Simpson’s Type Inequalities for s-Convex Functions via a Generalized Proportional Fractional Integral

Abstract

In this paper, we give new Simpson’s type integral inequalities for the class of functions whose derivatives of absolute values are s-convex via generalized proportional fractional integrals. Some results in the literature are particular cases of our results.

1. Introduction and Preliminaries

Simpson’s inequality is given by

$$\left|\frac{1}{3}\left[\frac{g\left(a\right)+g\left(b\right)}{2}+2g\left(\frac{a+b}{2}\right)\right]-\frac{1}{b-a}{\int }_a^{b}g\left(x\right)dx\right|\le \frac{{(b-a)}^4}{2880}{\parallel g^4\parallel }_{\infty },$$

where $g:[a,b]\to \mathbb{R}$ is a four times continuously differentiable function on $(a,b)$ and $\parallel g^{\left(4\right)}{\parallel }_{\infty }=\underset{x\in (a,b)}{sup}\left|g^{\left(4\right)}\right|<\infty$. This inequality has been studied and generalized by many scholars see for instance [1,2,3,4,5,6,7] and references cited therein.

Definition 1.

The function $g:[0,\infty )\to \mathbb{R}$ is a convex function if

$$g\left(\lambda x+(1-\lambda )y\right)\le \lambda g\left(x\right)+(1-\lambda )g\left(y\right)$$

holds for ever $x,y\in [0,\infty ]and\lambda \in [0,1]$. If the inequality in Definition 1 is reversed, then g is a concave function.

Definition 2.

The function $g:[0,\infty )\to \mathbb{R}$ is s-convex function (in the second sense) if

$$g\left(\lambda x+(1-\lambda )y\right)\le {\lambda }^{s}g\left(x\right)+{(1-\lambda )}^{s}g\left(y\right)$$

for ever $x,y\in [0,\infty )$ and $\lambda \in [0,1]$.

Remark 1.

Definition 2 reduces to Definition 1 when $s=1$. In the current paper $I^0$ denote the interior of an interval I and $L_1\left([a,b]\right)$ represent all integrable functions.

Theorem 1

([8]). Suppose that $g:[0,\infty )\to [0,\infty )$ is an s-convex function in the second sense, where $s\in (0,1)$ and let $a,b\in [0,\infty ),a<b$. If $g\in L_1\left([a,b]\right)$, then we have:

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

(1)

Definition 3

([9]). For an integrable function g on $[a,b]$, $a\ge 0$ and $\beta >0$, the right- and left-sided Riemann–Liouville fractional integral of order β are respectively given by

$${\mathcal{J}}_{a^{+}}^{\beta }g\left(x\right)=\frac{1}{\mathsf{\Gamma }\left(\beta \right)}{\int }_a^x{(x-t)}^{\beta -1}g\left(t\right)dt,x>a,$$

(2)

and

$${\mathcal{J}}_{b^{-}}^{\beta }g\left(x\right)=\frac{1}{\mathsf{\Gamma }\left(\beta \right)}{\int }_x^b{(t-x)}^{\beta -1}g\left(t\right)dt,t<b,$$

(3)

where ${\displaystyle \mathsf{\Gamma }\left(\beta \right)={\int }_0^{\infty }{e}^{-x}{x}^{\beta -1}dx}$ is the gamma function.

Definition 4

([10]). Suppose that the function g is integrable on $[a,b]$ and $\delta \in (0,1]$. Then for all $\beta \in \mathbb{C},Re\left(\beta \right)\ge 0$

$$\left({\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\right)\left(y\right)=\frac{1}{{\delta }^{\beta }\mathsf{\Gamma }\left(\beta \right)}{\int }_a^y{(y-u)}^{\beta -1}g\left(u\right)du,y>a,$$

(4)

and

$$\left({\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\right)\left(y\right)=\frac{1}{{\delta }^{\beta }\mathsf{\Gamma }\left(\beta \right)}{\int }_y^b{(u-y)}^{\beta -1}g\left(u\right)du,y<b.$$

(5)

The notations $\left({\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\right)\left(y\right)$ and $\left({\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\right)\left(y\right)$ are called respectively left- and right-sided generalized proportional fractional integral operators of order $\beta$.

Remark 2.

Definition 4 becomes the Riemann–Liouville fractional integrals given in Definition 3 for $\delta =1$.

For Riemann–Liouville fractional integrals, Chen and Huang in [11] obtained the following Simpson’s type inequality for s-convex functions.

Theorem 2.

Let $g:I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^0$ such that $g^{\prime }\in L_1\left([a,b]\right)$, where $a,b\in I^0$ with $a<b$. If $|g^{\prime }|$ is s-convex on $[a,b]$, for some fixed $s\in (0,1]$, then the following inequality holds:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta }g\left(\frac{a+b}{2}\right)\right]\right|\hfill \\ & \le \frac{b-a}{2^{s+1}}{M}_1(\beta ,s)\left[|g^{\prime }\left(a\right)|+|g^{\prime }\left(b\right)|\right],\hfill \end{array}\end{array}$$

(6)

where

$$M_1(\beta ,s)={\int }_0^1\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|\left[{(1+t)}^s+{(1-t)}^s\right]dt.$$

Theorem 3.

Let $g:I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^0$ such that $g^{\prime }\in L_1\left([a,b]\right)$, where $a,b\in I^0$ with $a<b$. If $|g^{\prime }{|}^q$ is s-convex on $[a,b]$, for some fixed $s\in (0,1]$ and $q>1$, then the following inequality holds:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta }g\left(\frac{a+b}{2}\right)\right]\right|\hfill \\ & \le \frac{b-a}{2{(s+1)}^{\frac{1}{q}}}{\left({\int }_0^1{\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|}^{p}dt\right)}^{\frac{1}{p}}\left[{\left(|g^{\prime }\left(b\right){|}^q+|g^{\prime }\left(\frac{a+b}{2}\right){|}^q\right)}^{\frac{1}{q}}+{\left(|g^{\prime }\left(a\right){|}^q+|g^{\prime }\left(\frac{a+b}{2}\right){|}^q\right)}^{\frac{1}{q}}\right],\hfill \end{array}\end{array}$$

(7)

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

Theorem 4.

Let $g:I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^0$ such that $g^{\prime }\in L_1\left([a,b]\right)$, where $a,b\in I^0$ with $a<b$. If $|g^{\prime }{|}^q$ is s-convex on $[a,b]$, for some fixed $s\in (0,1]$ and $q>1$, then the following inequality holds:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta }g\left(\frac{a+b}{2}\right)\right]\right|\hfill \\ & \le \frac{b-a}{2}{\left({\int }_0^1{\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|}^{p}dt\right)}^{\frac{1}{p}}\left[{\left(\frac{(2^{s+1}-1)|g^{\prime }\left(b\right){|}^q+{\left|g^{\prime }\left(a\right)\right|}^q}{2^s(s+1)}\right)}^{\frac{1}{q}}+{\left(\frac{(2^{s+1}-1)|g^{\prime }\left(a\right){|}^q+{\left|g^{\prime }\left(b\right)\right|}^q}{2^s(s+1)}\right)}^{\frac{1}{q}}\right],\hfill \end{array}\end{array}$$

(8)

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

Theorem 5.

Let $g:I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^0$ such that $g^{\prime }\in L_1\left([a,b]\right)$, where $a,b\in I^0$ with $a<b$. If $|g^{\prime }{|}^q$ is s-convex on $[a,b]$, for some fixed $s\in (0,1]$ and $q>1$, then the following inequality holds:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta }g\left(\frac{a+b}{2}\right)\right]\right|\hfill \\ & \le \frac{b-a}{2}{M}_2\left(\beta \right)\left[M_3(\beta ,s)+M_4(\beta ,s)\right],\hfill \end{array}\end{array}$$

(9)

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

$$\begin{array}{cc}\hfill M_2\left(\beta \right)& ={\left({\int }_0^1\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|dt\right)}^{\frac{1}{p}},\hfill \\ \hfill M_3(\beta ,s)& ={\int }_0^1\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|\left[{\left(\frac{1+t}{2}\right)}^s|g^{\prime }{\left(b\right)|}^q+{\left(\frac{1-t}{2}\right)}^s{\left|g^{\prime }\left(a\right)\right|}^q\right]dt,\hfill \\ \hfill M_4(\beta ,s)& ={\int }_0^1\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|\left[{\left(\frac{1+t}{2}\right)}^s|g^{\prime }{\left(a\right)|}^q+{\left(\frac{1-t}{2}\right)}^s{\left|g^{\prime }\left(b\right)\right|}^q\right]dt.\hfill \end{array}$$

In this paper, we introduce new Simpson’s inequalities for s-convex function in the second sense via a generalized proportional fractional integral which is the generalization of the result obtained by Chen and Huang [11]. These types of inequalities can be used to estimate the bounds of both regular and fractional integrals. The paper is organize as follows: In Section 2, we state our main results on inequalities of Simpson’s type for s-convex functions via generalized proportional fractional integral. Finally, Section 3 is devoted to the conclusion of our work.

2. Main Results

The following Lemma is required to prove our main results.

Lemma 1.

Let $g:I\to \mathbb{R}$ be an absolutely continuous mapping on $I^0$ such that $g^{\prime }\in L_1\left([a,b]\right)$, where $a,b\in I^0$ with $a<b$. Then we have the following equality:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)\right]\hfill \\ & =\frac{b-a}{2}{\int }_0^1\left[\left(\frac{t^{\beta }}{2}-\frac{1}{3}\right)g^{\prime }\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)+\left(\frac{1}{3}-\frac{t^{\beta }}{2}\right)g^{\prime }\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\right]dt.\hfill \end{array}\end{array}$$

(10)

Proof. 

Let

$$\begin{array}{cc}\hfill M& ={\int }_0^1\left[\left(\frac{t^{\beta }}{2}-\frac{1}{3}\right)g^{\prime }\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)+\left(\frac{1}{3}-\frac{t^{\beta }}{2}\right)g^{\prime }\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\right]dt\hfill \\ \hfill & ={\int }_0^1\left(\frac{1}{3}-\frac{t^{\beta }}{2}\right)g^{\prime }\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)dt+{\int }_0^1\left(\frac{t^{\beta }}{2}-\frac{1}{3}\right)g^{\prime }\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)dt\hfill \\ \hfill & =M_5+M_6.\hfill \end{array}$$

We compute $M_5$ and $M_6$ by using integration by parts and by change of variable. For this, we get

$$\begin{array}{cc}\hfill M_5& ={\int }_0^1\left(\frac{1}{3}-\frac{t^{\beta }}{2}\right)g^{\prime }\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)dt\hfill \\ \hfill & =\frac{2}{a-b}{\int }_0^1\left(\frac{1}{3}-\frac{t^{\beta }}{2}\right)d\left(g\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\right)\hfill \\ \hfill & =\frac{2}{b-a}\left[\frac{1}{6}g\left(a\right)+\frac{1}{3}g\left(\frac{a+b}{2}\right)-\frac{\beta }{2}{\left(\frac{2}{b-a}\right)}^{\beta }{\int }_a^{\frac{a+b}{2}}{\left(\left(\frac{a+b}{2}\right)-x\right)}^{\beta -1}g\left(x\right)dx\right]\hfill \\ \hfill & =\frac{2}{b-a}\left[\frac{1}{6}g\left(a\right)+\frac{1}{3}g\left(\frac{a+b}{2}\right)-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\frac{1}{\mathsf{\Gamma }\left(\beta \right){\delta }^{\beta }}{\int }_a^{\frac{a+b}{2}}{\left(\left(\frac{a+b}{2}\right)-x\right)}^{\beta -1}g\left(x\right)dx\right]\hfill \\ \hfill & =\frac{2}{b-a}\left[\frac{1}{6}g\left(a\right)+\frac{1}{3}g\left(\frac{a+b}{2}\right)-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}{\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)\right].\hfill \end{array}$$

Using similar argument as outlined above, we obtain:

$$\begin{array}{cc}\hfill M_6& ={\int }_0^1\left(\frac{t^{\beta }}{2}-\frac{1}{3}\right)g^{\prime }\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)dt\hfill \\ \hfill & =\frac{2}{b-a}\left[\frac{1}{6}g\left(b\right)+\frac{1}{3}g\left(\frac{a+b}{2}\right)-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}{\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)\right].\hfill \end{array}$$

By adding $M_5$ and $M_6$, we get the desired identity. □

Theorem 6.

Let $\delta ,\beta >0$ and let $g:I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^0$ such that $g^{\prime }\in L_1\left([a,b]\right)$, where $a,b\in I^0$ with $a<b$. If $|g^{\prime }|$ is s-convex on $[a,b]$, for $s\in (0,1]$, then the following inequalities hold:

$$\begin{array}{l}\left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)\right]\right|\\ \le \frac{b-a}{2^{s+1}}{M}_7(\beta ,s)\left[|g^{\prime }\left(a\right)|+|g^{\prime }\left(b\right)|\right]\\ \le \frac{b-a}{3(s+1)}\left[|g^{\prime }\left(a\right)|+|g^{\prime }\left(b\right)|\right],\end{array}$$

(11)

where

$$M_7(\beta ,s)={\int }_0^1\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|\left[{(1+t)}^s+{(1-t)}^s\right]dt.$$

Proof. 

Using Lemma 1 and s-convexity of $|g^{\prime }|$, we have:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)\right]\right|\hfill \\ & \le \frac{b-a}{2}{\int }_0^1\left[\left|\left(\frac{t^{\beta }}{2}-\frac{1}{3}\right)\right|\left|g^{\prime }\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)\right|+\left|\left(\frac{1}{3}-\frac{t^{\beta }}{2}\right)\right|\left|g^{\prime }\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\right|\right]dt\hfill \\ \hfill & =\frac{b-a}{2}{\int }_0^1\left|\left(\frac{t^{\beta }}{2}-\frac{1}{3}\right)\right|\left[\left|g^{\prime }\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)\right|+\left|g^{\prime }\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\right|\right]dt\hfill \\ \hfill & \le \frac{b-a}{2}{\int }_0^1\left[\left|\left(\frac{t^{\beta }}{2}-\frac{1}{3}\right)\right|\left[{\left(\frac{1+t}{2}\right)}^s|g^{\prime }\left(b\right)|+{\left(1-\frac{1+t}{2}\right)}^s|g^{\prime }\left(a\right)|+{\left(\frac{1+t}{2}\right)}^s|g^{\prime }\left(a\right)|+{\left(1-\frac{1+t}{2}\right)}^s\left|g^{\prime }\left(b\right)\right|\right]\right]dt\hfill \\ & =\frac{b-a}{2^{s+1}}{\int }_0^1\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|\left({(1+t)}^s+{(1-t)}^s\right)\left[|g^{\prime }\left(a\right)|+|g^{\prime }\left(b\right)|\right].\hfill \end{array}\end{array}$$

This complete the proof of the first inequality of (11). The second inequality of (11) follows since $\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|\le \frac{1}{3}$ for ever $t\in [0,1]$ and

$$\begin{array}{c}\hfill M_7(\beta ,s)\le \frac{1}{3}{\int }_0^1\left[{(1+t)}^s+{(1-t)}^s\right]dt=\frac{2^{s+1}}{3(s+1)},\end{array}$$

(12)

which proves the second inequality. □

Corollary 1.

Let $\delta ,\beta >0$ and let $g:I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^0$ such that $g^{\prime }\in L_1\left([a,b]\right)$, where $a,b\in I^0$ with $a<b$. If $|g^{\prime }|$ is convex on $[a,b]$, then the following inequalities hold:

$$\begin{array}{l}\left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)\right]\right|\\ \le \frac{b-a}{4}{M}_7(\beta ,1)\left[|g^{\prime }\left(a\right)|+|g^{\prime }\left(b\right)|\right]\\ \le \frac{b-a}{6}\left[|g^{\prime }\left(a\right)|+|g^{\prime }\left(b\right)|\right].\end{array}$$

(13)

Proof. 

Taking $s=1$ in Theorem 6 we have the result. □

Remark 3.

In Theorem 6 if we put

a.

$\delta =1$, then the first inequality (11) coincide with Theorem 2 and the second inequality coincide with Corollary 8 in [12].

b.

$\delta =\beta =1$, then the first inequality (11) coincide with Theorem 7 in [13].

Theorem 7.

Let $\delta ,\beta >0$ and let $g:I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^0$ such that $g^{\prime }\in L_1\left([a,b]\right)$, where $a,b\in I^0$ with $a<b$. If $|g^{\prime }{|}^q$ is s-convex on $[a,b]$ for $s\in (0,1]$ and $q>1$, then the following inequality holds:

$$\begin{array}{l}\left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)\right]\right|\\ \le \frac{b-a}{2}{\left({\int }_0^1{\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|}^{p}dt\right)}^{\frac{1}{p}}\left[{\left(\frac{|g^{\prime }{\left(b\right)|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{s+1}\right)}^{\frac{1}{q}}+{\left(\frac{|g^{\prime }{\left(a\right)|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{s+1}\right)}^{\frac{1}{q}}\right],\end{array}$$

(14)

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

Proof. 

Using H$\ddot{o}$lder’s inequality and Lemma 1, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)\right]\right|\hfill \\ & \le \frac{b-a}{2}{\int }_0^1\left[\left|\left(\frac{t^{\beta }}{2}-\frac{1}{3}\right)\right|\left|g^{\prime }\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)\right|+\left|\left(\frac{1}{3}-\frac{t^{\beta }}{2}\right)\right|\left|g^{\prime }\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\right|\right]dt\hfill \\ \hfill & =\frac{b-a}{2}{\int }_0^1\left|\left(\frac{t^{\beta }}{2}-\frac{1}{3}\right)\right|\left[\left|g^{\prime }\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)\right|+\left|g^{\prime }\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\right|\right]dt\hfill \\ \hfill & \le \frac{b-a}{2}{\left({\int }_0^1{\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|}^{p}dt\right)}^{\frac{1}{p}}\left[{\left({\int }_0^1{\left|g^{\prime }\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)\right|}^{q}dt\right)}^{\frac{1}{q}}+{\left({\int }_0^1{\left|g^{\prime }\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\right|}^{q}dt\right)}^{\frac{1}{q}}\right].\hfill \end{array}\end{array}$$

Since $|g^{\prime }{|}^q$ is s-convex, by using change of variable and Equation (1), we have

$${\int }_0^1{\left|g^{\prime }\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)\right|}^{q}dt=\frac{2}{b-a}{\int }_{\frac{a+b}{2}}^b{\left|g^{\prime }\left(x\right)\right|}^{q}dx\le \left[\frac{|g^{\prime }{\left(b\right)|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{s+1}\right]$$

(15)

and

$${\int }_0^1{\left|g^{\prime }\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\right|}^{q}dt=\frac{2}{b-a}{\int }_a^{\frac{a+b}{2}}{\left|g^{\prime }\left(x\right)\right|}^{q}dx\le \left[\frac{|g^{\prime }{\left(a\right)|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{s+1}\right].$$

(16)

Hence by using inequalities (15) and (16) we obtain

$$\begin{array}{cc}\hfill & \left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)\right]\right|\hfill \\ & \le \frac{b-a}{2}{\left({\int }_0^1{\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|}^{p}dt\right)}^{\frac{1}{p}}\left[{\left(\frac{|g^{\prime }{\left(b\right)|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{s+1}\right)}^{\frac{1}{q}}+{\left(\frac{|g^{\prime }{\left(a\right)|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{s+1}\right)}^{\frac{1}{q}}\right].\hfill \end{array}$$

Thus, the proof is complete. □

Corollary 2.

Let $\delta ,\beta >0$ and let $g:I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^0$ such that $g^{\prime }\in L_1\left([a,b]\right)$, where $a,b\in I^0$ with $a<b$. If $|g^{\prime }{|}^q$ is convex on $[a,b]$ and $q>1$, then the following inequality holds:

$$\begin{array}{cc}\hfill & \left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)\right]\right|\hfill \\ & \le \frac{b-a}{2}{\left({\int }_0^1{\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|}^{p}dt\right)}^{\frac{1}{p}}\left[{\left(\frac{|g^{\prime }{\left(b\right)|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{2}\right)}^{\frac{1}{q}}+{\left(\frac{|g^{\prime }{\left(a\right)|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{2}\right)}^{\frac{1}{q}}\right],\hfill \end{array}$$

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

Proof. 

We have the result by taking $s=1$ in Theorem 7. □

Remark 4.

In Theorem 7 if we put

a.

$\delta =1$, then the inequality (14) coincide with Theorem 3.

b.

$\delta =\beta =1$, then the inequality (14) coincide with Theorem 8 in [13].

Theorem 8.

Let $\delta ,\beta >0$ and let $g:I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^0$ such that $g^{\prime }\in L_1\left([a,b]\right)$, where $a,b\in I^0$ with $a<b$. If $|g^{\prime }{|}^q$ is $s-$convex on $[a,b]$, for $s\in (0,1]$ and $q>1$, then the following inequalities hold:

$$\begin{array}{l}\left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)\right]\right|\\ \le \frac{b-a}{2}{\left({\int }_0^1{\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|}^{p}dt\right)}^{\frac{1}{p}}\left[{\left(\frac{(2^{s+1}-1)|g^{\prime }\left(b\right){|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{2^s(s+1)}\right)}^{\frac{1}{q}}+{\left(\frac{(2^{s+1}-1)|g^{\prime }\left(a\right){|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{2^s(s+1)}\right)}^{\frac{1}{q}}\right]\\ \le \frac{b-a}{6}\left[{\left(\frac{(2^{s+1}-1)|g^{\prime }\left(b\right){|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{2^s(s+1)}\right)}^{\frac{1}{q}}+{\left(\frac{(2^{s+1}-1)|g^{\prime }\left(a\right){|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{2^s(s+1)}\right)}^{\frac{1}{q}}\right],\end{array}$$

(17)

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

Proof. 

By H$\ddot{o}$lder’s inequality, s-convexity of $|g^{\prime }{|}^q$ and Lemma 1, we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)\right]\right|\hfill \\ & \le \frac{b-a}{2}{\int }_0^1\left[\left|\left(\frac{t^{\beta }}{2}-\frac{1}{3}\right)\right|\left|g^{\prime }\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)\right|+\left|\left(\frac{1}{3}-\frac{t^{\beta }}{2}\right)\right|\left|g^{\prime }\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\right|\right]dt\hfill \\ \hfill & =\frac{b-a}{2}{\int }_0^1\left|\left(\frac{t^{\beta }}{2}-\frac{1}{3}\right)\right|\left[\left|g^{\prime }\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)\right|+\left|g^{\prime }\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\right|\right]dt\hfill \\ \hfill & \le \frac{b-a}{2}{\left({\int }_0^1{\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|}^{p}dt\right)}^{\frac{1}{p}}\left[{\left({\int }_0^1{\left|g^{\prime }\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)\right|}^{q}dt\right)}^{\frac{1}{q}}+{\left({\int }_0^1{\left|g^{\prime }\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\right|}^{q}dt\right)}^{\frac{1}{q}}\right]\hfill \\ \hfill & \le \frac{b-a}{2}{\left({\int }_0^1{\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|}^{p}dt\right)}^{\frac{1}{p}}[{\left({\int }_0^1{\left(\frac{1+t}{2}\right)}^s|g^{\prime }{\left(b\right)|}^q+{\left(1-\frac{1+t}{2}\right)}^s{\left|g^{\prime }\left(a\right)\right|}^{q}dt\right)}^{\frac{1}{q}}+\hfill \\ & {\int }_0^1{\left({\left(\frac{1+t}{2}\right)}^s|g^{\prime }{\left(a\right)|}^q+{\left(1-\frac{1+t}{2}\right)}^s{\left|g^{\prime }\left(b\right)\right|}^{q}dt\right)}^{\frac{1}{q}}]\hfill \\ \hfill & =\frac{b-a}{2}{\left({\int }_0^1{\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|}^{p}dt\right)}^{\frac{1}{p}}\left[{\left(\frac{(2^{s+1}-1)|g^{\prime }\left(b\right){|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{2^s(s+1)}\right)}^{\frac{1}{q}}+{\left(\frac{(2^{s+1}-1)|g^{\prime }\left(a\right){|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{2^s(s+1)}\right)}^{\frac{1}{q}}\right],\hfill \end{array}\end{array}$$

which is the desired first inequality. We get the second inequality from the first inequality due to the fact that $\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|\le \frac{1}{3}$ for all $t\in [0,1]$ and ${\displaystyle {\int }_0^1{(1+t)}^{s}dt=\frac{2^{s+1}-1}{(s+1)}}$ and ${\displaystyle {\int }_0^1{(1-t)}^{s}dt=\frac{1}{(s+1)}}$.

That concludes the proof. □

Corollary 3.

Let $\delta ,\beta >0$ and let $g:I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^0$ such that $g^{\prime }\in L_1\left([a,b]\right)$, where $a,b\in I^0$ with $a<b$. If $|g^{\prime }{|}^q$ is convex on $[a,b]$ and $q>1$, then the following inequalities hold:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)\right]\right|\hfill \\ & \le \frac{b-a}{2}{\left({\int }_0^1{\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|}^{p}dt\right)}^{\frac{1}{p}}\left[{\left(\frac{3|g^{\prime }{\left(b\right)|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{4}\right)}^{\frac{1}{q}}+{\left(\frac{3|g^{\prime }{\left(a\right)|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{4}\right)}^{\frac{1}{q}}\right]\hfill \\ \hfill & \le \frac{b-a}{6}\left[{\left(\frac{\left(3\right)|g^{\prime }{\left(b\right)|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{4}\right)}^{\frac{1}{q}}+{\left(\frac{3|g^{\prime }{\left(a\right)|}^q+{\left|g^{\prime }\left(\frac{a+b}{2}\right)\right|}^q}{4}\right)}^{\frac{1}{q}}\right],\hfill \end{array}\end{array}$$

(18)

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

Proof. 

Taking $s=1$ in Theorem 8 we get the result. □

Remark 5.

In Theorem 8 if we put

a.

$\delta =1$, then the first inequality (17) coincide with Theorem 4.

b.

$\delta =\beta =1$, then the first inequality (17) coincide with Theorem 9 in [13].

Theorem 9.

Let $\delta ,\beta >0$ and let $g:I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^0$ such that $g^{\prime }\in L_1\left([a,b]\right)$, where $a,b\in I^0$ with $a<b$. If $|g^{\prime }{|}^q$ is s-convex on $[a,b]$ for $s\in (0,1]$ and $q>1$, then the following inequalities hold:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)\right]\right|\hfill \\ \hfill & \le \frac{b-a}{2^{1+\frac{s}{q}}}{M}_8\left(\beta \right)\left[{\left(M_9(\beta ,s)|g^{\prime }{\left(b\right)|}^q+M_{10}(\beta ,s){\left|g^{\prime }\left(a\right)\right|}^q\right)}^{\frac{1}{q}}+{\left(M_9(\beta ,s)|g^{\prime }{\left(a\right)|}^q+M_{10}(\beta ,s){\left|g^{\prime }\left(b\right)\right|}^q\right)}^{\frac{1}{q}}\right]\hfill \\ \hfill & \le \frac{b-a}{2^{1+\frac{s}{q}}}{\left(\frac{1}{3}\right)}^{\frac{1}{p}}\left[{\left(\frac{2^{s+1}-1}{3(s+1)}|g^{\prime }{\left(b\right)|}^q+\frac{1}{3(s+1)}{\left|g^{\prime }\left(a\right)\right|}^q\right)}^{\frac{1}{q}}+{\left(\frac{2^{s+1}-1}{3(s+1)}|g^{\prime }{\left(a\right)|}^q+\frac{1}{3(s+1)}{\left|g^{\prime }\left(b\right)\right|}^q\right)}^{\frac{1}{q}}\right],\hfill \end{array}\end{array}$$

(19)

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

$$\begin{array}{cc}\hfill M_8\left(\beta \right)& ={\left({\int }_0^1{\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|}^{p}dt\right)}^{\frac{1}{p}},\hfill \\ \hfill M_9(\beta ,s)& ={\int }_0^1\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|{(1+t)}^{s}dt,\hfill \\ \hfill M_{10}(\beta ,s)& ={\int }_0^1\left|\frac{1}{3}-\frac{t^{\beta }}{2}\right|{(1-t)}^{s}dt.\hfill \end{array}$$

Proof. 

Using Lemma 1, s-convexity of $|g^{\prime }{|}^q$, the power mean inequality and H$\ddot{o}$lder’s inequality, we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)\right]\right|\hfill \\ \hfill & \le \frac{b-a}{2}{\int }_0^1\left[\left|\left(\frac{t^{\beta }}{2}-\frac{1}{3}\right)\right|\left|g^{\prime }\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)\right|+\left|\left(\frac{1}{3}-\frac{t^{\beta }}{2}\right)\right|\left|g^{\prime }\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\right|\right]dt\hfill \\ \hfill & \le \frac{b-a}{2}[{\left({\int }_0^1\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|dt\right)}^{\frac{1}{p}}{\left({\int }_0^1\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|{\left|g^{\prime }\left(\frac{1+t}{2}b+\frac{1-t}{2}a\right)\right|}^{q}dt\right)}^{\frac{1}{q}}\hfill \\ & +{\left({\int }_0^1\left|\frac{1}{3}-\frac{t^{\beta }}{2}\right|dt\right)}^{\frac{1}{p}}{\left({\int }_0^1\left|\frac{1}{3}-\frac{t^{\beta }}{2}\right|{\left|g^{\prime }\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)\right|}^{q}dt\right)}^{\frac{1}{q}}]\hfill \\ \hfill & \le \frac{b-a}{2}[{\left({\int }_0^1\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|dt\right)}^{\frac{1}{p}}{\left({\int }_0^1\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|\left[{\left(\frac{1+t}{2}\right)}^s|g^{\prime }{\left(b\right)|}^q+{\left(1-\frac{1+t}{2}\right)}^s{\left|g^{\prime }\left(a\right)\right|}^{q}dt\right]\right)}^{\frac{1}{q}}+\hfill \\ & {\left({\int }_0^1\left|\frac{1}{3}-\frac{t^{\beta }}{2}\right|dt\right)}^{\frac{1}{p}}{\left({\int }_0^1\left|\frac{1}{3}-\frac{t^{\beta }}{2}\right|\left[{\left(\frac{1+t}{2}\right)}^s|g^{\prime }{\left(a\right)|}^q+{\left(1-\frac{1+t}{2}\right)}^s{\left|g^{\prime }\left(b\right)\right|}^{q}dt\right]\right)}^{\frac{1}{q}}]\hfill \\ \hfill & =\frac{b-a}{2^{1+\frac{s}{q}}}[{\left({\int }_0^1\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|dt\right)}^{\frac{1}{p}}{\left({\int }_0^1\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|\left[{(1+t)}^s|g^{\prime }{\left(b\right)|}^q+{(1-t)}^s{\left|g^{\prime }\left(a\right)\right|}^{q}dt\right]\right)}^{\frac{1}{q}}+\hfill \\ & {\left({\int }_0^1\left|\frac{1}{3}-\frac{t^{\beta }}{2}\right|dt\right)}^{\frac{1}{p}}{\left({\int }_0^1\left|\frac{1}{3}-\frac{t^{\beta }}{2}\right|\left[{(1+t)}^s|g^{\prime }{\left(a\right)|}^q+{(1-t)}^s{\left|g^{\prime }\left(b\right)\right|}^{q}dt\right]\right)}^{\frac{1}{q}}],\hfill \end{array}\end{array}$$

which completes the first inequality. We get the second inequality from the first inequality due to the facts that:

$$\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|<\frac{1}{3}$$

for all $t\in [0,1],$

$${\displaystyle {\int }_0^1\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|{(1+t)}^{s}dt\le \frac{2^{s+1}-1}{3(s+1)}}$$

and

$${\displaystyle {\int }_0^1\left|\frac{t^{\beta }}{2}-\frac{1}{3}\right|{(1-t)}^{s}dt\le \frac{1}{3(s+1)}.}$$

This concludes the proof. □

Corollary 4.

Let $\delta ,\beta >0$ and let $g:I\subset [0,\infty )\to \mathbb{R}$ be a differentiable mapping on $I^0$ such that $g^{\prime }\in L_1\left([a,b]\right)$, where $a,b\in I^0$ with $a<b$. If $|g^{\prime }{|}^q$ is convex on $[a,b]$ and $q>1$, then the following inequalities hold:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \left|\frac{1}{6}\left[g\left(a\right)+4g\left(\frac{a+b}{2}\right)+g\left(b\right)\right]-\frac{2^{\beta -1}{\delta }^{\beta }\mathsf{\Gamma }(\beta +1)}{{(b-a)}^{\beta }}\left[{\mathcal{J}}_{a^{+}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)+{\mathcal{J}}_{b^{-}}^{\beta ,\delta }g\left(\frac{a+b}{2}\right)\right]\right|\hfill \\ & \le \frac{b-a}{2^{1+\frac{s}{q}}}{M}_8\left(\beta \right)\left[{\left(M_9(\beta ,1)|g^{\prime }{\left(b\right)|}^q+M_{10}(\beta ,1){\left|g^{\prime }\left(a\right)\right|}^q\right)}^{\frac{1}{q}}+{\left(M_9(\beta ,1)|g^{\prime }{\left(a\right)|}^q+M_{10}(\beta ,1){\left|g^{\prime }\left(b\right)\right|}^q\right)}^{\frac{1}{q}}\right]\hfill \\ \hfill & \le \frac{b-a}{2^{1+\frac{1}{q}}}{\left(\frac{1}{3}\right)}^{\frac{1}{p}}\left[{\left(\frac{1}{2}|g^{\prime }{\left(b\right)|}^q+\frac{1}{6}{\left|g^{\prime }\left(a\right)\right|}^q\right)}^{\frac{1}{q}}+{\left(\frac{1}{2}|g^{\prime }{\left(a\right)|}^q+\frac{1}{6}{\left|g^{\prime }\left(b\right)\right|}^q\right)}^{\frac{1}{q}}\right],\hfill \end{array}\end{array}$$

(20)

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

Proof. 

By taking $s=1$ in Theorem 9 we have the desired result. □

Remark 6.

In Theorem 9 if we put

a.

$\delta =1$, then the first inequality (19) coincide with Theorem 5.

b.

$\delta =\beta =1$, then the first inequality (19) coincide with Theorem 10 in [13].

3. Conclusions

Our results have introduced a new integral inequality of Simpson’s type integral inequalities using s-convexity via generalized proportional fractional integrals. The inequalities obtained are generalizations of Simpson’s type inequality that are given for the Riemann–Liouville fractional integrals in [13]. Similar inequalities could possibly be established for more generalized fractional integrals such as Riemann–Liouville fractional integrals of a function with respect to another generalized function and to a proportional fractional integral of a function with respect to another function.