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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.

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:

  1. (i)

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

  2. (ii)

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

  3. (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].

  4. (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, 1618, 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:

  1. (i)

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

  2. (ii)

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

  3. (iii)

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

  4. (iv)

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

  5. (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.

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

  1. (i)

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

  2. (ii)

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

  3. (iii)

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

  4. (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)

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.

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Acknowledgements

The research work of the first author is supported by the National Key R & D Program of China (Grant No. 2019YFA0706402). The research work of second author is supported by the Higher Education Commission of Pakistan under NRPU 2016, Project No. 5421. The research work of the third author is supported by the Ministry of Education and Science of the Russian Federation (the Agreement No. 02.a03.21.0008).

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Qiang, X., Farid, G., Pečarić, J. et al. Generalized fractional integral inequalities for exponentially \((s,m)\)-convex functions. J Inequal Appl 2020, 70 (2020). https://doi.org/10.1186/s13660-020-02335-7

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Keywords

  • Convex function
  • \((s,m)\)-convex function
  • Mittag-Leffler function
  • Fractional integral operators
  • Boundedness