Fractional Error Bounds for Lobatto Quadrature: A Convexity Approach via Riemann–Liouville Integrals

Abstract

In this paper, we establish a new fractional integral identity linked to the 4-point Lobatto quadrature rule within the Riemann–Liouville fractional calculus framework. Building on this identity, we derive several Lobatto-type inequalities under convexity assumptions, yielding error bounds that involve only first-order derivatives, thereby improving practical applicability. A numerical example with graphical illustration confirms the theoretical findings and demonstrates their accuracy. We also present applications to special means, highlighting the utility of the obtained inequalities. The integration of fractional analysis, quadrature theory, and numerical validation provides a robust methodology for refining and analyzing high-order integration rules.

1. Introduction

Numerical quadrature formulas are methods used to approximate definite integrals of the form $\underset{\mathfrak{a}}{\stackrel{\mathfrak{b}}{\int }}\mathfrak{f}\left(\mathfrak{u}\right)\mathrm{d}\mathfrak{u}$. Common approaches, such as the trapezoidal or Simpson’s rule, evaluate the function at equally spaced points. However, for higher accuracy with fewer evaluation points, Gaussian quadrature rules are often preferred. These methods optimally choose both the nodes and weights to maximize the degree of polynomial exactness.

Standard Gaussian quadrature rules, such as Gauss–Legendre, select n internal nodes in the interval $[-1,1]$ to achieve exactness for polynomials of degree up to $2n-1$. Among the variants of Gaussian quadrature, the Gauss–Lobatto rules are distinguished by including the endpoints of the interval as fixed nodes. This is particularly useful in applications where function values at the boundaries are required, such as in spectral methods or finite element schemes. Practical uses of the Gauss–Lobatto formula are illustrated in several recent studies [1,2,3], highlighting its effectiveness in real-world problems.

The 4-point Lobatto quadrature rule uses the endpoints $-1$ and 1, along with two interior points. It is exact for polynomials up to degree 5 and uses specially chosen weights and nodes—based on Legendre polynomials and their derivatives—to achieve high accuracy. The formula is given as follows:

$$\underset{-1}{\stackrel{1}{\int }}\mathfrak{f}\left(\mathfrak{u}\right)\mathrm{d}\mathfrak{u}\approx \frac{1}{6}\left(\mathfrak{f}\left(-1\right)+5\mathfrak{f}\left(-{\displaystyle \frac{\sqrt{5}}{5}}\right)+5\mathfrak{f}\left(\frac{\sqrt{5}}{5}\right)+\mathfrak{f}\left(1\right)\right).$$

The error of the 4-point Lobatto quadrature rule for a sufficiently smooth function $\mathfrak{f}$ on $[-1,1]$ is proportional to the sixth derivative of $\mathfrak{f}$. Specifically, the error expression is given by [4]

$$\left|\underset{-1}{\stackrel{1}{\int }}\mathfrak{f}\left(\mathfrak{u}\right)\mathrm{d}\mathfrak{u}-\frac{1}{6}\left(\mathfrak{f}\left(-1\right)+5\mathfrak{f}\left(-{\displaystyle \frac{\sqrt{5}}{5}}\right)+5\mathfrak{f}\left(\frac{\sqrt{5}}{5}\right)+\mathfrak{f}\left(1\right)\right)\right|\le {\displaystyle \frac{2}{23625}}{∥{\mathfrak{f}}^{\left(6\right)}∥}_{\infty },$$

where ${∥{\mathfrak{f}}^{\left(6\right)}∥}_{\infty }=\underset{u\in [-1,1]}{\sup }\left|{\mathfrak{f}}^{\left(6\right)}\left(\mathfrak{u}\right)\right|$.

To apply the 4-point Lobatto quadrature on a general interval $[\mathfrak{a},\mathfrak{b}]$, a change of variable is used to map $[-1,1]$ to $[\mathfrak{a},\mathfrak{b}]$. Let

$$\mathfrak{v}={\displaystyle \frac{\mathfrak{b}-\mathfrak{a}}{2}}\mathfrak{u}+{\displaystyle \frac{\mathfrak{a}+\mathfrak{b}}{2}},$$

so that $\mathfrak{u}\in [-1,1]$ corresponds to $\mathfrak{v}\in [\mathfrak{a},\mathfrak{b}]$. The quadrature formula becomes

$$\underset{\mathfrak{a}}{\stackrel{\mathfrak{b}}{\int }}\mathfrak{f}\left(\mathfrak{v}\right)\mathrm{d}\mathfrak{v}\approx {\displaystyle \frac{\mathfrak{b}-\mathfrak{a}}{12}}\left[\mathfrak{f}\left(\mathfrak{a}\right)+5\mathfrak{f}\left(\frac{5+\sqrt{5}}{10}\mathfrak{a}+\frac{5-\sqrt{5}}{10}\mathfrak{b}\right)+5\mathfrak{f}\left(\frac{5-\sqrt{5}}{10}\mathfrak{a}+\frac{5+\sqrt{5}}{10}\mathfrak{b}\right)+\mathfrak{f}\left(\mathfrak{b}\right)\right].$$

The error formula, which involves the sixth derivative of $\mathfrak{f}$, highlights a key limitation of the 4-point Lobatto rule: its error estimate is only meaningful for functions that are sufficiently smooth (at least six times continuously differentiable). For less regular functions, such as those arising in practical applications with limited differentiability, this high-order derivative may not exist or be difficult to bound, making the error estimate impractical.

This is where the concept of convexity becomes valuable. Under assumptions of convexity or generalized convexity (e.g., ${\mathfrak{f}}^{\prime }$ being convex), it is possible to derive alternative error bounds that depend only on lower-order derivatives, such as the first derivative, which are more accessible and easier to estimate. These refined inequalities exploit the sign properties and monotonicity of derivatives, allowing for computable error estimates even when full smoothness is not guaranteed. Thus, convexity-based analysis provides a powerful tool to enhance the practical applicability of high-order quadrature rules like Lobatto’s, especially in cases where classical remainder terms fail to provide useful information.

We recall that a function $\mathfrak{f}:I\subset \mathbb{R}\to \mathbb{R}$ is said to be convex, if

$$\mathfrak{f}\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\le \wp \mathfrak{f}\left(\mathfrak{a}\right)+\left(1-\wp \right)\mathfrak{f}\left(\mathfrak{b}\right)$$

holds for all $\mathfrak{a},\mathfrak{b}\in I$ and all $\wp \in [0,1]$ (see [5,6]).

It should be noted that while the notion of convexity has been widely used to establish various Newton–Cotes-type integral inequalities [7,8,9,10,11], the literature still lacks a comprehensive treatment of Gauss-type integral inequalities. Despite the contributions in [12,13,14,15], where the authors studied such inequalities in different settings, there remains a significant gap in the development of integral inequalities of Gauss-type.

Fractional calculus extends the classical notions of differentiation and integration to non-integer orders, providing a powerful framework for modeling systems with memory, hereditary properties, and anomalous behavior. Unlike standard calculus, where derivatives and integrals are defined only for integer orders, fractional calculus allows for the consideration of operators of arbitrary real or complex order. Among the most fundamental and widely used definitions in this theory are the Riemann–Liouville fractional integrals. These operators generalize the repeated integration process to fractional orders through a convolution with a power-law kernel and serve as the foundation for defining corresponding fractional derivatives. Due to their rigorous mathematical structure and applicability in diverse fields such as physics, engineering, and finance, Riemann–Liouville integrals play a central role in the development of fractional differential equations [16,17] and integral inequalities [18,19].

Definition 1

([20]). Assume that $\mathfrak{f}\in L^1[\mathfrak{a},\mathfrak{b}]$. For order $\alpha >0$ with $\mathfrak{a}\ge 0$, the left and right Riemann–Liouville fractional integrals are defined by

$$\begin{array}{cc}\hfill {\mathcal{I}}_{\mathfrak{a}}^{\alpha }\mathfrak{f}\left(s\right)=& \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}\stackrel{s}{\underset{\mathfrak{a}}{\int }}{\left(s-\wp \right)}^{\alpha -1}\mathfrak{f}(\wp )\mathrm{d}\wp ,s>\mathfrak{a},\hfill \\ \hfill {\mathcal{I}}_{\mathfrak{b}}^{\alpha }\mathfrak{f}\left(s\right)=& \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}\stackrel{\mathfrak{b}}{\underset{s}{\int }}{\left(\wp -s\right)}^{\alpha -1}\mathfrak{f}(\wp )\mathrm{d}\wp ,\mathfrak{b}>s,\hfill \end{array}$$

respectively, where $\mathsf{\Gamma }\left(\alpha \right)=\stackrel{\infty }{\underset{0}{\int }}$ $e^{-s}{s}^{\alpha -1}ds$ represents the Gamma function with ${\mathcal{I}}_{\mathfrak{a}}^0\mathfrak{f}\left(s\right)={\mathcal{I}}_{\mathfrak{b}}^0\mathfrak{f}\left(s\right)=\mathfrak{f}\left(s\right)$.

In recent years, a number of studies have been devoted to establishing integral inequalities using various types of fractional integrals. Notable contributions include works involving Riemann–Liouville [21,22], Caputo [23,24], Atangana–Baleanu [25], $\Psi$-Hilfer fractional operators [26], and generalized fractional integrals [27], among others. Inspired by these works, as well as recent studies on Gauss-type inequalities, in this paper, we establish new Lobatto-type inequalities via Riemann–Liouville fractional integrals. Our approach relies on the convexity of the first derivative of the involved function, which allows us to derive sharp and computable bounds without requiring higher-order smoothness.

The rest of this paper is structured as follows. Section 2 introduces a new fractional integral identity related to the 4-point Lobatto quadrature rule, which will serve as the main tool in our analysis. In Section 3, we establish several Lobatto-type inequalities by combining this identity with convexity and fractional calculus techniques. Section 4 presents a numerical example with graphical illustrations to validate the theoretical results and demonstrate their accuracy. Applications to special means are discussed in Section 5, highlighting the practical relevance of the inequalities. Finally, Section 6 concludes the paper with a brief summary and potential directions for future research.

2. Lobatto-Type Identity

We begin by establishing a key fractional integral identity that will serve as the foundation for our main results.

Lemma 1.

Let $\mathfrak{f}:\left[\mathfrak{a},\mathfrak{b}\right]\to \mathbb{R}$ be a differentiable function on $\left[\mathfrak{a},\mathfrak{b}\right]$ with $\mathfrak{a}<\mathfrak{b}$ and ${\mathfrak{f}}^{\prime }\in L^1\left[\mathfrak{a},\mathfrak{b}\right]$, then the following equality holds for $\alpha >0$

$$\begin{array}{cc}\hfill & \mathcal{L}\left(\mathfrak{a},\mathfrak{b},\mathfrak{f}\right)-\frac{\mathsf{\Gamma }\left(\alpha +1\right)}{2{\left(\mathfrak{b}-\mathfrak{a}\right)}^{\alpha }}\left({\mathcal{I}}_{{\mathfrak{b}}^{-}}^{\alpha }\mathfrak{f}\left(\mathfrak{a}\right)+{\mathcal{I}}_{{\mathfrak{a}}^{+}}^{\alpha }\mathfrak{f}\left(\mathfrak{b}\right)\right)\hfill \\ \hfill =& \frac{\mathfrak{b}-\mathfrak{a}}{2}\left(\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}\left({\wp }^{\alpha }-\frac{1}{12}\right)\left({\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)-{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right)\mathrm{d}\wp \right.\hfill \\ \hfill & +\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}\left({\wp }^{\alpha }-\frac{1}{2}\right)\left({\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)-{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right)\mathrm{d}\wp \hfill \\ \hfill & +\left.\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}\left({\wp }^{\alpha }-\frac{11}{12}\right)\left({\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)-{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right)\mathrm{d}\wp \right),\hfill \end{array}$$

(1)

where

$$\mathcal{L}\left(\mathfrak{a},\mathfrak{b},\mathfrak{f}\right)=\frac{1}{12}\left(\mathfrak{f}\left(\mathfrak{a}\right)+5\mathfrak{f}\left(\frac{5+\sqrt{5}}{10}\mathfrak{a}+\frac{5-\sqrt{5}}{10}\mathfrak{b}\right)+5\mathfrak{f}\left(\frac{5-\sqrt{5}}{10}\mathfrak{a}+\frac{5+\sqrt{5}}{10}\mathfrak{b}\right)+\mathfrak{f}\left(\mathfrak{b}\right)\right).$$

(2)

Proof. 

Let

$${\mathcal{N}}_1=\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}\left({\wp }^{\alpha }-\frac{1}{12}\right)\left({\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)-{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right)\mathrm{d}\wp ,$$

$${\mathcal{N}}_2=\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}\left({\wp }^{\alpha }-\frac{1}{2}\right)\left({\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)-{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right)\mathrm{d}\wp$$

and

$${\mathcal{N}}_3=\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}\left({\wp }^{\alpha }-\frac{11}{12}\right)\left({\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)-{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right)\mathrm{d}\wp .$$

Integrating by parts ${\mathcal{N}}_1$, we obtain

$$\begin{array}{cc}\hfill {\mathcal{N}}_1=& \stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}\left({\wp }^{\alpha }-\frac{1}{12}\right)\left({\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)-{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right)\mathrm{d}\wp \hfill \\ \hfill =& {\left.\frac{1}{\mathfrak{b}-\mathfrak{a}}\left({\wp }^{\alpha }-\frac{1}{12}\right)\left(\mathfrak{f}\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)+\mathfrak{f}\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right)\right|}_0^{{\displaystyle \frac{5-\sqrt{5}}{10}}}\hfill \\ \hfill & -\frac{\alpha }{\mathfrak{b}-\mathfrak{a}}\left(\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}{\wp }^{\alpha -1}\left(\mathfrak{f}\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)+\mathfrak{f}\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right)\mathrm{d}\wp \right)\hfill \\ \hfill =& \frac{1}{\mathfrak{b}-\mathfrak{a}}\left({\left(\frac{5-\sqrt{5}}{10}\right)}^{\alpha }-\frac{1}{12}\right)\left(\mathfrak{f}\left(\frac{5+\sqrt{5}}{10}\mathfrak{a}+\frac{5-\sqrt{5}}{10}\mathfrak{b}\right)+\mathfrak{f}\left(\frac{5-\sqrt{5}}{10}\mathfrak{a}+\frac{5+\sqrt{5}}{10}\mathfrak{b}\right)\right)\hfill \\ \hfill & +\frac{1}{\mathfrak{b}-\mathfrak{a}}\left(\frac{1}{12}\right)\left(\mathfrak{f}\left(\mathfrak{a}\right)+\mathfrak{f}\left(\mathfrak{b}\right)\right)\hfill \\ \hfill & -\frac{\alpha }{\mathfrak{b}-\mathfrak{a}}\left(\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}{\wp }^{\alpha -1}\mathfrak{f}\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\mathrm{d}\wp +\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}{\wp }^{\alpha -1}\mathfrak{f}\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\mathrm{d}\wp \right).\hfill \end{array}$$

(3)

Similarly, we obtain

$$\begin{array}{cc}\hfill {\mathcal{N}}_2=& \stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}\left({\wp }^{\alpha }-\frac{1}{2}\right)\left({\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)-{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right)\mathrm{d}\wp \hfill \\ \hfill =& {\left.\frac{1}{\mathfrak{b}-\mathfrak{a}}\left({\wp }^{\alpha }-\frac{1}{2}\right)\left(\mathfrak{f}\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)+\mathfrak{f}\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right)\right|}_{{\displaystyle \frac{5-\sqrt{5}}{10}}}^{{\displaystyle \frac{5+\sqrt{5}}{10}}}\hfill \\ \hfill & -\frac{\alpha }{\mathfrak{b}-\mathfrak{a}}\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}{\wp }^{\alpha -1}\left(\mathfrak{f}\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)+\mathfrak{f}\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right)\mathrm{d}\wp \hfill \\ \hfill =& \frac{1}{\mathfrak{b}-\mathfrak{a}}\left({\left(\frac{5+\sqrt{5}}{10}\right)}^{\alpha }-\frac{1}{2}\right)\left(\mathfrak{f}\left(\frac{5-\sqrt{5}}{10}\mathfrak{a}+\frac{5+\sqrt{5}}{10}\mathfrak{b}\right)+\mathfrak{f}\left(\frac{5+\sqrt{5}}{10}\mathfrak{a}+\frac{5-\sqrt{5}}{10}\mathfrak{b}\right)\right)\hfill \\ \hfill & -\frac{1}{\mathfrak{b}-\mathfrak{a}}\left({\left(\frac{5-\sqrt{5}}{10}\right)}^{\alpha }-\frac{1}{2}\right)\left(\mathfrak{f}\left(\frac{5+\sqrt{5}}{10}\mathfrak{a}+\frac{5-\sqrt{5}}{10}\mathfrak{b}\right)+\mathfrak{f}\left(\frac{5-\sqrt{5}}{10}\mathfrak{a}+\frac{5+\sqrt{5}}{10}\mathfrak{b}\right)\right)\hfill \\ \hfill & -\frac{\alpha }{\mathfrak{b}-\mathfrak{a}}\left(\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}{\wp }^{\alpha -1}\mathfrak{f}\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\mathrm{d}\wp +\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}{\wp }^{\alpha -1}\mathfrak{f}\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\mathrm{d}\wp \right)\hfill \end{array}$$

(4)

and

$$\begin{array}{cc}\hfill {\mathcal{N}}_3=& \stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}\left({\wp }^{\alpha }-\frac{11}{12}\right)\left({\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)-{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right)\mathrm{d}\wp \hfill \\ \hfill =& {\left.\frac{1}{\mathfrak{b}-\mathfrak{a}}\left({\wp }^{\alpha }-\frac{11}{12}\right)\left(\mathfrak{f}\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)+\mathfrak{f}\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right)\right|}_{{\displaystyle \frac{5+\sqrt{5}}{10}}}^1\hfill \\ \hfill & -\frac{\alpha }{\mathfrak{b}-\mathfrak{a}}\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}{\wp }^{\alpha -1}\left(\mathfrak{f}\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)+\mathfrak{f}\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right)\mathrm{d}\wp \hfill \\ \hfill =& \frac{1}{\mathfrak{b}-\mathfrak{a}}\left(1-\frac{11}{12}\right)\left(\mathfrak{f}\left(\mathfrak{b}\right)+\mathfrak{f}\left(\mathfrak{a}\right)\right)\hfill \\ \hfill & -\frac{1}{\mathfrak{b}-\mathfrak{a}}\left({\left(\frac{5+\sqrt{5}}{10}\right)}^{\alpha }-\frac{11}{12}\right)\left(\mathfrak{f}\left(\frac{5-\sqrt{5}}{10}\mathfrak{a}+\frac{5+\sqrt{5}}{10}\mathfrak{b}\right)+\mathfrak{f}\left(\frac{5+\sqrt{5}}{10}\mathfrak{a}+\frac{5-\sqrt{5}}{10}\mathfrak{b}\right)\right)\hfill \\ \hfill & -\frac{\alpha }{\mathfrak{b}-\mathfrak{a}}\left(\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}{\wp }^{\alpha -1}\mathfrak{f}\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\mathrm{d}\wp +\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}{\wp }^{\alpha -1}\mathfrak{f}\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\mathrm{d}\wp \right).\hfill \end{array}$$

(5)

Adding (3)–(5), we get

$$\begin{array}{cc}\hfill & {\mathcal{N}}_1+{\mathcal{N}}_2+{\mathcal{N}}_3\hfill \\ \hfill =& \frac{1}{6\left(\mathfrak{b}-\mathfrak{a}\right)}\left(\mathfrak{f}\left(\mathfrak{a}\right)+5\mathfrak{f}\left(\frac{5+\sqrt{5}}{10}\mathfrak{a}+\frac{5-\sqrt{5}}{10}\mathfrak{b}\right)+5\mathfrak{f}\left(\frac{5-\sqrt{5}}{10}\mathfrak{a}+\frac{5+\sqrt{5}}{10}\mathfrak{b}\right)+\mathfrak{f}\left(\mathfrak{b}\right)\right)\hfill \\ \hfill & -\frac{\alpha }{\mathfrak{b}-\mathfrak{a}}\left(\stackrel{1}{\underset{0}{\int }}{\wp }^{\alpha -1}\mathfrak{f}\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\mathrm{d}\wp +\stackrel{1}{\underset{0}{\int }}{\wp }^{\alpha -1}\mathfrak{f}\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\mathrm{d}\wp \right)\hfill \\ \hfill =& \frac{1}{6\left(\mathfrak{b}-\mathfrak{a}\right)}\left(\mathfrak{f}\left(\mathfrak{a}\right)+5\mathfrak{f}\left(\frac{5+\sqrt{5}}{10}\mathfrak{a}+\frac{5-\sqrt{5}}{10}\mathfrak{b}\right)+5\mathfrak{f}\left(\frac{5-\sqrt{5}}{10}\mathfrak{a}+\frac{5+\sqrt{5}}{10}\mathfrak{b}\right)+\mathfrak{f}\left(\mathfrak{b}\right)\right)\hfill \\ \hfill & -\frac{\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(\mathfrak{b}-\mathfrak{a}\right)}^{\alpha +1}}\left(\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}\stackrel{\mathfrak{b}}{\underset{\mathfrak{a}}{\int }}{\left(\mathfrak{u}-\mathfrak{a}\right)}^{\alpha -1}\mathfrak{f}\left(\mathfrak{u}\right)\mathrm{d}\mathfrak{u}+\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}\stackrel{\mathfrak{b}}{\underset{\mathfrak{a}}{\int }}{\left(\mathfrak{b}-\mathfrak{u}\right)}^{\alpha -1}\mathfrak{f}\left(\mathfrak{u}\right)\mathrm{d}\mathfrak{u}\right)\hfill \\ \hfill =& \frac{1}{6\left(\mathfrak{b}-\mathfrak{a}\right)}\left(\mathfrak{f}\left(\mathfrak{a}\right)+5\mathfrak{f}\left(\frac{5+\sqrt{5}}{10}\mathfrak{a}+\frac{5-\sqrt{5}}{10}\mathfrak{b}\right)+5\mathfrak{f}\left(\frac{5-\sqrt{5}}{10}\mathfrak{a}+\frac{5+\sqrt{5}}{10}\mathfrak{b}\right)+\mathfrak{f}\left(\mathfrak{b}\right)\right)\hfill \\ \hfill & -\frac{\mathsf{\Gamma }\left(\alpha +1\right)}{{\left(\mathfrak{b}-\mathfrak{a}\right)}^{\alpha +1}}\left({\mathcal{I}}_{{\mathfrak{b}}^{-}}^{\alpha }\mathfrak{f}\left(\mathfrak{a}\right)+{\mathcal{I}}_{{\mathfrak{a}}^{+}}^{\alpha }\mathfrak{f}\left(\mathfrak{b}\right)\right).\hfill \end{array}$$

(6)

Multiplying both sides of (6) by ${\displaystyle \frac{\mathfrak{b}-\mathfrak{a}}{2}}$, we get the desired result. □

3. Lobatto-Type Inequalities

Based on the identity established in the previous section, we now derive several Lobatto-type inequalities using convexity and fractional calculus tools.

Theorem 1.

Let $\mathfrak{f}$ be as in Lemma 1. If $\left|{\mathfrak{f}}^{\prime }\right|$ is convex, then we have

$$\begin{array}{cc}\hfill & \left|\mathcal{L}\left(\mathfrak{a},\mathfrak{b},\mathfrak{f}\right)-\frac{\mathsf{\Gamma }\left(\alpha +1\right)}{2{\left(\mathfrak{b}-\mathfrak{a}\right)}^{\alpha }}\left({\mathcal{I}}_{{\mathfrak{b}}^{-}}^{\alpha }\mathfrak{f}\left(\mathfrak{a}\right)+{\mathcal{I}}_{{\mathfrak{a}}^{+}}^{\alpha }\mathfrak{f}\left(\mathfrak{b}\right)\right)\right|\hfill \\ \hfill \le & \frac{\mathfrak{b}-\mathfrak{a}}{2}\left({\mathcal{C}}_1\left(\alpha \right)+{\mathcal{C}}_2\left(\alpha \right)+{\mathcal{C}}_3\left(\alpha \right)\right)\left(\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|+\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|\right),\hfill \end{array}$$

where $\mathcal{L}$ is defined as in (2), and ${\mathcal{C}}_1\left(\alpha \right),{\mathcal{C}}_2\left(\alpha \right)$ and ${\mathcal{C}}_3\left(\alpha \right)$ are given by

$$\begin{array}{cc}\hfill {\mathcal{C}}_1\left(\alpha \right)=& \stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}\left|{\wp }^{\alpha }-\frac{1}{12}\right|\mathrm{d}\wp \hfill \\ \hfill =& \left\{\begin{array}{ccc}{\displaystyle \frac{2\alpha }{\alpha +1}}{\left({\displaystyle \frac{1}{12}}\right)}^{1+{\displaystyle \frac{1}{\alpha }}}+{\displaystyle \frac{1}{\alpha +1}}{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{\alpha +1}-{\displaystyle \frac{5-\sqrt{5}}{120}}\hfill & \mathit{if}& \alpha <\frac{ln12}{ln10-ln\left(5-\sqrt{5}\right)},\\ {\displaystyle \frac{5-\sqrt{5}}{120}}-{\displaystyle \frac{1}{\alpha +1}}{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{\alpha +1}\hfill & \mathit{if}& \alpha \ge \frac{ln12}{ln10-ln\left(5-\sqrt{5}\right)},\end{array}\right.\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill {\mathcal{C}}_2\left(\alpha \right)=& \stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{1}{2}\right|\mathrm{d}\wp \hfill \\ \hfill =& \left\{\begin{array}{ccc}{\displaystyle \frac{1}{\alpha +1}}\left({\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{\alpha +1}-{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{\alpha +1}\right)-{\displaystyle \frac{\sqrt{5}}{10}}\hfill & \mathit{if}& 0\le \alpha <\frac{ln2}{ln10-ln\left(5-\sqrt{5}\right)},\\ {\displaystyle \frac{\alpha }{\alpha +1}}{\left({\displaystyle \frac{1}{2}}\right)}^{{\displaystyle \frac{1}{\alpha }}}-\frac{1}{2}+{\displaystyle \frac{1}{\alpha +1}}\left({\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{\alpha +1}+{\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{\alpha +1}\right)\hfill & \mathit{if}& \frac{ln2}{ln10-ln\left(5-\sqrt{5}\right)}\le \alpha \le \frac{ln2}{ln10-ln\left(5+\sqrt{5}\right)},\\ {\displaystyle \frac{\sqrt{5}}{10}}-{\displaystyle \frac{1}{\alpha +1}}\left({\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{\alpha +1}-{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{\alpha +1}\right)\hfill & \mathit{if}& \alpha >\frac{ln2}{ln10-ln\left(5+\sqrt{5}\right)},\end{array}\right.\hfill \end{array}$$

(8)

and

$$\begin{array}{cc}\hfill {\mathcal{C}}_3\left(\alpha \right)=& \stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{11}{12}\right|\mathrm{d}\wp \hfill \\ \hfill =& \left\{\begin{array}{ccc}{\displaystyle \frac{1-11\alpha }{12\left(\alpha +1\right)}}-{\displaystyle \frac{1}{\alpha +1}}{\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{\alpha +1}+{\displaystyle \frac{55+11\sqrt{5}}{120}}\hfill & \mathit{if}& \alpha <\frac{ln12-ln11}{ln10-ln\left(5+\sqrt{5}\right)},\\ {\displaystyle \frac{2\alpha }{\alpha +1}}{\left(\frac{11}{12}\right)}^{1+{\displaystyle \frac{1}{\alpha }}}+{\displaystyle \frac{1}{\alpha +1}}{\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{\alpha +1}-{\displaystyle \frac{55+11\sqrt{5}}{120}}+{\displaystyle \frac{1-11\alpha }{12\left(\alpha +1\right)}}\hfill & \mathit{if}& \alpha \ge \frac{ln12-ln11}{ln10-ln\left(5+\sqrt{5}\right)}.\end{array}\right.\hfill \end{array}$$

(9)

Proof. 

Using the absolute value on both sides of (1) and the convexity of $\left|{\mathfrak{f}}^{\prime }\right|$ yields

$$\begin{array}{cc}\hfill & \left|\mathcal{L}\left(\mathfrak{a},\mathfrak{b},\mathfrak{f}\right)-\frac{\mathsf{\Gamma }\left(\alpha +1\right)}{2{\left(\mathfrak{b}-\mathfrak{a}\right)}^{\alpha }}\left({\mathcal{I}}_{{\mathfrak{b}}^{-}}^{\alpha }\mathfrak{f}\left(\mathfrak{a}\right)+{\mathcal{I}}_{{\mathfrak{a}}^{+}}^{\alpha }\mathfrak{f}\left(\mathfrak{b}\right)\right)\right|\hfill \\ \hfill \le & \frac{\mathfrak{b}-\mathfrak{a}}{2}\left(\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}\left|{\wp }^{\alpha }-\frac{1}{12}\right|\left(\left|{\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\right|+\left|{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right|\right)\mathrm{d}\wp \right.\hfill \\ \hfill & +\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{1}{2}\right|\left(\left|{\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\right|+\left|{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right|\right)\mathrm{d}\wp \hfill \\ \hfill & +\left.\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{11}{12}\right|\left(\left|{\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\right|+\left|{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right|\right)\mathrm{d}\wp \right)\hfill \\ \hfill \le & \frac{\mathfrak{b}-\mathfrak{a}}{2}\left(\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}\left|{\wp }^{\alpha }-\frac{1}{12}\right|\mathrm{d}\wp +\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{1}{2}\right|\mathrm{d}\wp +\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{11}{12}\right|\mathrm{d}\wp \right)\left(\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|+\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|\right)\hfill \\ \hfill =& \frac{\mathfrak{b}-\mathfrak{a}}{2}\left({\mathcal{C}}_1\left(\alpha \right)+{\mathcal{C}}_2\left(\alpha \right)+{\mathcal{C}}_3\left(\alpha \right)\right)\left(\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|+\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|\right),\hfill \end{array}$$

where we have used (7)–(9).

The proof is finished. □

Corollary 1.

If we attempt to take $\alpha =1$, Theorem 1 yields the following Lobatto-type inequality

$$\begin{array}{cc}\hfill & \left|\frac{1}{12}\left(\mathfrak{f}\left(\mathfrak{a}\right)+5\mathfrak{f}\left(\frac{5+\sqrt{5}}{10}\mathfrak{a}+\frac{5-\sqrt{5}}{10}\mathfrak{b}\right)+5\mathfrak{f}\left(\frac{5-\sqrt{5}}{10}\mathfrak{a}+\frac{5+\sqrt{5}}{10}\mathfrak{b}\right)+\mathfrak{f}\left(\mathfrak{b}\right)\right)-\frac{1}{\mathfrak{b}-\mathfrak{a}}\underset{\mathfrak{a}}{\stackrel{\mathfrak{b}}{\int }}\mathfrak{f}\left(\mathfrak{u}\right)\mathrm{d}\mathfrak{u}\right|\hfill \\ \hfill \le & \frac{\left(101-30\sqrt{5}\right)\left(\mathfrak{b}-\mathfrak{a}\right)}{720}\left(\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|+\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|\right).\hfill \end{array}$$

Theorem 2.

Let $\mathfrak{f}$ be as in Lemma 1. If ${\left|{\mathfrak{f}}^{\prime }\right|}^q$ is convex, where $q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$, then we have

$$\begin{array}{cc}\hfill & \left|\mathcal{L}\left(\mathfrak{a},\mathfrak{b},\mathfrak{f}\right)-\frac{\mathsf{\Gamma }\left(\alpha +1\right)}{2{\left(\mathfrak{b}-\mathfrak{a}\right)}^{\alpha }}\left({\mathcal{I}}_{{\mathfrak{b}}^{-}}^{\alpha }\mathfrak{f}\left(\mathfrak{a}\right)+{\mathcal{I}}_{{\mathfrak{a}}^{+}}^{\alpha }\mathfrak{f}\left(\mathfrak{b}\right)\right)\right|\hfill \\ \hfill \le & \frac{\mathfrak{b}-\mathfrak{a}}{2}\left[\left({\left({\mathcal{D}}_1\left(\alpha ,p\right)\right)}^{{\displaystyle \frac{1}{p}}}+{\left({\mathcal{D}}_3\left(\alpha ,p\right)\right)}^{{\displaystyle \frac{1}{p}}}\right){\left(\frac{5-\sqrt{5}}{10}\right)}^{{\displaystyle \frac{1}{q}}}\right.\left({\left(\frac{\left(15+\sqrt{5}\right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+\left(5-\sqrt{5}\right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q}{20}\right)}^{{\displaystyle \frac{1}{q}}}+{\left(\frac{\left(5-\sqrt{5}\right){\left|{\mathfrak{f}}^{\prime }\left(a\right)\right|}^q+\left(15+\sqrt{5}\right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q}{20}\right)}^{{\displaystyle \frac{1}{q}}}\right)\hfill \\ \hfill & +\left.2{\left({\mathcal{D}}_2\left(\alpha ,p\right)\right)}^{{\displaystyle \frac{1}{p}}}{\left(\frac{\sqrt{5}}{10}\right)}^{{\displaystyle \frac{1}{q}}}{\left({\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+{\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q\right)}^{{\displaystyle \frac{1}{q}}}\right],\hfill \end{array}$$

where

$${\mathcal{D}}_1\left(\alpha ,p\right)=\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}{\left|{\wp }^{\alpha }-\frac{1}{12}\right|}^p\mathrm{d}\wp ,$$

(10)

$${\mathcal{D}}_2\left(\alpha ,p\right)=\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}{\left|{\wp }^{\alpha }-\frac{1}{2}\right|}^p\mathrm{d}\wp$$

(11)

and

$${\mathcal{D}}_3\left(\alpha ,p\right)=\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}{\left|{\wp }^{\alpha }-\frac{11}{12}\right|}^p\mathrm{d}\wp .$$

(12)

Proof. 

Using the absolute value on both sides of (1), then applying Hölder’s inequality along with convexity of ${\left|{\mathfrak{f}}^{\prime }\right|}^q$, yields

$$\begin{array}{cc}\hfill & \left|\mathcal{L}\left(\mathfrak{a},\mathfrak{b},\mathfrak{f}\right)-\frac{\mathsf{\Gamma }\left(\alpha +1\right)}{2{\left(\mathfrak{b}-\mathfrak{a}\right)}^{\alpha }}\left({\mathcal{I}}_{{\mathfrak{b}}^{-}}^{\alpha }\mathfrak{f}\left(\mathfrak{a}\right)+{\mathcal{I}}_{{\mathfrak{a}}^{+}}^{\alpha }\mathfrak{f}\left(\mathfrak{b}\right)\right)\right|\hfill \\ \hfill \le & \frac{\mathfrak{b}-\mathfrak{a}}{2}\left[{\left(\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}{\left|{\wp }^{\alpha }-\frac{1}{12}\right|}^p\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{p}}}\right.\left({\left(\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}{\left|{\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\right|}^q\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{q}}}+{\left(\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}{\left|{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right|}^q\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{q}}}\right)\hfill \\ \hfill & +{\left(\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}{\left|{\wp }^{\alpha }-\frac{1}{2}\right|}^p\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{p}}}\left({\left(\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}{\left|{\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\right|}^q\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{q}}}+{\left(\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}{\left|{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right|}^q\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{q}}}\right)\hfill \\ \hfill & +{\left(\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}{\left|{\wp }^{\alpha }-\frac{11}{12}\right|}^p\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{p}}}\left.\left({\left(\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}{\left|{\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\right|}^q\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{q}}}+{\left(\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}{\left|{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right|}^q\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{q}}}\right)\right]\hfill \\ \hfill \le & \frac{\mathfrak{b}-\mathfrak{a}}{2}\left[{\left(\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}{\left|{\wp }^{\alpha }-\frac{1}{12}\right|}^p\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{p}}}\left({\left(\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}\left(\left(1-\wp \right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+\wp {\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q\right)\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{q}}}\right.+{\left(\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}\left(\wp {\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+\left(1-\wp \right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q\right)\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{q}}}\right)\hfill \\ \hfill & +{\left(\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}{\left|{\wp }^{\alpha }-\frac{1}{2}\right|}^p\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{p}}}\left({\left(\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}\left(\left(1-\wp \right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+\wp {\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q\right)\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{q}}}+{\left(\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}\left(\wp {\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+\left(1-\wp \right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q\right)\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{q}}}\right)\hfill \\ \hfill & +{\left(\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}{\left|{\wp }^{\alpha }-\frac{11}{12}\right|}^p\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{p}}}\left({\left(\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}\left(\left(1-\wp \right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+\wp {\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q\right)\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{q}}}+\left.{\left(\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}\left(\wp {\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+\left(1-\wp \right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q\right)\mathrm{d}\wp \right)}^{{\displaystyle \frac{1}{q}}}\right)\right]\hfill \\ \hfill =& \frac{\mathfrak{b}-\mathfrak{a}}{2}\left[\left({\left({\mathcal{D}}_1\left(\alpha ,p\right)\right)}^{{\displaystyle \frac{1}{p}}}+{\left({\mathcal{D}}_3\left(\alpha ,p\right)\right)}^{{\displaystyle \frac{1}{p}}}\right){\left(\frac{5-\sqrt{5}}{10}\right)}^{{\displaystyle \frac{1}{q}}}\right.\left({\left(\frac{\left(15+\sqrt{5}\right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+\left(5-\sqrt{5}\right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q}{20}\right)}^{{\displaystyle \frac{1}{q}}}+{\left(\frac{\left(5-\sqrt{5}\right){\left|{\mathfrak{f}}^{\prime }\left(a\right)\right|}^q+\left(15+\sqrt{5}\right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q}{20}\right)}^{{\displaystyle \frac{1}{q}}}\right)\hfill \\ \hfill & +\left.2{\left({\mathcal{D}}_2\left(\alpha ,p\right)\right)}^{{\displaystyle \frac{1}{p}}}{\left(\frac{\sqrt{5}}{10}\right)}^{{\displaystyle \frac{1}{q}}}{\left({\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+{\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q\right)}^{{\displaystyle \frac{1}{q}}}\right],\hfill \end{array}$$

where ${\mathcal{D}}_1\left(\alpha ,p\right),{\mathcal{D}}_2\left(\alpha ,p\right)$ and ${\mathcal{D}}_3\left(\alpha ,p\right)$ are defined by (10), (11), and (12), respectively. The proof is achieved. □

Corollary 2.

By setting $\alpha =1$, Theorem 2 yields the following Lobatto-type inequality

$$\begin{array}{cc}\hfill & \left|\frac{1}{12}\left(\mathfrak{f}\left(\mathfrak{a}\right)+5\mathfrak{f}\left(\frac{5+\sqrt{5}}{10}\mathfrak{a}+\frac{5-\sqrt{5}}{10}\mathfrak{b}\right)+5\mathfrak{f}\left(\frac{5-\sqrt{5}}{10}\mathfrak{a}+\frac{5+\sqrt{5}}{10}\mathfrak{b}\right)+\mathfrak{f}\left(\mathfrak{b}\right)\right)-\frac{1}{\mathfrak{b}-\mathfrak{a}}\underset{\mathfrak{a}}{\stackrel{\mathfrak{b}}{\int }}\mathfrak{f}\left(\mathfrak{u}\right)\mathrm{d}\mathfrak{u}\right|\hfill \\ \hfill \le & \frac{\mathfrak{b}-\mathfrak{a}}{60{\left(p+1\right)}^{\frac{1}{p}}}\left({\left(\frac{5^{p+1}+{\left(25-6\sqrt{5}\right)}^{p+1}}{60}\right)}^{{\displaystyle \frac{1}{p}}}{\left(\frac{5-\sqrt{5}}{10}\right)}^{{\displaystyle \frac{1}{q}}}\right.\left({\left(\frac{\left(15+\sqrt{5}\right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+\left(5-\sqrt{5}\right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q}{20}\right)}^{{\displaystyle \frac{1}{q}}}+{\left(\frac{\left(5-\sqrt{5}\right){\left|{\mathfrak{f}}^{\prime }\left(a\right)\right|}^q+\left(15+\sqrt{5}\right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q}{20}\right)}^{{\displaystyle \frac{1}{q}}}\right)\hfill \\ \hfill & +\left.6{\left(\frac{{\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+{\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q}{2}\right)}^{{\displaystyle \frac{1}{q}}}\right).\hfill \end{array}$$

Theorem 3.

Let $\mathfrak{f}$ be as in Lemma 1. If ${\left|{\mathfrak{f}}^{\prime }\right|}^q$ is convex, where $q>0$, then we have

$$\left|\mathcal{L}\left(\mathfrak{a},\mathfrak{b},\mathfrak{f}\right)-\frac{\mathsf{\Gamma }\left(\alpha +1\right)}{2{\left(\mathfrak{b}-\mathfrak{a}\right)}^{\alpha }}\left({\mathcal{I}}_{{\mathfrak{b}}^{-}}^{\alpha }\mathfrak{f}\left(\mathfrak{a}\right)+{\mathcal{I}}_{{\mathfrak{a}}^{+}}^{\alpha }\mathfrak{f}\left(\mathfrak{b}\right)\right)\right|\le \frac{\mathfrak{b}-\mathfrak{a}}{2}c\left(\frac{1}{q}\right)\left(\stackrel{6}{\sum _{i=1}}{\mathcal{E}}_i\left(\alpha ,q\right)\right)\left(\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|+\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|\right),$$

where

$$c\left(\frac{1}{q}\right)=\left\{\begin{array}{ccc}{2}^{\frac{1}{q}-1}\hfill & \mathit{if}& q\le 1,\\ 1\hfill & \mathit{if}& q>1,\end{array}\right.$$

and ${\left\{{\mathcal{E}}_i\left(\alpha ,q\right)\right\}}_{i=1,2,\dots 6}$ are defined by

$$\begin{array}{cc}\hfill {\mathcal{E}}_1\left(\alpha ,q\right)=& \stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}\left|{\wp }^{\alpha }-\frac{1}{12}\right|{\left(1-\wp \right)}^{{\displaystyle \frac{1}{q}}}\mathrm{d}\wp \hfill \\ \hfill =& \left\{\begin{array}{ccc}\frac{q}{12\left(1+q\right)}\left(1-2{\left(1-{\left({\displaystyle \frac{1}{12}}\right)}^{{\displaystyle \frac{1}{\alpha }}}\right)}^{{\displaystyle \frac{1+q}{q}}}\right)-2{\mathcal{B}}_{{\left({\displaystyle \frac{1}{12}}\right)}^{{\displaystyle \frac{1}{\alpha }}}}\left(\alpha +1,{\displaystyle \frac{1+q}{q}}\right)\hfill & & \\ +{\mathcal{B}}_{{\displaystyle \frac{5-\sqrt{5}}{10}}}\left(\alpha +1,{\displaystyle \frac{1+q}{q}}\right)+\frac{q}{12\left(1+q\right)}{\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}\hfill & \mathit{if}& \alpha <{\displaystyle \frac{ln12}{ln10-ln\left(5-\sqrt{5}\right)}},\\ \left(\frac{q}{12\left(1+q\right)}\left(1-{\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}\right)-{\mathcal{B}}_{{\displaystyle \frac{5-\sqrt{5}}{10}}}\left(\alpha +1,{\displaystyle \frac{1+q}{q}}\right)\right)\hfill & \mathit{if}& \alpha \ge {\displaystyle \frac{ln12}{ln10-ln\left(5-\sqrt{5}\right)}},\end{array}\right.\hfill \end{array}$$

(13)

$$\begin{array}{cc}\hfill {\mathcal{E}}_2\left(\alpha ,q\right)=& \stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}\left|{\wp }^{\alpha }-\frac{1}{12}\right|{\wp }^{{\displaystyle \frac{1}{q}}}\mathrm{d}\wp \hfill \\ \hfill =& \left\{\begin{array}{ccc}\frac{2\alpha q^2}{\left(1+q\right)\left(\left(\alpha +1\right)q+1\right)}{\left(\frac{1}{12}\right)}^{{\displaystyle \frac{\left(\alpha +1\right)q+1}{\alpha q}}}+\frac{q}{\left(\alpha +1\right)q+1}{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{\left(\alpha +1\right)q+1}{q}}}\hfill & & \\ -\frac{q}{12\left(1+q\right)}{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}\hfill & \mathit{if}& \alpha <{\displaystyle \frac{ln12}{ln10-ln\left(5-\sqrt{5}\right)}},\\ \frac{q}{12\left(1+q\right)}{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}-\frac{q}{\left(\alpha +1\right)q+1}{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{\left(\alpha +1\right)q+1}{q}}}\hfill & \mathit{if}& \alpha \ge {\displaystyle \frac{ln12}{ln10-ln\left(5-\sqrt{5}\right)}},\end{array}\right.\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill {\mathcal{E}}_3\left(\alpha ,q\right)=& \stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{1}{2}\right|{\left(1-\wp \right)}^{{\displaystyle \frac{1}{q}}}\mathrm{d}\wp \hfill \\ \hfill =& \left\{\begin{array}{c}{\mathcal{B}}_{{\displaystyle \frac{5+\sqrt{5}}{10}}}\left(\alpha +1,{\displaystyle \frac{1+q}{q}}\right)-{\mathcal{B}}_{{\displaystyle \frac{5-\sqrt{5}}{10}}}\left(\alpha +1,{\displaystyle \frac{1+q}{q}}\right)\hfill \\ -\frac{q}{2\left(1+q\right)}\left({\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}-{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}\right)\hfill & \mathit{if}& \alpha <\frac{ln2}{ln10-ln\left(5-\sqrt{5}\right)}\\ \frac{q}{1+q}\left({\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}-{\left(1-{\left(\frac{1}{2}\right)}^{{\displaystyle \frac{1}{\alpha }}}\right)}^{{\displaystyle \frac{1+q}{q}}}\right)-2{\mathcal{B}}_{{\left({\displaystyle \frac{1}{2}}\right)}^{{\displaystyle \frac{1}{\alpha }}}}\left(\alpha +1,{\displaystyle \frac{1+q}{q}}\right)\hfill & & \\ +{\mathcal{B}}_{{\displaystyle \frac{5-\sqrt{5}}{10}}}\left(\alpha +1,{\displaystyle \frac{1+q}{q}}\right)+{\mathcal{B}}_{{\displaystyle \frac{5+\sqrt{5}}{10}}}\left(\alpha +1,{\displaystyle \frac{1+q}{q}}\right)\hfill & \mathit{if}& \frac{ln2}{ln10-ln\left(5-\sqrt{5}\right)}\le \alpha \le \frac{ln2}{ln10-ln\left(5+\sqrt{5}\right)}\\ \frac{q}{2\left(1+q\right)}\left({\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}-{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}\right)\hfill & & \\ -{\mathcal{B}}_{{\displaystyle \frac{5+\sqrt{5}}{10}}}\left(\alpha +1,{\displaystyle \frac{1+q}{q}}\right)+{\mathcal{B}}_{{\displaystyle \frac{5-\sqrt{5}}{10}}}\left(\alpha +1,{\displaystyle \frac{1+q}{q}}\right)\hfill & \mathit{if}& \alpha >\frac{ln2}{ln10-ln\left(5+\sqrt{5}\right)}\end{array}\right.\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill {\mathcal{E}}_4\left(\alpha ,q\right)=& \stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{1}{2}\right|{\left(1-\wp \right)}^{{\displaystyle \frac{1}{q}}}\mathrm{d}\wp \hfill \\ \hfill =& \left\{\begin{array}{ccc}\frac{q}{\left(\alpha +1\right)q+1}\left({\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{\left(\alpha +1\right)q+1}{q}}}-{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{\left(\alpha +1\right)q+1}{q}}}\right)\hfill & & \\ -\frac{q}{2\left(1+q\right)}\left({\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}-{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}\right)\hfill & \mathit{if}& \alpha <\frac{ln2}{ln10-ln\left(5-\sqrt{5}\right)},\\ \frac{2q}{2\left(1+q\right)}{\left(\frac{1}{2}\right)}^{{\displaystyle \frac{1+q}{\alpha q}}}-\frac{2q}{\left(\alpha +1\right)q+1}{\left(\frac{1}{2}\right)}^{{\displaystyle \frac{\left(\alpha +1\right)q+1}{\alpha q}}}\hfill & & \\ -\frac{q}{2\left(1+q\right)}\left({\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}-{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}\right)\hfill & & \\ +\frac{q}{\left(\alpha +1\right)q+1}\left({\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{\left(\alpha +1\right)q+1}{q}}}+{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{\left(\alpha +1\right)q+1}{q}}}\right)\hfill & \mathit{if}& \frac{ln2}{ln10-ln\left(5-\sqrt{5}\right)}\le \alpha \le \frac{ln2}{ln10-ln\left(5+\sqrt{5}\right)},\\ \frac{q}{2\left(1+q\right)}\left({\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}-{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}\right)\hfill & & \\ -\frac{q}{\left(\alpha +1\right)q+1}\left({\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{\left(\alpha +1\right)q+1}{q}}}-{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{\left(\alpha +1\right)q+1}{q}}}\right)\hfill & \mathit{if}& \alpha >\frac{ln2}{ln10-ln\left(5+\sqrt{5}\right)},\end{array}\right.\hfill \end{array}$$

(16)

$$\begin{array}{cc}\hfill {\mathcal{E}}_5\left(\alpha ,q\right)=& \stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{11}{12}\right|{\left(1-\wp \right)}^{{\displaystyle \frac{1}{q}}}\mathrm{d}\wp \hfill \\ \hfill =& \left\{\begin{array}{ccc}\mathcal{B}\left(\alpha +1,{\displaystyle \frac{1+q}{q}}\right)-{\mathcal{B}}_{{\displaystyle \frac{5+\sqrt{5}}{10}}}\left(\alpha +1,{\displaystyle \frac{1+q}{q}}\right)-{\displaystyle \frac{11q}{12\left(1+q\right)}}{\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}\hfill & \mathit{if}& \alpha \le \frac{ln12-ln11}{ln10-ln\left(5+\sqrt{5}\right)},\\ {\displaystyle \frac{11q}{12\left(1+q\right)}}\left({\left({\displaystyle \frac{5-\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}-2{\left(1-{\left(\frac{11}{12}\right)}^{{\displaystyle \frac{1}{\alpha }}}\right)}^{{\displaystyle \frac{1+q}{q}}}\right)-2{\mathcal{B}}_{{\left(\frac{11}{12}\right)}^{{\displaystyle \frac{1}{\alpha }}}}\left(\alpha +1,{\displaystyle \frac{1+q}{q}}\right)\hfill & & \\ +\mathcal{B}\left(\alpha +1,{\displaystyle \frac{1+q}{q}}\right)+{\mathcal{B}}_{{\displaystyle \frac{5+\sqrt{5}}{10}}}\left(\alpha +1,{\displaystyle \frac{1+q}{q}}\right)\hfill & \mathit{if}& \alpha >\frac{ln12-ln11}{ln10-ln\left(5+\sqrt{5}\right)},\end{array}\right.\hfill \end{array}$$

(17)

and

$$\begin{array}{cc}\hfill {\mathcal{E}}_6\left(\alpha ,q\right)=& \stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{11}{12}\right|{\wp }^{{\displaystyle \frac{1}{q}}}\mathrm{d}\wp \hfill \\ \hfill =& \left\{\begin{array}{ccc}{\displaystyle \frac{q}{\left(\alpha +1\right)q+1}}\left(1-{\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{\left(\alpha +1\right)q+1}{q}}}\right)-\frac{11q}{12\left(1+q\right)}\left(1-{\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}\right)\hfill & \mathit{if}& \alpha \le \frac{ln12-ln11}{ln10-ln\left(5+\sqrt{5}\right)},\\ {\displaystyle \frac{q}{\left(\alpha +1\right)q+1}}\left(1+{\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{\left(\alpha +1\right)q+1}{q}}}-2{\left(\frac{11}{12}\right)}^{{\displaystyle \frac{\left(\alpha +1\right)q+1}{\alpha q}}}\right)\hfill & & \\ -\frac{11q}{12\left(1+q\right)}\left(1+{\left({\displaystyle \frac{5+\sqrt{5}}{10}}\right)}^{{\displaystyle \frac{1+q}{q}}}-2{\left(\frac{11}{12}\right)}^{{\displaystyle \frac{1+q}{\alpha q}}}\right)\hfill & \mathit{if}& \alpha >\frac{ln12-ln11}{ln10-ln\left(5+\sqrt{5}\right)}.\end{array}\right.\hfill \end{array}$$

(18)

Proof. 

Using the absolute value of both sides of (1), we deduce

$$\begin{array}{cc}\hfill & \left|\mathcal{L}\left(\mathfrak{a},\mathfrak{b},\mathfrak{f}\right)-\frac{\mathsf{\Gamma }\left(\alpha +1\right)}{2{\left(\mathfrak{b}-\mathfrak{a}\right)}^{\alpha }}\left({\mathcal{I}}_{{\mathfrak{b}}^{-}}^{\alpha }\mathfrak{f}\left(\mathfrak{a}\right)+{\mathcal{I}}_{{\mathfrak{a}}^{+}}^{\alpha }\mathfrak{f}\left(\mathfrak{b}\right)\right)\right|\hfill \\ \hfill \le & \frac{\mathfrak{b}-\mathfrak{a}}{2}\left(\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}\left|{\wp }^{\alpha }-\frac{1}{12}\right|\left|{\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\right|\mathrm{d}\wp +\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}\left|{\wp }^{\alpha }-\frac{1}{12}\right|\left|{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right|\mathrm{d}\wp \right.\hfill \\ \hfill & +\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{1}{2}\right|\left|{\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\right|\mathrm{d}\wp +\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{1}{2}\right|\left|{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right|\mathrm{d}\wp \hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill & +\left.\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{11}{12}\right|\left|{\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\right|\mathrm{d}\wp +\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{11}{12}\right|\left|{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right|\mathrm{d}\wp \right).\hfill \end{array}$$

(20)

Since ${\left|{\mathfrak{f}}^{\prime }\right|}^q$ is convex, then we have

$${\left|{\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\right|}^q\le \left(1-\wp \right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+\wp {\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q.$$

(21)

Obviously, (21) gives

$$\left|{\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\right|\le {\left(\left(1-\wp \right){\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+\wp {\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q\right)}^{{\displaystyle \frac{1}{q}}}.$$

(22)

Using the following algebraic inequality ${\left(u+v\right)}^r\le c\left(r\right)\left(u^r+v^r\right)$ for $u,v\ge 0$ with $c\left(r\right)=\left\{\begin{array}{ccc}{2}^{r-1}\hfill & \mathrm{if}& r\ge 1\\ 1\hfill & \mathrm{if}& 0<r<1\end{array}\right.$, (22) becomes

$$\left|{\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\right|\le c\left(\frac{1}{q}\right)\left({\left(1-\wp \right)}^{{\displaystyle \frac{1}{q}}}\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|+{\wp }^{{\displaystyle \frac{1}{q}}}\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|\right).$$

(23)

Similarly, we have

$$\left|{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right|\le c\left(\frac{1}{q}\right)\left({\wp }^{{\displaystyle \frac{1}{q}}}\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|+{\left(1-\wp \right)}^{{\displaystyle \frac{1}{q}}}\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|\right).$$

(24)

Substituting (23) and (24) in (19), we obtain

$$\begin{array}{cc}\hfill & \left|\mathcal{L}\left(\mathfrak{a},\mathfrak{b},\mathfrak{f}\right)-\frac{\mathsf{\Gamma }\left(\alpha +1\right)}{2{\left(\mathfrak{b}-\mathfrak{a}\right)}^{\alpha }}\left({\mathcal{I}}_{{\mathfrak{b}}^{-}}^{\alpha }\mathfrak{f}\left(\mathfrak{a}\right)+{\mathcal{I}}_{{\mathfrak{a}}^{+}}^{\alpha }\mathfrak{f}\left(\mathfrak{b}\right)\right)\right|\hfill \\ \hfill \le & \frac{\mathfrak{b}-\mathfrak{a}}{2}\left(\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|+\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|\right)c\left(\frac{1}{q}\right)\left(\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}\left|{\wp }^{\alpha }-\frac{1}{12}\right|\left({\left(1-\wp \right)}^{{\displaystyle \frac{1}{q}}}+{\wp }^{{\displaystyle \frac{1}{q}}}\right)\mathrm{d}\wp \right.\hfill \\ \hfill & +\left.\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{1}{2}\right|\left({\left(1-\wp \right)}^{{\displaystyle \frac{1}{q}}}+{\wp }^{{\displaystyle \frac{1}{q}}}\right)\mathrm{d}\wp +\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{11}{12}\right|\left({\left(1-\wp \right)}^{{\displaystyle \frac{1}{q}}}+{\wp }^{{\displaystyle \frac{1}{q}}}\right)\mathrm{d}\wp \right)\hfill \\ \hfill =& \frac{\mathfrak{b}-\mathfrak{a}}{2}\left(\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|+\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|\right)c\left(\frac{1}{q}\right)\left(\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}\left|{\wp }^{\alpha }-\frac{1}{12}\right|{\left(1-\wp \right)}^{{\displaystyle \frac{1}{q}}}\mathrm{d}\wp \right.\hfill \\ \hfill & +\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}\left|{\wp }^{\alpha }-\frac{1}{12}\right|{\wp }^{{\displaystyle \frac{1}{q}}}\mathrm{d}\wp +\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{1}{2}\right|{\left(1-\wp \right)}^{{\displaystyle \frac{1}{q}}}\mathrm{d}\wp +\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{1}{2}\right|{\wp }^{{\displaystyle \frac{1}{q}}}\mathrm{d}\wp \hfill \\ \hfill & +\left.\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{11}{12}\right|{\left(1-\wp \right)}^{{\displaystyle \frac{1}{q}}}\mathrm{d}\wp +\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}\left|{\wp }^{\alpha }-\frac{11}{12}\right|{\wp }^{{\displaystyle \frac{1}{q}}}\mathrm{d}\wp \right)\hfill \\ \hfill =& \frac{\mathfrak{b}-\mathfrak{a}}{2}c\left(\frac{1}{q}\right)\left(\stackrel{6}{\sum _{i=1}}{\mathcal{E}}_i\left(\alpha ,q\right)\right)\left(\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|+\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|\right),\hfill \end{array}$$

where we have used (13)–(18).

The proof is completed. □

Corollary 3.

If we attempt to set $\alpha =1$, Theorem 3 gives

$$\begin{array}{cc}\hfill & \left|\frac{1}{12}\left(\mathfrak{f}\left(\mathfrak{a}\right)+5\mathfrak{f}\left(\frac{5+\sqrt{5}}{10}\mathfrak{a}+\frac{5-\sqrt{5}}{10}\mathfrak{b}\right)+5\mathfrak{f}\left(\frac{5-\sqrt{5}}{10}\mathfrak{a}+\frac{5+\sqrt{5}}{10}\mathfrak{b}\right)+\mathfrak{f}\left(\mathfrak{b}\right)\right)-\frac{1}{\mathfrak{b}-\mathfrak{a}}\underset{\mathfrak{a}}{\stackrel{\mathfrak{b}}{\int }}\mathfrak{f}\left(\mathfrak{u}\right)\mathrm{d}\mathfrak{u}\right|\hfill \\ \hfill \le & \frac{\mathfrak{b}-\mathfrak{a}}{\left(1+q\right)\left(1+2q\right)}c\left(\frac{1}{q}\right)\left(\frac{q-10q^2}{12}+2q^2\left({\left(\frac{1}{12}\right)}^{{\displaystyle \frac{1+2q}{q}}}+{\left(\frac{1}{2}\right)}^{{\displaystyle \frac{1+2q}{q}}}+{\left(\frac{11}{12}\right)}^{{\displaystyle \frac{1+2q}{q}}}\right)-\frac{17q\left(1+2q\right)}{12}{\left(\frac{5+\sqrt{5}}{10}\right)}^{{\displaystyle \frac{1+q}{q}}}\right.\hfill \\ \hfill & -\left.\frac{7q\left(1+2q\right)}{12}{\left(\frac{5-\sqrt{5}}{10}\right)}^{{\displaystyle \frac{1+q}{q}}}+2q\left(1+q\right)\left({\left(\frac{5+\sqrt{5}}{10}\right)}^{{\displaystyle \frac{1+2q}{q}}}+{\left(\frac{5-\sqrt{5}}{10}\right)}^{{\displaystyle \frac{1+2q}{q}}}\right)\right)\left(\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|+\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|\right).\hfill \end{array}$$

Theorem 4.

Let $\mathfrak{f}$ be as in Lemma 1. If ${\left|{\mathfrak{f}}^{\prime }\right|}^q$ is convex, where $q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$, then we have

$$\begin{array}{cc}\hfill & \left|\mathcal{L}\left(\mathfrak{a},\mathfrak{b},\mathfrak{f}\right)-\frac{\mathsf{\Gamma }\left(\alpha +1\right)}{2{\left(\mathfrak{b}-\mathfrak{a}\right)}^{\alpha }}\left({\mathcal{I}}_{{\mathfrak{b}}^{-}}^{\alpha }\mathfrak{f}\left(\mathfrak{a}\right)+{\mathcal{I}}_{{\mathfrak{a}}^{+}}^{\alpha }\mathfrak{f}\left(\mathfrak{b}\right)\right)\right|\hfill \\ \hfill \le & \frac{\mathfrak{b}-\mathfrak{a}}{2}\left(\frac{2}{p}\left({\mathcal{D}}_1\left(\alpha ,p\right)+{\mathcal{D}}_2\left(\alpha ,p\right)+{\mathcal{D}}_3\left(\alpha ,p\right)\right)+\frac{1}{q}\left({\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+{\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q\right)\right),\hfill \end{array}$$

where ${\mathcal{D}}_1\left(\alpha ,p\right),{\mathcal{D}}_2\left(\alpha ,p\right)$ and ${\mathcal{D}}_3\left(\alpha ,p\right)$ are defined by (10), (11), and (12), respectively.

Proof. 

Using the absolute value of both sides of (1), then applying Young’s inequality along with the convexity of ${\left|{\mathfrak{f}}^{\prime }\right|}^q$, we deduce

$$\begin{array}{cc}\hfill & \left|\mathcal{L}\left(\mathfrak{a},\mathfrak{b},\mathfrak{f}\right)-\frac{\mathsf{\Gamma }\left(\alpha +1\right)}{2{\left(\mathfrak{b}-\mathfrak{a}\right)}^{\alpha }}\left({\mathcal{I}}_{{\mathfrak{b}}^{-}}^{\alpha }\mathfrak{f}\left(\mathfrak{a}\right)+{\mathcal{I}}_{{\mathfrak{a}}^{+}}^{\alpha }\mathfrak{f}\left(\mathfrak{b}\right)\right)\right|\hfill \\ \hfill \le & \frac{\mathfrak{b}-\mathfrak{a}}{2}\left(\frac{1}{p}\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}{\left|{\wp }^{\alpha }-\frac{1}{12}\right|}^p\mathrm{d}\wp +\frac{1}{q}\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}{\left|{\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\right|}^q\mathrm{d}\wp \right.+\frac{1}{p}\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}{\left|{\wp }^{\alpha }-\frac{1}{12}\right|}^p\mathrm{d}\wp +\frac{1}{q}\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}{\left|{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right|}^q\mathrm{d}\wp \hfill \\ \hfill & +\frac{1}{p}\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}{\left|{\wp }^{\alpha }-\frac{1}{2}\right|}^p\mathrm{d}\wp +\frac{1}{q}\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}{\left|{\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\right|}^q\mathrm{d}\wp +\frac{1}{p}\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}{\left|{\wp }^{\alpha }-\frac{1}{2}\right|}^p\mathrm{d}\wp +\frac{1}{q}\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}{\left|{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right|}^q\mathrm{d}\wp \hfill \\ \hfill & +\frac{1}{p}\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}{\left|{\wp }^{\alpha }-\frac{11}{12}\right|}^p\mathrm{d}\wp +\frac{1}{q}\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}{\left|{\mathfrak{f}}^{\prime }\left(\left(1-\wp \right)\mathfrak{a}+\wp \mathfrak{b}\right)\right|}^q\mathrm{d}\wp +\left.\frac{1}{p}\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}{\left|{\wp }^{\alpha }-\frac{11}{12}\right|}^p\mathrm{d}\wp +\frac{1}{q}\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}{\left|{\mathfrak{f}}^{\prime }\left(\wp \mathfrak{a}+\left(1-\wp \right)\mathfrak{b}\right)\right|}^q\mathrm{d}\wp \right)\hfill \\ \hfill \le & \frac{\mathfrak{b}-\mathfrak{a}}{2}\left(\frac{2}{p}\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}{\left|{\wp }^{\alpha }-\frac{1}{12}\right|}^p\mathrm{d}\wp +\frac{1}{q}\left({\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+{\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q\right)\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}\mathrm{d}\wp \right.+\frac{2}{p}\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}{\left|{\wp }^{\alpha }-\frac{1}{2}\right|}^p\mathrm{d}\wp \hfill \\ \hfill & +\frac{1}{q}\left({\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+{\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q\right)\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}\mathrm{d}\wp +\left.\frac{2}{p}\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}{\left|{\wp }^{\alpha }-\frac{11}{12}\right|}^p\mathrm{d}\wp +\frac{1}{q}\left({\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+{\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q\right)\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}\mathrm{d}\wp \right)\hfill \\ \hfill =& \frac{\mathfrak{b}-\mathfrak{a}}{2}\left(\frac{2}{p}\left(\stackrel{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\underset{0}{\int }}{\left|{\wp }^{\alpha }-\frac{1}{12}\right|}^p\mathrm{d}\wp +\stackrel{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\underset{{\displaystyle \frac{5-\sqrt{5}}{10}}}{\int }}{\left|{\wp }^{\alpha }-\frac{1}{2}\right|}^p\mathrm{d}\wp +\stackrel{1}{\underset{{\displaystyle \frac{5+\sqrt{5}}{10}}}{\int }}{\left|{\wp }^{\alpha }-\frac{11}{12}\right|}^p\mathrm{d}\wp \right)+\frac{1}{q}\left({\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+{\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q\right)\right)\hfill \\ \hfill =& \frac{\mathfrak{b}-\mathfrak{a}}{2}\left(\frac{2}{p}\left({\mathcal{D}}_1\left(\alpha ,p\right)+{\mathcal{D}}_2\left(\alpha ,p\right)+{\mathcal{D}}_3\left(\alpha ,p\right)\right)+\frac{1}{q}\left({\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+{\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q\right)\right),\hfill \end{array}$$

where ${\mathcal{D}}_1\left(\alpha ,p\right),{\mathcal{D}}_2\left(\alpha ,p\right)$ and ${\mathcal{D}}_3\left(\alpha ,p\right)$ are defined by (10), (11), and (12), respectively. The proof is achieved. □

Corollary 4.

By setting $\alpha =1$, Theorem 4 becomes

$$\begin{array}{cc}\hfill & \left|\frac{1}{12}\left(\mathfrak{f}\left(\mathfrak{a}\right)+5\mathfrak{f}\left(\frac{5+\sqrt{5}}{10}\mathfrak{a}+\frac{5-\sqrt{5}}{10}\mathfrak{b}\right)+5\mathfrak{f}\left(\frac{5-\sqrt{5}}{10}\mathfrak{a}+\frac{5+\sqrt{5}}{10}\mathfrak{b}\right)+\mathfrak{f}\left(\mathfrak{b}\right)\right)-\frac{1}{\mathfrak{b}-\mathfrak{a}}\underset{\mathfrak{a}}{\stackrel{\mathfrak{b}}{\int }}\mathfrak{f}\left(\mathfrak{u}\right)\mathrm{d}\mathfrak{u}\right|\hfill \\ \hfill \le & \frac{\mathfrak{b}-\mathfrak{a}}{2}\left(\frac{4}{p\left(p+1\right)}\left(\frac{5^{p+1}+{\left(25-6\sqrt{5}\right)}^{p+1}+{\left(6\sqrt{5}\right)}^{p+1}}{{60}^{p+1}}\right)+\frac{1}{q}\left({\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{a}\right)\right|}^q+{\left|{\mathfrak{f}}^{\prime }\left(\mathfrak{b}\right)\right|}^q\right)\right).\hfill \end{array}$$

4. Numerical Validation and Graphical Illustration

To confirm the accuracy and effectiveness of the derived inequalities, we present a numerical example with visual representation of the error bounds and convergence behavior.

Example 1.

Let $\mathfrak{f}$ be a function defined on the interval $[\mathfrak{a},\mathfrak{b}]=[0,1]$ by $\mathfrak{f}\left(\mathfrak{u}\right)={\mathfrak{u}}^2$. The first-order derivative ${\mathfrak{f}}^{\prime }\left(\mathfrak{u}\right)=2\mathfrak{u}$ satisfies the condition of Theorem 1, which gives

$$\left|{\displaystyle \frac{1}{3}}-\mathsf{\Gamma }(\alpha +1)\left({\displaystyle \frac{1}{\mathsf{\Gamma }(\alpha +3)}}+{\displaystyle \frac{1}{2(\alpha +2)\mathsf{\Gamma }\left(\alpha \right)}}\right)\right|\le {\mathcal{C}}_1\left(\alpha \right)+{\mathcal{C}}_2\left(\alpha \right)+{\mathcal{C}}_3\left(\alpha \right),$$

(25)

where ${\mathcal{C}}_1\left(\alpha \right)$, ${\mathcal{C}}_2\left(\alpha \right)$, and ${\mathcal{C}}_3\left(\alpha \right)$ are defined by (7)–(9), respectively.

The left- and right-hand sides of inequality (25) are plotted in Figure 1 for $\alpha \in (0,4]$, while

Table 1. Values of both sides of (25) for selected $\alpha$.

$\mathit{\alpha }$	LHS	RHS
$0.100$	$1.234\times {10}^{-1}$	$4.091\times {10}^{-1}$
$0.500$	$3.333\times {10}^{-2}$	$1.747\times {10}^{-1}$
$1.000$	$0.000\times {10}^0$	$9.422\times {10}^{-2}$
$1.500$	$4.762\times {10}^{-3}$	$1.205\times {10}^{-1}$
$2.000$	$0.000\times {10}^0$	$1.706\times {10}^{-1}$
$2.500$	$7.937\times {10}^{-3}$	$2.171\times {10}^{-1}$
$3.000$	$1.667\times {10}^{-2}$	$2.524\times {10}^{-1}$
$3.500$	$2.525\times {10}^{-2}$	$2.798\times {10}^{-1}$
$4.000$	$3.333\times {10}^{-2}$	$3.018\times {10}^{-1}$

lists the values of both sides of the inequality for selected values of α within the same interval.

5. Applications

To illustrate the applicability of our results, we present several applications to special means of real numbers. We first recall the following special means:

The formula for the arithmetic mean is $\mathcal{A}\left(\mathfrak{a},\mathfrak{b}\right)={\displaystyle \frac{\mathfrak{a}+\mathfrak{b}}{2}}$.

The geometric mean is $\mathcal{G}\left(\mathfrak{a},\mathfrak{b}\right)=\sqrt{\mathfrak{a}\mathfrak{b}},\mathfrak{a},\mathfrak{b}>0$.

The p-logarithmic mean is ${\mathcal{L}}_p\left(\mathfrak{a},\mathfrak{b}\right)={\left({\displaystyle \frac{\mathfrak{b}-\mathfrak{a}}{\left(p+1\right)\left(\mathfrak{b}-\mathfrak{a}\right)}}\right)}^{{\displaystyle \frac{1}{p}}}$, $\mathfrak{a},\mathfrak{b}>0,\mathfrak{a}\ne \mathfrak{b}$ and $p\in \mathbb{R}\setminus \left\{-1,0\right\}$.

Proposition 1.

For real numbers $\mathfrak{a},\mathfrak{b}$, such that $0<\mathfrak{a}<\mathfrak{b}$, we have

$$\begin{array}{cc}\hfill & \left|A\left({\mathfrak{a}}^3,{\mathfrak{b}}^3\right)+15{\mathcal{G}}^2\left(\mathfrak{a},\mathfrak{b}\right)\mathcal{A}\left(\mathfrak{a},\mathfrak{b}\right)-3{\mathcal{L}}_3^3\left(\mathfrak{a},\mathfrak{b}\right)\right|\hfill \\ \hfill \le & \frac{\left(101-30\sqrt{5}\right)\left(\mathfrak{b}-\mathfrak{a}\right)}{80}\left({\mathfrak{a}}^2+{\mathfrak{b}}^2\right).\hfill \end{array}$$

Proof. 

Applying Corollary 1 to the function $\mathfrak{f}\left(\mathfrak{u}\right)=u^3$, yields the statement. □

Proposition 2.

For real numbers $\mathfrak{a},\mathfrak{b}$, such that $0<\mathfrak{a}<\mathfrak{b}$, we have

$$\begin{array}{cc}\hfill & \left|A\left({\mathfrak{a}}^3,{\mathfrak{b}}^3\right)+15{\mathcal{G}}^2\left(\mathfrak{a},\mathfrak{b}\right)\mathcal{A}\left(\mathfrak{a},\mathfrak{b}\right)-3{\mathcal{L}}_3^3\left(\mathfrak{a},\mathfrak{b}\right)\right|\hfill \\ \hfill \le & \frac{\mathfrak{b}-\mathfrak{a}}{4}\left(\frac{5849-2250\sqrt{5}}{10800}+27\left({\mathfrak{a}}^4+{\mathfrak{b}}^4\right)\right).\hfill \end{array}$$

Proof. 

Applying Corollary 4 with $q=2$ to the function $\mathfrak{f}\left(\mathfrak{u}\right)=u^3$, yields the statement. □

6. Conclusions and Future Directions

We have introduced a new fractional integral identity for the 4-point Lobatto quadrature rule and used it to derive practical error inequalities under convexity assumptions. These bounds depend on lower-order derivatives, making them more applicable than classical estimates. A numerical example with graphical validation confirms the accuracy of the theoretical results, and applications to special means illustrate their utility. This work highlights the value of combining fractional calculus with quadrature analysis. Future research may extend these ideas to other quadrature rules, multidimensional integrals, or stochastic settings.