Some New Approaches to Fractional Euler–Maclaurin-Type Inequalities via Various Function Classes

Abstract

This paper aims to examine an approach that studies many Euler–Maclaurin-type inequalities for various function classes applying Riemann–Liouville fractional integrals. Afterwards, our results are provided by using special cases of obtained theorems and examples. Moreover, several Euler–Maclaurin-type inequalities are presented for bounded functions by fractional integrals. Some fractional Euler–Maclaurin-type inequalities are established for Lipschitzian functions. Finally, several Euler–Maclaurin-type inequalities are constructed by fractional integrals of bounded variation.

1. Introduction

Inequality theory is well known and still a fascinating field of research with a wide range of applications in several areas of mathematics. In mathematical analysis, convex functions are significant in the study of inequalities due to their specific geometric and analytical properties. Afterwards, mathematicians have been interested in fractional calculus due to its fundamental properties and its applications.

In [1], Dragomir proved an estimate of the remainder for Simpson’s quadratic formula for functions of bounded variation and gave some applications for special means. Budak et al. [2] presented some variations of Simpson-type inequalities using generalized fractional integrals in the context of differentiable convex functions. For additional information on Simpson-type inequalities and other features associated with Riemann–Liouville fractional integrals, readers can see [3,4,5] and its references.

In the literature, evaluations for three-step quadratic kernels are frequently referred to as Newton-type results, because the three-point Newton–Cotes quadrature is a rule of Simpson’s second rule. In recent years, a number of mathematicians have focused on Newton-type inequalities. For example, in [6], Erden et al. established some Newton-type integral inequalities for functions whose first derivative is arithmetically–harmonically convex in absolute value at a given power. Moreover, in [7], Sitthiwirattham et al. offered some Newton-type inequalities for Riemann–Liouville fractional integrals by using convex functions and functions of bounded variation. Please refer to [8,9,10] and its references for more details on the Newton-type inequality, which includes convex differentiable functions.

Dedić et al. [11] constructed a set of inequalities using the Euler–Maclaurin-type inequalities, and the results were used to produce specific error estimates in the case of the Maclaurin quadrature rules. In [12], the results were applied to provide some error estimates for the Simpson $3/8$ quadrature rules. In paper [13], several Euler–Maclaurin-type inequalities were considered for the case of differentiable convex functions. Moreover, in paper [14], several Euler–Maclaurin-type inequalities were established using the Riemann–Liouville fractional integrals. The reader is referred to [15,16,17,18] and the references therein for further information on such types of inequalities.

The purpose of this paper is to use Riemann–Liouville fractional integrals to obtain Euler–Maclaurin-type inequalities for the different function classes. This paper is organized according to the following plan: In Section 3, an integral equality is proved that is essential in order to establish the main findings. In Section 4, using the Riemann–Liouville fractional integrals, some Euler–Maclaurin-type inequalities are constructed for differentiable convex functions. In Section 5, we give several Euler–Maclaurin-type inequalities for bounded functions by fractional integrals. In Section 6, some fractional Euler–Maclaurin-type inequalities will be proved for Lipschitzian functions. Moreover, in Section 7, the Euler–Maclaurin-type inequalities are constructed by fractional integrals of bounded variation. Furthermore, in Section 8, we provide several graphical examples in order to demonstrate the accuracy of the newly established inequalities. Finally, in Section 9, the summary and concluding remarks are noted.

2. Preliminaries

The Riemann–Liouville integrals  $J_{\sigma +}^{\alpha }\mathcal{F}$ and $J_{\delta -}^{\alpha }\mathcal{F}$ of order $\alpha >0$ with $\sigma \ge 0$ are given by

$$J_{\sigma +}^{\alpha }\mathcal{F}\left(x\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_{\sigma }^x{\left(x-\mu \right)}^{\alpha -1}\mathcal{F}\left(\mu \right)d\mu ,x>\sigma$$

and

$$J_{\delta -}^{\alpha }\mathcal{F}\left(x\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_x^{\delta }{\left(\mu -x\right)}^{\alpha -1}\mathcal{F}\left(\mu \right)d\mu ,x<\delta ,$$

respectively [19,20]. Here, $\mathcal{F}$ belongs to $L_1[\sigma ,\delta ]$, and $\Gamma \left(\alpha \right)$ denotes the Gamma function, which is defined as

$$\Gamma \left(\alpha \right):={\int }_0^{\infty }{e}^{-u}{u}^{\alpha -1}du.$$

The fractional integral coincides with the classical integral for the case of $\alpha =1.$

The formula for Simpson’s quadrature, commonly referred as Simpson’s $1/3$ rule, is as follows:

$${\int }_{\sigma }^{\delta }\mathcal{F}\left(x\right)dx\approx \frac{\delta -\sigma }{6}\left[\mathcal{F}\left(\sigma \right)+4\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+\mathcal{F}\left(\delta \right)\right].$$

The most popular Newton–Cotes quadrature using a three-point Simpson-type inequality is as follows:

Theorem 1. 

Let $\mathcal{F}:\left[\sigma ,\delta \right]\to \mathbb{R}$ denote a four times differentiable and continuous function on $\left(\sigma ,\delta \right)$, and let ${∥{\mathcal{F}}^{\left(4\right)}∥}_{\infty }={\displaystyle \underset{x\in \left(\sigma ,\delta \right)}{sup}}\left|{\mathcal{F}}^{\left(4\right)}\left(x\right)\right|<\infty .$ Then, the following inequality holds:

$$\left|\frac{1}{6}\left[\mathcal{F}\left(\sigma \right)+4\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+\mathcal{F}\left(\delta \right)\right]-\frac{1}{\delta -\sigma }{\int }_{\sigma }^{\delta }\mathcal{F}\left(x\right)dx\right|\le \frac{1}{2880}{∥{\mathcal{F}}^{\left(4\right)}∥}_{\infty }{\left(\delta -\sigma \right)}^4.$$

The Newton–Cotes quadrature formula, frequently referred to as Simpson’s second formula (also known as Simpson’s $3/8$ rule; see [18]), is defined as follows:

$${\int }_{\sigma }^{\delta }\mathcal{F}\left(x\right)dx\approx \frac{\delta -\sigma }{8}\left[\mathcal{F}\left(\sigma \right)+3\mathcal{F}\left(\frac{2\sigma +\delta }{3}\right)+3\mathcal{F}\left(\frac{\sigma +2\delta }{3}\right)+\mathcal{F}\left(\delta \right)\right].$$

The Simpson $3/8$ rule, a classical closed-type quadrature rule based on the Simpson $3/8$ inequality, approximates this.

Theorem 2. 

If $\mathcal{F}:\left[\sigma ,\delta \right]\to \mathbb{R}$ is a four times differentiable and continuous function on $\left(\sigma ,\delta \right),$ and ${∥{\mathcal{F}}^{\left(4\right)}∥}_{\infty }={\displaystyle \underset{x\in \left(\sigma ,\delta \right)}{sup}}\left|{\mathcal{F}}^{\left(4\right)}\left(x\right)\right|<\infty ,$ then one has the inequality

$$\left|\frac{1}{8}\left[\mathcal{F}\left(\sigma \right)+3\mathcal{F}\left(\frac{2\sigma +\delta }{3}\right)+3\mathcal{F}\left(\frac{\sigma +2\delta }{3}\right)+\mathcal{F}\left(\delta \right)\right]-\frac{1}{\delta -\sigma }{\int }_{\sigma }^{\delta }\mathcal{F}\left(x\right)dx\right|\le \frac{1}{6480}{∥{\mathcal{F}}^{\left(4\right)}∥}_{\infty }{\left(\delta -\sigma \right)}^4.$$

The Maclaurin rule, which is derived from the Maclaurin formula (see [18]), is equivalent to the corresponding dual Simpson’s $3/8$ formula:

$${\int }_{\sigma }^{\delta }\mathcal{F}\left(x\right)dx\approx \frac{\delta -\sigma }{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right].$$

The Maclaurin rule, which is derived from the Maclaurin inequality, is equivalent to the corresponding dual Simpson’s $3/8$ formula:

Theorem 3. 

Let $\mathcal{F}:\left[\sigma ,\delta \right]\to \mathbb{R}$ be a four times differentiable and continuous function on $\left(\sigma ,\delta \right)$, and let ${∥{\mathcal{F}}^{\left(4\right)}∥}_{\infty }={\displaystyle \underset{x\in \left(\sigma ,\delta \right)}{sup}}\left|{\mathcal{F}}^{\left(4\right)}\left(x\right)\right|<\infty .$ Then, the following inequality holds:

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{1}{\delta -\sigma }{\int }_{\sigma }^{\delta }\mathcal{F}\left(x\right)dx\right|\hfill \\ \hfill & \le \frac{7}{51840}{∥{\mathcal{F}}^{\left(4\right)}∥}_{\infty }{\left(\delta -\sigma \right)}^4.\hfill \end{array}$$

3. A Crucial Equality

In this section, we express integral equality in order to demonstrate the main results of the study.

Lemma 1. 

If $\mathcal{F}:[\sigma ,\delta ]\to \mathbb{R}$ is an absolutely continuous function $(\sigma ,\delta )$ such that ${\mathcal{F}}^{\prime }\in L_1\left[\sigma ,\delta \right]$, then the equality

$$\begin{array}{cc}\hfill & \frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\hfill \\ \hfill & =\frac{\delta -\sigma }{4}\left[I_1+I_2\right].\hfill \end{array}$$

is valid. Here,

$$\left\{\begin{array}{c}{I}_1=\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha }\left[{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)-{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right]d\mu ,\hfill \\ I_2=\underset{\frac{1}{3}}{\overset{1}{\int }}\left({\mu }^{\alpha }-\frac{3}{4}\right)\left[{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)-{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right]d\mu .\hfill \end{array}\right.$$

Proof. 

If we first use the integration by parts, then one can obtain

$$\begin{array}{cc}\hfill I_1& =\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha }\left[{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)-{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right]d\mu \hfill \\ \hfill & {\left.=\frac{2}{\delta -\sigma }{\mu }^{\alpha }\left[\mathcal{F}\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)+\mathcal{F}\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right]\right|}_0^{\frac{1}{3}}\hfill \\ \hfill & -\frac{2\alpha }{\delta -\sigma }\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha -1}\left[\mathcal{F}\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)+\mathcal{F}\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right]d\mu \hfill \\ \hfill & =\frac{2}{\delta -\sigma }{\left(\frac{1}{3}\right)}^{\alpha }\left[\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]\hfill \\ \hfill & -\frac{2\alpha }{\delta -\sigma }\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha -1}\left[\mathcal{F}\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)+\mathcal{F}\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right]d\mu .\hfill \end{array}$$

(1)

In a similar manner, we have

$$\begin{array}{cc}\hfill I_2& =\underset{\frac{1}{3}}{\overset{1}{\int }}\left({\mu }^{\alpha }-\frac{3}{4}\right)\left[{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)-{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right]d\mu \hfill \\ \hfill & =\frac{1}{\delta -\sigma }\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)-\frac{2}{\delta -\sigma }\left({\left(\frac{1}{3}\right)}^{\alpha }-\frac{3}{4}\right)\left[\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]\hfill \\ \hfill & -\frac{2\alpha }{\delta -\sigma }\underset{\frac{1}{3}}{\overset{1}{\int }}{\mu }^{\alpha -1}\left[\mathcal{F}\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)+\mathcal{F}\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right]d\mu .\hfill \end{array}$$

(2)

Combining (1) and (2) allows us to easily obtain

$$\begin{array}{cc}\hfill I_1+I_2& =\frac{1}{2\left(\delta -\sigma \right)}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]\hfill \\ \hfill & -\frac{2\alpha }{\delta -\sigma }\underset{0}{\overset{1}{\int }}{\mu }^{\alpha -1}\left[\mathcal{F}\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)+\mathcal{F}\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right].\hfill \end{array}$$

(3)

If we use the change of the variable $x=\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma$ and $y=\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta$ for $\mu \in \left[0,1\right]$, then the equality (3) can be rewritten as follows

$$\begin{array}{cc}\hfill I_1+I_2& =\frac{1}{2\left(\delta -\sigma \right)}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]\hfill \\ \hfill & -\frac{2^{\alpha +1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha +1}}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right].\hfill \end{array}$$

(4)

Finally, if we multiply both sides of (4) by $\frac{\delta -\sigma }{4}$, then we conclude the proof of Lemma 1. □

4. Convex Functions: Euler–Maclaurin-Type Inequalities with Fractional Integrals

In this section, we obtain several Euler–Maclaurin-type inequalities for differentiable convex functions by using the Riemann–Liouville fractional integrals.

Theorem 4. 

Suppose that Lemma 1 holds, and the function $\left|{\mathcal{F}}^{\prime }\right|$ is convex on the interval $[\sigma ,\delta ]$. Then, one can prove the following fractional Euler–Maclaurin-type inequality:

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & \le \frac{\delta -\sigma }{4}\left({\Omega }_1\left(\alpha \right)+{\Omega }_2\left(\alpha \right)\right)\left[\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|+\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|\right],\hfill \end{array}$$

(5)

where

$${\Omega }_1\left(\alpha \right)=\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha }d\mu =\frac{1}{\alpha +1}{\left(\frac{1}{3}\right)}^{\alpha +1}$$

and

$${\Omega }_2\left(\alpha \right)=\underset{\frac{1}{3}}{\overset{1}{\int }}\left|{\mu }^{\alpha }-\frac{3}{4}\right|d\mu =\left\{\begin{array}{cc}\frac{1}{\alpha +1}\left(1-{\left(\frac{1}{3}\right)}^{\alpha +1}\right)-\frac{1}{2},\hfill & 0<\alpha \le \frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)},\hfill \\ \hfill & \\ \frac{2\alpha }{\alpha +1}{\left(\frac{3}{4}\right)}^{1+\frac{1}{\alpha }}+\frac{1}{\alpha +1}{\left(\frac{1}{3}\right)}^{\alpha +1}+\frac{1}{\alpha +1}-1,\hfill & \frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}<\alpha .\hfill \end{array}\right.$$

Proof. 

By taking into account the absolute value of Lemma 1, one may directly obtain

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & \le \frac{\delta -\sigma }{4}\left\{\underset{0}{\overset{\frac{1}{3}}{\int }}\left|{\mu }^{\alpha }\right|\left[\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)\right|+\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right|\right]d\mu \right.\hfill \\ \hfill & \left.+\underset{\frac{1}{3}}{\overset{1}{\int }}\left|{\mu }^{\alpha }-\frac{3}{4}\right|\left[\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)\right|+\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right|\right]d\mu \right\}.\hfill \end{array}$$

(6)

Since $\left|{\mathcal{F}}^{\prime }\right|$ is convex, it yields

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & \le \frac{\delta -\sigma }{4}\left\{\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha }\left[\frac{\mu }{2}\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|+\frac{2-\mu }{2}\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|+\frac{\mu }{2}\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|+\frac{2-\mu }{2}\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|\right]d\mu \right.\hfill \\ \hfill & \left.+\underset{\frac{1}{3}}{\overset{1}{\int }}\left|{\mu }^{\alpha }-\frac{3}{4}\right|\left[\frac{\mu }{2}\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|+\frac{2-\mu }{2}\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|+\frac{\mu }{2}\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|+\frac{2-\mu }{2}\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|\right]d\mu \right\}\hfill \\ \hfill & =\frac{\delta -\sigma }{4}\left({\Omega }_1\left(\alpha \right)+{\Omega }_2\left(\alpha \right)\right)\left[\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|+\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|\right],\hfill \end{array}$$

which completes the proof of Theorem 1. □

Remark 1. 

If we assign $\alpha =1$ in Theorem 4, then we can obtain the following Euler–Maclaurin-type inequality:

$$\left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{1}{\delta -\sigma }\underset{\sigma }{\overset{\delta }{\int }}\mathcal{F}\left(\mu \right)d\mu \right|\le \frac{25\left(\delta -\sigma \right)}{576}\left[\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|+\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|\right],$$

which is given in (Corollary 1, [13]).

Theorem 5. 

Let us consider the assumptions in Lemma 1, and the function ${\left|{\mathcal{F}}^{\prime }\right|}^q$, $q>1$ is convex on $[\sigma ,\delta ]$. Then, the following Euler–Maclaurin-type inequality holds:

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & \le \frac{\delta -\sigma }{4}\left\{{\left(\frac{1}{\left(\alpha p+1\right)}{\left(\frac{1}{3}\right)}^{\alpha p+1}\right)}^{\frac{1}{p}}\left[{\left(\frac{{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q+11{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q}{36}\right)}^{\frac{1}{q}}+{\left(\frac{{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q+11{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q}{36}\right)}^{\frac{1}{q}}\right]\right.\hfill \\ \hfill & \left.+{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}{\left|{\mu }^{\alpha }-\frac{3}{4}\right|}^{p}d\mu \right)}^{\frac{1}{p}}\left[{\left(\frac{2{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q+4{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q}{9}\right)}^{\frac{1}{q}}+{\left(\frac{2{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q+4{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q}{9}\right)}^{\frac{1}{q}}\right]\right\}.\hfill \end{array}$$

(7)

Here, $\frac{1}{p}+\frac{1}{q}=1$.

Proof. 

If we apply Hölder’s inequality to (6), then we obtain

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & \le \frac{\delta -\sigma }{4}\left\{{\left(\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha p}d\mu \right)}^{\frac{1}{p}}{\left(\underset{0}{\overset{\frac{1}{3}}{\int }}{\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)\right|}^{q}d\mu \right)}^{\frac{1}{q}}\right.\hfill \\ \hfill & +{\left(\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha p}d\mu \right)}^{\frac{1}{p}}{\left(\underset{0}{\overset{\frac{1}{3}}{\int }}{\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right|}^{q}d\mu \right)}^{\frac{1}{q}}\hfill \\ \hfill & +{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}{\left|{\mu }^{\alpha }-\frac{3}{4}\right|}^{p}d\mu \right)}^{\frac{1}{p}}{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}{\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)\right|}^{q}d\mu \right)}^{\frac{1}{q}}\hfill \\ \hfill & \left.+{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}{\left|{\mu }^{\alpha }-\frac{3}{4}\right|}^{p}d\mu \right)}^{\frac{1}{p}}{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}{\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right|}^{q}d\mu \right)}^{\frac{1}{q}}\right\}.\hfill \end{array}$$

Taking advantage of the convexity of ${\left|{\mathcal{F}}^{\prime }\right|}^q$, we can easily obtain

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & \le \frac{\delta -\sigma }{4}\left\{{\left(\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha p}d\mu \right)}^{\frac{1}{p}}{\left(\underset{0}{\overset{\frac{1}{3}}{\int }}\left(\frac{\mu }{2}{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q+\frac{2-\mu }{2}{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q\right)d\mu \right)}^{\frac{1}{q}}\right.\hfill \\ \hfill & +{\left(\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha p}d\mu \right)}^{\frac{1}{p}}{\left(\underset{0}{\overset{\frac{1}{3}}{\int }}\left(\frac{\mu }{2}{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q+\frac{2-\mu }{2}{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q\right)d\mu \right)}^{\frac{1}{q}}\hfill \\ \hfill & +{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}{\left|{\mu }^{\alpha }-\frac{3}{4}\right|}^{p}d\mu \right)}^{\frac{1}{p}}{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}\left(\frac{\mu }{2}{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q+\frac{2-\mu }{2}{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q\right)d\mu \right)}^{\frac{1}{q}}\hfill \\ \hfill & \left.+{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}{\left|{\mu }^{\alpha }-\frac{3}{4}\right|}^{p}d\mu \right)}^{\frac{1}{p}}{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}\left(\frac{\mu }{2}{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q+\frac{2-\mu }{2}{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q\right)d\mu \right)}^{\frac{1}{q}}\right\}\hfill \\ \hfill & =\frac{\delta -\sigma }{4}\left\{{\left(\frac{1}{\left(\alpha p+1\right)}{\left(\frac{1}{3}\right)}^{\alpha p+1}\right)}^{\frac{1}{p}}\left[{\left(\frac{{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q+11{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q}{36}\right)}^{\frac{1}{q}}+{\left(\frac{{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q+11{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q}{36}\right)}^{\frac{1}{q}}\right]\right.\hfill \\ \hfill & \left.+{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}{\left|{\mu }^{\alpha }-\frac{3}{4}\right|}^{p}d\mu \right)}^{\frac{1}{p}}\left[{\left(\frac{2{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q+4{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q}{9}\right)}^{\frac{1}{q}}+{\left(\frac{2{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q+4{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q}{9}\right)}^{\frac{1}{q}}\right]\right\}.\hfill \end{array}$$

This ends the proof of Theorem 5. □

Corollary 1. 

If we select $\alpha =1$ in Theorem 5, then the following inequality holds:

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{1}{\delta -\sigma }\underset{\sigma }{\overset{\delta }{\int }}\mathcal{F}\left(\mu \right)d\mu \right|\hfill \\ \hfill & \le \frac{\delta -\sigma }{4}\left\{{\left(\frac{1}{p+1}\left[{\left(\frac{1}{4}\right)}^{p+1}+{\left(\frac{5}{12}\right)}^{p+1}\right]\right)}^{\frac{1}{p}}\right.\hfill \\ \hfill & \times \left[{\left(\frac{4{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q+2{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q}{9}\right)}^{\frac{1}{q}}+{\left(\frac{4{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q+2{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q}{9}\right)}^{\frac{1}{q}}\right]\hfill \\ \hfill & \left.+{\left(\frac{1}{p+1}{\left(\frac{1}{3}\right)}^{p+1}\right)}^{\frac{1}{p}}\left[{\left(\frac{11{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q+{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q}{36}\right)}^{\frac{1}{q}}+{\left(\frac{11{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q+{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q}{36}\right)}^{\frac{1}{q}}\right]\right\}.\hfill \end{array}$$

Theorem 6. 

Assume that the assumptions of Lemma 1 are satisfied, and the function ${\left|{\mathcal{F}}^{\prime }\right|}^q$, $q\ge 1$ is convex on $[\sigma ,\delta ].$ Then, we obtain the following Euler–Maclaurin-type inequality:

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & \le \frac{\delta -\sigma }{4}\left\{{\left({\Omega }_1\left(\alpha \right)\right)}^{1-\frac{1}{q}}\left[{\left[{\Omega }_3\left(\alpha \right){\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q+{\Omega }_4\left(\alpha \right){\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q\right]}^{\frac{1}{q}}\right.\right.\hfill \\ \hfill & +\left.{\left[{\Omega }_3\left(\alpha \right){\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q+{\Omega }_4\left(\alpha \right){\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q\right]}^{\frac{1}{q}}\right]\hfill \\ \hfill & +{\left({\Omega }_2\left(\alpha \right)\right)}^{1-\frac{1}{q}}\left[{\left[{\Omega }_5\left(\alpha \right){\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q+{\Omega }_6\left(\alpha \right){\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q\right]}^{\frac{1}{q}}\right.\hfill \\ \hfill & \left.\left.+{\left[{\Omega }_5\left(\alpha \right){\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q+{\Omega }_6\left(\alpha \right){\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q\right]}^{\frac{1}{q}}\right]\right\}.\hfill \end{array}$$

(8)

Here, ${\Omega }_1\left(\alpha \right)$ and ${\Omega }_2\left(\alpha \right)$ are specified in Theorem 4, and

$${\Omega }_3\left(\alpha \right)=\underset{0}{\overset{\frac{1}{3}}{\int }}\frac{{\mu }^{\alpha +1}}{2}d\mu =\frac{1}{2\left(\alpha +2\right)}{\left(\frac{1}{3}\right)}^{\alpha +2},$$

$${\Omega }_4\left(\alpha \right)=\underset{0}{\overset{\frac{1}{3}}{\int }}\left(\frac{2-\mu }{2}\right){\mu }^{\alpha }d\mu =\frac{5\alpha +11}{2\left(\alpha +1\right)\left(\alpha +2\right)}{\left(\frac{1}{3}\right)}^{\alpha +2},$$

$${\Omega }_5\left(\alpha \right)=\underset{\frac{1}{3}}{\overset{1}{\int }}\frac{\mu }{2}\left|{\mu }^{\alpha }-\frac{3}{4}\right|d\mu =\left\{\begin{array}{cc}\frac{1}{2\left(\alpha +2\right)}\left(1-{\left(\frac{1}{3}\right)}^{\alpha +2}\right)-\frac{1}{6},\hfill & 0<\alpha \le \frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)},\hfill \\ \hfill & \\ \frac{1}{2\left(\alpha +2\right)}\left[\alpha {\left(\frac{3}{4}\right)}^{1+\frac{2}{\alpha }}+{\left(\frac{1}{3}\right)}^{\alpha +2}+1\right]-\frac{5}{24},\hfill & \frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}<\alpha ,\hfill \end{array}\right.$$

$${\Omega }_6\left(\alpha \right)=\underset{\frac{1}{3}}{\overset{1}{\int }}\left(\frac{2-\mu }{2}\right)\left|{\mu }^{\alpha }-\frac{3}{4}\right|d\mu =\left\{\begin{array}{cc}\frac{\alpha +3}{2\left(\alpha +1\right)\left(\alpha +2\right)}-\frac{1}{\alpha +1}{\left(\frac{1}{3}\right)}^{\alpha +1}+\frac{1}{2\left(\alpha +2\right)}{\left(\frac{1}{3}\right)}^{\alpha +2}-\frac{1}{3},\hfill & 0<\alpha \le \frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)},\hfill \\ \begin{array}{c}\frac{2\alpha }{\alpha +1}{\left(\frac{3}{4}\right)}^{1+\frac{1}{\alpha }}+\frac{1}{\alpha +1}{\left(\frac{1}{3}\right)}^{\alpha +1}+\frac{1}{\alpha +1}-1\hfill \\ -\frac{1}{2\left(\alpha +2\right)}\left[\alpha {\left(\frac{3}{4}\right)}^{1+\frac{2}{\alpha }}+{\left(\frac{1}{3}\right)}^{\alpha +2}+1\right]+\frac{5}{24},\hfill \end{array}\hfill & \frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}<\alpha .\hfill \end{array}\right.$$

Proof. 

When we apply (6) to the power mean inequality, we have

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & \le \frac{\delta -\sigma }{4}\left\{{\left(\underset{0}{\overset{\frac{1}{3}}{\int }}\left|{\mu }^{\alpha }\right|d\mu \right)}^{1-\frac{1}{q}}{\left(\underset{0}{\overset{\frac{1}{3}}{\int }}\left|{\mu }^{\alpha }\right|{\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)\right|}^{q}d\mu \right)}^{\frac{1}{q}}\right.\hfill \\ \hfill & +{\left(\underset{0}{\overset{\frac{1}{3}}{\int }}\left|{\mu }^{\alpha }\right|d\mu \right)}^{1-\frac{1}{q}}{\left(\underset{0}{\overset{\frac{1}{3}}{\int }}\left|{\mu }^{\alpha }\right|{\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right|}^{q}d\mu \right)}^{\frac{1}{q}}\hfill \\ \hfill & +{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}\left|{\mu }^{\alpha }-\frac{3}{4}\right|d\mu \right)}^{1-\frac{1}{q}}{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}\left|{\mu }^{\alpha }-\frac{3}{4}\right|{\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)\right|}^{q}d\mu \right)}^{\frac{1}{q}}\hfill \\ \hfill & \left.+{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}\left|{\mu }^{\alpha }-\frac{3}{4}\right|d\mu \right)}^{1-\frac{1}{q}}{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}\left|{\mu }^{\alpha }-\frac{3}{4}\right|{\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right|}^{q}d\mu \right)}^{\frac{1}{q}}\right\}.\hfill \end{array}$$

By using the convexity of ${\left|{\mathcal{F}}^{\prime }\right|}^q$, it follows that

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & \le \frac{\delta -\sigma }{4}\left\{{\left(\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha }d\mu \right)}^{1-\frac{1}{q}}{\left(\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha }\left[\frac{\mu }{2}{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q+\frac{2-\mu }{2}{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q\right]d\mu \right)}^{\frac{1}{q}}\right.\hfill \\ \hfill & +{\left(\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha }d\mu \right)}^{1-\frac{1}{q}}{\left(\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha }\left[\frac{\mu }{2}{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q+\frac{2-\mu }{2}{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q\right]d\mu \right)}^{\frac{1}{q}}\hfill \\ \hfill & +{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}\left|{\mu }^{\alpha }-\frac{3}{4}\right|d\mu \right)}^{1-\frac{1}{q}}{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}\left|{\mu }^{\alpha }-\frac{3}{4}\right|\left[\frac{\mu }{2}{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q+\frac{2-\mu }{2}{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q\right]d\mu \right)}^{\frac{1}{q}}\hfill \\ \hfill & \left.+{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}\left|{\mu }^{\alpha }-\frac{3}{4}\right|d\mu \right)}^{1-\frac{1}{q}}{\left(\underset{\frac{1}{3}}{\overset{1}{\int }}\left|{\mu }^{\alpha }-\frac{3}{4}\right|\left[\frac{\mu }{2}{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q+\frac{2-\mu }{2}{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q\right]d\mu \right)}^{\frac{1}{q}}\right\}.\hfill \end{array}$$

This finishes the proof of Theorem 6. □

Remark 2. 

If we choose $\alpha =1$ in Theorem 6, then we obtain the following Euler–Maclaurin-type inequality:

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{1}{\delta -\sigma }\underset{\sigma }{\overset{\delta }{\int }}\mathcal{F}\left(\mu \right)d\mu \right|\hfill \\ \hfill & \le \frac{\delta -\sigma }{72}\left\{\left[{\left(\frac{{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q+8{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q}{9}\right)}^{\frac{1}{q}}+{\left(\frac{{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q+8{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q}{9}\right)}^{\frac{1}{q}}\right]\right.\hfill \\ \hfill & \left.+{\left(\frac{17}{8}\right)}^{1-\frac{1}{q}}\left[{\left(\frac{361{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q+863{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q}{576}\right)}^{\frac{1}{q}}+{\left(\frac{361{\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|}^q+863{\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|}^q}{576}\right)}^{\frac{1}{q}}\right]\right\},\hfill \end{array}$$

which is established in (Corollary 3, [13]).

5. Bounded Functions: Euler–Maclaurin-Type Inequalities with Fractional Integrals

In this section, we consider some Euler–Maclaurin-type inequalities for bounded functions by fractional integrals.

Theorem 7. 

Note that the conditions of Lemma 1 hold. If there exists $m,M\in \mathbb{R}$ so that $m\le {\mathcal{F}}^{\prime }\left(\mu \right)\le M$ for $\mu \in \left[\sigma ,\delta \right],$ then we have

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & \le \frac{\delta -\sigma }{4}\left\{{\Omega }_1\left(\alpha \right)+{\Omega }_2\left(\alpha \right)\right\}\left(M-m\right).\hfill \end{array}$$

(9)

Proof. 

By using the Lemma 1, we obtain

$$\begin{array}{cc}\hfill & \frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\hfill \\ \hfill & =\frac{\delta -\sigma }{4}\left\{\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha }\left[{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)-\frac{m+M}{2}\right]d\mu \right.\hfill \\ \hfill & +\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha }\left[\frac{m+M}{2}-{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right]d\mu \hfill \\ \hfill & +\underset{\frac{1}{3}}{\overset{1}{\int }}\left({\mu }^{\alpha }-\frac{3}{4}\right)\left[{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)-\frac{m+M}{2}\right]d\mu \hfill \\ \hfill & \left.+\underset{\frac{1}{3}}{\overset{1}{\int }}\left({\mu }^{\alpha }-\frac{3}{4}\right)\left[\frac{m+M}{2}-{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right]d\mu \right\}.\hfill \end{array}$$

(10)

Let us take the absolute value of (10). Then, it follows that

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & \le \frac{\delta -\sigma }{4}\left\{\underset{0}{\overset{\frac{1}{3}}{\int }}\left|{\mu }^{\alpha }\right|\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)-\frac{m+M}{2}\right|d\mu \right.\hfill \\ \hfill & +\underset{0}{\overset{\frac{1}{3}}{\int }}\left|{\mu }^{\alpha }\right|\left|\frac{m+M}{2}-{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right|d\mu \hfill \\ \hfill & +\underset{\frac{1}{3}}{\overset{1}{\int }}\left|{\mu }^{\alpha }-\frac{3}{4}\right|\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)-\frac{m+M}{2}\right|d\mu \hfill \\ \hfill & \left.+\underset{\frac{1}{3}}{\overset{1}{\int }}\left|{\mu }^{\alpha }-\frac{3}{4}\right|\left|\frac{m+M}{2}-{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right|d\mu \right\}.\hfill \end{array}$$

It is known that $m\le {\mathcal{F}}^{\prime }\left(\mu \right)\le M$ for $\mu \in \left[\sigma ,\delta \right].$ Then, we readily obtain

$$\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)-\frac{m+M}{2}\right|\le \frac{M-m}{2},$$

(11)

$$\left|\frac{m+M}{2}-{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right|\le \frac{M-m}{2}.$$

(12)

With the aid of the (11) and (12), we have

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & \le \frac{\delta -\sigma }{4}\left(M-m\right)\left\{\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha }d\mu +\underset{\frac{1}{3}}{\overset{1}{\int }}\left|{\mu }^{\alpha }-\frac{3}{4}\right|d\mu \right\}=\frac{\delta -\sigma }{4}\left\{{\Omega }_1\left(\alpha \right)+{\Omega }_2\left(\alpha \right)\right\}\left(M-m\right).\hfill \end{array}$$

□

Corollary 2. 

If we consider $\alpha =1$ in Theorem 7, then the following inequality holds:

$$\left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{1}{\delta -\sigma }{\int }_{\sigma }^{\delta }\mathcal{F}\left(\mu \right)d\mu \right|\le \frac{25\left(\delta -\sigma \right)}{576}\left(M-m\right).$$

Corollary 3. 

Under the assumption of Theorem 7, if there exist $M\in {\mathbb{R}}^{+}$ such that $\left|{\mathcal{F}}^{\prime }\left(\mu \right)\right|\le M$ for all $\mu \in \left[\sigma ,\delta \right],$ then we have

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & \le \frac{\delta -\sigma }{2}\left\{{\Omega }_1\left(\alpha \right)+{\Omega }_2\left(\alpha \right)\right\}M.\hfill \end{array}$$

Corollary 4. 

For $\alpha =1$ in Corollary 3, the following inequality holds:

$$\left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{1}{\delta -\sigma }{\int }_{\sigma }^{\delta }\mathcal{F}\left(\mu \right)d\mu \right|\le \frac{25\left(\delta -\sigma \right)}{288}M.$$

6. Lipschitzian Functions: Euler–Maclaurin-Type Inequalities with Fractional Integrals

In this section, we prove some fractional Euler–Maclaurin-type inequalities for Lipschitzian functions.

Theorem 8. 

Consider that the assumptions of Lemma 1 are valid. If ${\mathcal{F}}^{\prime }$ is a L-Lipschitzian function on $\left[\sigma ,\delta \right],$ then the following inequality holds:

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & \le \frac{{\left(\delta -\sigma \right)}^2}{4}L\left\{{\Omega }_7\left(\alpha \right)+{\Omega }_8\left(\alpha \right)\right\},\hfill \end{array}$$

where

$${\Omega }_7\left(\alpha \right)=\underset{0}{\overset{\frac{1}{3}}{\int }}\left(1-\mu \right){\mu }^{\alpha }d\mu =\frac{1}{\alpha +1}{\left(\frac{1}{3}\right)}^{\alpha +1}-\frac{1}{\alpha +2}{\left(\frac{1}{3}\right)}^{\alpha +2}$$

and

$$\begin{array}{ccc}\hfill {\Omega }_8\left(\alpha \right)& =& \underset{\frac{1}{3}}{\overset{1}{\int }}\left(1-\mu \right)\left|{\mu }^{\alpha }-\frac{3}{4}\right|d\mu \hfill \\ & & \\ & =& \left\{\begin{array}{cc}\frac{1}{\alpha +1}\left(1-{\left(\frac{1}{3}\right)}^{\alpha +1}\right)-\frac{1}{\alpha +2}\left(1-{\left(\frac{1}{3}\right)}^{\alpha +2}\right)-\frac{1}{6},\hfill & 0<\alpha \le \frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)},\hfill \\ \begin{array}{c}\frac{2\alpha }{\alpha +1}{\left(\frac{3}{4}\right)}^{1+\frac{1}{\alpha }}+\frac{1}{\alpha +1}{\left(\frac{1}{3}\right)}^{\alpha +1}+\frac{1}{\alpha +1}\hfill \\ -\frac{1}{\alpha +2}\left[\alpha {\left(\frac{3}{4}\right)}^{1+\frac{2}{\alpha }}+{\left(\frac{1}{3}\right)}^{\alpha +2}+1\right]-\frac{7}{12},\hfill \end{array}\hfill & \frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}<\alpha .\hfill \end{array}\right.\hfill \end{array}$$

Proof. 

By using Lemma 1, and since ${\mathcal{F}}^{\prime }$ is the L-Lipschitzian function, we easily have

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & =\left|\frac{\delta -\sigma }{4}\left\{\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha }\left[{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)-{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right]d\mu \right.\right.\hfill \\ \hfill & \left.\left.+\underset{\frac{1}{3}}{\overset{1}{\int }}\left({\mu }^{\alpha }-\frac{3}{4}\right)\left[{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)-{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right]d\mu \right\}\right|\hfill \\ \hfill & \le \frac{\delta -\sigma }{4}\left\{\underset{0}{\overset{\frac{1}{3}}{\int }}\left|{\mu }^{\alpha }\right|\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)-{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right|d\mu \right.\hfill \\ \hfill & \left.+\underset{\frac{1}{3}}{\overset{1}{\int }}\left|{\mu }^{\alpha }-\frac{3}{4}\right|\left|{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\delta +\frac{2-\mu }{2}\sigma \right)-{\mathcal{F}}^{\prime }\left(\frac{\mu }{2}\sigma +\frac{2-\mu }{2}\delta \right)\right|d\mu \right\}\hfill \\ \hfill & \le \frac{\delta -\sigma }{4}\left\{\underset{0}{\overset{\frac{1}{3}}{\int }}{\mu }^{\alpha }L\left(1-\mu \right)\left(\delta -\sigma \right)d\mu +\underset{\frac{1}{3}}{\overset{1}{\int }}\left|{\mu }^{\alpha }-\frac{3}{4}\right|L\left(1-\mu \right)\left(\delta -\sigma \right)d\mu \right\}\hfill \\ \hfill & =\frac{{\left(\delta -\sigma \right)}^2}{4}L\left\{{\Omega }_7\left(\alpha \right)+{\Omega }_8\left(\alpha \right)\right\}.\hfill \end{array}$$

□

Corollary 5. 

Note that $\alpha =1$ in Theorem 8. Then, the following Euler–Maclaurin-type inequality holds:

$$\left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{1}{\delta -\sigma }{\int }_{\sigma }^{\delta }\mathcal{F}\left(\mu \right)d\mu \right|\le \frac{475{\left(\delta -\sigma \right)}^2}{20736}L.$$

7. Functions of Bounded Variation: Euler–Maclaurin-Type Inequalities via Fractional Integrals

In this section, we establish some Euler–Maclaurin-type inequalities by fractional integrals of bounded variation.

Theorem 9. 

If $\mathcal{F}:\left[\sigma ,\delta \right]\to \mathbb{R}$ is a function of bounded variation on $\left[\sigma ,\delta \right],$ then we obtain

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & \le \frac{1}{2}max\left\{{\left(\frac{1}{3}\right)}^{\alpha },\left|{\left(\frac{1}{3}\right)}^{\alpha }-\frac{3}{4}\right|,\frac{1}{4}\right\}\underset{\sigma }{\overset{\delta }{\bigvee }}\left(\mathcal{F}\right),\hfill \end{array}$$

where $\underset{\sigma }{\overset{\delta }{\bigvee }}\left(\mathcal{F}\right)$ denotes the total variation of $\mathcal{F}$ on $\left[\sigma ,\delta \right].$

Proof. 

Define the function $K_{\alpha }\left(x\right)$ as

$$K_{\alpha }\left(x\right)=\left\{\begin{array}{cc}{\left(x-\sigma \right)}^{\alpha },\hfill & \sigma \le x<\frac{5\sigma +\delta }{6},\hfill \\ {\left(x-\sigma \right)}^{\alpha }-\frac{3}{4}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha },\hfill & \frac{5\sigma +\delta }{6}\le x<\frac{\sigma +\delta }{2},\hfill \\ \frac{3}{4}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha }-{\left(\delta -x\right)}^{\alpha },\hfill & \frac{\sigma +\delta }{2}\le x<\frac{\sigma +5\delta }{6},\hfill \\ -{\left(\delta -x\right)}^{\alpha },\hfill & \frac{\sigma +5\delta }{6}\le x\le \delta .\hfill \end{array}\right.$$

With the help of integrating by parts, we have

$$\begin{array}{cc}\hfill & \underset{\sigma }{\overset{\delta }{\int }}{K}_{\alpha }\left(x\right)d\mathcal{F}\left(x\right)\hfill \\ \hfill & =\underset{\sigma }{\overset{\frac{5\sigma +\delta }{6}}{\int }}{\left(x-\sigma \right)}^{\alpha }d\mathcal{F}\left(x\right)+\underset{\frac{5\sigma +\delta }{6}}{\overset{\frac{\sigma +\delta }{2}}{\int }}\left[{\left(x-\sigma \right)}^{\alpha }-\frac{3}{4}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha }\right]d\mathcal{F}\left(x\right)\hfill \\ \hfill & +\underset{\frac{\sigma +\delta }{2}}{\overset{\frac{\sigma +5\delta }{6}}{\int }}\left[\frac{3}{4}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha }-{\left(\delta -x\right)}^{\alpha }\right]d\mathcal{F}\left(x\right)-\underset{\frac{\sigma +5\delta }{6}}{\overset{\delta }{\int }}{\left(\delta -x\right)}^{\alpha }d\mathcal{F}\left(x\right)\hfill \\ \hfill & ={\left.{\left(x-\sigma \right)}^{\alpha }\mathcal{F}\left(x\right)\right|}_{\sigma }^{\frac{5\sigma +\delta }{6}}-\alpha \underset{\sigma }{\overset{\frac{5\sigma +\delta }{6}}{\int }}{\left(x-\sigma \right)}^{\alpha -1}\mathcal{F}\left(x\right)dx\hfill \\ \hfill & +{\left.\left[{\left(x-\sigma \right)}^{\alpha }-\frac{3}{4}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha }\right]\mathcal{F}\left(x\right)\right|}_{\frac{5\sigma +\delta }{6}}^{\frac{\sigma +\delta }{2}}-\alpha \underset{\frac{5\sigma +\delta }{6}}{\overset{\frac{\sigma +\delta }{2}}{\int }}{\left(x-\sigma \right)}^{\alpha -1}\mathcal{F}\left(x\right)dx\hfill \end{array}$$

$$\begin{array}{cc}\hfill & +{\left.\left[\frac{3}{4}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha }-{\left(\delta -x\right)}^{\alpha }\right]\mathcal{F}\left(x\right)\right|}_{\frac{\sigma +\delta }{2}}^{\frac{\sigma +5\delta }{6}}-\alpha \underset{\frac{\sigma +\delta }{2}}{\overset{\frac{\sigma +5\delta }{6}}{\int }}{\left(\delta -x\right)}^{\alpha -1}\mathcal{F}\left(x\right)dx\hfill \\ \hfill & -{\left.{\left(\delta -x\right)}^{\alpha }\mathcal{F}\left(x\right)\right|}_{\frac{\sigma +5\delta }{6}}^{\delta }-\alpha \underset{\frac{\sigma +5\delta }{6}}{\overset{\delta }{\int }}{\left(\delta -x\right)}^{\alpha -1}\mathcal{F}\left(x\right)dx\hfill \\ \hfill & =\frac{3}{4}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha }\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+\frac{3}{4}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha }\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)+\frac{1}{2}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha }\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)\hfill \\ \hfill & -\alpha \underset{\sigma }{\overset{\frac{\sigma +\delta }{2}}{\int }}{\left(x-\sigma \right)}^{\alpha -1}\mathcal{F}\left(x\right)dx-\alpha \underset{\frac{\sigma +\delta }{2}}{\overset{\delta }{\int }}{\left(\delta -x\right)}^{\alpha -1}\mathcal{F}\left(x\right)dx\hfill \\ \hfill & =\frac{{\left(\delta -\sigma \right)}^{\alpha }}{2^{\alpha -1}}\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\Gamma \left(\alpha +1\right)\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right].\hfill \end{array}$$

Namely, we have

$$\begin{array}{cc}\hfill & \frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\hfill \\ \hfill & =\frac{2^{\alpha -1}}{{\left(\delta -\sigma \right)}^{\alpha }}\underset{\sigma }{\overset{\delta }{\int }}{K}_{\alpha }\left(x\right)d\mathcal{F}\left(x\right).\hfill \end{array}$$

It is known that if $g,\mathcal{F}:\left[\sigma ,\delta \right]\to \mathbb{R}$ are such that g is continuous on $\left[\sigma ,\delta \right]$ and $\mathcal{F}$ is of bounded variation on $\left[\sigma ,\delta \right]$, then $\underset{\sigma }{\overset{\delta }{\int }}g\left(\mu \right)d\mathcal{F}\left(\mu \right)$ exists, and

$$\left|\underset{\sigma }{\overset{\delta }{\int }}g\left(\mu \right)d\mathcal{F}\left(\mu \right)\right|\le \underset{\mu \in \left[\sigma ,\delta \right]}{sup}\left|g\left(\mu \right)\right|\underset{\sigma }{\overset{\delta }{\bigvee }}\left(\mathcal{F}\right).$$

(13)

By using (13), it follows that

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\sigma +}^{\alpha }\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+J_{\delta -}^{\alpha }\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)\right]\right|\hfill \\ \hfill & =\frac{2^{\alpha -1}}{{\left(\delta -\sigma \right)}^{\alpha }}\left|\underset{\sigma }{\overset{\delta }{\int }}{K}_{\alpha }\left(x\right)d\mathcal{F}\left(x\right)\right|\hfill \\ \hfill & \le \frac{2^{\alpha -1}}{{\left(\delta -\sigma \right)}^{\alpha }}\left\{\left|\underset{\sigma }{\overset{\frac{5\sigma +\delta }{6}}{\int }}{\left(x-\sigma \right)}^{\alpha }d\mathcal{F}\left(x\right)\right|\right.\hfill \\ \hfill & +\left|\underset{\frac{5\sigma +\delta }{6}}{\overset{\frac{\sigma +\delta }{2}}{\int }}\left[{\left(x-\sigma \right)}^{\alpha }-\frac{3}{4}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha }\right]d\mathcal{F}\left(x\right)\right|\hfill \end{array}$$

$$\begin{array}{cc}\hfill & +\left|\underset{\frac{\sigma +\delta }{2}}{\overset{\frac{\sigma +5\delta }{6}}{\int }}\left[\frac{3}{4}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha }-{\left(\delta -x\right)}^{\alpha }\right]d\mathcal{F}\left(x\right)\right|\hfill \\ \hfill & \left.+\left|\underset{\frac{\sigma +5\delta }{6}}{\overset{\delta }{\int }}\left[-{\left(\delta -x\right)}^{\alpha }\right]d\mathcal{F}\left(x\right)\right|\right\}\hfill \\ \hfill & \le \frac{2^{\alpha -1}}{{\left(\delta -\sigma \right)}^{\alpha }}\left\{\underset{x\in \left[\sigma ,\frac{5\sigma +\delta }{6}\right]}{sup}\left|{\left(x-\sigma \right)}^{\alpha }\right|\underset{\sigma }{\overset{\frac{5\sigma +\delta }{6}}{\bigvee }}\left(\mathcal{F}\right)\right.\hfill \\ \hfill & +\underset{x\in \left[\frac{5\sigma +\delta }{6},\frac{\sigma +\delta }{2}\right]}{sup}\left|{\left(x-\sigma \right)}^{\alpha }-\frac{3}{4}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha }\right|\underset{\frac{5\sigma +\delta }{6}}{\overset{\frac{\sigma +\delta }{2}}{\bigvee }}\left(\mathcal{F}\right)\hfill \\ \hfill & +\underset{x\in \left[\frac{\sigma +\delta }{2},\frac{\sigma +5\delta }{6}\right]}{sup}\left|\frac{3}{4}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha }-{\left(\delta -x\right)}^{\alpha }\right|\underset{\frac{\sigma +\delta }{2}}{\overset{\frac{\sigma +5\delta }{6}}{\bigvee }}\left(\mathcal{F}\right)\hfill \\ \hfill & \left.+\underset{x\in \left[\frac{\sigma +5\delta }{6},\delta \right]}{sup}\left|-{\left(\delta -x\right)}^{\alpha }\right|\underset{\frac{\sigma +5\delta }{6}}{\overset{\delta }{\bigvee }}\left(\mathcal{F}\right)\right\}\hfill \\ \hfill & =\frac{2^{\alpha -1}}{{\left(\delta -\sigma \right)}^{\alpha }}\left\{{\left(\frac{\delta -\sigma }{6}\right)}^{\alpha }\underset{\sigma }{\overset{\frac{5\sigma +\delta }{6}}{\bigvee }}\left(\mathcal{F}\right)\right.\hfill \\ \hfill & +max\left[\frac{1}{4}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha },\left|{\left(\frac{\delta -\sigma }{6}\right)}^{\alpha }-\frac{3}{4}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha }\right|\right]\underset{\frac{5\sigma +\delta }{6}}{\overset{\frac{\sigma +\delta }{2}}{\bigvee }}\left(\mathcal{F}\right)\hfill \\ \hfill & +max\left[\left|\frac{3}{4}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha }-{\left(\frac{\delta -\sigma }{6}\right)}^{\alpha }\right|,\frac{1}{4}{\left(\frac{\delta -\sigma }{2}\right)}^{\alpha }\right]\underset{\frac{\sigma +\delta }{2}}{\overset{\frac{\sigma +5\delta }{6}}{\bigvee }}\left(\mathcal{F}\right)\hfill \\ \hfill & \left.+{\left(\frac{\delta -\sigma }{6}\right)}^{\alpha }\underset{\frac{\sigma +5\delta }{6}}{\overset{\delta }{\bigvee }}\left(\mathcal{F}\right)\right\}\hfill \\ \hfill & =\frac{1}{2}\left\{{\left(\frac{1}{3}\right)}^{\alpha }\underset{\sigma }{\overset{\frac{5\sigma +\delta }{6}}{\bigvee }}\left(\mathcal{F}\right)+max\left[\frac{1}{4},\left|{\left(\frac{1}{3}\right)}^{\alpha }-\frac{3}{4}\right|\right]\underset{\frac{5\sigma +\delta }{6}}{\overset{\frac{\sigma +\delta }{2}}{\bigvee }}\left(\mathcal{F}\right)\right.\hfill \\ \hfill & \left.+max\left[\left|{\left(\frac{1}{3}\right)}^{\alpha }-\frac{3}{4}\right|,\frac{1}{4}\right]\underset{\frac{\sigma +\delta }{2}}{\overset{\frac{\sigma +5\delta }{6}}{\bigvee }}\left(\mathcal{F}\right)+{\left(\frac{1}{3}\right)}^{\alpha }\underset{\frac{\sigma +5\delta }{6}}{\overset{\delta }{\bigvee }}\left(\mathcal{F}\right)\right\}\hfill \\ \hfill & \le \frac{1}{2}max\left\{{\left(\frac{1}{3}\right)}^{\alpha },\left|{\left(\frac{1}{3}\right)}^{\alpha }-\frac{3}{4}\right|,\frac{1}{4}\right\}\underset{\sigma }{\overset{\delta }{\bigvee }}\left(\mathcal{F}\right).\hfill \end{array}$$

This finishes the proof of Theorem 9. □

Corollary 6. 

Let us consider $\alpha =1$ in Theorem 9. Then, the following inequality holds:

$$\left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{1}{\delta -\sigma }{\int }_{\sigma }^{\delta }\mathcal{F}\left(\mu \right)d\mu \right|\le \frac{5}{24}\underset{\sigma }{\overset{\delta }{\bigvee }}\left(\mathcal{F}\right).$$

8. Examples of Main Results

Example 1. 

Let us consider that a function $\mathcal{F}:[\sigma ,\delta ]=[0,3]\to \mathbb{R}$ is defined by $\mathcal{F}\left(x\right)=\frac{x^2}{2}$ with $\alpha \in (0,10]$. Then, the left-hand side of (5) coincides with

$$\begin{array}{cc}\hfill & \left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{5\sigma +\delta }{6}\right)+2\mathcal{F}\left(\frac{\sigma +\delta }{2}\right)+3\mathcal{F}\left(\frac{\sigma +5\delta }{6}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{{\left(\delta -\sigma \right)}^{\alpha }}\left[J_{\frac{\sigma +\delta }{2}-}^{\alpha }\mathcal{F}\left(\sigma \right)+J_{\frac{\sigma +\delta }{2}+}^{\alpha }\mathcal{F}\left(\delta \right)\right]\right|\hfill \\ \hfill & =\left|\frac{1}{8}\left[3\mathcal{F}\left(\frac{1}{2}\right)+2\mathcal{F}\left(\frac{3}{2}\right)+3\mathcal{F}\left(\frac{5}{2}\right)\right]-\frac{2^{\alpha -1}\Gamma \left(\alpha +1\right)}{3^{\alpha }}\left[J_{\frac{3}{2}-}^{\alpha }\mathcal{F}\left(0\right)+J_{\frac{3}{2}+}^{\alpha }\mathcal{F}\left(3\right)\right]\right|\hfill \\ \hfill & =\left|\frac{3}{2}-\frac{9\left({\alpha }^2+3\alpha +4\right)}{8\left(\alpha +1\right)\left(\alpha +2\right)}\right|.\hfill \end{array}$$

(14)

The right-hand side of (5) becomes

$$\begin{array}{cc}\hfill & \frac{\delta -\sigma }{4}\left({\Omega }_1\left(\alpha \right)+{\Omega }_2\left(\alpha \right)\right)\left[\left|{\mathcal{F}}^{\prime }\left(\sigma \right)\right|+\left|{\mathcal{F}}^{\prime }\left(\delta \right)\right|\right]=\frac{9}{4}\left({\Omega }_1\left(\alpha \right)+{\Omega }_2\left(\alpha \right)\right)\hfill \\ \hfill & =\left\{\begin{array}{cc}\frac{9\left(1-\alpha \right)}{8\left(\alpha +1\right)},\hfill & 0<\alpha \le \frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)},\hfill \\ \frac{9}{4}\left[\frac{2\alpha }{\alpha +1}{\left(\frac{3}{4}\right)}^{1+\frac{1}{\alpha }}+\frac{2}{\alpha +1}{\left(\frac{1}{3}\right)}^{\alpha +1}+\frac{1}{\alpha +1}-1\right],\hfill & \frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}<\alpha \le 10.\hfill \end{array}\right.\hfill \end{array}$$

Consequently, we have

$$\left\{\begin{array}{cc}\left|\frac{3}{2}-\frac{9\left({\alpha }^2+3\alpha +4\right)}{8\left(\alpha +1\right)\left(\alpha +2\right)}\right|\le \frac{9\left(1-\alpha \right)}{8\left(\alpha +1\right)},\hfill & 0<\alpha \le \frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)},\hfill \\ \hfill & \\ \left|\frac{3}{2}-\frac{9\left({\alpha }^2+3\alpha +4\right)}{8\left(\alpha +1\right)\left(\alpha +2\right)}\right|\le \frac{9}{4}\left[\frac{2\alpha }{\alpha +1}{\left(\frac{3}{4}\right)}^{1+\frac{1}{\alpha }}+\frac{2}{\alpha +1}{\left(\frac{1}{3}\right)}^{\alpha +1}+\frac{1}{\alpha +1}-1\right],\hfill & \frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}<\alpha \le 10.\hfill \end{array}\right.$$

One can see the correctness of the inequalities (15) in Figure 1.

Figure 1. For all values of $\alpha \in (0,10]$, the left-hand side of (5) is consistently below the right-hand side of (5) in Example 1. (a) The specified values of $\alpha$ from 0 to $\frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}$. (b) The specified values of $\alpha$ from $\frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}$ to 10.

Figure 1. For all values of $\alpha \in (0,10]$, the left-hand side of (5) is consistently below the right-hand side of (5) in Example 1. (a) The specified values of $\alpha$ from 0 to $\frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}$. (b) The specified values of $\alpha$ from $\frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}$ to 10.

Example 2. 

Note that function $\mathcal{F}:[\sigma ,\delta ]=[0,3]\to \mathbb{R}$ is presented by $\mathcal{F}\left(x\right)=\frac{x^2}{2}$. From Theorem 5 with $\alpha \in (0,10]$ and $p=q=2$, the left-hand side of (7) reduces to equality (14), and the right-hand side of (7) is equal to

$$\begin{array}{cc}\hfill & \frac{3}{4}\left\{{\left[\frac{1}{2\alpha +1}{\left(\frac{1}{3}\right)}^{2\alpha +1}\right]}^{\frac{1}{2}}\left(\frac{1+\sqrt{11}}{2}\right)+\left[\frac{3}{2\left(\alpha +1\right)}{\left(\frac{1}{3}\right)}^{\alpha +1}\right.\right.\hfill \\ \hfill & \left.{\left.-\frac{1}{2\alpha +1}{\left(\frac{1}{3}\right)}^{2\alpha +1}+\frac{1}{2\alpha +1}-\frac{3}{2\left(\alpha +1\right)}+\frac{3}{8}\right]}^{\frac{1}{2}}\left(2+\sqrt{2}\right)\right\}.\hfill \end{array}$$

Consequently, we have the inequality

$$\begin{array}{cc}\hfill \left|\frac{3}{2}-\frac{9\left({\alpha }^2+3\alpha +4\right)}{8\left(\alpha +1\right)\left(\alpha +2\right)}\right|& \le \frac{3}{4}\left\{{\left[\frac{1}{2\alpha +1}{\left(\frac{1}{3}\right)}^{2\alpha +1}\right]}^{\frac{1}{2}}\left(\frac{1+\sqrt{11}}{2}\right)+\left[\frac{3}{2\left(\alpha +1\right)}{\left(\frac{1}{3}\right)}^{\alpha +1}\right.\right.\hfill \\ \hfill & \left.{\left.-\frac{1}{2\alpha +1}{\left(\frac{1}{3}\right)}^{2\alpha +1}+\frac{1}{2\alpha +1}-\frac{3}{2\left(\alpha +1\right)}+\frac{3}{8}\right]}^{\frac{1}{2}}\left(2+\sqrt{2}\right)\right\}.\hfill \end{array}$$

One can see the correctness of the inequality (16) in Figure 2.

Figure 2. In Example 2, depending on $\alpha \in (0,10]$, MATLAB has been used to compute and plot the graph of both sides of (7).

Figure 2. In Example 2, depending on $\alpha \in (0,10]$, MATLAB has been used to compute and plot the graph of both sides of (7).

Example 3. 

A function $\mathcal{F}:[\sigma ,\delta ]=[0,3]\to \mathbb{R}$ is given by $\mathcal{F}\left(x\right)=\frac{x^2}{2}$. From Theorem 6, with $\alpha \in (0,10]$ and $q=2$, the left-hand side of (8) becomes equality (14), and the right-hand side of (8) coincides with

$$\frac{9}{4}\left\{{\left({\Omega }_1\left(\alpha \right)\right)}^{\frac{1}{2}}\left[{\left[{\Omega }_3\left(\alpha \right)\right]}^{\frac{1}{2}}+{\left[{\Omega }_4\left(\alpha \right)\right]}^{\frac{1}{2}}\right]+{\left({\Omega }_2\left(\alpha \right)\right)}^{\frac{1}{2}}\left[{\left[{\Omega }_5\left(\alpha \right)\right]}^{\frac{1}{2}}+{\left[{\Omega }_6\left(\alpha \right)\right]}^{\frac{1}{2}}\right]\right\}.$$

Finally, we have the inequality

$$\begin{array}{cc}\hfill \left|\frac{3}{2}-\frac{9\left({\alpha }^2+3\alpha +4\right)}{8\left(\alpha +1\right)\left(\alpha +2\right)}\right|& \le \frac{9}{4}\left\{{\left({\Omega }_1\left(\alpha \right)\right)}^{\frac{1}{2}}\left[{\left[{\Omega }_3\left(\alpha \right)\right]}^{\frac{1}{2}}+{\left[{\Omega }_4\left(\alpha \right)\right]}^{\frac{1}{2}}\right]\right.\hfill \\ \hfill & \left.+{\left({\Omega }_2\left(\alpha \right)\right)}^{\frac{1}{2}}\left[{\left[{\Omega }_5\left(\alpha \right)\right]}^{\frac{1}{2}}+{\left[{\Omega }_6\left(\alpha \right)\right]}^{\frac{1}{2}}\right]\right\}.\hfill \end{array}$$

One can see the correctness of the inequalities (17) in Figure 3.

Figure 3. In Example 3, using MATLAB software, it is clear that the left-hand side of (8) constantly stays below the right-hand side. (a) The specified values of $\alpha$ from 0 to $\frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}$. (b) The specified values of $\alpha$ from $\frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}$ to 10.

Figure 3. In Example 3, using MATLAB software, it is clear that the left-hand side of (8) constantly stays below the right-hand side. (a) The specified values of $\alpha$ from 0 to $\frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}$. (b) The specified values of $\alpha$ from $\frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}$ to 10.

Example 4. 

Consider that a function $\mathcal{F}:[\sigma ,\delta ]=[0,3]\to \mathbb{R}$ is presented by $\mathcal{F}\left(x\right)=\frac{x^2}{2}$. From Theorem 7, with $\alpha \in (0,10]$ and $0\le {\mathcal{F}}^{\prime }\left(\mu \right)\le 3$, the left-hand side of (9) coincides with equality (14), and the right-hand side of (9) is

$$\begin{array}{cc}\hfill & \frac{9}{4}\left\{{\Omega }_1\left(\alpha \right)+{\Omega }_2\left(\alpha \right)\right\}\hfill \\ \hfill & =\left\{\begin{array}{cc}\frac{9\left(1-\alpha \right)}{8\left(\alpha +1\right)},\hfill & 0<\alpha \le \frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)},\hfill \\ \frac{9}{4}\left[\frac{2\alpha }{\alpha +1}{\left(\frac{3}{4}\right)}^{1+\frac{1}{\alpha }}+\frac{2}{\alpha +1}{\left(\frac{1}{3}\right)}^{\alpha +1}+\frac{1}{\alpha +1}-1\right],\hfill & \frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}<\alpha \le 10.\hfill \end{array}\right.\hfill \end{array}$$

One can see the correctness of the inequalities (18) in Figure 4.

Figure 4. Example 4 shows that the left side of (9) consistently remains lower than the right side when evaluated with MATLAB software. (a) The specified values of $\alpha$ from 0 to $\frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}$. (b) The specified values of $\alpha$ from $\frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}$ to 10.

Figure 4. Example 4 shows that the left side of (9) consistently remains lower than the right side when evaluated with MATLAB software. (a) The specified values of $\alpha$ from 0 to $\frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}$. (b) The specified values of $\alpha$ from $\frac{ln\left(\frac{3}{4}\right)}{ln\left(\frac{1}{3}\right)}$ to 10.

9. Concluding Remarks and Future Works

In this paper, we proved some Euler–Maclaurin-type inequalities for various function classes by using Riemann–Liouville fractional integrals. First of all, we gave an integral equality that is necessary in order to prove the main findings of the paper. Subsequently, several Euler–Maclaurin-type inequalities were investigated for differentiable convex functions by using the Riemann–Liouville fractional integrals. In addition to this, we gave some Euler–Maclaurin-type for bounded functions by fractional integrals. Moreover, some fractional Euler–Maclaurin-type inequalities were established for Lipschitzian functions. Furthermore, some Euler–Maclaurin-type inequalities were proved by fractional integrals of bounded variation. Finally, we gave several examples using graphs in order to show that our main result is correct.

In future papers, the ideas and strategies for our results about Euler–Maclaurin-type inequalities by Riemann–Liouville fractional integrals may open new avenues for further research in this field. Improvements or generalizations of our results can be investigated by using different kinds of convex function classes or other types of fractional integral operators. In addition, one can obtain some Euler–Maclaurin-type inequalities for various function classes with the help of the quantum calculus.