\frac{a+mb}{2}\right)C_{\alpha ,{(\frac{a+mb}{2})}^{+}}(mb;p)\\ & \le \left({ϵ}_{\mu ,\alpha ,l,{\omega }^{\prime },{(\frac{a+mb}{2})}^{+}}^{\gamma ,\delta ,k,c}f\right)(mb;p)+m^{\alpha +1}\left({ϵ}_{\mu ,\alpha ,l,m^{\mu }{\omega }^{\prime },{(\frac{a+mb}{2m})}^{-}}^{\gamma ,\delta ,k,c}f\right)(\frac{a}{m};p)\\ & \le \frac{1}{mb-a}(f(a)-m^{2}f\left(\frac{a}{m^2}\right))C_{\alpha +1,{(\frac{a+mb}{2})}^{+}}(mb;p)\\ & +m^{\alpha +1}(f(b)+mf\left(\frac{a}{m^2}\right))C_{\alpha ,{(\frac{a+mb}{2m})}^{-}}(\frac{a}{m};p),\end{array}$$ 23

where ${\omega }^{\prime }=\frac{2^{\mu }\omega }{{(mb-a)}^{\mu }}$.

Proof

Since f is an m-convex function, we have

$$2f\left(\frac{a+mb}{2}\right)\le f(\frac{t}{2}a+\frac{2-t}{2}mb)+mf(\frac{2-t}{2m}a+\frac{t}{2}b).$$ 24

Also from m-convexity of f, we have

$$\begin{array}{rl}& f(\frac{t}{2}a+m\frac{2-t}{2}b)+mf(\frac{2-t}{2m}a+\frac{t}{2}b)\\ & \le \frac{t}{2}(f(a)-m^{2}f\left(\frac{a}{m^2}\right))+m(f(b)+mf\left(\frac{a}{m^2}\right)).\end{array}$$ 25

Multiplying (24) by $t^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)$ on both sides and then integrating over $[0,1]$, we have

$$\begin{array}{rl}& 2f\left(\frac{a+mb}{2}\right){\int }_0^1{t}^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)dt\\ & \le {\int }_0^1{t}^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(\frac{t}{2}a+\frac{2-t}{2}mb)dt\\ & +m{\int }_0^1{t}^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(\frac{2-t}{2m}a+\frac{t}{2}b)dt.\end{array}$$ 26

Putting $u=\frac{t}{2}a+\frac{2-t}{2}mb$ and $v=\frac{2-t}{2m}a+\frac{t}{2}b$ in (26), we have

$$\begin{array}{rl}& 2f\left(\frac{a+mb}{2}\right){\int }_{\frac{a+mb}{2}}^{mb}{(mb-u)}^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}({\omega }^{\prime }{(mb-u)}^{\mu };p)du\\ & \le {\int }_{\frac{a+mb}{2}}^{mb}{(mb-u)}^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}({\omega }^{\prime }{(mb-u)}^{\mu };p)f(u)du\\ & +m^{\alpha +1}{\int }_{\frac{a}{m}}^{\frac{a+mb}{2m}}{(v-\frac{a}{m})}^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(m^{\mu }{\omega }^{\prime }{(v-\frac{a}{m})}^{\mu }:p)f(v)dv.\end{array}$$

By using (3), (4), and (5) we get the first inequality of (23).

Now multiplying (25) by $t^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)$ on both sides and then integrating over $[0,1]$, we have

$$\begin{array}{rl}& {\int }_0^1{t}^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(\frac{t}{2}a+m\frac{2-t}{2}b)dt\\ & +m{\int }_0^1{t}^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)f(\frac{2-t}{2m}a+\frac{t}{2}b)\\ & \le \frac{1}{2}(f(a)-m^{2}f\left(\frac{a}{m^2}\right)){\int }_0^1{t}^{\alpha }{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)dt\\ & +m(f(b)+mf\left(\frac{a}{m^2}\right)){\int }_0^1{t}^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(\omega t^{\mu };p)dt.\end{array}$$ 27

Putting $u=\frac{t}{2}a+m\frac{2-t}{2}b$ and $v=\frac{2-t}{2m}a+\frac{t}{2}b$ in (27), we have

$$\begin{array}{rl}& {\int }_{\frac{a+mb}{2}}^{mb}{(mb-u)}^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}({\omega }^{\prime }{(mb-u)}^{\mu };p)f(u)du\\ & +{\int }_{\frac{a}{m}}^{\frac{a+mb}{2m}}{(v-\frac{a}{m})}^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(m^{\mu }{\omega }^{\prime }{(v-\frac{a}{m})}^{\mu };p)f(v)dv\\ & \le \frac{1}{2}(f(a)-m^{2}f\left(\frac{a}{m^2}\right)){\int }_{\frac{a+mb}{2}}^{mb}{(mb-u)}^{\alpha }{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}({\omega }^{\prime }{(mb-u)}^{\mu };p)dt\\ & +m^{\alpha +1}(f(b)+mf\left(\frac{a}{m^2}\right))\\ & \times {\int }_{\frac{a}{m}}^{\frac{a+mb}{2m}}{(v-\frac{a}{m})}^{\alpha -1}{E}_{\mu ,\alpha ,l}^{\gamma ,\delta ,k,c}(m^{\mu }{\omega }^{\prime }{(v-\frac{a}{m})}^{\mu };p)dt.\end{array}$$ 28

By using (3), (4), and (6), we get the second inequality of (23). □

If we take $m=1$ in (23), then we get the following Hadamard type inequality for a convex function.

Corollary 2.12

Let $f:[a,b]\subseteq [0,\mathrm{\infty })\to \mathbb{R}$ be a function such that $f\in L_1[a,b]$ with $a<b$. If f is convex on $[a,b]$, then the following inequalities for extended generalized fractional integral operators hold:

$$\begin{array}{rl}& f\left(\frac{a+b}{2}\right)C_{\alpha ,{(\frac{a+b}{2})}^{+}}(b;p)\\ & \le [\left({ϵ}_{\mu ,\alpha ,l,{\omega }^{\prime },{(\frac{a+b}{2})}^{+}}^{\gamma ,\delta ,k,c}f\right)(b;p)+\left({ϵ}_{\mu ,\alpha ,l,{\omega }^{\prime },{(\frac{a+b}{2})}_{-}}^{\gamma ,\delta ,k,c}f\right)(a;p)]\\ & \le \frac{f(a)+f(b)}{2}{C}_{\alpha ,{(\frac{a+b}{2})}^{-}}(a;p),\end{array}$$ 29

where ${\omega }^{\prime }=\frac{2^{\mu }\omega }{{(b-a)}^{\mu }}$.

Remark 2.13

In Theorem 2.11.

• (i)If we put $p=0$, then we get [6, Theorem 3.10].
• (ii)If we put $\omega =p=0$, $m=1$, and $\alpha =1$, then we get the classical Hadamard inequality.

Remark 2.14

In Corollary 2.12.

• (i)If we put $p=0$, then we get [1, Theorem 3.9].
• (ii)If we put $\omega =p=0$ and $\alpha =1$, then we get the classical Hadamard inequality.
• (iii)If we take $\omega =p=0$, then we get [14, Theorem 4].

Concluding remarks

We have investigated more general fractional integral inequalities. By selecting specific values of parameters quite interesting results can be obtained. For example selecting $p=0$, fractional integral inequalities for fractional integral operators defined by Salim and Faraj in [12], selecting $l=\delta =1$, fractional integral inequalities for fractional integral operators defined by Rahman et al. in [11], selecting $p=0$ and $l=\delta =1$, fractional integral inequalities for fractional integral operators defined by Shukla and Prajapati in [15] (see also [16]), selecting $p=0$ and $l=\delta =k=1$, fractional integral inequalities for fractional integral operators defined by Prabhakar in [10], selecting $p=\omega =0$, fractional integral inequalities for Riemann–Liouville fractional integral operators.

Authors’ contributions

All authors have equal contribution in this article. All authors read and approved the final manuscript.

Funding

Not applicable.

Competing interests

It is declared that authors have no competing interests.