Generalized fractional integral inequalities for exponentially $(s,m)$-convex functions

Abstract

In this paper we have derived the fractional integral inequalities by defining exponentially $(s,m)$-convex functions. These inequalities provide upper bounds, boundedness, continuity, and Hadamard type inequality for fractional integrals containing an extended Mittag-Leffler function. The results about fractional integral operators for s-convex, m-convex, $(s,m)$-convex, exponentially convex, exponentially s-convex, and convex functions are direct consequences of presented results.

1 Introduction

Convex functions are very useful in mathematical analysis due to their fascinating properties and convenient characterizations.

Definition 1

A function $f: I\rightarrow \mathbb{R}$ is said to be convex function if the following inequality holds:

$$ f\bigl(ta+(1-t)b\bigr)\leq tf(a)+(1-t)f(b) $$

(1.1)

for all $a,b \in I$ and $t\in [0, 1 ]$. If inequality (1.1) holds in reverse order, then the function f is called concave function.

A graphical interpretation of a convex function f over an interval $[a,b]$ provides at a glance the following well-known Hadamard inequality:

$$ f \biggl(\frac{a+b}{2} \biggr)\leq \frac{1}{b-a} \int _{a}^{b}f(x)\,dx \leq \frac{f(a)+f(b)}{2}. $$

(1.2)

This inequality has been studied extensively, and a lot of its versions have been published by defining new functions obtained from inequality (1.1). Next we define some of these definitions.

Definition 2

([10])

Let $s\in [0, 1 ]$. A function $f:[0,\infty ) \rightarrow \mathbb{R}$ is said to be s-convex function in the second sense if

$$ f\bigl(ta+(1-t)b\bigr)\leq t^{s}f(a)+(1-t)^{s}f(b) $$

holds for all $a,b \in [0,\infty )$ and $t\in [0, 1 ]$.

In [22], Toader gave the following definition of m-convex function.

Definition 3

A function $f:[0,b]\rightarrow \mathbb{R}$, $b>0$, is said to be m-convex if

$$ f \bigl(tx+m(1-t)y \bigr)\leq tf(x)+m(1-t)f(y) $$

holds, where $m \in [0,1]$, $x,y\in [0,b]$, and $t\in [0,1]$.

In [4], Awan et al. gave the following definition of exponentially convex function.

Definition 4

A function $f:K \rightarrow \mathbb{R}$, where K is an interval, is said to be an exponentially convex function if

$$ f\bigl(ta+(1-t)b\bigr)\leq t\frac{f(a)}{e^{\alpha {a}}}+(1-t) \frac{f(b)}{e^{\alpha {b}}} $$

(1.3)

holds for all $a,b \in {K}$, $t\in [0, 1 ]$, and $\alpha \in \mathbb{R}$. If the inequality in (1.3) is reversed, then f is called exponentially concave.

In [12], Mehreen and Anwar gave the following definition of exponentially s-convex function.

Definition 5

([12])

Let $s\in (0, 1 ]$ and $K\subseteq [0,\infty )$ be an interval. A function $f:K\rightarrow \mathbb{R}$ is said to be exponentially s-convex in the second sense if

$$ f\bigl(ta+(1-t)b\bigr)\leq t^{s}\frac{f(a)}{e^{\alpha {a}}}+(1-t)^{s} \frac{f(b)}{e^{\alpha {b}}} $$

(1.4)

holds for all $a,b \in {K}$, $t\in [0, 1 ]$, and $\alpha \in \mathbb{R}$. If the inequality in (1.4) is reversed, then f is called exponentially s-concave function.

In [1], Anastassiou gave the following definition of $(s,m)$-convex function.

Definition 6

([1])

A function $f:[0,b] \rightarrow \mathbb{R}$ is said to be an $(s,m)$-convex function, where $(s,m)\in [0,1 ]^{2}$ and $b>0$, if for every $x,y\in [0,b]$ and $t\in [0,1]$ we have

$$ f\bigl(ta+m(1-t)b\bigr)\leq t^{s}f(a)+m\bigl(1-t^{s} \bigr)f(b). $$

The aim of this paper is to define a further generalization named exponentially $(s,m)$-convex function (Definition 9) and explore the bounds of generalized fractional integral operators containing Mittag-Leffler functions in their kernels. The Mittag-Leffler function $E_{\sigma }(t)$ was introduced by Gosta [13] in 1903:

$$ E_{\sigma }(t)=\sum_{n=0}^{\infty } \frac{t^{n}}{\varGamma (\sigma n + 1)}, $$

where $t,\sigma \in \mathbb{C}, \Re {(\sigma )}>0$ and $\varGamma (\cdot)$ is the gamma function.

The Mittag-Leffler function is a direct generalization of the exponential function to which it reduces for $\sigma =1$. In the solution of fractional integral equations and fractional differential equations, the Mittag-Leffler function arises naturally. Due to its importance, the Mittag-Leffler function has been further generalized and extended by many researchers, we refer the reader to [3, 9, 19, 20]. Recently in [2], Andrić et al. introduced a generalized Mittag-Leffler function defined as follows.

Definition 7

Let $\mu,\sigma,l,\gamma,c\in \mathbb{C}$, $\Re (\mu ),\Re (\sigma ),\Re (l)>0$, $\Re (c)>\Re (\gamma )>0$ with $p\geq 0$, $\delta >0$, and $0< k\leq \delta +\Re (\mu )$. Then the extended generalized Mittag-Leffler function is defined by

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

(1.5)

where $\beta _{p}$ is the generalized beta function defined as follows:

$$ \beta _{p}(x,y) = \int _{0}^{1}t^{x-1}(1-t)^{y-1}e^{-\frac{p}{t(1-t)}} \,dt $$

and $(c)_{nk}$ is the Pochhammer symbol defined by $(c)_{nk}=\frac{\varGamma (c+nk)}{\varGamma (c)}$.

Remark 1

The function given in (1.5) is a generalization of the following Mittag-Leffler functions:

• (i) If $p=0$ in (1.5), then it reduces to the Salim–Faraj function defined in [19].

• (ii) If $l=\delta =1$ in (1.5), then it reduces to the function defined by Rahman et al. in [15].

• (iii) If $p=0$ and $l=\delta =1$ in (1.5), then it reduces to the Shukla–Prajapati function defined in [20], see also [21].

• (iv) If $p=0$ and $l=\delta =k=1$ in (1.5), then it reduces to the Prabhakar function defined in [14].

Derivative property of the generalized Mittag-Leffler function is given in following lemma.

Lemma 1

([2])

If$m\in \mathbb{N},\omega,\mu,\sigma,l,\gamma,c\in \mathbb{C},\Re ( \mu ),\Re (\sigma ),\Re (l)>0,\Re (c)>\Re (\gamma )>0$with$p\geq 0,\delta >0$and$0< k<\delta +\Re (\mu )$, then

$$\begin{aligned} \biggl(\frac{d}{dt} \biggr)^{m} \bigl[t^{\sigma -1}E_{\mu,\sigma,l}^{ \gamma,\delta,k,c}\bigl(\omega t^{\mu };p\bigr)\bigr]=t^{\sigma -m-1}E_{\mu,\sigma -m,l}^{ \gamma,\delta,k,c} \bigl(\omega t^{\mu };p\bigr),\quad \Re (\sigma )>m. \end{aligned}$$

(1.6)

Fractional integral operators are very useful in advancement of mathematical inequalities. Many researchers have established fractional integral inequalities due to different kinds of fractional and conformable integral operators, see [1, 2, 5, 6, 8, 11, 16–18, 23]. The Mittag-Leffler function is used to define generalized fractional integral operators. The left-sided and right-sided fractional integral operators containing Mittag-Leffler function (1.5) are defined as follows.

Definition 8

([2])

Let $\omega,\mu,\sigma,l,\gamma,c\in \mathbb{C}$, $\Re (\mu ),\Re (\sigma ),\Re (l)>0$, $\Re (c)>\Re (\gamma )>0$ with $p\geq 0$, $\delta >0$ and $0< k\leq \delta +\Re (\mu )$. Let $f\in L_{1}[a,b]$ and $x\in [a,b]$. Then the generalized fractional integral operators containing Mittag-Leffler function are defined by

$$\begin{aligned} &\bigl( \epsilon _{\mu,\sigma,l,\omega,a^{+}}^{\gamma,\delta,k,c}f \bigr) (x;p)= \int _{a}^{x}(x-t)^{\sigma -1}E_{\mu,\sigma,l}^{ \gamma,\delta,k,c} \bigl(\omega (x-t)^{\mu };p\bigr)f(t)\,dt, \end{aligned}$$

(1.7)

$$\begin{aligned} & \bigl( \epsilon _{\mu,\sigma,l,\omega,b^{-}}^{\gamma,\delta,k,c}f \bigr) (x;p)= \int _{x}^{b}(t-x)^{\sigma -1}E_{\mu,\sigma,l}^{ \gamma,\delta,k,c} \bigl(\omega (t-x)^{\mu };p\bigr)f(t)\,dt, \end{aligned}$$

(1.8)

where $E_{\mu,\sigma,l}^{\gamma,\delta,k,c}(\cdot)$ is the Mittag-Leffler function given in (1.5).

Remark 2

Integral operators given in (1.7) and (1.8) are the generalization of the following fractional integral operators containing Mittag-Leffler function:

• (i) If we take $p=0$, it reduces to the fractional integral operators defined by Salim and Faraj in [19].

• (ii) If we take $l=\delta =1$, it reduces to the fractional integral operators defined by Rahman et al. in [15].

• (iii) If we take $p=0$ and $l=\delta =1$, it reduces to the fractional integral operators defined by Srivastava and Tomovski in [21].

• (iv) If we take $p=0$ and $l=\delta =k=1$, it reduces to the fractional integral operators defined by Prabhakar in [14].

• (v) If we take $p=\omega =0$, it reduces to the right-sided and left-sided Riemann–Liouville fractional integrals.

In [8], Farid et al. proved that

$$ \bigl(\epsilon _{\mu,\sigma,l,\omega,a^{+}}^{\gamma,\delta,k,c}1 \bigr) (x;p)=(x-a)^{\sigma } E_{\mu,\sigma +1,l}^{\gamma,\delta,k,c}\bigl(w(x-a)^{ \mu };p \bigr) $$

(1.9)

and

$$ \bigl(\epsilon _{\mu,\tau,l,\omega,b^{-}}^{\gamma,\delta,k,c}1 \bigr) (x;p)=(b-x)^{\tau } E_{\mu,\tau +1,l}^{\gamma,\delta,k,c}\bigl(w(b-x)^{ \mu };p \bigr). $$

(1.10)

We will follow the upcoming notations in the main results:

$$\begin{aligned} &D_{\sigma,\omega,a^{+}}(x;p)= \bigl(\epsilon _{\mu,\sigma,l, \omega,a}^{\gamma,\delta,k,c}1 \bigr) (x;p), \end{aligned}$$

(1.11)

$$\begin{aligned} &D_{\tau,\omega,b^{-}}(x;p)= \bigl(\epsilon _{\mu,\tau,l,\omega,b^{-}}^{ \gamma,\delta,k,c}1 \bigr) (x;p). \end{aligned}$$

(1.12)

In the upcoming section we define a new definition named exponentially $(s,m)$-convex function which generalizes convex, s-convex, m-convex, exponentially convex, and exponentially s-convex functions. Further this definition is used to establish the upper bounds of left-sided and right-sided generalized fractional integral operators (1.7) and (1.8). The upper bounds provide the continuity of these operators. A modulus inequality is obtained for differentiable functions which in absolute value are exponentially $(s,m)$-convex. Furthermore a fractional version of the Hadamard inequality is proved.

2 Main results

Definition 9

Let $s\in [0, 1 ]$ and $K\subseteq [0,\infty )$ be an interval. A function $f:K \rightarrow \mathbb{R}$ is said to be exponentially $(s,m)$-convex function in the second sense if

$$ f\bigl(ta+m(1-t)b\bigr)\leq t^{s}\frac{f(a)}{e^{\alpha {a}}}+m(1-t)^{s} \frac{f(b)}{e^{\alpha {b}}} $$

holds for all $a,b \in K$, $m\in [0, 1 ]$, and $\alpha \in \mathbb{R}$.

Remark 3

• (i) For $m=1$, one can get an exponentially s-convex function.

• (ii) For $\alpha =0$, one can get an $(s,m)$-convex function.

• (iii) For $\alpha =0$, $m=1$, one can get an s-convex function in the second sense.

• (iv) For $\alpha =0$, $s=1$, $m=1$, one can get a convex function.

Theorem 1

Let$f:K\subseteq [0,\infty )\longrightarrow \mathbb{R}$be a real-valued function. Iffis positive and exponentially$(s,m)$-convex, then for$a,b\in K,a< b$, and$\sigma,\tau \geq 1$, the following fractional integral inequality for generalized integral operators (1.7) and (1.8) holds:

$$\begin{aligned} & \bigl(\epsilon _{\mu,\sigma,l,\omega,a^{+}}^{\gamma,\delta,k,c}f \bigr) (x;p)+ \bigl(\epsilon _{\mu,\tau,l,\omega,b^{-}}^{\gamma, \delta,k,c}f \bigr) (x;p) \\ & \quad \leq \biggl(\frac{f(a)}{e^{\alpha {a}}}+ \frac{mf(\frac{x}{m})}{e^{\frac{\alpha x}{m}}} \biggr) \frac{(x-a)D_{\sigma -1,a^{+}}(x;p)}{s+1} \\ &\qquad{}+ \biggl(\frac{f(b)}{e^{\beta {b}}}+ \frac{mf(\frac{x}{m})}{e^{\frac{\beta x}{m}}} \biggr) \frac{(b-x)D_{\tau -1,b^{-}}(x;p)}{s+1},\quad x\in [a,b] \alpha, \beta \in \mathbb{R}. \end{aligned}$$

(2.1)

Proof

Let $x\in [a, b]$. Then, for $t\in [a,x)$ and $\sigma \geq 1$, one can have the following inequality:

$$ (x-t)^{\sigma -1}E_{\mu,\sigma,l}^{\gamma,\delta,k,c}\bigl(\omega (x-t)^{ \mu };p\bigr) \leq (x-a)^{\sigma -1}E_{\mu,\sigma,l}^{\gamma,\delta,k,c} \bigl( \omega (x-a)^{\mu };p\bigr). $$

(2.2)

As f is exponentially $(s,m)$-convex, therefore one can obtain

$$ f(t) \leq \biggl(\frac{x-t}{x-a} \biggr)^{s} \frac{f(a)}{e^{\alpha {a}}}+m \biggl(\frac{t-a}{x-a} \biggr)^{s} \frac{f(\frac{x}{m})}{e^{\frac{\alpha x}{m}}}, \quad\alpha \in \mathbb{R}. $$

(2.3)

By multiplying (2.2) and (2.3) and then integrating over $[a,x]$, we get

$$\begin{aligned} & \int _{a}^{x} (x-t)^{\sigma -1}E_{\mu,\alpha,l}^{\gamma,\delta,k,c} \bigl( \omega (x-t)^{\mu };p\bigr) f(t) \,dt \\ &\quad\leq \frac{(x-a)^{\alpha -1}E_{\mu,\sigma,l}^{\gamma,\delta,k,c}(\omega (x-a)^{\mu };p)}{(x-a)^{s}} \\ &\qquad{}\times \biggl(\frac{f(a)}{e^{\alpha {a}}} \int _{a}^{x}(x-t)^{s}\,dt+ \frac{mf(\frac{x}{m})}{e^{\frac{\alpha x}{m}}} \int _{a}^{x}(t-a)^{s}\,dt \biggr), \end{aligned}$$

that is, the left integral operator satisfies the following inequality:

$$ \bigl(\epsilon _{\mu,\sigma,l,\omega,a^{+}}^{\gamma,\delta,k,c}f \bigr) (x;p) \leq \frac{(x-a)D_{\sigma -1,a^{+}}(x;p)}{s+1} \biggl( \frac{f(a)}{e^{\alpha {a}}}+m \frac{f(\frac{x}{m})}{e^{\frac{\alpha x}{m}}} \biggr). $$

(2.4)

On the other hand, for $t\in (x,b]$ and $\tau \geq 1$, one can have the following inequality:

$$ (t-x)^{\tau -1}E_{\mu,\tau,l}^{\gamma,\delta,k,c}\bigl(\omega (t-x)^{ \mu };p\bigr) \leq (b-x)^{\tau -1}E_{\mu,\tau,l}^{\gamma,\delta,k,c} \bigl( \omega (b-x)^{\mu };p\bigr). $$

(2.5)

Again from exponential $(s,m)$-convexity of f, we have

$$ f(t) \leq \biggl(\frac{t-x}{b-x} \biggr)^{s} \frac{f(b)}{e^{\beta {b}}}+m \biggl(\frac{b-t}{b-x} \biggr)^{s} \frac{f(\frac{x}{m})}{e^{\frac{\beta x}{m}}},\quad \beta \in \mathbb{R}. $$

(2.6)

By multiplying (2.5) and (2.6) and then integrating over $[x,b]$, we get

$$\begin{aligned} & \int _{x}^{b} (t-x)^{\tau -1}E_{\mu,\tau,l}^{\gamma,\delta,k,c} \bigl( \omega (t-x)^{\mu };p\bigr) f(t) \,dt \\ &\quad\leq \frac{(b-x)^{\tau -1}E_{\mu,\tau,l}^{\gamma,\delta,k,c}(\omega (b-x)^{\mu };p)}{(b-x)^{s}} \\ & \qquad{}\times \biggl(\frac{f(b)}{e^{\beta {b}}} \int _{x}^{b}(t-x)^{s}\,dt+ \frac{mf(\frac{x}{m})}{e^{\frac{\beta x}{m}}} \int _{x}^{b}(b-t)^{s}\,dt \biggr), \end{aligned}$$

that is, the right integral operator satisfies the following inequality:

$$ \bigl(\epsilon _{\mu,\tau,l,\omega,b^{-}}^{\gamma,\delta,k,c}f \bigr) (x;p) \leq \frac{(b-x)D_{\tau -1,b^{-}}(x;p)}{s+1} \biggl( \frac{f(b)}{e^{\beta {b}}}+ \frac{mf(\frac{x}{m})}{e^{\frac{\beta x}{m}}} \biggr). $$

(2.7)

By adding (2.4) and (2.7), the required inequality (2.1) can be obtained. □

The following special cases are considered.

Corollary 1

If we set$\sigma =\tau$in (2.1), then the following inequality is obtained:

$$\begin{aligned} & \bigl(\epsilon _{\mu,\sigma,l,\omega,a^{+}}^{\gamma,\delta,k,c}f \bigr) (x;p)+ \bigl(\epsilon _{\mu,\sigma,l,\omega,b^{-}}^{ \gamma,\delta,k,c}f \bigr) (x;p) \\ &\quad \leq \biggl(\frac{f(a)}{e^{\alpha {a}}}+ \frac{mf(\frac{x}{m})}{e^{\frac{\alpha x}{m}}} \biggr) \frac{(x-a)D_{\sigma -1,a^{+}}(x;p)}{s+1} \\ &\qquad{}+ \biggl(\frac{f(b)}{e^{\beta {b}}}+ \frac{mf(\frac{x}{m})}{e^{\frac{\beta x}{m}}} \biggr) \frac{(b-x)D_{\sigma -1,b^{-}}(x;p)}{s+1},\quad x\in [a,b]. \end{aligned}$$

(2.8)

Corollary 2

Along with the assumption of Theorem 1, if$f\in L_{\infty }[a,b]$, then the following inequality is obtained:

$$\begin{aligned} & \bigl(\epsilon _{\mu,\sigma,l,\omega,a^{+}}^{\gamma,\delta,k,c}f \bigr) (x;p)+ \bigl(\epsilon _{\mu,\tau,l,\omega,b^{-}}^{\gamma, \delta,k,c}f \bigr) (x;p) \\ &\quad \leq \frac{ \Vert f \Vert _{\infty }}{s+1} \biggl( \biggl( \frac{1}{e^{\alpha {a}}}+ \frac{m}{e^{\frac{\alpha x}{m}}} \biggr) (x-a)D_{ \sigma -1,a^{+}}(x;p) \\ &\qquad{}+ \biggl(\frac{1}{e^{\beta {b}}}+ \frac{m}{e^{\frac{\beta x}{m}}} \biggr){(b-x)D_{\tau -1,b^{-}}(x;p)} \biggr). \end{aligned}$$

(2.9)

Corollary 3

For$\sigma =\tau$in (2.9), we get the following result:

$$\begin{aligned} & \bigl(\epsilon _{\mu,\sigma,l,\omega,a^{+}}^{\gamma,\delta,k,c}f \bigr) (x;p)+ \bigl(\epsilon _{\mu,\sigma,l,\omega,b^{-}}^{ \gamma,\delta,k,c}f \bigr) (x;p) \\ &\quad \leq \frac{ \Vert f \Vert _{\infty }}{s+1} \biggl( \biggl( \frac{1}{e^{\alpha {a}}}+ \frac{m}{e^{\frac{\alpha x}{m}}} \biggr) (x-a)D_{ \sigma -1,a^{+}}(x;p) \\ &\qquad{}+ \biggl(\frac{1}{e^{\beta {b}}}+ \frac{m}{e^{\frac{\beta x}{m}}} \biggr){(b-x)D_{\sigma -1,b^{-}}(x;p)} \biggr). \end{aligned}$$

(2.10)

Corollary 4

For$s=1$in (2.9), we get the following result:

$$\begin{aligned} & \bigl(\epsilon _{\mu,\sigma,l,\omega,a^{+}}^{\gamma,\delta,k,c}f \bigr) (x;p)+ \bigl(\epsilon _{\mu,\tau,l,\omega,b^{-}}^{\gamma, \delta,k,c}f \bigr) (x;p) \\ &\quad \leq \frac{ \Vert f \Vert _{\infty }}{2} \biggl( \biggl( \frac{1}{e^{\alpha {a}}}+ \frac{m}{e^{\frac{\alpha x}{m}}} \biggr) (x-a)D_{ \alpha -1,a^{+}}(x;p) \\ &\qquad{} + \biggl(\frac{1}{e^{\beta {b}}}+ \frac{m}{e^{\frac{\beta x}{m}}} \biggr) (b-x)D_{\beta -1,b^{-}}(x;p) \biggr). \end{aligned}$$

(2.11)

Theorem 2

With the assumptions of Theorem 1, if$f\in L_{\infty }[a,b]$, then operators defined in (1.7) and (1.8) are continuous.

Proof

If $f\in L_{\infty }[a,b]$, then from (2.4) we have

$$\begin{aligned} \bigl\vert \bigl(\epsilon _{\mu,\sigma,l,\omega,a^{+}}^{\gamma, \delta,k,c}f \bigr) (x;p) \bigr\vert & \leq \frac{2 \Vert f \Vert _{\infty }(x-a)D_{\sigma -1,a^{+}}(x;p)}{s+1} \biggl( \frac{1}{e^{\alpha {a}}}+ \frac{m}{e^{\frac{\alpha x}{m}}} \biggr) \\ &\leq \frac{2(b-a)D_{\sigma -1,a^{+}}(b;p)}{s+1} \biggl( \frac{1}{e^{\alpha {a}}}+\frac{m}{e^{\frac{\alpha a}{m}}} \biggr) \Vert f \Vert _{ \infty }, \end{aligned}$$

(2.12)

that is, $\vert (\epsilon _{\mu,\sigma,l,\omega,a^{+}}^{\gamma, \delta,k,c}f )(x;p) \vert \leq M \Vert f \Vert _{\infty }$, where $M=\frac{2(b-a)D_{\sigma -1,a^{+}}(b;p)}{s+1} ( \frac{1}{e^{\alpha {a}}}+\frac{m}{e^{\frac{\alpha a}{m}}} )$. Therefore $(\epsilon _{\mu,\sigma,l,\omega,a^{+}}^{\gamma,\delta,k,c}f )(x;p)$ is bounded, also it is easy to see that it is linear, hence this is a continuous operator. On the other hand, from (2.7) one can obtain

$$ \bigl\vert \bigl(\epsilon _{\mu,\tau,l,\omega,b^{-}}^{\gamma,\delta,k,c}f \bigr) (x;p) \bigr\vert \leq K \Vert f \Vert _{\infty }, $$

where $K=\frac{2(b-a)D_{\tau -1,b^{-}}(a;p)}{s+1} ( \frac{1}{e^{\beta {a}}}+\frac{m}{e^{\frac{\beta a}{m}}} )$. Therefore $(\epsilon _{\mu,\tau,l,\omega,b^{-}}^{\gamma,\delta,k,c}f )(x;p)$ is bounded, also it is linear, hence continuous. □

The next result provides the boundedness of a sum of left and right integrals at an arbitrary point for functions whose derivatives in absolute values are exponentially $(s,m)$-convex.

Theorem 3

Let$f:K\subseteq [0,\infty )\longrightarrow \mathbb{R}$be a real-valued function. Iffis differentiable and$|f'|$is exponentially$(s,m)$-convex, then for$a,b\in K,a< b$, and$\sigma,\tau \geq 1$, the following fractional integral inequality for generalized integral operators (1.7) and (1.8) holds:

$$\begin{aligned} & \bigl\vert \bigl(\epsilon _{\mu,\sigma +1,l,\omega,a^{+}}^{\gamma, \delta,k,c}f \bigr) (x;p)+ \bigl(\epsilon _{\mu,\tau +1,l,\omega,b^{-}}^{ \gamma,\delta,k,c}f \bigr) (x;p) \\ &\qquad{} - \bigl(D_{\sigma -1,a^{+}}(x;p)f(a)+D_{\tau -1,b^{-}}(x;p)f(b) \bigr) \bigr\vert \\ &\quad \leq \biggl(\frac{ \vert f'(a) \vert }{e^{\alpha {a}}}+ \frac{m \vert f'(\frac{x}{m}) \vert }{e^{\frac{\alpha x}{m}}} \biggr) \frac{(x-a)D_{\sigma -1,a^{+}}(x;p)}{s+1} \\ &\qquad{}+ \biggl(\frac{ \vert f'(b) \vert }{e^{\beta {b}}}+ \frac{m \vert f'(\frac{x}{m}) \vert }{e^{\frac{\beta x}{m}}} \biggr) \frac{(b-x)D_{\tau -1,b^{-}}(x;p)}{s+1} , \quad x\in [a,b],\alpha,\beta \in \mathbb{R}. \end{aligned}$$

(2.13)

Proof

Let $x\in [a,b]$ and $t\in [a,x)$, by using exponential $(s,m)$-convexity of $|f'|$, we have

$$ \bigl\vert f'(t) \bigr\vert \leq \biggl( \frac{x-t}{x-a} \biggr)^{s} \frac{ \vert f'(a) \vert }{e^{\alpha {a}}}+m \biggl( \frac{t-a}{x-a} \biggr)^{s} \frac{ \vert f'(\frac{x}{m}) \vert }{e^{\frac{\alpha x}{m}}}. $$

(2.14)

From (2.14), one can have

$$ f'(t) \leq \biggl(\frac{x-t}{x-a} \biggr)^{s} \frac{ \vert f'(a) \vert }{e^{\alpha {a}}}+m \biggl(\frac{t-a}{x-a} \biggr)^{s} \frac{ \vert f'(\frac{x}{m}) \vert }{e^{\frac{\alpha x}{m}}}. $$

(2.15)

The product of (2.2) and (2.15) gives the following inequality:

$$\begin{aligned} &(x-t)^{\sigma -1}E_{\mu,\sigma,l}^{\gamma,\delta,k,c}\bigl(\omega (x-t)^{ \mu };p\bigr)f'(t)\,dt \\ &\quad\leq (x-a)^{\sigma -1-s}E_{\mu,\sigma,l}^{\gamma,\delta,k,c}\bigl( \omega (x-a)^{\mu };p\bigr) \biggl(\frac{ \vert f'(a) \vert }{e^{\alpha {a}}}(x-t)^{s}+ \frac{m \vert f'(\frac{x}{m}) \vert }{e^{\frac{\alpha x}{m}}}(t-a)^{s} \biggr). \end{aligned}$$

(2.16)

After integrating the above inequality over $[a,x]$, we get

$$\begin{aligned} & \int _{a}^{x}(x-t)^{\sigma -1}E_{\mu,\sigma,l}^{\gamma,\delta,k,c} \bigl( \omega (x-t)^{\mu };p\bigr)f'(t)\,dt \\ &\quad \leq (x-a)^{\sigma -1-s}E_{\mu,\sigma,l}^{\gamma, \delta,k,c}\bigl(\omega (x-a)^{\mu };p\bigr) \\ &\qquad{}\times \biggl(\frac{ \vert f'(a) \vert }{e^{\alpha {a}}} \int _{a}^{x}(x-t)^{s}\,dt+ \frac{m \vert f'(\frac{x}{m}) \vert }{e^{\frac{\alpha x}{m}}} \int _{a}^{x}(t-a)^{s}\,dt \biggr) \\ &\quad = \frac{(x-a)^{\sigma }E_{\mu,\sigma,l}^{\gamma,\delta,k,c}(\omega (x-t)^{\mu };p)}{s+1} \biggl(\frac{ \vert f'(a) \vert }{e^{\alpha {a}}}+ \frac{m \vert f'(\frac{x}{m}) \vert }{e^{\frac{\alpha x}{m}}} \biggr). \end{aligned}$$

(2.17)

The left-hand side of (2.17) is calculated as follows:

$$ \int _{a}^{x}(x-t)^{\sigma -1}E_{\mu,\sigma,l}^{\gamma,\delta,k,c} \bigl( \omega (x-t)^{\mu };p\bigr)f'(t)\,dt. $$

(2.18)

Put $x-t=z$, that is, $t=x-z$, also using the derivative property (1.6) of Mittag-Leffler function, we have

$$\begin{aligned} & \int _{0}^{x-a}z^{\sigma -1}E_{\mu,\sigma,l}^{\gamma,\delta,k,c} \bigl( \omega z^{\mu };p\bigr)f'(x-z)\,dz \\ &\quad=(x-a)^{\sigma -1}E_{\mu,\sigma,l}^{\gamma,\delta,k,c}\bigl(\omega (x-a)^{ \mu };p\bigr)f(a)- \int _{0}^{x-a} z^{\sigma -2}E_{\mu,\sigma,l}^{ \gamma,\delta,k,c} \bigl(\omega z^{\mu };p\bigr)f(x-z)\,dz. \end{aligned}$$

Now putting $x-z=t$ in the second term of the right-hand side of the above equation and then using (1.7), we get

$$\begin{aligned} & \int _{0}^{x-a}z^{\sigma -1}E_{\mu,\sigma,l}^{\gamma,\delta,k,c} \bigl( \omega z^{\mu };p\bigr)f'(x-z)\,dz \\ &\quad=(x-a)^{\sigma -1}E_{\mu,\sigma,l}^{\gamma,\delta,k,c}\bigl(\omega (x-a)^{ \mu };p\bigr)f(a)- \bigl(\epsilon _{\mu,\sigma +1,l,\omega,a^{+}}^{ \gamma,\delta,k,c}f \bigr) (x;p). \end{aligned}$$

Therefore (2.17) takes the following form:

$$\begin{aligned} & \bigl(D_{\sigma -1,a^{+}}(x;p) \bigr)f(a)- \bigl(\epsilon _{\mu, \sigma +1,l,\omega,a^{+}}^{\gamma,\delta,k,c}f \bigr) (x;p) \\ &\quad\leq \frac{(x-a)D_{\sigma -1,a^{+}}(x;p)}{s+1} \biggl( \frac{ \vert f'(a) \vert }{e^{\alpha {a}}}+ \frac{m \vert f'(\frac{x}{m}) \vert }{e^{\frac{\alpha x}{m}}} \biggr). \end{aligned}$$

(2.19)

Also from (2.14) one can have

$$ f'(t)\geq - \biggl( \biggl(\frac{x-t}{x-a} \biggr)^{s} \frac{ \vert f'(a) \vert }{e^{\alpha {a}}}+m \biggl(\frac{t-a}{x-a} \biggr)^{s} \frac{ \vert f'(\frac{x}{m}) \vert }{e^{\frac{\alpha x}{m}}} \biggr). $$

(2.20)

Following the same procedure as we did for (2.15), one can obtain

$$\begin{aligned} & \bigl(\epsilon _{\mu,\sigma +1,l,\omega,a^{+}}^{\gamma,\delta,k,c}f \bigr) (x;p)-D_{\sigma -1,a^{+}}(x;p)f(a) \\ &\quad\leq \frac{(x-a)D_{\sigma -1,a^{+}}(x;p)}{s+1} \biggl( \frac{ \vert f'(a) \vert }{e^{\alpha {a}}}+ \frac{m \vert f'(\frac{x}{m}) \vert }{e^{\frac{\alpha x}{m}}} \biggr). \end{aligned}$$

(2.21)

From (2.19) and (2.21), we get

$$\begin{aligned} & \bigl\vert \bigl(\epsilon _{\mu,\sigma +1,l,\omega,a^{+}}^{\gamma, \delta,k,c}f \bigr) (x;p)-D_{\sigma -1,a^{+}}(x;p)f(a) \bigr\vert \\ &\quad\leq \frac{(x-a)D_{\sigma -1,a^{+}}(x;p)}{s+1} \biggl( \frac{ \vert f'(a) \vert }{e^{\alpha {a}}}+ \frac{m \vert f'(\frac{x}{m}) \vert }{e^{\frac{\alpha x}{m}}} \biggr). \end{aligned}$$

(2.22)

Now we let $x\in [a,b]$ and $t\in (x,b]$. Then, by exponential $(s,m)$-convexity of $|f'|$, we have

$$ \bigl\vert f'(t) \bigr\vert \leq \biggl( \frac{t-x}{b-x} \biggr)^{s} \frac{ \vert f'(b) \vert }{e^{\beta {b}}}+m \biggl( \frac{b-t}{b-x} \biggr)^{s} \frac{ \vert f'(\frac{x}{m}) \vert }{e^{\frac{\beta x}{m}}},\quad \beta \in \mathbb{R}. $$

(2.23)

On the same lines as we have done for (2.2), (2.15), and (2.20), one can get from (2.5) and (2.23) the following inequality:

$$\begin{aligned} & \bigl\vert \bigl(\epsilon _{\mu,\tau +1,l,\omega,b^{-}}^{\gamma, \delta,k,c}f \bigr) (x;p)-D_{\tau -1,b^{-}}(x;p)f(b) \bigr\vert \\ &\quad\leq \frac{(b-x)D_{\tau -1,b^{-}}(x;p)}{s+1} \biggl( \frac{ \vert f'(b) \vert }{e^{\beta {b}}}+ \frac{m \vert f'(\frac{x}{m}) \vert }{e^{\frac{\beta x}{m}}} \biggr). \end{aligned}$$

(2.24)

From inequalities (2.22) and (2.24) via the triangular inequality, (2.13) can be obtained. □

Corollary 5

If we put$\sigma =\tau$in (2.13), then the following inequality is obtained:

$$\begin{aligned} & \bigl\vert \bigl(\epsilon _{\mu,\sigma +1,l,\omega,a^{+}}^{\gamma, \delta,k,c}f \bigr) (x;p)+ \bigl(\epsilon _{\mu,\sigma +1,l,\omega,b^{-}}^{ \gamma,\delta,k,c}f \bigr) (x;p) \\ &\qquad{} - \bigl(D_{\sigma -1,a^{+}}(x;p)f(a)+D_{\sigma -1,b^{-}}(x;p)f(b) \bigr) \bigr\vert \\ &\quad \leq \biggl(\frac{ \vert f'(a) \vert }{e^{\alpha {a}}}+ \frac{m \vert f'(\frac{x}{m}) \vert }{e^{\frac{\alpha x}{m}}} \biggr) \frac{(x-a)D_{\sigma -1,a^{+}}(x;p)}{s+1} \\ &\qquad{}+ \biggl(\frac{ \vert f'(b) \vert }{e^{\beta {b}}}+ \frac{m \vert f'(\frac{x}{m}) \vert }{e^{\frac{\beta x}{m}}} \biggr) \frac{(b-x)D_{\sigma -1,b^{-}}(x;p)}{s+1} ,\quad x\in [a,b],\alpha,\beta \in \mathbb{R}. \end{aligned}$$

(2.25)

Definition 10

Let $f: [a,b]\rightarrow \mathbb{R}$ be a function, we will say that f is exponentially m-symmetric about $\frac{a+b}{2}$ if

$$\begin{aligned} \frac{f(x)}{e^{\alpha {x}}}= \frac{f(\frac{a+b-x}{m})}{e^{\alpha {(\frac{a+b-x}{m})}}}, \quad\alpha \in \mathbb{R}. \end{aligned}$$

(2.26)

It is required to give the following lemma which will be helpful to produce Hadamard type estimations for the generalized fractional integral operators.

Lemma 2

Let$f:K\subseteq [0,\infty )\longrightarrow \mathbb{R}, a,b\in K,a< b$, be an exponentially$(s,m)$-convex function. Iffis exponentiallym-symmetric about$\frac{a+b}{2}$, then the following inequality holds:

$$\begin{aligned} f \biggl(\frac{a+b}{2} \biggr)\leq (1+m) \frac{f(x)}{2^{s}e^{\alpha {x}}},\quad \alpha \in \mathbb{R}. \end{aligned}$$

(2.27)

Proof

Since f is exponentially $(s,m)$-convex, so

$$ f \biggl(\frac{a+b}{2} \biggr) \leq \frac{f(at+(1-t)b)}{2^{s}e^{\alpha ({at+(1-t)b})}}+ \frac{mf(\frac{a(1-t)+bt}{m})}{2^{s}e^{\alpha (\frac{a(1-t)+bt}{m})}},\quad t \in [0,1]. $$

(2.28)

Let $x=at+(1-t)b$, where $x\in [a,b]$. Then we have $a+b-x=bt+(1-t)a$, and we get

$$ f \biggl(\frac{a+b}{2} \biggr) \leq \frac{f(x)}{2^{s}e^{\alpha {x}}}+m \frac{f(\frac{a+b-x}{m})}{2^{s}e^{\alpha (\frac{a+b-x}{m})}}. $$

(2.29)

Now, using that f is exponentially m-symmetric, we will get (2.27). □

Theorem 4

Let$f:K\subseteq [0,\infty )\longrightarrow \mathbb{R}, a,b\in K,a< b$, be a real-valued function. Iffis positive, exponentially$(s,m)$-convex and exponentiallym-symmetric about$\frac{a+b}{2}$, then for$\sigma,\tau >0$, the following fractional integral inequality for generalized integral operators (1.7) and (1.8) holds:

$$\begin{aligned} &\frac{2^{s}h(\alpha )}{1+m}f \biggl(\frac{a+b}{2} \biggr) \bigl[D_{ \tau +1,b^{-}}(a;p)+D_{\sigma +1,a^{+}}(b;p) \bigr] \\ &\quad\leq \bigl(\epsilon _{\mu,\tau +1,l,\omega,b^{-}}^{\gamma, \delta,k,c}f \bigr) (a;p)+ \bigl( \epsilon _{\mu,\sigma +1,l,\omega,a^{+}}^{ \gamma,\delta,k,c}f \bigr) (b;p) \\ &\quad\leq \bigl[D_{\tau -1,b^{-}}(a;p)+D_{\sigma -1,a^{+}}(b;p) \bigr] \frac{(b-a)^{2}}{s+1} \biggl( \frac{f(\frac{a}{m})}{e^{\frac{\alpha a}{m}}}+ \frac{f(b)}{e^{\beta {b}}} \biggr) ,\quad \alpha,\beta \in \mathbb{R}, \end{aligned}$$

(2.30)

where$h(\alpha ) = e^{\alpha b}$for$\alpha <0$and$h(\alpha ) = e^{\alpha a}$for$\alpha \geq 0$.

Proof

For $x\in [a,b]$, we have

$$ (x-a)^{\tau }E_{\mu,\tau,l}^{\gamma,\delta,k,c}\bigl(\omega (x-a)^{ \mu };p\bigr)\leq (b-a)^{\tau }E_{\mu,\tau,l}^{\gamma,\delta,k,c} \bigl( \omega (b-a)^{\mu };p\bigr),\quad \tau >0. $$

(2.31)

As f is exponentially $(s,m)$-convex, so for $x\in [a,b]$, we have

$$ f(x) \leq \biggl(\frac{x-a}{b-a} \biggr)^{s} \frac{f(b)}{e^{\alpha {b}}}+m \biggl(\frac{b-x}{b-a} \biggr)^{s} \frac{f(\frac{a}{m})}{e^{\frac{\alpha a}{m}}},\quad \alpha \in \mathbb{R}. $$

(2.32)

By multiplying (2.31) and (2.32) and then integrating over $[a,b]$, we get

$$\begin{aligned} & \int _{a}^{b}(x-a)^{\tau }E_{\mu,\tau,l}^{\gamma,\delta,k,c} \bigl( \omega (x-a)^{\mu };p\bigr)f(x)\,dx \\ &\quad\leq {(b-a)^{\tau -s}E_{\mu,\tau,l}^{\gamma,\delta,k,c}\bigl(\omega (b-a)^{ \mu };p\bigr)} \biggl(\frac{f(b)}{e^{\alpha {b}}} \int _{a}^{b} (x-a)^{s}\,dx+ \frac{mf(\frac{a}{m})}{e^{\frac{\alpha a}{m}}} \int _{a}^{b} (b-x)^{s}\,dx \biggr), \end{aligned}$$

from which we have

$$\begin{aligned} &\bigl(\epsilon _{\mu,\tau +1,l,\omega,b^{-}}^{\gamma,\delta,k,c}f \bigr) (a;p)\leq \frac{(b-a)^{\tau +1}E_{\mu,\tau,l}^{\gamma,\delta,k,c}(\omega (b-a)^{\mu };p)}{s+1} \biggl(\frac{f(b)}{e^{\alpha {b}}}+ \frac{mf(\frac{a}{m})}{e^{\frac{\alpha a}{m}}} \biggr), \end{aligned}$$

(2.33)

$$\begin{aligned} & \bigl(\epsilon _{\mu,\tau +1,l,\omega,b^{-}}^{\gamma,\delta,k,c}f \bigr) (a;p)\leq \frac{(b-a)^{2}}{s+1}D_{\tau -1,b^{-}}(a;p) \biggl( \frac{f(b)}{e^{\alpha {b}}}+ \frac{mf(\frac{a}{m})}{e^{\frac{\alpha a}{m}}} \biggr). \end{aligned}$$

(2.34)

On the other hand, for $x\in [a,b]$, we have

$$ (b-x)^{\sigma }E_{\mu,\sigma,l}^{\gamma,\delta,k,c}\bigl(\omega (b-x)^{ \mu };p\bigr)\leq (b-a)^{\sigma }E_{\mu,\sigma,l}^{\gamma,\delta,k,c} \bigl( \omega (b-a)^{\mu };p\bigr),\quad \alpha >0. $$

(2.35)

By multiplying (2.32) and (2.35) and then integrating over $[a,b]$, we get

$$\begin{aligned} & \int _{a}^{b}(b-x)^{\sigma }E_{\mu,\sigma,l}^{\gamma,\delta,k,c} \bigl( \omega (b-x)^{\mu };p\bigr)f(x)\,dx \\ &\quad\leq {(b-a)^{\sigma -s}E_{\mu,\sigma,l}^{\gamma,\delta,k,c}\bigl( \omega (b-a)^{\mu };p\bigr)} \biggl(\frac{f(b)}{e^{\alpha {b}}} \int _{a}^{b} (x-a)^{s}\,dx+ \frac{mf(\frac{a}{m})}{e^{\frac{\alpha a}{m}}} \int _{a}^{b} (b-x)^{s}\,dx \biggr), \end{aligned}$$

from which we have

$$\begin{aligned} & \bigl(\epsilon _{\mu,\sigma +1,l,\omega,a^{+}}^{\gamma,\delta,k,c}f \bigr) (b;p)\leq \frac{(b-a)^{\sigma +1}E_{\mu,\sigma,l}^{\gamma,\delta,k,c}(\omega (b-a)^{\mu };p)}{s+1} \biggl(\frac{f(b)}{e^{\alpha {b}}}+ \frac{mf(\frac{a}{m})}{e^{\frac{\alpha a}{m}}} \biggr), \end{aligned}$$

(2.36)

$$\begin{aligned} &\bigl(\epsilon _{\mu,\sigma +1,l,\omega,a^{+}}^{\gamma,\delta,k,c}f \bigr) (b;p)\leq \frac{(b-a)^{2}}{s+1}D_{\sigma -1,a^{+}}(b;p) \biggl( \frac{f(b)}{e^{\alpha {b}}}+ \frac{mf(\frac{a}{m})}{e^{\frac{\alpha a}{m}}} \biggr). \end{aligned}$$

(2.37)

Adding (2.34) and (2.37), we get

$$\begin{aligned} & \bigl(\epsilon _{\mu,\tau +1,l,\omega,b^{-}}^{\gamma,\delta,k,c}f \bigr) (a;p)+ \bigl(\epsilon _{\mu,\sigma +1,l,\omega,a^{+}}^{ \gamma,\delta,k,c}f \bigr) (b;p) \\ &\quad\leq \bigl[D_{\tau -1,b^{-}}(a;p)+D_{\sigma -1,a^{+}}(b;p) \bigr] \frac{(b-a)^{2}}{s+1} \biggl(\frac{f(b)}{e^{\alpha {b}}}+ \frac{mf(\frac{a}{m})}{e^{\frac{\alpha a}{m}}} \biggr). \end{aligned}$$

(2.38)

Multiplying (2.27) with $(x-a)^{\tau }E_{\mu,\tau,l}^{\gamma,\delta,k,c}(\omega (x-a)^{ \mu };p)$ and integrating over $[a,b]$, we get

$$\begin{aligned} &f \biggl(\frac{a+b}{2} \biggr) \int _{a}^{b}(x-a)^{\tau }E_{\mu,\tau,l}^{ \gamma,\delta,k,c} \bigl(\omega (x-a)^{\mu };p\bigr)\,dx \\ &\quad \leq \frac{m+1}{2^{s}} \int _{a}^{b}(x-a)^{\tau }E_{\mu,\tau,l}^{ \gamma,\delta,k,c} \bigl(\omega (x-a)^{\mu };p\bigr)\frac{f(x)}{e^{\alpha {x}}}\,dx. \end{aligned}$$

(2.39)

By using (1.8) and (1.12), we get

$$\begin{aligned} f \biggl(\frac{a+b}{2} \biggr)D_{\tau +1,b^{-}}(a;p)\leq \frac{m+1}{2^{s}e^{\alpha {x}}} \bigl(\epsilon _{\mu,\tau +1,l, \omega,b^{-}}^{\gamma,\delta,k,c}f \bigr) (a;p). \end{aligned}$$

(2.40)

Multiplying (2.27) with $(b-x)^{\sigma }E_{\mu,\sigma,l}^{\gamma,\delta,k,c}(\omega (b-x)^{ \mu };p)$ and integrating over $[a,b]$, also using (1.7) and(1.11), we get

$$\begin{aligned} f \biggl(\frac{a+b}{2} \biggr)D_{\sigma +1,a^{+}}(b;p)\leq \frac{m+1}{2^{s}h(\alpha )} \bigl(\epsilon _{\mu,\sigma +1,l,\omega,a^{+}}^{ \gamma,\delta,k,c}f \bigr) (b;p). \end{aligned}$$

(2.41)

By adding (2.40) and (2.41), we get

$$\begin{aligned} &\frac{2^{s}h(\alpha )}{1+m}f \biggl(\frac{a+b}{2} \biggr) \bigl[D_{ \tau +1,b^{-}}(a;p)+D_{\sigma +1,a^{+}}(b;p) \bigr] \\ &\quad\leq \bigl(\epsilon _{\mu,\tau +1,l,\omega,b^{-}}^{\gamma, \delta,k,c}f \bigr) (a;p)+ \bigl( \epsilon _{\mu,\sigma +1,l,\omega,a^{+}}^{ \gamma,\delta,k,c}f \bigr) (b;p). \end{aligned}$$

(2.42)

By combining (2.38) and (2.42), inequality (2.30) can be obtained. □

Corollary 6

If we put$\sigma =\tau$in (2.30), then the following inequality is obtained:

$$\begin{aligned} &\frac{2^{s}e^{\alpha {x}}}{1+m}f \biggl(\frac{a+b}{2} \biggr) \bigl[D_{ \sigma +1,b^{-}}(a;p)+D_{\sigma +1,a^{+}}(b;p) \bigr] \\ &\quad\leq \bigl(\epsilon _{\mu,\sigma +1,l,\omega,b^{-}}^{\gamma, \delta,k,c}f \bigr) (a;p)+ \bigl( \epsilon _{\mu,\sigma +1,l,\omega,a^{+}}^{ \gamma,\delta,k,c}f \bigr) (b;p) \\ &\quad\leq \bigl(D_{\sigma -1,b^{-}}(a;p)+D_{\sigma -1,a^{+}}(b;p) \bigr) \frac{(b-a)^{2}}{s+1} \biggl(\frac{f(b)}{e^{\alpha {b}}}+ \frac{mf(\frac{a}{m})}{e^{\frac{\alpha a}{m}}} \biggr). \end{aligned}$$

(2.43)

3 Concluding remarks

This paper has investigated generalized fractional integral inequalities which provide the bounds of fractional integral operators containing Mittag-Leffler functions in their kernels. By setting different values of parameters involved in the Mittag-Leffler function, the results for various known fractional operators can be obtained. For example, by setting $p=0$, fractional integral inequalities for fractional operators defined by Salim and Faraj in [19] can be obtained; by setting $l=\delta =1$, fractional integral inequalities for fractional operators defined by Rahman et al. in [15] can be deduced, by setting $p=0$ and $l=\delta =1$, fractional integral inequalities for fractional operators defined by Shukla and Prajapati in [20] (see also [21]) can be deduced, by setting $p=0$ and $l=\delta =k=1$, fractional integral inequalities for fractional operators defined by Prabhakar in [14] can be deduced, by setting $p=\omega =0$ fractional integral inequalities for Riemann–Liouville fractional integrals can be deduced. Also all the results of this paper hold for s-convex, m-convex, exponentially convex, exponentially s-convex, and convex functions. In particular results for $(s,m)$-convex functions, which are proved in [7], can be obtained.