Fractional Simpson-like Inequalities with Parameter for Differential s-tgs-Convex Functions

Abstract

In this paper, we first prove a new parameterized identity. Based on this identity we establish some parametrized Simpson-like type symmetric inequalities, for functions whose first derivatives are s-$tgs$-convex via Reimann–Liouville frational operators. Some special cases are discussed. Applications to numerical quadrature are provided.

1. Introduction

Symmetric inequalities often arise in various branches of mathematics, such as algebra, analysis, and optimization. They have numerous applications and play a crucial role in proving theorems and solving problems in areas like number theory, combinatorics, and inequalities themselves. In the study of symmetric inequalities, techniques such as rearrangement inequality, Cauchy–Schwarz inequality, and the method of Lagrange multipliers are commonly employed to establish the validity of the inequalities and find optimal solutions [1,2,3].

Fractional calculus is a field of applied mathematics that deals with derivatives and integrals of arbitrary orders and their applications in different fields of sciences and engineering. In real life, fractional calculus is generated from various fractional operators such as Riemann–Liouville, Caputo, Hadamard, and so on; due to its widespread use in different fields, this calculus has attracted many researchers. The most-used operator is that of Riemann-Liouville given by the following definition

Definition 1

([4]). For any integrable function $\mathcal{L}$ on $[\mathit{k},g]$ with $\mathit{k}\ge 0$, $I_{{\mathit{k}}^{+}}^{\varsigma }\mathcal{L}$ and $I_{g^{-}}^{\varsigma }\mathcal{L}$ are the Riemann–Liouville fractional integrals of order $\varsigma >0$ given by

$$\begin{array}{ccc}\hfill I_{{\mathit{k}}^{+}}^{\varsigma }\mathcal{L}\left(x\right)& =& \frac{1}{\mathsf{\Gamma }\left(\varsigma \right)}\stackrel{x}{\underset{\mathit{k}}{\int }}{\left(x-\mathsf{\Lambda }\right)}^{\varsigma -1}\mathcal{L}\left(\mathsf{\Lambda }\right)d\mathsf{\Lambda },x>\mathit{k},\hfill \\ \hfill I_{g^{-}}^{\varsigma }\mathcal{L}\left(x\right)& =& \frac{1}{\mathsf{\Gamma }\left(\varsigma \right)}\stackrel{g}{\underset{x}{\int }}{\left(\mathsf{\Lambda }-x\right)}^{\varsigma -1}\mathcal{L}\left(\mathsf{\Lambda }\right)d\mathsf{\Lambda },g>x,\hfill \end{array}$$

respectively, where $\mathsf{\Gamma }\left(\varsigma \right)=\stackrel{\infty }{\underset{0}{\int }}$$e^{-\mathsf{\Lambda }}{\mathsf{\Lambda }}^{\varsigma -1}d\mathsf{\Lambda }$ is the gamma function and

$$I_{{\mathit{k}}^{+}}^0\mathcal{L}\left(x\right)=I_{g^{-}}^0\mathcal{L}\left(x\right)=\mathcal{L}\left(x\right).$$

The theory of convexity plays a central and attractive role in many fields of research. This theory provides us with a powerful tool for solving a large class of problems that arise in pure and applied mathematics, defined as follows:

Definition 2

([5]). A function $\mathcal{L}:I\to \mathbb{R}$ is said to be convex, if

$$\mathcal{L}\left(\mathsf{\Lambda }x+\left(1-\mathsf{\Lambda }\right)y\right)\le \mathsf{\Lambda }\mathcal{L}\left(x\right)+\left(1-\mathsf{\Lambda }\right)\mathcal{L}\left(y\right)$$

holds for all $x,y\in I$ and all $\mathsf{\Lambda }\in [0,1]$.

In [6], Awan et al. introduced the class of s-$tgs$-convex functions.

Definition 3

([6]). We say that a function $\mathcal{L}:I\subset \mathbb{R}\to \mathbb{R}$ is s-$tgs$-convex on I, if

$$\mathcal{L}\left(\mathsf{\Lambda }x+\left(1-\mathsf{\Lambda }\right)y\right)\le {\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^s\left(\mathcal{L}\left(x\right)+\mathcal{L}\left(y\right)\right)$$

holds for all $x,y\in I$ and $\mathsf{\Lambda }\in [0,1]$, with $s\in \left[0,1\right]$.

Convexity has a close relation in the development of the theory of inequalities, of which it plays an important role in the study of qualitative properties of solutions of ordinary, partial, and integral differential equations as well as in numerical analysis, which is used for establishing the estimates of the errors for quadrature rules; see [7,8,9,10,11,12,13,14,15,16,17,18].

The following Newton–Cotes inequality involving four points is known in the literature as the $3/8$-Simpson inequality

$$\left|\frac{1}{8}\left(\mathcal{L}\left(\mathit{k}\right)+3\mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+3\mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{1}{g-\mathit{k}}\underset{\mathit{k}}{\stackrel{g}{\int }}\mathcal{L}\left(w\right)dw\right|\le \frac{{\left(g-\mathit{k}\right)}^4}{6480}{∥{\mathcal{L}}^{\left(4\right)}∥}_{\infty },$$

(1)

where $\mathcal{L}$ is a four-times continuously differentiable function on $\left[\mathit{k},g\right]$ and

$${∥{\mathcal{L}}^{\left(4\right)}∥}_{\infty }=\underset{x\in \left[\mathit{k},g\right]}{sup}\left|{\mathcal{L}}^{\left(4\right)}\left(x\right)\right|.$$

Recently, Mahmoudi and Meftah [19] discussed more general inequalities of four points and gave the following results

Theorem 1.

Let $\mathcal{L}:\left[\mathit{k},g\right]\to \mathbb{R}$ be a differentiable function on $\left[\mathit{k},g\right]$ such that ${\mathcal{L}}^{\prime }\in L^1\left[\mathit{k},g\right]$ with $0\le \mathit{k}<g$. If $\left|{\mathcal{L}}^{\prime }\right|$ is s-convex in the second sense for some fixed $s\in \left(0,1\right]$, then we have

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\rho }\left(\mathcal{L}\left(\mathit{k}\right)+\rho \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\rho \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{1}{g-\mathit{k}}\underset{\mathit{k}}{\stackrel{g}{\int }}\mathcal{L}\left(w\right)dw\right|\hfill \\ & \le & \frac{g-\mathit{k}}{9\left(s+1\right)\left(s+2\right)}\left(\left(\frac{3s+4-2\rho }{2+2\rho }+2{\left(\frac{2\rho -1}{2+2\rho }\right)}^{s+2}\right)\left(\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(g\right)\right|\right)\right.\hfill \\ & & +\left.\left(\frac{3\rho s+\left(2\rho -4\right)}{2+2\rho }+{\left(\frac{1}{2}\right)}^{s+1}+2{\left(\frac{3}{2+2\rho }\right)}^{s+2}\right)\left(\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|\right)\right),\hfill \end{array}$$

where ρ is a positive number.

Theorem 2.

Let $\mathcal{L}:\left[\mathit{k},g\right]\to \mathbb{R}$ be a differentiable function on $\left[\mathit{k},g\right]$ such that ${\mathcal{L}}^{\prime }\in L^1\left[\mathit{k},g\right]$ with $0\le \mathit{k}<g$. If ${\left|{\mathcal{L}}^{\prime }\right|}^q$ is s-convex in the second sense for some fixed $s\in \left(0,1\right]$, then we have

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\rho }\left(\mathcal{L}\left(\mathit{k}\right)+\rho \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\rho \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{1}{g-\mathit{k}}\underset{\mathit{k}}{\stackrel{g}{\int }}\mathcal{L}\left(w\right)dw\right|\hfill \\ & \le & \frac{g-\mathit{k}}{18{\left(p+1\right)}^{p+1}}\left({\left(\frac{3^{p+1}+{\left(2\rho -1\right)}^{p+1}}{2{\left(1+\rho \right)}^{p+1}}\right)}^{\frac{1}{p}}{\left(\frac{{\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^q+{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^q}{s+1}\right)}^{\frac{1}{q}}\right.\hfill \\ & & +\left.{\left(\frac{{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^q+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^q}{s+1}\right)}^{\frac{1}{q}}+{\left(\frac{3^{p+1}+{\left(2\rho -1\right)}^{p+1}}{2{\left(1+\rho \right)}^{p+1}}\right)}^{\frac{1}{p}}{\left(\frac{{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^q+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^q}{s+1}\right)}^{\frac{1}{q}}\right),\hfill \end{array}$$

where ρ is a positive number and $q>1$ with $\frac{1}{p}+\frac{1}{q}=1$.

Theorem 3.

Let $\mathcal{L}:\left[\mathit{k},g\right]\to \mathbb{R}$ be a differentiable function on $\left[\mathit{k},g\right]$ such that ${\mathcal{L}}^{\prime }\in L^1\left[\mathit{k},g\right]$ with $0\le \mathit{k}<g$. If ${\left|{\mathcal{L}}^{\prime }\right|}^q$ is s-convex in the second sense for some fixed $s\in \left(0,1\right]$, then we have

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\rho }\left(\mathcal{L}\left(\mathit{k}\right)+\rho \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\rho \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{1}{g-\mathit{k}}\underset{\mathit{k}}{\stackrel{g}{\int }}\mathcal{L}\left(w\right)dw\right|\hfill \\ & \le & \frac{g-\mathit{k}}{9{\left(\left(s+1\right)\left(s+2\right)\right)}^{\frac{1}{q}}}\left({\left(\frac{9+{\left(2\rho -1\right)}^2}{8{\left(1+\rho \right)}^2}\right)}^{1-\frac{1}{q}}\left(\left(\frac{3s+4-2\rho }{2+2\rho }+2{\left(\frac{2\rho -1}{2+2\rho }\right)}^{s+2}\right){\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^q\right.\right.\hfill \\ & & +{\left.\left(\frac{\left(2\rho -1\right)s+\left(2\rho -4\right)}{2+2\rho }+2{\left(\frac{3}{2+2\rho }\right)}^{s+2}\right){\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^q\right)}^{\frac{1}{q}}\hfill \\ & & +\frac{1}{4}{\left(2s+{\left(\frac{1}{2}\right)}^{s-1}\right)}^{\frac{1}{q}}{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^q+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^q\right)}^{\frac{1}{q}}\hfill \\ & & +{\left(\frac{9+{\left(2\rho -1\right)}^2}{8{\left(1+\rho \right)}^2}\right)}^{1-\frac{1}{q}}\left(\left(\frac{\left(2\rho -1\right)s+\left(2\rho -4\right)}{2+2\rho }+2{\left(\frac{3}{2+2\rho }\right)}^{s+2}\right){\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^q\right.\hfill \\ & & +\left.{\left.\left(\frac{3s+4-2\rho }{2+2\rho }+2{\left(\frac{2\rho -1}{2+2\rho }\right)}^{s+2}\right){\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^q\right)}^{\frac{1}{q}}\right),\hfill \end{array}$$

where ρ is a positive number and $q\ge 1$.

Motivated by the above results, in this paper we first prove a new parameterized identity. Based on this identity, we establish some new fractional Simpson-like type inequalities for functions whose first derivatives are s-$tgs$-convex. We end this work with some applications.

2. Main Results

Let us first recall some special functions (see [4]).

The incomplete beta function is given by

$$B_m\left({\xi }_1,{\xi }_2\right)=\stackrel{m}{\underset{0}{\int }}{\mathsf{\Lambda }}^{{\xi }_1-1}{\left(1-\mathsf{\Lambda }\right)}^{{\xi }_2-1}d\mathsf{\Lambda },$$

where ${\xi }_1,{\xi }_2\in \mathbb{C}$ such that $\Re \left({\xi }_1\right)>0$, $\Re \left({\xi }_2\right)>0$ and $0\le m<1$. The case where $m=1$ gives the classical beta function, i.e.,

$$B\left({\xi }_1,{\xi }_2\right)=\stackrel{1}{\underset{0}{\int }}{\mathsf{\Lambda }}^{{\xi }_1-1}{\left(1-\mathsf{\Lambda }\right)}^{{\xi }_2-1}d\mathsf{\Lambda }.$$

The hypergeometric function is defined for $\Re \left(c\right)>\Re \left(b\right)>0$ and $\left|z\right|<1$, as follows

$${}_2{F}_1\left(a,b,c;z\right)=\frac{1}{B\left(b,c-b\right)}\underset{0}{\stackrel{1}{\int }}{\mathsf{\Lambda }}^{b-1}{\left(1-\mathsf{\Lambda }\right)}^{c-b-1}{\left(1-z\mathsf{\Lambda }\right)}^{-a}d\mathsf{\Lambda },$$

where $B\left(·,·\right)$ is the beta function.

Lemma 1.

Let $\mathcal{L}:\left[\mathit{k},g\right]\to \mathbb{R}$ be a differentiable function on $\left[\mathit{k},g\right]$ such that ${\mathcal{L}}^{\prime }\in L^1\left[\mathit{k},g\right]$, then the following equality holds

$$\begin{array}{ccc}& & \frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{3^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha }}\mathcal{R}\left(\alpha ,\mathcal{L}\right)\hfill \\ & =& \frac{g-\mathit{k}}{9}\left(\underset{0}{\stackrel{1}{\int }}\left({\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right){\mathcal{L}}^{\prime }\left(\left(1-\mathsf{\Lambda }\right)\mathit{k}+\mathsf{\Lambda }\frac{2\mathit{k}+g}{3}\right)d\mathsf{\Lambda }\right.\hfill \\ & & +\underset{0}{\stackrel{1}{\int }}\left(\frac{1}{2}-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right){\mathcal{L}}^{\prime }\left(\left(1-\mathsf{\Lambda }\right)\frac{2\mathit{k}+g}{3}+\mathsf{\Lambda }\frac{\mathit{k}+2g}{3}\right)d\mathsf{\Lambda }\hfill \\ & & +\left.\underset{0}{\stackrel{1}{\int }}\left(\frac{3}{2+2\theta }-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right){\mathcal{L}}^{\prime }\left(\left(1-\mathsf{\Lambda }\right)\frac{\mathit{k}+2g}{3}+\mathsf{\Lambda }g\right)d\mathsf{\Lambda }\right),\hfill \end{array}$$

where θ is positive number, $0<\alpha \le 1$, and

$$\mathcal{R}\left(\alpha ,\mathcal{L}\right)=I_{{\left(\frac{2\mathit{k}+g}{3}\right)}^{-}}^{\alpha }\mathcal{L}\left(\mathit{k}\right)+I_{{\left(\frac{2\mathit{k}+g}{3}\right)}^{+}}^{\alpha }\mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+I_{{\left(\frac{\mathit{k}+2g}{3}\right)}^{+}}^{\alpha }\mathcal{L}\left(g\right).$$

(2)

Proof. 

Let

$$\begin{array}{ccc}\hfill I_1& =& \underset{0}{\stackrel{1}{\int }}\left({\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right){\mathcal{L}}^{\prime }\left(\left(1-\mathsf{\Lambda }\right)\mathit{k}+\mathsf{\Lambda }\frac{2\mathit{k}+g}{3}\right)d\mathsf{\Lambda },\hfill \\ \hfill I_2& =& \underset{0}{\stackrel{1}{\int }}\left(\frac{1}{2}-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right){\mathcal{L}}^{\prime }\left(\left(1-\mathsf{\Lambda }\right)\frac{2\mathit{k}+g}{3}+\mathsf{\Lambda }\frac{\mathit{k}+2g}{3}\right)d\mathsf{\Lambda }\hfill \end{array}$$

and

$$I_3=\underset{0}{\stackrel{1}{\int }}\left(\frac{3}{2+2\theta }-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right){\mathcal{L}}^{\prime }\left(\left(1-\mathsf{\Lambda }\right)\frac{\mathit{k}+2g}{3}+\mathsf{\Lambda }g\right)d\mathsf{\Lambda }.$$

By using integration by parts in $I_1$, we obtain

$$\begin{array}{ccc}\hfill I_1& =& {\left.\frac{3}{g-\mathit{k}}\left({\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right)\mathcal{L}\left(\left(1-\mathsf{\Lambda }\right)\mathit{k}+\mathsf{\Lambda }\frac{2\mathit{k}+g}{3}\right)\right|}_{\mathsf{\Lambda }=0}^{\mathsf{\Lambda }=1}\hfill \\ & & -\frac{3\alpha }{g-\mathit{k}}\underset{0}{\stackrel{1}{\int }}{\mathsf{\Lambda }}^{\alpha -1}\mathcal{L}\left(\left(1-\mathsf{\Lambda }\right)a+\mathsf{\Lambda }\frac{2\mathit{k}+g}{3}\right)d\mathsf{\Lambda }\hfill \\ & =& \frac{6\theta -3}{\left(g-\mathit{k}\right)\left(2+2\theta \right)}\mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\frac{9}{\left(g-\mathit{k}\right)\left(2+2\theta \right)}\mathcal{L}\left(\mathit{k}\right)\hfill \\ & & -\frac{3^{\alpha +1}\alpha }{{\left(g-\mathit{k}\right)}^{\alpha +1}}\underset{\mathit{k}}{\stackrel{\frac{2\mathit{k}+g}{3}}{\int }}{\left(u-a\right)}^{\alpha -1}\mathcal{L}\left(u\right)du\hfill \\ & =& \frac{6\theta -3}{\left(g-\mathit{k}\right)\left(2+2\theta \right)}\mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\frac{9}{\left(g-\mathit{k}\right)\left(2+2\theta \right)}\mathcal{L}\left(\mathit{k}\right)-\frac{3^{\alpha +1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha +1}}{I}_{{\frac{2\mathit{k}+g}{3}}^{-}}^{\alpha }\mathcal{L}\left(\mathit{k}\right).\hfill \end{array}$$

(3)

Similarly, we obtain

$$\begin{array}{ccc}\hfill I_2& =& {\left.\frac{3}{g-\mathit{k}}\left(\frac{1}{2}-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right)\mathcal{L}\left(\left(1-\mathsf{\Lambda }\right)\frac{2\mathit{k}+g}{3}+\mathsf{\Lambda }\frac{\mathit{k}+2g}{3}\right)\right|}_{\mathsf{\Lambda }=0}^{\mathsf{\Lambda }=1}\hfill \\ & & -\frac{3\alpha }{g-\mathit{k}}\underset{0}{\stackrel{1}{\int }}{\left(1-\mathsf{\Lambda }\right)}^{\alpha -1}\mathcal{L}\left(\left(1-\mathsf{\Lambda }\right)\frac{2\mathit{k}+g}{3}+\mathsf{\Lambda }\frac{\mathit{k}+2g}{3}\right)d\mathsf{\Lambda }\hfill \\ & =& \frac{3}{2\left(g-\mathit{k}\right)}\mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\frac{3}{2\left(g-\mathit{k}\right)}\mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)\hfill \\ & & -\frac{3^{\alpha +1}\alpha }{{\left(g-\mathit{k}\right)}^{\alpha +1}}\underset{\frac{2\mathit{k}+g}{3}}{\stackrel{\frac{\mathit{k}+2g}{3}}{\int }}{\left(\frac{\mathit{k}+2g}{3}-u\right)}^{\alpha -1}\mathcal{L}\left(u\right)du\hfill \\ & =& \frac{3}{2\left(g-\mathit{k}\right)}\mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\frac{3}{2\left(g-\mathit{k}\right)}\mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)-\frac{3^{\alpha +1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha +1}}{I}_{{\left(\frac{2\mathit{k}+g}{3}\right)}^{+}}^{\alpha }\mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)\hfill \end{array}$$

(4)

and

$$\begin{array}{ccc}\hfill I_3& =& {\left.\frac{3}{g-\mathit{k}}\left(\frac{3}{2+2\theta }-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right)\mathcal{L}\left(\left(1-\mathsf{\Lambda }\right)\frac{\mathit{k}+2g}{3}+\mathsf{\Lambda }g\right)\right|}_{\mathsf{\Lambda }=0}^{\mathsf{\Lambda }=1}\hfill \\ & & -\frac{3\alpha }{g-\mathit{k}}\underset{0}{\stackrel{1}{\int }}{\left(1-\mathsf{\Lambda }\right)}^{\alpha -1}\mathcal{L}\left(\left(1-\mathsf{\Lambda }\right)\frac{\mathit{k}+2g}{3}+\mathsf{\Lambda }g\right)dt\hfill \\ & =& \frac{9}{\left(2+2\theta \right)\left(g-\mathit{k}\right)}\mathcal{L}\left(g\right)+\frac{6\theta -3}{\left(g-\mathit{k}\right)\left(2+2\theta \right)}\mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)\hfill \\ & & -\frac{3^{\alpha +1}\alpha }{{\left(g-\mathit{k}\right)}^{\alpha +1}}\underset{\frac{\mathit{k}+2g}{3}}{\stackrel{g}{\int }}{\left(g-u\right)}^{\alpha -1}\mathcal{L}\left(u\right)du\hfill \\ & =& \frac{9}{\left(g-\mathit{k}\right)\left(2+2\theta \right)}\mathcal{L}\left(g\right)+\frac{6\theta -3}{\left(g-\mathit{k}\right)\left(2+2\theta \right)}\mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)-\frac{3^{\alpha +1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha +1}}{I}_{{\left(\frac{\mathit{k}+2g}{3}\right)}^{+}}^{\alpha }\mathcal{L}\left(g\right).\hfill \end{array}$$

(5)

Summing (3)–(5), and then multiplying the resulting equality by $\frac{g-\mathit{k}}{9}$, we obtain the desired result. □

Theorem 4.

Assume that all the assumptions of Lemma 1 are satisfied. Moreover, if $\left|{\mathcal{L}}^{\prime }\right|$ is s-$tgs$-convex, then the following inequality holds

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{3^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha }}\mathcal{R}\left(\alpha ,\mathcal{L}\right)\right|\hfill \\ & \le & \frac{g-\mathit{k}}{9}\left({\phi }_{\theta }\left(s+1,s+1;\alpha \right)\left(\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(g\right)\right|\right)\right.\hfill \\ & & +\left.\left({\phi }_2\left(s+1,s+1;\alpha \right)+{\phi }_{\theta }\left(s+1,s+1;\alpha \right)\right)\left(\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|\right)\right),\hfill \end{array}$$

where $s\in \left[0,1\right],\mathcal{R}\left(\alpha ,\mathcal{L}\right)$ is defined by (2),

$${\varkappa }_{\theta }\left(x,y;\alpha \right)=B_{{\left(\frac{3}{2+2\theta }\right)}^{\frac{1}{\alpha }}}\left(x,y\right)-B_{1-{\left(\frac{3}{2+2\theta }\right)}^{\frac{1}{\alpha }}}\left(y,x\right)$$

(6)

and

$${\phi }_{\theta }\left(x,y;\alpha \right)=\frac{3}{2+2\theta }{\varkappa }_{\theta }\left(x,y;\alpha \right)-{\varkappa }_{\theta }\left(x+\alpha ,y;\alpha \right).$$

(7)

Proof. 

By Lemma 1, modulus, and the s-$tgs$-convexity of $\left|{\mathcal{L}}^{\prime }\right|$, we determine

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{3^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha }}\mathcal{R}\left(\alpha ,\mathcal{L}\right)\right|\hfill \\ & \le & \frac{g-\mathit{k}}{9}\left(\underset{0}{\stackrel{1}{\int }}\left|{\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right|\left|{\mathcal{L}}^{\prime }\left(\left(1-\mathsf{\Lambda }\right)\mathit{k}+\mathsf{\Lambda }\frac{2\mathit{k}+g}{3}\right)\right|d\mathsf{\Lambda }\right.\hfill \\ & & +\underset{0}{\stackrel{1}{\int }}\left|\frac{1}{2}-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|\left|{\mathcal{L}}^{\prime }\left(\left(1-\mathsf{\Lambda }\right)\frac{2\mathit{k}+g}{3}+\mathsf{\Lambda }\frac{\mathit{k}+2g}{3}\right)\right|d\mathsf{\Lambda }\hfill \\ & & +\left.\underset{0}{\stackrel{1}{\int }}\left|\frac{3}{2+2\theta }-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|\left|{\mathcal{L}}^{\prime }\left(\left(1-\mathsf{\Lambda }\right)\frac{\mathit{k}+2g}{3}+\mathsf{\Lambda }g\right)\right|d\mathsf{\Lambda }\right)\hfill \\ & \le & \frac{g-\mathit{k}}{9}\left(\underset{0}{\stackrel{1}{\int }}\left|{\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^s\left(\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|\right)d\mathsf{\Lambda }\right.\hfill \\ & & +\underset{0}{\stackrel{1}{\int }}\left|\frac{1}{2}-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^s\left(\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|\right)d\mathsf{\Lambda }\hfill \\ & & +\left.\underset{0}{\stackrel{1}{\int }}\left|\frac{3}{2+2\theta }-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^s\left(\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(g\right)\right|\right)d\mathsf{\Lambda }\right)\hfill \\ & =& \frac{g-\mathit{k}}{9}\left(\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|\underset{0}{\stackrel{1}{\int }}\left|{\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^{s}dt\right.\hfill \\ & & +\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|\hfill \\ & & \times \left(\underset{0}{\stackrel{1}{\int }}\left|{\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^{s}d\mathsf{\Lambda }+\underset{0}{\stackrel{1}{\int }}\left|\frac{1}{2}-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^{s}d\mathsf{\Lambda }\right)\hfill \\ & & +\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|\hfill \\ & & \times \left(\underset{0}{\stackrel{1}{\int }}\left|\frac{1}{2}-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^{s}dt+\underset{0}{\stackrel{1}{\int }}\left|\frac{3}{2+2\theta }-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^{s}d\mathsf{\Lambda }\right)\hfill \\ & & +\left.\left|{\mathcal{L}}^{\prime }\left(g\right)\right|\underset{0}{\stackrel{1}{\int }}\left|\frac{3}{2+2\theta }-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^{s}d\mathsf{\Lambda }\right)\hfill \\ & =& \frac{g-\mathit{k}}{9}\left({\phi }_{\theta }\left(s+1,s+1;\alpha \right)\left(\left|f^{\prime }\left(\mathit{k}\right)\right|+\left|f^{\prime }\left(g\right)\right|\right)\right.\hfill \\ & & +\left.\left({\phi }_2\left(s+1,s+1;\alpha \right)+{\phi }_{\theta }\left(s+1,s+1;\alpha \right)\right)\left(\left|f^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|+\left|f^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|\right)\right),\hfill \end{array}$$

where ${\varkappa }_{\theta }$ and ${\phi }_{\theta }$ are defined as in (6) and (7), respectively, and we have used the fact that

$$\begin{array}{ccc}& & \underset{0}{\stackrel{1}{\int }}\left|{\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^{s}d\mathsf{\Lambda }=\left|\frac{3}{2+2\theta }-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^{s}d\mathsf{\Lambda }\hfill \\ & =& \frac{3}{2+2\theta }\left(B_{{\left(\frac{3}{2+2\theta }\right)}^{\frac{1}{\alpha }}}\left(s+1,s+1\right)-B_{1-{\left(\frac{3}{2+2\theta }\right)}^{\frac{1}{\alpha }}}\left(s+1,s+1\right)\right)\hfill \\ & & -\left(B_{{\left(\frac{3}{2+2\theta }\right)}^{\frac{1}{\alpha }}}\left(s+\alpha +1,s+1\right)-B_{1-{\left(\frac{3}{2+2\theta }\right)}^{\frac{1}{\alpha }}}\left(s+1,s+\alpha +1\right)\right)\hfill \\ & =& \frac{3}{2+2\theta }{\varkappa }_{\theta }\left(s+1,s+1;\alpha \right)-{\varkappa }_{\theta }\left(s+\alpha +1,s+1;\alpha \right)\hfill \\ & =& {\phi }_{\theta }\left(s+1,s+1;\alpha \right)\hfill \end{array}$$

(8)

and

$$\begin{array}{ccc}\hfill \underset{0}{\stackrel{1}{\int }}\left|\frac{1}{2}-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^{s}d\mathsf{\Lambda }& =& \underset{0}{\stackrel{1}{\int }}\left|{\mathsf{\Lambda }}^{\alpha }-\frac{1}{2}\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^{s}d\mathsf{\Lambda }\hfill \\ & =& {\phi }_2\left(s+1,s+1;\alpha \right),\hfill \end{array}$$

(9)

which ends the proof. □

Corollary 1.

Taking $\alpha =1$, Theorem 4 becomes

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{1}{g-\mathit{k}}\underset{\mathit{k}}{\stackrel{g}{\int }}\mathcal{L}\left(u\right)du\right|\hfill \\ & \le & \frac{g-\mathit{k}}{9}\left({\phi }_{\theta }\left(s+1,s+1;1\right)\left(\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(g\right)\right|\right)\right.\hfill \\ & & +\left.\left({\phi }_2\left(s+1,s+1;1\right)+{\phi }_{\theta }\left(s+1,s+1;1\right)\right)\left(\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|\right)\right),\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill {\phi }_{\theta }\left(s+1,s+1;1\right)& =& \frac{3}{2+2\theta }{B}_{\frac{3}{2+2\theta }}\left(s+1,s+1\right)-B_{\frac{3}{2+2\theta }}\left(s+2,s+1\right)\hfill \\ & & +\frac{2\theta -1}{2+2\theta }{B}_{\frac{2\theta -1}{2+2\theta }}\left(s+1,s+1\right)-B_{\frac{2\theta -1}{2+2\theta }}\left(s+2,s+1\right)\hfill \end{array}$$

and

$${\phi }_2\left(s+1,s+1;1\right)=B_{\frac{1}{2}}\left(s+1,s+1\right)-2B_{\frac{1}{2}}\left(s+2,s+1\right).$$

Corollary 2.

Under the assumption of Theorem 4, and if $\left|f^{\prime }\right|$ is P-function, then we have

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{3^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha }}\mathcal{R}\left(\alpha ,\mathcal{L}\right)\right|\hfill \\ & \le & \frac{g-\mathit{k}}{36}\left(\frac{\left(2{\theta }^2-2\theta +5\right)\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|+\left(3{\theta }^2+6\right)\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|+\left(3{\theta }^2+6\right)\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|+\left(2{\theta }^2-2\theta +5\right)\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}{{\left(1+\theta \right)}^2}\right).\hfill \end{array}$$

Proof. 

Just replace s with 0 in Theorem 4. □

Remark 1.

In Corollary 2, choosing $\theta =3$, we obtain the fractional Simpson’s $3/8$ formula

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left(\mathcal{L}\left(\mathit{k}\right)+3\mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+3\mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{3^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha }}\mathcal{R}\left(\alpha ,\mathcal{L}\right)\right|\hfill \\ \hfill & \le \frac{25\left(g-\mathit{k}\right)}{144}\left(\frac{17\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|+33\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|+33\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|+17\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}{100}\right).\hfill \end{array}$$

Remark 2.

In Corollary 2, choosing $\theta =\frac{27}{13}$, we obtain the fractional corrected Simpson’s $3/8$ formula

$$\begin{array}{cc}\hfill & \left|\frac{1}{80}\left(\mathcal{L}\left(\mathit{k}\right)+27\mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+27\mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{3^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha }}\mathcal{R}\left(\alpha ,\mathcal{L}\right)\right|\hfill \\ \hfill & \le \frac{2401\left(g-\mathit{k}\right)}{14400}\left(\frac{1601\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|+3201\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|+3201\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|+1601\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}{9604}\right).\hfill \end{array}$$

Corollary 3.

In Corollary 2, if we take $\alpha =1$, then we have

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{1}{g-\mathit{k}}\underset{\mathit{k}}{\stackrel{g}{\int }}\mathcal{L}\left(u\right)du\right|\hfill \\ & \le & \frac{g-\mathit{k}}{36}\left(\frac{\left(2{\theta }^2-2\theta +5\right)\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|+\left(3{\theta }^2+6\right)\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|+\left(3{\theta }^2+6\right)\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|+\left(2{\theta }^2-2\theta +5\right)\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}{{\left(1+\theta \right)}^2}\right).\hfill \end{array}$$

Corollary 4.

Under the assumptions of Theorem 4, and if $\left|{\mathcal{L}}^{\prime }\right|$ is $tgs$-function, then we have

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{3^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha }}\mathcal{R}\left(\alpha ,\mathcal{L}\right)\right|\hfill \\ & \le & \frac{g-\mathit{k}}{1728}\left(\frac{27\left(1+4\theta \right)+{\left(2\theta -1\right)}^4}{{\left(1+\theta \right)}^4}\left(\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(g\right)\right|\right)\right.\hfill \\ & & +\left.\left(\frac{4{\left(1+\theta \right)}^4+27\left(1+4\theta \right)+{\left(2\theta -1\right)}^4}{{\left(1+\theta \right)}^4}\right)\left(\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|\right)\right).\hfill \end{array}$$

Proof. 

Just replace s with 1 in Theorem 4. □

Corollary 5.

Taking $\alpha =1$ in Corollary 4 gives

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{1}{g-\mathit{k}}\underset{\mathit{k}}{\stackrel{g}{\int }}\mathcal{L}\left(u\right)du\right|\hfill \\ & \le & \frac{g-\mathit{k}}{1728}\left(\frac{27\left(1+4\theta \right)+{\left(2\theta -1\right)}^4}{{\left(1+\theta \right)}^4}\left(\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(g\right)\right|\right)\right.\hfill \\ & & +\left.\left(\frac{4{\left(1+\theta \right)}^4+27\left(1+4\theta \right)+{\left(2\theta -1\right)}^4}{{\left(1+\theta \right)}^4}\right)\left(\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|\right)\right).\hfill \end{array}$$

Theorem 5.

Assume that all the assumptions of Theorem 4 are satisfied. Moreover, if ${\left|{\mathcal{L}}^{\prime }\right|}^{\zeta }$ is s-$tgs$-convex, where $\zeta >1$ with $\frac{1}{\tau }+\frac{1}{\zeta }=1$, then the following inequality

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{3^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha }}\mathcal{R}\left(\alpha ,\mathcal{L}\right)\right|\hfill \\ & \le & \frac{g-\mathit{k}}{9}{\left(B\left(s+1,s+1\right)\right)}^{\frac{1}{\zeta }}\hfill \\ & & \times \left({\left(\frac{B\left(\frac{1}{\alpha },\tau +1\right)}{\alpha }{\left(\frac{3}{2+2\theta }\right)}^{\tau +\frac{1}{\alpha }}+\frac{{}_2{F}_1\left(\frac{\alpha -1}{\alpha },1,\tau +2;\frac{2\theta -1}{2+2\theta }\right)}{\alpha \left(\tau +1\right)}{\left(\frac{2\theta -1}{2+2\theta }\right)}^{\tau +\frac{1}{\alpha }}\right)}^{\frac{1}{\tau }}\right.\hfill \\ & & \times \left({\left({\left|f^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|f^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|f^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|f^{\prime }\left(g\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.{\left(\frac{B\left(\frac{1}{\alpha },\tau +1\right)}{2^{\tau +\frac{1}{\alpha }}\alpha }+\frac{{}_2{F}_1\left(\frac{\alpha -1}{\alpha },1,\tau +2;\frac{1}{2}\right)}{2^{\tau +1}\alpha \left(\tau +1\right)}\right)}^{\frac{1}{\tau }}{\left({\left|f^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|f^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \end{array}$$

holds, where $\mathcal{R}\left(\alpha ,\mathcal{L}\right)$ is defined by (2) B and ${}_2{F}_1$ are beta and hypergeometric functions, respectively.

Proof. 

By Lemma 1, modulus, Hölder’s inequality, and the s-$tgs$-convexity of ${\left|{\mathcal{L}}^{\prime }\right|}^{\zeta }$, we obtain

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{3^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha }}\mathcal{R}\left(\alpha ,\mathcal{L}\right)\right|\hfill \\ & \le & \frac{g-\mathit{k}}{9}\left({\left(\underset{0}{\stackrel{1}{\int }}{\left|{\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right|}^{\tau }d\mathsf{\Lambda }\right)}^{\frac{1}{\tau }}{\left(\underset{0}{\stackrel{1}{\int }}{\left|{\mathcal{L}}^{\prime }\left(\left(1-\mathsf{\Lambda }\right)\mathit{k}+\mathsf{\Lambda }\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }d\mathsf{\Lambda }\right)}^{\frac{1}{\zeta }}\right.\hfill \\ & & +{\left(\underset{0}{\stackrel{1}{\int }}{\left|\frac{1}{2}-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|}^{\tau }d\mathsf{\Lambda }\right)}^{\frac{1}{\tau }}\left({\left(\underset{0}{\stackrel{1}{\int }}{\left|{\mathcal{L}}^{\prime }\left(\left(1-\mathsf{\Lambda }\right)\frac{2\mathit{k}+g}{3}+\mathsf{\Lambda }\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }d\mathsf{\Lambda }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.{\left(\underset{0}{\stackrel{1}{\int }}{\left|\frac{3}{2+2\theta }-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|}^{\tau }d\mathsf{\Lambda }\right)}^{\frac{1}{\tau }}{\left(\underset{0}{\stackrel{1}{\int }}{\left|{\mathcal{L}}^{\prime }\left(\left(1-\mathsf{\Lambda }\right)\frac{\mathit{k}+2g}{3}+\mathsf{\Lambda }g\right)\right|}^{\zeta }d\mathsf{\Lambda }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & \le & \frac{g-\mathit{k}}{9}\left({\left(\underset{0}{\stackrel{1}{\int }}{\left|{\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right|}^{\tau }d\mathsf{\Lambda }\right)}^{\frac{1}{\tau }}{\left(\underset{0}{\stackrel{1}{\int }}{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^s\left({\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)d\mathsf{\Lambda }\right)}^{\frac{1}{\zeta }}\right.\hfill \\ & & +{\left(\underset{0}{\stackrel{1}{\int }}{\left|\frac{1}{2}-{\mathsf{\Lambda }}^{\alpha }\right|}^{\tau }d\mathsf{\Lambda }\right)}^{\frac{1}{\tau }}{\left(\underset{0}{\stackrel{1}{\int }}{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^s\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)d\mathsf{\Lambda }\right)}^{\frac{1}{\zeta }}\hfill \\ & & +\left.{\left(\underset{0}{\stackrel{1}{\int }}{\left|{\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right|}^{\tau }d\mathsf{\Lambda }\right)}^{\frac{1}{\tau }}{\left(\underset{0}{\stackrel{1}{\int }}{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^s\left({\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\right)d\mathsf{\Lambda }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & =& \frac{g-\mathit{k}}{9}{\left(B\left(s+1,s+1\right)\right)}^{\frac{1}{\zeta }}\hfill \\ & & \times \left({\left(\frac{1}{\alpha }{\left(\frac{3}{2+2\theta }\right)}^{\tau +\frac{1}{\alpha }}B\left(\frac{1}{\alpha },\tau +1\right)+\frac{{}_2{F}_1\left(\frac{\alpha -1}{\alpha },1,\tau +2;\frac{2\theta -1}{2+2\theta }\right)}{\alpha \left(\tau +1\right)}{\left(\frac{2\theta -1}{2+2\theta }\right)}^{\tau +\frac{1}{\alpha }}\right)}^{\frac{1}{\tau }}\right.\hfill \\ & & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.{\left(\frac{B\left(\frac{1}{\alpha },\tau +1\right)}{2^{\tau +\frac{1}{\alpha }}\alpha }+\frac{{}_2{F}_1\left(\frac{\alpha -1}{\alpha },1,\tau +2;\frac{1}{2}\right)}{2^{\tau +1}\alpha \left(\tau +1\right)}\right)}^{\frac{1}{\tau }}{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right),\hfill \end{array}$$

where we used

$$\begin{array}{ccc}& & \underset{0}{\stackrel{1}{\int }}{\left|{\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right|}^{\tau }d\mathsf{\Lambda }\hfill \\ & =& \underset{0}{\stackrel{{\left(\frac{3}{2+2\theta }\right)}^{\frac{1}{\alpha }}}{\int }}{\left(\frac{3}{2+2\theta }-{\mathsf{\Lambda }}^{\alpha }\right)}^{\tau }d\mathsf{\Lambda }+\underset{{\left(\frac{3}{2+2\theta }\right)}^{\frac{1}{\alpha }}}{\stackrel{1}{\int }}{\left({\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right)}^{\tau }d\mathsf{\Lambda }\hfill \\ & =& \frac{1}{\alpha }{\left(\frac{3}{2+2\theta }\right)}^{\tau +\frac{1}{\alpha }}\underset{0}{\stackrel{1}{\int }}{\vartheta }^{\frac{1}{\alpha }-1}{\left(1-\vartheta \right)}^{\tau }d\vartheta +\frac{1}{\alpha }{\left(\frac{2\theta -1}{2+2\theta }\right)}^{\tau +\frac{1}{\alpha }}\underset{0}{\stackrel{1}{\int }}{\left(1-\vartheta \right)}^{\tau }{\left(1-\frac{2\theta -1}{2+2\theta }\vartheta \right)}^{\frac{1}{\alpha }-1}d\vartheta \hfill \\ & =& \frac{1}{\alpha }{\left(\frac{3}{2+2\theta }\right)}^{\tau +\frac{1}{\alpha }}B\left(\frac{1}{\alpha },\tau +1\right)+\frac{1}{\alpha \left(\tau +1\right)}{\left(\frac{2\theta -1}{2+2\theta }\right)}^{\tau +\frac{1}{\alpha }}{.}_2{F}_1\left(\frac{\alpha -1}{\alpha },1,\tau +2;\frac{2\theta -1}{2+2\theta }\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}& & \underset{0}{\stackrel{1}{\int }}{\left|\frac{1}{2}-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|}^{\tau }d\mathsf{\Lambda }\hfill \\ & =& \underset{0}{\stackrel{1-{\left(\frac{1}{2}\right)}^{\frac{1}{\alpha }}}{\int }}{\left({\left(1-\mathsf{\Lambda }\right)}^{\alpha }-\frac{1}{2}\right)}^{\tau }d\mathsf{\Lambda }+\underset{1-{\left(\frac{1}{2}\right)}^{\frac{1}{\alpha }}}{\stackrel{1}{\int }}{\left(\frac{1}{2}-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right)}^{\tau }d\mathsf{\Lambda }\hfill \\ & =& \frac{1}{2^{\tau +1}\alpha }\underset{0}{\stackrel{1}{\int }}{\left(1-x\right)}^{\tau }{\left(1-\frac{1}{2}x\right)}^{\frac{1}{\alpha }-1}dx+\frac{1}{2^{\tau +\frac{1}{\alpha }}\alpha }\underset{0}{\stackrel{1}{\int }}{x}^{\frac{1}{\alpha }-1}{\left(1-x\right)}^{\tau }dx.\hfill \\ & =& \frac{1}{2^{\tau +1}\alpha \left(\tau +1\right)}{.}_2{F}_1\left(\frac{\alpha -1}{\alpha },1,\tau +2;\frac{1}{2}\right)+\frac{1}{2^{\tau +\frac{1}{\alpha }}\alpha }B\left(\frac{1}{\alpha },\tau +1\right),\hfill \end{array}$$

which ends the proof. □

Corollary 6.

Taking $\alpha =1$ in Theorem 5, it yields

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{1}{g-\mathit{k}}\underset{\mathit{k}}{\stackrel{g}{\int }}\mathcal{L}\left(u\right)du\right|\hfill \\ & \le & \frac{g-\mathit{k}}{18}{\left(B\left(s+1,s+1\right)\right)}^{\frac{1}{\zeta }}{\left(\frac{1}{\tau +1}\right)}^{\frac{1}{\tau }}\left({\left(\frac{3^{\tau +1}+{\left(2\theta -1\right)}^{\tau +1}}{2{\left(1+\theta \right)}^{\tau +1}}\right)}^{\frac{1}{\tau }}\right.\hfill \\ & & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right).\hfill \end{array}$$

Corollary 7.

Under the assumptions of Theorem 5, and if ${\left|{\mathcal{L}}^{\prime }\right|}^{\zeta }$ is P-function, then we have

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{3^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha }}\mathcal{R}\left(\alpha ,\mathcal{L}\right)\right|\hfill \\ & \le & \frac{g-\mathit{k}}{9}\left({\left(\frac{B\left(\frac{1}{\alpha },\tau +1\right)}{\alpha }{\left(\frac{3}{2+2\theta }\right)}^{\tau +\frac{1}{\alpha }}+\frac{{}_2{F}_1\left(\frac{\alpha -1}{\alpha },1,\tau +2;\frac{2\theta -1}{2+2\theta }\right)}{\alpha \left(\tau +1\right)}{\left(\frac{2\theta -1}{2+2\theta }\right)}^{\tau +\frac{1}{\alpha }}\right)}^{\frac{1}{\tau }}\right.\hfill \\ & & \times \left({\left({\left|f^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|f^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.{\left(\frac{B\left(\frac{1}{\alpha },\tau +1\right)}{2^{\tau +\frac{1}{\alpha }}\alpha }+\frac{{}_2{F}_1\left(\frac{\alpha -1}{\alpha },1,\tau +2;\frac{1}{2}\right)}{2^{\tau +1}\alpha \left(\tau +1\right)}\right)}^{\frac{1}{\tau }}{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right).\hfill \end{array}$$

Proof. 

Just replace s with 0 in Theorem 5. □

Corollary 8.

In Corollary 7, if we take $\alpha =1$, then we have

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{1}{g-\mathit{k}}\underset{\mathit{k}}{\stackrel{g}{\int }}\mathcal{L}\left(u\right)du\right|\hfill \\ & \le & \frac{g-\mathit{k}}{18}{\left(\frac{1}{\tau +1}\right)}^{\frac{1}{\tau }}\left({\left(\frac{3^{\tau +1}+{\left(2\theta -1\right)}^{\tau +1}}{2{\left(1+\theta \right)}^{\tau +1}}\right)}^{\frac{1}{\tau }}\right.\hfill \\ & & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right).\hfill \end{array}$$

Corollary 9.

Under the assumptions of Theorem 5, and if ${\left|{\mathcal{L}}^{\prime }\right|}^{\zeta }$ is $tgs$-function, then we have

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{3^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha }}\mathcal{R}\left(\alpha ,\mathcal{L}\right)\right|\hfill \\ & \le & \frac{g-\mathit{k}}{9}{\left(\frac{1}{6}\right)}^{\frac{1}{\zeta }}\left({\left(\frac{B\left(\frac{1}{\alpha },\tau +1\right)}{\alpha }{\left(\frac{3}{2+2\theta }\right)}^{\tau +\frac{1}{\alpha }}+\frac{{}_2{F}_1\left(\frac{\alpha -1}{\alpha },1,\tau +2;\frac{2\theta -1}{2+2\theta }\right)}{\alpha \left(\tau +1\right)}{\left(\frac{2\theta -1}{2+2\theta }\right)}^{\tau +\frac{1}{\alpha }}\right)}^{\frac{1}{\tau }}\right.\hfill \\ & & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.{\left(\frac{B\left(\frac{1}{\alpha },\tau +1\right)}{2^{\tau +\frac{1}{\alpha }}\alpha }+\frac{{}_2{F}_1\left(\frac{\alpha -1}{\alpha },1,\tau +2;\frac{1}{2}\right)}{2^{\tau +1}\alpha \left(\tau +1\right)}\right)}^{\frac{1}{\tau }}{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right).\hfill \end{array}$$

Proof. 

Just replace s with 1 in Theorem 5. □

Corollary 10.

In Corollary 9, if we take $\alpha =1$, then we have

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{1}{g-\mathit{k}}\underset{\mathit{k}}{\stackrel{g}{\int }}\mathcal{L}\left(u\right)du\right|\hfill \\ & \le & \frac{g-\mathit{k}}{18}{\left(\frac{1}{6}\right)}^{\frac{1}{\zeta }}{\left(\frac{1}{\tau +1}\right)}^{\frac{1}{\tau }}\left({\left(\frac{3^{\tau +1}+{\left(2\theta -1\right)}^{\tau +1}}{2{\left(1+\theta \right)}^{\tau +1}}\right)}^{\frac{1}{\tau }}\right.\hfill \\ & & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right).\hfill \end{array}$$

Remark 3.

In Corollary 10, choosing $\theta =3$, we obtain Simpson’s $3/8$ formula

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left(\mathcal{L}\left(\mathit{k}\right)+3\mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+3\mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{1}{g-\mathit{k}}\underset{\mathit{k}}{\stackrel{g}{\int }}\mathcal{L}\left(u\right)du\right|\hfill \\ \hfill & \le \frac{g-\mathit{k}}{72}{\left(\frac{1}{6}\right)}^{\frac{1}{\zeta }}{\left(\frac{1}{\tau +1}\right)}^{\frac{1}{\tau }}\left({\left(\frac{3^{\tau +1}+5^{\tau +1}}{8}\right)}^{\frac{1}{\tau }}\right.\hfill \\ \hfill & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ \hfill & +\left.{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right).\hfill \end{array}$$

Remark 4.

In Corollary 10, choosing $\theta =\frac{27}{13}$, we obtain the corrected Simpson’s $3/8$ formula

$$\begin{array}{cc}\hfill & \left|\frac{1}{80}\left(\mathcal{L}\left(\mathit{k}\right)+27\mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+27\mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{1}{g-\mathit{k}}\underset{\mathit{k}}{\stackrel{g}{\int }}\mathcal{L}\left(u\right)du\right|\hfill \\ \hfill & \le \frac{g-\mathit{k}}{18}{\left(\frac{1}{12}\right)}^{\frac{1}{\zeta }}{\left(\frac{1}{\tau +1}\right)}^{\frac{1}{\tau }}\left({\left(\frac{{\left(39\right)}^{\tau +1}+{\left(41\right)}^{\tau +1}}{{\left(40\right)}^{\tau +1}}\right)}^{\frac{1}{\tau }}\right.\hfill \\ \hfill & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ \hfill & +\left.{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right).\hfill \end{array}$$

Theorem 6.

Assume that all the assumptions of Theorem 5 are satisfied. Moreover, if ${\left|{\mathcal{L}}^{\prime }\right|}^{\zeta }$ is s-$tgs$-convex, where $\zeta \ge 1$, then the following inequality

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{3^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha }}\mathcal{R}\left(\alpha ,\mathcal{L}\right)\right|\hfill \\ & \le & \frac{g-\mathit{k}}{9}\left({\left(\frac{2\theta -3\alpha -1}{\left(2+2\theta \right)\left(\alpha +1\right)}+\frac{2\alpha }{\alpha +1}{\left(\frac{3}{2+2\theta }\right)}^{1+\frac{1}{\alpha }}\right)}^{1-\frac{1}{\zeta }}{\left({\phi }_{\theta }\left(s+1,s+1;\alpha \right)\right)}^{\frac{1}{\zeta }}\right.\hfill \\ & & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.{\left(\frac{1-\alpha +\alpha 2^{1-\frac{1}{\alpha }}}{2\left(\alpha +1\right)}\right)}^{1-\frac{1}{\zeta }}{\left({\phi }_2\left(s+1,s+1;\alpha \right)\right)}^{\frac{1}{\zeta }}{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right),\hfill \end{array}$$

holds, where $\mathcal{R}\left(\alpha ,\mathcal{L}\right)$ and ${\phi }_{\theta }$ are defined by (2) and (7), respectively.

Proof. 

By Lemma 1, modulus, power mean inequality, and the s-$tgs$-convexity of ${\left|{\mathcal{L}}^{\prime }\right|}^{\zeta }$, we obtain

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{3^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha }}\mathcal{R}\left(\alpha ,\mathcal{L}\right)\right|\hfill \\ & \le & \frac{g-\mathit{k}}{9}\left({\left(\underset{0}{\stackrel{1}{\int }}\left|{\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right|d\mathsf{\Lambda }\right)}^{1-\frac{1}{\zeta }}{\left(\underset{0}{\stackrel{1}{\int }}\left|{\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right|{\left|{\mathcal{L}}^{\prime }\left(\left(1-\mathsf{\Lambda }\right)\mathit{k}+\mathsf{\Lambda }\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }d\mathsf{\Lambda }\right)}^{\frac{1}{\zeta }}\right.\hfill \\ & & +{\left(\underset{0}{\stackrel{1}{\int }}\left|\frac{1}{2}-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|d\mathsf{\Lambda }\right)}^{1-\frac{1}{\zeta }}\hfill \\ & & \times {\left(\underset{0}{\stackrel{1}{\int }}\left|\frac{1}{2}-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|{\left|{\mathcal{L}}^{\prime }\left(\left(1-\mathsf{\Lambda }\right)\frac{2\mathit{k}+g}{3}+\mathsf{\Lambda }\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }d\mathsf{\Lambda }\right)}^{\frac{1}{\zeta }}\hfill \\ & & +{\left(\underset{0}{\stackrel{1}{\int }}\left|\frac{3}{2+2\theta }-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|d\mathsf{\Lambda }\right)}^{1-\frac{1}{\zeta }}\hfill \\ & & \times \left.{\left(\underset{0}{\stackrel{1}{\int }}\left|\frac{3}{2+2\theta }-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|{\left|{\mathcal{L}}^{\prime }\left(\left(1-\mathsf{\Lambda }\right)\frac{\mathit{k}+2g}{3}+\mathsf{\Lambda }g\right)\right|}^{\zeta }d\mathsf{\Lambda }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & \le & \frac{g-\mathit{k}}{9}\left({\left(\underset{0}{\stackrel{1}{\int }}\left|{\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right|d\mathsf{\Lambda }\right)}^{1-\frac{1}{\zeta }}\left({\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }\underset{0}{\stackrel{1}{\int }}\left|{\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^{s}d\mathsf{\Lambda }\right.\right.\hfill \\ & & +{\left.{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\underset{0}{\stackrel{1}{\int }}\left|{\mathsf{\Lambda }}^{\alpha }-\frac{3}{2+2\theta }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^{s}d\mathsf{\Lambda }\right)}^{\frac{1}{\zeta }}\hfill \\ & & +{\left(\underset{0}{\stackrel{1}{\int }}\left|\frac{1}{2}-{\mathsf{\Lambda }}^{\alpha }\right|d\mathsf{\Lambda }\right)}^{1-\frac{1}{\zeta }}\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\underset{0}{\stackrel{1}{\int }}\left|\frac{1}{2}-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^{s}d\mathsf{\Lambda }\right.\hfill \\ & & +{\left.{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\underset{0}{\stackrel{1}{\int }}\left|\frac{1}{2}-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^{s}d\mathsf{\Lambda }\right)}^{\frac{1}{\zeta }}\hfill \\ & & +{\left(\underset{0}{\stackrel{1}{\int }}\left|\frac{3}{2+2\theta }-{\mathsf{\Lambda }}^{\alpha }\right|d\mathsf{\Lambda }\right)}^{1-\frac{1}{\zeta }}\left({\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\underset{0}{\stackrel{1}{\int }}\left|\frac{3}{2+2\theta }-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^{s}d\mathsf{\Lambda }\right.\hfill \\ & & +\left.{\left.{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\underset{0}{\stackrel{1}{\int }}\left|\frac{3}{2+2\theta }-{\left(1-\mathsf{\Lambda }\right)}^{\alpha }\right|{\mathsf{\Lambda }}^s{\left(1-\mathsf{\Lambda }\right)}^{s}d\mathsf{\Lambda }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & =& \frac{g-\mathit{k}}{9}\left({\left(\frac{2\theta -3\alpha -1}{\left(2+2\theta \right)\left(\alpha +1\right)}+\frac{2\alpha }{\alpha +1}{\left(\frac{3}{2+2\theta }\right)}^{1+\frac{1}{\alpha }}\right)}^{1-\frac{1}{\zeta }}{\left({\phi }_{\theta }\left(s+1,s+1;\alpha \right)\right)}^{\frac{1}{\zeta }}\right.\hfill \\ & & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.{\left(\frac{1-\alpha +\alpha 2^{1-\frac{1}{\alpha }}}{2\left(\alpha +1\right)}\right)}^{1-\frac{1}{\zeta }}{\left({\phi }_2\left(s+1,s+1;\alpha \right)\right)}^{\frac{1}{\zeta }}{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right),\hfill \end{array}$$

where we have considered (8) and (9). The proof is achieved. □

Corollary 11.

In Theorem 6, if we take $\alpha =1$, then we have

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{1}{g-\mathit{k}}\underset{\mathit{k}}{\stackrel{g}{\int }}\mathcal{L}\left(u\right)du\right|\hfill \\ & \le & \frac{g-\mathit{k}}{9}\left({\left(\frac{2{\theta }^2-2\theta +5}{{\left(2+2\theta \right)}^2}\right)}^{1-\frac{1}{\zeta }}{\left({\phi }_{\theta }\left(s+1,s+1;1\right)\right)}^{\frac{1}{\zeta }}\right.\hfill \\ & & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.{\left(\frac{1}{4}\right)}^{1-\frac{1}{\zeta }}{\left({\phi }_2\left(s+1,s+1;1\right)\right)}^{\frac{1}{\zeta }}{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right).\hfill \end{array}$$

Corollary 12.

Under the assumptions of Theorem 6, and if ${\left|{\mathcal{L}}^{\prime }\right|}^{\zeta }$ is P-function, then we have

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{3^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha }}\mathcal{R}\left(\alpha ,\mathcal{L}\right)\right|\hfill \\ & \le & \frac{g-\mathit{k}}{9}\left(\left(\frac{2\theta -3\alpha -1}{\left(2+2\theta \right)\left(\alpha +1\right)}+\frac{2\alpha }{\alpha +1}{\left(\frac{3}{2+2\theta }\right)}^{1+\frac{1}{\alpha }}\right)\right.\hfill \\ & & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.\frac{1-\alpha +\alpha 2^{1-\frac{1}{\alpha }}}{2\left(\alpha +1\right)}{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right).\hfill \end{array}$$

Proof. 

Just replace s with 0 in Theorem 6. □

Corollary 13.

In Corollary 12, if we take $\alpha =1$, then we have

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{1}{g-\mathit{k}}\underset{\mathit{k}}{\stackrel{g}{\int }}\mathcal{L}\left(u\right)du\right|\hfill \\ & \le & \frac{g-\mathit{k}}{36}\left(\frac{2{\theta }^2-2\theta +5}{{\left(1+\theta \right)}^2}\left({\left({\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\right.\hfill \\ & & +\left.{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right).\hfill \end{array}$$

Corollary 14.

Under the assumptions of Theorem 6, and if ${\left|{\mathcal{L}}^{\prime }\right|}^{\zeta }$ is $tgs$-function, then we have

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{3^{\alpha -1}\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(g-\mathit{k}\right)}^{\alpha }}\mathcal{R}\left(\alpha ,\mathcal{L}\right)\right|\hfill \\ & \le & \frac{g-\mathit{k}}{9}\left({\left(\frac{2\theta -3\alpha -1}{\left(2+2\theta \right)\left(\alpha +1\right)}+\frac{2\alpha }{\alpha +1}{\left(\frac{3}{2+2\theta }\right)}^{1+\frac{1}{\alpha }}\right)}^{1-\frac{1}{\zeta }}\right.\hfill \\ & & \times {\left(\frac{4\theta -{\alpha }^2-5\alpha -2}{4\left(1+\theta \right)\left(\alpha +2\right)\left(\alpha +3\right)}+\frac{\alpha }{\alpha +2}{\left(\frac{3}{2+2\theta }\right)}^{1+\frac{2}{\alpha }}-\frac{2\alpha }{3\left(\alpha +3\right)}{\left(\frac{3}{2+2\theta }\right)}^{1+\frac{3}{\alpha }}\right)}^{\frac{1}{\zeta }}\hfill \\ & & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +{\left(\frac{1-\alpha +\alpha 2^{1-\frac{1}{\alpha }}}{2\left(\alpha +1\right)}\right)}^{1-\frac{1}{\zeta }}{\left(\frac{6-5\alpha -{\alpha }^2}{12\left(\alpha +2\right)\left(\alpha +3\right)}+\frac{\alpha }{2\left(\alpha +2\right)}{\left(\frac{1}{2}\right)}^{\frac{2}{\alpha }}-\frac{\alpha }{3\left(\alpha +3\right)}{\left(\frac{1}{2}\right)}^{\frac{3}{\alpha }}\right)}^{\frac{1}{\zeta }}\hfill \\ & & \times \left.{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right).\hfill \end{array}$$

Proof. 

Just replace s with 1 in Theorem 6. □

Corollary 15.

In Corollary 14, if we take $\alpha =1$, then we have

$$\begin{array}{ccc}& & \left|\frac{1}{2+2\theta }\left(\mathcal{L}\left(\mathit{k}\right)+\theta \mathcal{L}\left(\frac{2\mathit{k}+g}{3}\right)+\theta \mathcal{L}\left(\frac{\mathit{k}+2g}{3}\right)+\mathcal{L}\left(g\right)\right)-\frac{1}{g-\mathit{k}}\underset{\mathit{k}}{\stackrel{g}{\int }}\mathcal{L}\left(u\right)du\right|\hfill \\ & \le & \frac{g-\mathit{k}}{36}\left(\frac{2{\theta }^2-2\theta +5}{{\left(1+\theta \right)}^2}{\left(\frac{\left(\theta -2\right){\left(1+\theta \right)}^2}{3\left(2{\theta }^2-2\theta +5\right)}+\frac{36\left(1+\theta \right)-27}{8\left(2{\theta }^2-2\theta +5\right){\left(1+\theta \right)}^2}\right)}^{\frac{1}{\zeta }}\right.\hfill \\ & & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left(\mathit{k}\right)\right|}^{\zeta }+{\left|f^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|f^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(g\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.{\left(\frac{1}{8}\right)}^{\frac{1}{\zeta }}{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2\mathit{k}+g}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{\mathit{k}+2g}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right).\hfill \end{array}$$

3. Applications

Let ${\rm Y}$ be the division of points $\mathit{k}={\psi }_0<{\psi }_1<$…$<{\psi }_n=g$ of $\left[\mathit{k},g\right]$, and consider the following formula for quadrature

$$\underset{\mathit{k}}{\stackrel{g}{\int }}\mathcal{L}\left(w\right)dw=\lambda \left(\mathcal{L},{\rm Y}\right)+R\left(\mathcal{L},{\rm Y}\right),$$

where

$$\lambda \left(\mathcal{L},{\rm Y}\right)=\underset{ϵ=0}{\sum ^{n-1}}\frac{{\psi }_{ϵ+1}-{\psi }_{ϵ}}{2+2\theta }\left(\mathcal{L}\left({\psi }_{ϵ}\right)+\theta \mathcal{L}\left(\frac{2{\psi }_{ϵ}+{\psi }_{ϵ+1}}{3}\right)+\theta \mathcal{L}\left(\frac{{\psi }_{ϵ}+2{\psi }_{ϵ+1}}{3}\right)+\mathcal{L}\left({\psi }_{ϵ+1}\right)\right)$$

and $R\left(\mathcal{L},{\rm Y}\right)$ represents the associated approximation error.

Proposition 1.

Let $\mathcal{L}$ be as in Lemma 1 and $n\in \mathbb{N}$. If $\left|{\mathcal{L}}^{\prime }\right|$ is P-function, we have

$$\begin{array}{ccc}& & \left|R\left(\mathcal{L},{\rm Y}\right)\right|\le \underset{ϵ=0}{\sum ^{n-1}}\frac{{\left({\psi }_{ϵ+1}-{\psi }_{ϵ}\right)}^2}{36}\left(\frac{2{\theta }^2-2\theta +5}{{\left(1+\theta \right)}^2}\left(\left|{\mathcal{L}}^{\prime }\left({\psi }_{ϵ}\right)\right|+\left|{\mathcal{L}}^{\prime }\left({\psi }_{ϵ+1}\right)\right|\right)\right.\hfill \\ & & +\left.\frac{3{\theta }^2+6}{{\left(1+\theta \right)}^2}\left(\left|{\mathcal{L}}^{\prime }\left(\frac{2{\psi }_{ϵ}+{\psi }_{ϵ+1}}{3}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(\frac{{\psi }_{ϵ}+2{\psi }_{ϵ+1}}{3}\right)\right|\right)\right).\hfill \end{array}$$

Proof. 

Applying Corollary 3 on $\left[{\psi }_{{}_{ϵ}},{\psi }_{{}_{ϵ}+1}\right]$$\left(ϵ=0,1,\dots ,n-1\right)$ of the partition ${\rm Y}$, we determine

$$\begin{array}{ccc}& & \left|\frac{\mathcal{L}\left({\psi }_{ϵ}\right)+\theta \mathcal{L}\left(\frac{2{\psi }_{ϵ}+{\psi }_{{}_{ϵ}+1}}{3}\right)+\theta \mathcal{L}\left(\frac{{\psi }_{{}_{ϵ}}+2{\psi }_{{}_{ϵ}+1}}{3}\right)+\mathcal{L}\left({\psi }_{{}_{ϵ}+1}\right)}{2+2\theta }-\frac{1}{{\psi }_{{}_{ϵ}+1}-{\psi }_{{}_{ϵ}}}\underset{{\psi }_{ϵ}}{\stackrel{{\psi }_{{}_{ϵ}+1}}{\int }}\mathcal{L}\left(w\right)dw\right|\hfill \\ & \le & \frac{{\psi }_{ϵ+1}-{\psi }_{ϵ}}{36}\left(\frac{2{\theta }^2-2\theta +5}{{\left(1+\theta \right)}^2}\left(\left|{\mathcal{L}}^{\prime }\left({\psi }_{ϵ}\right)\right|+\left|{\mathcal{L}}^{\prime }\left({\psi }_{ϵ+1}\right)\right|\right)\right.\hfill \\ & & +\left.\frac{3{\theta }^2+6}{{\left(1+\theta \right)}^2}\left(\left|{\mathcal{L}}^{\prime }\left(\frac{2{\psi }_{ϵ}+{\psi }_{ϵ+1}}{3}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(\frac{{\psi }_{ϵ}+2{\psi }_{ϵ+1}}{3}\right)\right|\right)\right).\hfill \end{array}$$

The desired inequality follows by multiplying the above inequality by $\left({\psi }_{ϵ+1}-{\psi }_{ϵ}\right)$, then adding the result over $ϵ=0,1,$…$,n-1$ and using the triangle inequality. □

Proposition 2.

Let $\mathcal{L}$ be as in Lemma 1 and $n\in \mathbb{N}$. If $\left|{\mathcal{L}}^{\prime }\right|$ is $tgs$-convex, we have

$$\begin{array}{ccc}& & \left|R\left(\mathcal{L},{\rm Y}\right)\right|\hfill \\ & \le & \underset{i=0}{\sum ^{n-1}}\frac{{\left({\psi }_{ϵ+1}-{\psi }_{ϵ}\right)}^2}{1728}\left(\frac{27\left(1+4\theta \right)+{\left(2\theta -1\right)}^4}{{\left(1+\theta \right)}^4}\left(\left|{\mathcal{L}}^{\prime }\left({\psi }_{ϵ}\right)\right|+\left|{\mathcal{L}}^{\prime }\left({\psi }_{ϵ+1}\right)\right|\right)\right.\hfill \\ & & +\left.\left(\frac{4{\left(1+\theta \right)}^4+27\left(1+4\theta \right)+{\left(2\theta -1\right)}^4}{{\left(1+\theta \right)}^4}\right)\left(\left|{\mathcal{L}}^{\prime }\left(\frac{2{\psi }_{ϵ}+{\psi }_{ϵ+1}}{3}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(\frac{{\psi }_{ϵ}+2{\psi }_{ϵ+1}}{3}\right)\right|\right)\right).\hfill \end{array}$$

Proof. 

Applying Corollary 5 on $\left[{\psi }_{{}_{ϵ}},{\psi }_{{}_{ϵ}+1}\right]$$\left(ϵ=0,1,\dots ,n-1\right)$ of the partition ${\rm Y}$, we obtain

$$\begin{array}{ccc}& & \left|\frac{\mathcal{L}\left({\psi }_{ϵ}\right)+\theta \mathcal{L}\left(\frac{2{\psi }_{ϵ}+{\psi }_{{}_{ϵ}+1}}{3}\right)+\theta \mathcal{L}\left(\frac{{\psi }_{{}_{ϵ}}+2{\psi }_{{}_{ϵ}+1}}{3}\right)+\mathcal{L}\left({\psi }_{{}_{ϵ}+1}\right)}{2+2\theta }-\frac{1}{{\psi }_{{}_{ϵ}+1}-{\psi }_{{}_{ϵ}}}\underset{{\psi }_{ϵ}}{\stackrel{{\psi }_{{}_{ϵ}+1}}{\int }}\mathcal{L}\left(w\right)dw\right|\hfill \\ & \le & \frac{{\psi }_{ϵ+1}-{\psi }_{ϵ}}{1728}\left(\frac{27\left(1+4\theta \right)+{\left(2\theta -1\right)}^4}{{\left(1+\theta \right)}^4}\left(\left|{\mathcal{L}}^{\prime }\left({\psi }_{ϵ}\right)\right|+\left|{\mathcal{L}}^{\prime }\left({\psi }_{ϵ+1}\right)\right|\right)\right.\hfill \\ & & +\left.\left(\frac{4{\left(1+\theta \right)}^4+27\left(1+4\theta \right)+{\left(2\theta -1\right)}^4}{{\left(1+\theta \right)}^4}\right)\left(\left|{\mathcal{L}}^{\prime }\left(\frac{2{\psi }_{ϵ}+{\psi }_{ϵ+1}}{3}\right)\right|+\left|{\mathcal{L}}^{\prime }\left(\frac{{\psi }_{ϵ}+2{\psi }_{ϵ+1}}{3}\right)\right|\right)\right).\hfill \end{array}$$

The desired inequality follows by multiplying the above inequality by $\left({\psi }_{ϵ+1}-{\psi }_{ϵ}\right)$, then adding the result over $ϵ$ =0, 1, …, $n-1$ and using the triangle inequality. □

Proposition 3.

Let $\mathcal{L}$ be as in Lemma 1 and $n\in \mathbb{N}$. If ${\left|{\mathcal{L}}^{\prime }\right|}^{\zeta }$ is P-convex where $\zeta ,\tau >1$ with $\frac{1}{\tau }+\frac{1}{\zeta }=1$, we have

$$\begin{array}{ccc}& & \left|R\left(\mathcal{L},{\rm Y}\right)\right|\hfill \\ & \le & \underset{i=0}{\sum ^{n-1}}\frac{{\left({\psi }_{ϵ+1}-{\psi }_{ϵ}\right)}^2}{18}{\left(\frac{1}{\tau +1}\right)}^{\frac{1}{\tau }}\left({\left(\frac{3^{\tau +1}+{\left(2\theta -1\right)}^{\tau +1}}{2{\left(1+\theta \right)}^{\tau +1}}\right)}^{\frac{1}{\tau }}\right.\hfill \\ & & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left({\psi }_{ϵ}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2{\psi }_{ϵ}+{\psi }_{ϵ+1}}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{{\psi }_{ϵ}+2{\psi }_{ϵ+1}}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left({\psi }_{ϵ+1}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2{\psi }_{ϵ}+{\psi }_{ϵ+1}}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{{\psi }_{ϵ}+2{\psi }_{ϵ+1}}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right).\hfill \end{array}$$

Proof. 

Applying Corollary 8 on $\left[{\psi }_{{}_{ϵ}},{\psi }_{{}_{ϵ}+1}\right]$$\left(ϵ=0,1,\dots ,n-1\right)$ of the partition $\mathrm{{\rm Y}}$, we determine

$$\begin{array}{ccc}& & \left|\frac{\mathcal{L}\left({\psi }_{ϵ}\right)+\theta \mathcal{L}\left(\frac{2{\psi }_{ϵ}+{\psi }_{{}_{ϵ}+1}}{3}\right)+\theta \mathcal{L}\left(\frac{{\psi }_{{}_{ϵ}}+2{\psi }_{{}_{ϵ}+1}}{3}\right)+\mathcal{L}\left({\psi }_{{}_{ϵ}+1}\right)}{2+2\theta }-\frac{1}{{\psi }_{{}_{ϵ}+1}-{\psi }_{{}_{ϵ}}}\underset{{\psi }_{ϵ}}{\stackrel{{\psi }_{{}_{ϵ}+1}}{\int }}\mathcal{L}\left(w\right)dw\right|\hfill \\ & \le & \frac{{\psi }_{ϵ+1}-{\psi }_{ϵ}}{18}{\left(\frac{1}{\tau +1}\right)}^{\frac{1}{\tau }}\left({\left(\frac{3^{\tau +1}+{\left(2\theta -1\right)}^{\tau +1}}{2{\left(1+\theta \right)}^{\tau +1}}\right)}^{\frac{1}{\tau }}\right.\hfill \\ & & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left({\psi }_{ϵ}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2{\psi }_{ϵ}+{\psi }_{ϵ+1}}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{{\psi }_{ϵ}+2{\psi }_{ϵ+1}}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left({\psi }_{ϵ+1}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2{\psi }_{ϵ}+{\psi }_{ϵ+1}}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{{\psi }_{ϵ}+2{\psi }_{ϵ+1}}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right).\hfill \end{array}$$

The desired inequality follows by multiplying the above inequality by $\left({\psi }_{ϵ+1}-{\psi }_{ϵ}\right)$, then adding the result over $ϵ=0,1,$…$,n-1$ and using the triangle inequality. □

Proposition 4.

Let $\mathcal{L}$ be as in Lemma 1 and $n\in \mathbb{N}$. If ${\left|{\mathcal{L}}^{\prime }\right|}^{\zeta }$ is $tgs$-convex where $\zeta ,\tau >1$ with $\frac{1}{\tau }+\frac{1}{\zeta }=1$, we have

$$\begin{array}{ccc}& & \left|R\left(f,\mathrm{{\rm Y}}\right)\right|\hfill \\ & \le & \underset{i=0}{\sum ^{n-1}}\frac{{\left({\psi }_{ϵ+1}-{\psi }_{ϵ}\right)}^2}{18}{\left(\frac{1}{6}\right)}^{\frac{1}{\zeta }}{\left(\frac{1}{\tau +1}\right)}^{\frac{1}{\tau }}\left({\left(\frac{3^{\tau +1}+{\left(2\theta -1\right)}^{\tau +1}}{2{\left(1+\theta \right)}^{\tau +1}}\right)}^{\frac{1}{\tau }}\right.\hfill \\ & & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left({\psi }_{ϵ}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2{\psi }_{ϵ}+{\psi }_{ϵ+1}}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{{\psi }_{ϵ}+2{\psi }_{ϵ+1}}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left({\psi }_{ϵ+1}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2{\psi }_{ϵ}+{\psi }_{ϵ+1}}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{{\psi }_{ϵ}+2{\psi }_{ϵ+1}}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right).\hfill \end{array}$$

Proof. 

Applying Corollary 10 on $\left[{\psi }_{{}_{ϵ}},{\psi }_{{}_{ϵ}+1}\right]$$\left(ϵ=0,1,\dots ,n-1\right)$ of the partition $\mathrm{{\rm Y}}$, we obtain

$$\begin{array}{ccc}& & \left|\frac{\mathcal{L}\left({\psi }_{ϵ}\right)+\theta \mathcal{L}\left(\frac{2{\psi }_{ϵ}+{\psi }_{{}_{ϵ}+1}}{3}\right)+\theta \mathcal{L}\left(\frac{{\psi }_{{}_{ϵ}}+2{\psi }_{{}_{ϵ}+1}}{3}\right)+\mathcal{L}\left({\psi }_{{}_{ϵ}+1}\right)}{2+2\theta }-\frac{1}{{\psi }_{{}_{ϵ}+1}-{\psi }_{{}_{ϵ}}}\underset{{\psi }_{ϵ}}{\stackrel{{\psi }_{{}_{ϵ}+1}}{\int }}\mathcal{L}\left(w\right)dw\right|\hfill \\ & \le & \frac{{\psi }_{ϵ+1}-{\psi }_{ϵ}}{18}{\left(\frac{1}{6}\right)}^{\frac{1}{\zeta }}{\left(\frac{1}{\tau +1}\right)}^{\frac{1}{\tau }}\left({\left(\frac{3^{\tau +1}+{\left(2\theta -1\right)}^{\tau +1}}{2{\left(1+\theta \right)}^{\tau +1}}\right)}^{\frac{1}{\tau }}\right.\hfill \\ & & \times \left({\left({\left|{\mathcal{L}}^{\prime }\left({\psi }_{ϵ}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{2{\psi }_{ϵ}+{\psi }_{ϵ+1}}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}+{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{{\psi }_{ϵ}+2{\psi }_{ϵ+1}}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left({\psi }_{ϵ+1}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right)\hfill \\ & & +\left.{\left({\left|{\mathcal{L}}^{\prime }\left(\frac{2{\psi }_{ϵ}+{\psi }_{ϵ+1}}{3}\right)\right|}^{\zeta }+{\left|{\mathcal{L}}^{\prime }\left(\frac{{\psi }_{ϵ}+2{\psi }_{ϵ+1}}{3}\right)\right|}^{\zeta }\right)}^{\frac{1}{\zeta }}\right).\hfill \end{array}$$

The desired inequality follows by multiplying the above inequality by $\left({\psi }_{ϵ+1}-{\psi }_{ϵ}\right)$, then adding the result over $ϵ=0,1,$…$,n-1$ and using the triangle inequality. □

Let us consider the following means for arbitrary real numbers $\mathit{k},g$

The Arithmetic mean: $A\left(\mathit{k},g,n\right)=\frac{\mathit{k}+g+n}{3}$.

The Harmonic mean: $H\left(\mathit{k},g,n\right)=\frac{3}{\frac{1}{\mathit{k}}+\frac{1}{g}+\frac{1}{n}}$

The Geometric means: $G\left(\mathit{k},g\right)=\sqrt{\mathit{k}g}$

The p-Logarithmic mean: $L_p\left(\mathit{k},g\right)={\left(\frac{g^{p+1}-{\mathit{k}}^{p+1}}{\left(p+1\right)\left(g-\mathit{k}\right)}\right)}^{\frac{1}{p}}$, $\mathit{k},g>0,\mathit{k}\ne g$, and $p\in \mathbb{R}‵\left\{-1,0\right\}$.

Proposition 5.

Let $\mathit{k},g\in \mathbb{R}$ with $0<\mathit{k}<g$, then we have

$$\begin{array}{cc}\hfill & \left|\frac{1}{4}\left(2A\left(g^{-3},{\mathit{k}}^{-3}\right)+H^{-1}\left(g,g,\mathit{k}\right)+H^{-1}\left(g,\mathit{k},\mathit{k}\right)\right)-4G^{-6}\left(\mathit{k},g\right)L_3^3\left(\mathit{k},g\right)\right|\hfill \\ \hfill & \le \frac{g-\mathit{k}}{12\mathit{k}g}\left(\frac{5}{g^2}+{\left(\frac{2\mathit{k}+g}{\mathit{k}g}\right)}^2+{\left(\frac{\mathit{k}+2g}{\mathit{k}g}\right)}^2+\frac{5}{{\mathit{k}}^2}\right).\hfill \end{array}$$

Proof. 

The assertion follows from Corollary 2 with $\alpha =1$ and $\theta =1$, applied to the function $\mathcal{L}\left(w\right)=w^3$ on $\left[\frac{1}{g},\frac{1}{\mathit{k}}\right]$. □

Proposition 6.

Let $\mathit{k},g\in \mathbb{R}$ with $0<\mathit{k}<g$, then we have

$$\begin{array}{cc}\hfill & \left|2A\left({\mathit{k}}^2,g^2\right)+3A^2\left(\mathit{k},\mathit{k},g\right)+3A^2\left(\mathit{k},g,g\right)-8L_2^2\left(\mathit{k},g\right)\right|\hfill \\ \hfill & \le \frac{\sqrt{57}\left(g-\mathit{k}\right)}{27}\left(\frac{{\left(13{\mathit{k}}^2+4\mathit{k}g+g^2\right)}^{\frac{1}{2}}+{\left({\mathit{k}}^2+4\mathit{k}g+13g^2\right)}^{\frac{1}{2}}}{12}+\frac{{\left(5{\mathit{k}}^2+8\mathit{k}g+5g^2\right)}^{\frac{1}{2}}}{3\sqrt{19}}\right).\hfill \end{array}$$

Proof. 

The assertion follows from Corollary 8 with $\theta =3$ and $\zeta =2$, applied to the function $\mathcal{L}\left(w\right)=\frac{1}{2}{w}^2$ on $\left[\mathit{k},g\right]$, in which ${\left|{\mathcal{L}}^{\prime }\left(w\right)\right|}^2=w^2$ is P-function. □

4. Conclusions

In this study, we have considered the fractional Newton–Cotes type integral inequalities involving four points via a Riemann–Liouville integral operator. We have established for the first time a novel parametrized integral identity. Based on this equality, we have derived several $3/8$ Simpson-like type inequalities for functions whose first derivatives belongs in the class of s-$tgs$-convex functions. Some special cases are discussed according to the values of parameters. Some applications to numerical quadratures are presented. The obtained results may lead to additional research in this fascinating field as well as generalizations in other types of calculations, including multiplicative calculus and quantum calculus.