*Article* **Some New Simpson's-Formula-Type Inequalities for Twice-Differentiable Convex Functions via Generalized Fractional Operators**

**Muhammad Aamir Ali 1,\*, Hasan Kara 2, Jessada Tariboon 3, Suphawat Asawasamrit 3, Hüseyin Budak <sup>2</sup> and Fatih Hezenci <sup>2</sup>**


**Abstract:** From the past to the present, various works have been dedicated to Simpson's inequality for differentiable convex functions. Simpson-type inequalities for twice-differentiable functions have been the subject of some research. In this paper, we establish a new generalized fractional integral identity involving twice-differentiable functions, then we use this result to prove some new Simpson'sformula-type inequalities for twice-differentiable convex functions. Furthermore, we examine a few special cases of newly established inequalities and obtain several new and old Simpson's-formulatype inequalities. These types of analytic inequalities, as well as the methodologies for solving them, have applications in a wide range of fields where symmetry is crucial.

**Keywords:** Simpson-type inequalities; convex function; fractional integrals

## **1. Introduction**

Simpson's inequality is widely used in many areas of mathematics. For four times continuously differentiable functions, the classical Simpson's inequality is expressed as follows:

**Theorem 1.** *Suppose that <sup>f</sup>* : [*a*, *<sup>b</sup>*] <sup>→</sup> <sup>R</sup> *is a four times continuously differentiable mapping on* (*a*, *b*), *and suppose also that* % % % *<sup>f</sup>* (4) % % % <sup>∞</sup> <sup>=</sup> sup *x*∈(*a*,*b*) *<sup>f</sup>* (4)(*x*) <sup>&</sup>lt; <sup>∞</sup>. *Then, one has the inequality*

$$\frac{1}{\beta} \left| \frac{f(a) + f(b)}{2} + 2f\left(\frac{a+b}{2}\right) \right| - \frac{1}{b-a} \int\_a^b f(\mathbf{x}) d\mathbf{x} \right| \le \frac{1}{2880} \left\| f^{(4)} \right\|\_{\infty} (b-a)^4.$$

Many researchers have studied various Simpson's inequalities. More precisely, some studies have focused on Simpson's type for the convex function, because this focus has been an effective and powerful way to solve many problems in inequality theory and other areas of mathematics. For example, Alomari et al. established some inequalities of Simpson's type for *s*-convex functions by using differentiable functions [1]. Subsequently, Sarikaya et al. established new variants of Simpson's-type inequalities based on differentiable convex functions in [2,3]. Additionally, some papers have listed Simpson's-type inequalities in various convex classes [4–8]. Moreover, in the papers [9,10], researchers extended the Simpson inequalities for differentiable functions to Riemann–Liouville fractional integrals. Thereupon, several mathematicians studied fractional Simpson inequalities for these kinds of fractional integral operators [11–19]. For more studies related to different

**Citation:** Ali, M.A.; Kara, H.; Tariboon, J.; Asawasamrit, S.; Budak, H.; Hezenci, H. Some New Simpson's-Formula-Type Inequalities for Twice-Differentiable Convex Functions via Generalized Fractional Operators. *Symmetry* **2021**, *13*, 2249. https://doi.org/10.3390/sym13122249

Academic Editor: Nicusor Minculete

Received: 29 October 2021 Accepted: 11 November 2021 Published: 25 November 2021

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**Copyright:** © 2021 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/).

integral operator inequalities, one can see [20–31]. In addition, Sarikaya et al. obtained several Simpson-type inequalities for mappings whose second derivatives are convex [32]. In this article, after giving the definition of the generalized fractional integral operators, we construct a new identity for twice-differentiable functions. Using this equality, we prove several Simpson-type inequalities for functions whose second derivatives are convex. Then, with the help of special choices, the main results in this paper are shown to generalize many studies. In addition to all these, new results for *k*-Riemann–Liouville fractional integrals are also obtained.

First of all, general definitions and theorems that are used throughout the article are presented.

**Definition 1.** *Let us consider <sup>f</sup>* <sup>∈</sup> *<sup>L</sup>*1[*a*, *<sup>b</sup>*]. *The Riemann–Liouville integrals <sup>J</sup><sup>α</sup> <sup>a</sup>*<sup>+</sup> *f and J<sup>α</sup> <sup>b</sup>*<sup>−</sup> *<sup>f</sup> of order α* > 0 *with a* ≥ 0 *are defined by*

$$J\_{a+}^{a}f(\mathbf{x}) = \frac{1}{\Gamma(\alpha)} \int\_{a}^{\mathbf{x}} (\mathbf{x} - t)^{a-1} f(t) dt, \ \mathbf{x} > a, \ \mathbf{x}$$

*and*

$$f\_{b-}^{\alpha}f(\mathbf{x}) = \frac{1}{\Gamma(\alpha)} \int\_{\mathbf{x}}^{b} (t-\mathbf{x})^{\alpha-1} f(t)dt, \ \mathbf{x} < b,$$

*respectively. Here,* Γ(*α*) *is the gamma function and J*<sup>0</sup> *<sup>a</sup>*<sup>+</sup> *f*(*x*) = *J*<sup>0</sup> *<sup>b</sup>*<sup>−</sup> *<sup>f</sup>*(*x*) = *<sup>f</sup>*(*x*).

For further information and several properties of Riemann–Liouville fractional integrals, please refer to [33–35].

In [36], Budak et al. prove the following identity for twice-differentiable functions and they also prove corresponding Simpson-type inequalities.

**Lemma 1** ([36])**.** *Let <sup>f</sup>* : [*a*, *<sup>b</sup>*] <sup>→</sup> <sup>R</sup> *be a twice-differentiable mapping* (*a*, *<sup>b</sup>*) *such that <sup>f</sup>* <sup>∈</sup> *L*1([*a*, *b*])*. Then, the following equality holds:*

$$\begin{aligned} &\frac{1}{6}\left[f(a) + 4f\left(\frac{a+b}{2}\right) + f(b)\right] - \frac{2^{a-1}\Gamma(a+1)}{(b-a)^a} \left[l^a\_{\left(\frac{a+b}{2}\right)+} f(b) + l^a\_{\left(\frac{a+b}{2}\right)-} f(a)\right] \\ &= \quad \frac{(b-a)^2}{6} \int\_0^1 w(t)f''(tb + (1-t)a)dt, \end{aligned}$$

*where*

$$w(t) = \begin{cases} \ t \left( 1 - \frac{3 \cdot 2^{\alpha}}{\alpha + 1} t^{\alpha} \right), & t \in \left[ 0, \frac{1}{2} \right], \\\\ (1 - t) \left( 1 - \frac{3 \cdot 2^{\alpha}}{\alpha + 1} (1 - t)^{\alpha} \right), & t \in \left( \frac{1}{2}, 1 \right]. \end{cases}$$

In [37], Hezenci et al. prove another version of the results given in [36].

However, the generalized fractional integrals were introduced by Sarikaya and Ertu ˘gral as follows:

**Definition 2** ([38])**.** *Let us note that a function ϕ* : [0, ∞) → [0, ∞) *satisfies the following condition:*

$$\int\_0^1 \frac{\wp(t)}{t} dt < \infty.$$

*We consider the following left-sided and right-sided generalized fractional integral operators*

$$I\_{a+}I\_{\emptyset}f(\mathbf{x}) = \int\_{a}^{\mathbf{x}} \frac{\varphi(\mathbf{x} - t)}{\mathbf{x} - t} f(t)dt, \quad \mathbf{x} > a,\tag{1}$$

*and*

$$I\_{b-}I\_{\emptyset}f(\mathbf{x}) = \int\_{\mathbf{x}}^{b} \frac{\varphi(t-\mathbf{x})}{t-\mathbf{x}} f(t)dt, \; \mathbf{x} < b,\tag{2}$$

*respectively.*

The most important feature of generalized fractional integrals is that they generalize some types of fractional integrals such as Riemann–Liouville fractional integrals, *k*-Riemann–Liouville fractional integrals, Hadamard fractional integrals, Katugampola fractional integrals, conformable fractional integrals, etc. These significant special cases of the integral operators (1) and (2) are used as follows:


In recent years, several papers have been devoted to obtaining inequalities for generalized fractional integrals; for some of them please refer to [39–45].

Inspired by the ongoing studies, we give the generalized fractional version of the inequalities proved by Budak et al. in [36] for twice-differentiable convex functions. The fundamental benefit of these inequalities is that they can be turned into classical integral inequalities of Simpson's type [32], Riemann–Liouville fractional integral inequalities of Simpson's type [36], and *k*-Riemann–Liouville fractional integral inequalities of Simpson's type without having to prove each one separately.

## **2. Simpson's-Type Inequalities for Twice-Differentiable Functions**

In this section, we prove some new inequalities of Simpson's type for twice-differentiable convex functions via the generalized fractional integrals. For brevity in the rest of the paper, we define

$$A(t) = \int\_0^t T(s)ds,$$

where

$$T(s) = \int\_0^s \frac{\wp((b-a)u)}{u} du.$$

**Lemma 2.** *Let <sup>f</sup>* : [*a*, *<sup>b</sup>*] <sup>→</sup> <sup>R</sup> *be a twice-differentiable mapping* (*a*, *<sup>b</sup>*) *such that <sup>f</sup>* <sup>∈</sup> *<sup>L</sup>*1([*a*, *<sup>b</sup>*])*. Then, the following equality for generalized fractional integrals holds:*

$$\begin{aligned} &\frac{1}{6}\left[f(a) + 4f\left(\frac{a+b}{2}\right) + f(b)\right] - \frac{1}{2T\left(\frac{1}{2}\right)}\left[\left(\frac{a+b}{2}\right) + {}^{I}\_{\Psi}f(b) + \left(\frac{a+b}{2}\right) - {}^{I}\_{\Psi}f(a)\right],\\ &= \quad \frac{(b-a)^2}{6}\int\_0^1 \varpi(t)f''(tb + (1-t)a)dt,\end{aligned}$$

*where*

$$\varpi(t) = \begin{cases} \begin{array}{cc} t - \frac{3A(t)}{T(1/2)}, & t \in \left[0, \frac{1}{2}\right], \\\\ 1 - t - \frac{3A(1-t)}{T(1/2)}, & t \in \left(\frac{1}{2}, 1\right]. \end{array} \end{cases}$$

**Proof.** Using integration by parts, we obtain

$$\begin{split} I\_{1} &= \int\_{0}^{1/2} \left( t - \frac{3A(t)}{T(1/2)} \right) f''(tb + (1-t)a) dt \\ &= \left. \left( t - \frac{3A(t)}{T(1/2)} \right) \frac{f'(tb + (1-t)a)}{b-a} \right|\_{0}^{\frac{1}{2}} \\ &\quad - \frac{1}{b-a} \int\_{0}^{1/2} \left( 1 - \frac{3T(t)}{T(1/2)} \right) f'(tb + (1-t)a) dt \\ &= \left. \frac{1}{b-a} \left( \frac{1}{2} - \frac{3A(1/2)}{T(1/2)} \right) f'\left(\frac{a+b}{2}\right) \right. \\ &\quad - \frac{1}{b-a} \left[ \left( 1 - \frac{3T(t)}{T(1/2)} \right) \frac{f(tb + (1-t)a)}{(b-a)} \right]\_{0}^{1/2} \\ &\quad + \frac{1}{b-a} \int\_{0}^{1/2} \frac{3}{T(1/2)} \frac{g((b-a)t)}{t} f(tb + (1-t)) dt \right] \\ &= \left. \frac{1}{b-a} \left( \frac{1}{2} - \frac{3A(1/2)}{T(1/2)} \right) f'\left(\frac{a+b}{2}\right) + \frac{2}{(b-a)^{2}} f\left(\frac{a+b}{2}\right) \right. \\ &\left. \frac{f(a)}{(b-a)^{2}} + \frac{3}{T(1/2)(b-a)^{2}} \left( \frac{a+b}{2} \right) - l\_{\Phi}f(a). \end{split}$$

Similarly, we have

$$\begin{aligned} I\_2 &= \int\_{1/2}^1 \left( 1 - t - \frac{3A(1-t)}{T(1/2)} \right) f''(tb + (1-t)a) dt \\ &= \quad - \frac{1}{b-a} \left( \frac{1}{2} - \frac{3A(1/2)}{T(1/2)} \right) f'\left(\frac{a+b}{2}\right) + \frac{2}{(b-a)^2} f\left(\frac{a+b}{2}\right) \\ &+ \frac{f(b)}{(b-a)^2} + \frac{3}{T(1/2)(b-a)^2} \left( \frac{a+b}{2} \right) + I\_\Phi f(b). \end{aligned}$$

If *<sup>I</sup>*<sup>1</sup> and *<sup>I</sup>*<sup>2</sup> are added and then multiplied by (*b*−*a*) 2 <sup>6</sup> , the desired result is obtained.

**Remark 1.** *If we take ϕ*(*t*) = *t in Lemma 2, then Lemma 2 reduces to [32] (Lemma 2.1).*

**Remark 2.** *Let us note that ϕ*(*t*) = *<sup>t</sup> α* <sup>Γ</sup>(*α*)*, α* > 0 *in Lemma 2, then Lemma 2 reduces to Lemma 1.*

**Corollary 1.** *If we choose ϕ*(*t*) = *<sup>t</sup> α k <sup>k</sup>*Γ*<sup>k</sup>* (*α*)*, <sup>α</sup>*, *<sup>k</sup>* <sup>&</sup>gt; <sup>0</sup> *in Lemma 2, then the following equality for k-Riemann–Liouville fractional integrals holds:*

$$\begin{aligned} &\frac{1}{6}\left[f(a) + 4f\left(\frac{a+b}{2}\right) + f(b)\right] - \frac{2^{\frac{a-b}{k}}\Gamma(a+k)}{(b-a)^{\frac{a}{k}}} \left[J\_{\left(\frac{a+b}{2}\right)+k}^{\alpha}f(b) + J\_{\left(\frac{a+b}{2}\right)-k}^{\alpha}f(a)\right],\\ &= \quad \frac{(b-a)^2}{6} \int\_0^1 m(t)f''(tb + (1-t)a)dt,\end{aligned}$$

*where*

$$m(t) = \begin{cases} \begin{array}{cc} t\left(1 - \frac{3k \cdot 2^{\frac{k}{2}}}{a+k} t^{\frac{a}{k}}\right), & t \in \left[0, \frac{1}{2}\right], \\\\ (1-t)\left(1 - \frac{3k \cdot 2^{\frac{k}{2}}}{a+k} (1-t)^{\frac{k}{2}}\right), & t \in \left(\frac{1}{2}, 1\right]. \end{array} \end{cases}$$

**Proof.** For *ϕ*(*τ*) = *<sup>τ</sup> α k <sup>k</sup>*Γ*<sup>k</sup>* (*α*), we have

$$\Lambda(\mathbf{s}) = \frac{(\kappa\_2 - \kappa\_1)^{\frac{\mathbf{f}}{\mathbf{f}}}}{\alpha \Gamma\_k(\alpha)} \mathbf{s}^{\frac{\mathbf{f}}{\mathbf{f}}} = \frac{(\kappa\_2 - \kappa\_1)^{\frac{\mathbf{f}}{\mathbf{f}}}}{\Gamma\_k(\alpha + k)} \mathbf{s}^{\frac{\mathbf{f}}{\mathbf{f}}} \tag{3}$$

$$\Lambda(1/2) = \frac{(\kappa\_2 - \kappa\_1)^{\frac{\theta}{k}}}{2^{\frac{\theta}{k}} \Gamma\_k(a+k)}\tag{4}$$

and

$$\Delta(\tau) = \frac{(\kappa\_2 - \kappa\_1)^{\frac{\mathfrak{g}}{k}}}{\Gamma\_k(\mathfrak{a} + k)} \int\_0^\tau s^{\frac{\mathfrak{g}}{k}} ds = \frac{k(\kappa\_2 - \kappa\_1)^{\frac{\mathfrak{g}}{k}}}{(\mathfrak{a} + k)\Gamma\_k(\mathfrak{a} + k)} \tau^{\frac{\mathfrak{g}}{k} + 1}. \tag{5}$$

Then it follows that

$$
\phi(\tau) = m(\tau).
$$

which completes the proof.

**Theorem 2.** *Assume that the assumptions of Lemma 2 hold. Assume also that the mapping* | *f* | *is convex on* [*a*, *b*]. *Then, we have the following Simpson-type inequality for generalized fractional integrals*

$$\begin{aligned} &\left|\frac{1}{6}\left[f(a) + 4f\left(\frac{a+b}{2}\right) + f(b)\right] - \frac{1}{2T\left(\frac{1}{2}\right)}\left[\left(\frac{a+b}{2}\right) + \left.I\_{\mathcal{G}}f(b) + \left.(\frac{a+b}{2}) - I\_{\mathcal{G}}f(a)\right|\right]\right| \\ &\leq \quad \frac{(b-a)^2}{6}\left(\int\_0^{\frac{1}{2}} \left|t - \frac{3A(t)}{T\left(1/2\right)}\right| dt\right) \left[\left|f''(a)\right| + \left|f''(b)\right|\right]. \end{aligned}$$

**Proof.** By taking the modulus in Lemma 2, we have

$$\begin{aligned} & \left| \frac{1}{6} \left[ f(a) + 4f\left(\frac{a+b}{2}\right) + f(b) \right] - \frac{1}{2T\left(\frac{1}{2}\right)} \left[ \left(\frac{a+b}{2}\right) + I\_{\theta}f(b) + \left(\frac{a+b}{2}\right) - I\_{\theta}f(a) \right] \right| \\ & \le \underbrace{\left( \frac{(b-a)^2}{6} \right)^2}\_{0} \int |\phi(t)| |f''(tb + (1-t)a)| dt \\ &= \underbrace{\left(\frac{(b-a)^2}{6}\right)^2}\_{0} \left[ \left| \frac{\frac{1}{2}}{\int 0 \right| t - \frac{3A(t)}{T(1/2)} \left| \left| f''(tb + (1-t)a) \right| dt} \right. \\ & \left. + \int \left| 1 - t - \frac{3A(1-t)}{T(1/2)} \right| \left| f''(tb + (1-t)a) \right| dt \right]. \end{aligned} (6)$$

With the help of the convexity of | *f* |, we obtain

$$\left| \frac{1}{6} \left[ f(a) + 4f\left(\frac{a+b}{2}\right) + f(b) \right] - \frac{1}{2T\left(\frac{1}{2}\right)} \left[ \left(\frac{a+b}{2}\right) + \left. I\_{\mathbb{Q}}f(b) + \left. \left(\frac{a+b}{2}\right) - I\_{\mathbb{Q}}f(a) \right| \right] \right| $$

<sup>≤</sup> (*<sup>b</sup>* <sup>−</sup> *<sup>a</sup>*) 2 6 ⎡ ⎢ ⎣ 1 2 0 *<sup>t</sup>* <sup>−</sup> <sup>3</sup>*A*(*t*) *T*(1/2) + *t f* (*b*) + (1 − *t*) *f* (*a*) , *dt* + 1 1 2 <sup>1</sup> <sup>−</sup> *<sup>t</sup>* <sup>−</sup> <sup>3</sup>*A*(<sup>1</sup> <sup>−</sup> *<sup>t</sup>*) *T*(1/2) + *t f* (*b*) + (1 − *t*) *f* (*a*) , *dt* ⎤ ⎥ ⎦ <sup>=</sup> (*<sup>b</sup>* <sup>−</sup> *<sup>a</sup>*) 2 6 ⎧ ⎪⎨ ⎪⎩ ⎡ ⎢ ⎣ 1 2 0 *t <sup>t</sup>* <sup>−</sup> <sup>3</sup>*A*(*t*) *T*(1/2) *dt* + 1 1 2 *t* <sup>1</sup> <sup>−</sup> *<sup>t</sup>* <sup>−</sup> <sup>3</sup>*A*(<sup>1</sup> <sup>−</sup> *<sup>t</sup>*) *T*(1/2) *dt* ⎤ ⎥ ⎦ *f* (*b*) + ⎡ ⎢ ⎣ 1 2 0 (1 − *t*) *<sup>t</sup>* <sup>−</sup> <sup>3</sup>*A*(*t*) *T*(1/2) *dt* + 1 1 2 (1 − *t*) <sup>1</sup> <sup>−</sup> *<sup>t</sup>* <sup>−</sup> <sup>3</sup>*A*(<sup>1</sup> <sup>−</sup> *<sup>t</sup>*) *T*(1/2) *dt* ⎤ ⎥ ⎦ *f* (*a*) ⎫ ⎪⎬ ⎪⎭ <sup>=</sup> (*<sup>b</sup>* <sup>−</sup> *<sup>a</sup>*) 2 6 ⎡ ⎢ ⎣ 1 2 0 *t <sup>t</sup>* <sup>−</sup> <sup>3</sup>*A*(*t*) *T*(1/2) *dt* + 1 1 2 *t* <sup>1</sup> <sup>−</sup> *<sup>t</sup>* <sup>−</sup> <sup>3</sup>*A*(<sup>1</sup> <sup>−</sup> *<sup>t</sup>*) *T*(1/2) *dt* ⎤ ⎥ ⎦ + *f* (*a*) <sup>+</sup> *f* (*b*) , <sup>=</sup> (*<sup>b</sup>* <sup>−</sup> *<sup>a</sup>*) 2 6 ⎛ ⎜⎝ 1 2 0 *<sup>t</sup>* <sup>−</sup> <sup>3</sup>*A*(*t*) *T*(1/2) *dt* ⎞ ⎟⎠+ *f* (*a*) <sup>+</sup> *f* (*b*) , .

This completes the proof of Theorem 2.

**Remark 3.** *Consider ϕ*(*t*) = *t in Theorem 2, then Theorem 2 reduces to [32] (Theorem 2.2).*

**Remark 4.** *If we assign ϕ*(*t*) = *<sup>t</sup> α* <sup>Γ</sup>(*α*), *α* > 0 *in Theorem 2, then we obtain the following Simpsontype inequality for Riemann–Liouville fractional integrals*

$$\begin{aligned} &\left|\frac{1}{6}\left[f(a)+4f\left(\frac{a+b}{2}\right)+f(b)\right]-\frac{2^{a-1}\Gamma(a+1)}{(b-a)^a}\left[l^a\_{\left(\frac{a+b}{2}\right)+}f(b)+l^a\_{\left(\frac{a+b}{2}\right)-}f(a)\right]\right|\\ &\leq \quad \frac{(b-a)^2}{6}\Theta(a)\left[|f'(a)|+|f''(b)|\right].\end{aligned}$$

*Here,*

$$\Theta(\alpha) = \frac{1}{4(\alpha+2)} \left( \alpha \left( \frac{\alpha+1}{3} \right)^{\frac{2}{\alpha}} + \frac{3}{\alpha+1} \right) - \frac{1}{8'} \tag{7}$$

*which is given by Budak et al. in [36].*

**Corollary 2.** *For ϕ*(*t*) = *<sup>t</sup> α k <sup>k</sup>*Γ*<sup>k</sup>* (*α*), *<sup>k</sup>*, *<sup>α</sup>* <sup>&</sup>gt; <sup>0</sup> *in Theorem 2, we have the following Simpson-type inequality for k-Riemann–Liouville fractional integrals*

$$\begin{aligned} &\left|\frac{1}{6}\left[f(a)+4f\left(\frac{a+b}{2}\right)+f(b)\right]-\frac{2^{\frac{a-k}{k}}\Gamma(a+k)}{(b-a)^{\frac{0}{k}}}\left[l^{\frac{a+k}{\left(\frac{a+k}{2}\right)}+k}\_{\left(\frac{a+k}{2}\right)}f(b)+l^{\frac{a}{\left(\frac{a+k}{2}\right)}-k}\_{\left(\frac{a+k}{2}\right)}f(a)\right]\right|\\ &\leq \quad\frac{(b-a)^{2}}{6}\Theta(a,k)\left[|f^{\prime\prime}(a)|+|f^{\prime\prime}(b)|\right],\end{aligned}$$

*where*

$$\Theta(\alpha, k) = \frac{k}{4(\alpha + 2k)} \left( \frac{a}{k} \left( \frac{a+k}{3k} \right)^{\frac{2k}{a}} + \frac{3k}{a+k} \right) - \frac{1}{8}. \tag{8}$$

**Proof.** Let *ϕ*(*τ*) = *<sup>τ</sup> α k <sup>k</sup>*Γ*<sup>k</sup>* (*α*). By the equalities (3)–(5), we have

$$\begin{aligned} \left. \int\_0^{\frac{1}{2}} \right| \tau - \frac{3\Delta(\tau)}{\Lambda(1/2)} \Big| d\tau &=& \int\_0^{\frac{1}{2}} \Big| \tau - \frac{3k \cdot 2^{\frac{k}{2}}}{a+k} \tau^{\frac{a}{k}} \Big| d\tau \\ &=& \frac{k}{4(a+2k)} \left( \frac{a}{k} \left( \frac{a+k}{3k} \right)^{\frac{2k}{a}} + \frac{3k}{a+k} \right) - \frac{1}{8} . \end{aligned}$$

This completes the proof.

**Theorem 3.** *Suppose that the assumptions of Lemma 2 hold. Suppose also that the mapping* | *f* | *q , q* > 1, *is convex on* [*a*, *b*]. *Then, the following Simpson-type inequality for generalized fractional integrals*

$$\begin{split} & \left| \frac{1}{6} \left[ f(a) + 4f\left(\frac{a+b}{2}\right) + f(b) \right] - \frac{1}{2T\left(\frac{1}{2}\right)} \left[ \left( \left(\frac{a+b}{2}\right) + \left. I\_{\mathbb{P}}f(b) + \left(\frac{a+b}{2}\right) - I\_{\mathbb{P}}f(a) \right| \right] \right| \\ & \leq \quad \frac{(b-a)^{2}}{6} \left( \int\_{0}^{\frac{1}{2}} \left| t - \frac{3A(t)}{T(1/2)} \right|^{p} dt \right)^{\frac{1}{p}} \\ & \qquad \times \left[ \left( \frac{|f''(b)|^{q} + 3|f''(a)|^{q}}{8} \right)^{\frac{1}{q}} + \left( \frac{3|f''(b)|^{q} + |f''(a)|^{q}}{8} \right)^{\frac{1}{q}} \right] \end{split}$$

*is valid. Here,* <sup>1</sup> *<sup>p</sup>* <sup>+</sup> <sup>1</sup> *<sup>q</sup>* = 1*.*

**Proof.** By applying the Hölder inequality in inequality (6), we obtain

$$\begin{split} & \left| \frac{1}{6} \left[ f(a) + 4f\left(\frac{a+b}{2}\right) + f(b) \right] - \frac{1}{2T\left(\frac{1}{2}\right)} \left[ \left(\frac{a+b}{2}\right) + I\_{q}f(b) + \left(\frac{a+b}{2}\right) - I\_{q}f(a) \right] \right| \\ & \leq \quad \frac{(b-a)^{2}}{6} \left[ \left( \int \left| t - \frac{3A(t)}{T(1/2)} \right|^{p} dt \right)^{\frac{1}{p}} \left( \int \left| f'(tb + (1-t)a) \right|^{q} dt \right)^{\frac{1}{q}} \\ & \quad + \left( \int \left| 1 - t - \frac{3A(1-t)}{T(1/2)} \right|^{p} dt \right)^{\frac{1}{p}} \left( \int \left| f'(tb + (1-t)a) \right|^{q} dt \right)^{\frac{1}{q}} \right]. \end{split}$$

By using the convexity of | *f* | *q* , we obtain

$$\begin{aligned} &\left|\frac{1}{6}\left[f(a) + 4f\left(\frac{a+b}{2}\right) + f(b)\right] - \frac{1}{2T\left(\frac{1}{2}\right)} \left[\left(\frac{a+b}{2}\right) + {}\_{\!0}I\_{\!p}f(b) + \left(\frac{a+b}{2}\right) - {}\_{\!0}I\_{\!p}f(a)\right]\right| \\ &\leq \quad \frac{(b-a)^2}{6} \left(\int\_0^{\frac{1}{2}} \left|t - \frac{3A(t)}{T(1/2)}\right|^p dt\right)^{\frac{1}{p}} \end{aligned}$$

$$\begin{split} & \quad \times \left[ \left( \int\_{0}^{\frac{1}{2}} \left[ t \left| f''(b) \right|^{q} + (1-t) \left| f''(a) \right|^{q} \right] dt \right)^{\frac{1}{q}} + \left( \int\_{\frac{1}{2}}^{\frac{1}{2}} \left[ t \left| f''(b) \right|^{q} + (1-t) \left| f''(a) \right|^{q} \right] dt \right)^{\frac{1}{q}} \right] \\ &= \quad \cdot \frac{(b-a)^{2}}{6} \left( \int\_{0}^{\frac{1}{2}} \left| t - \frac{3A(t)}{T(1/2)} \right|^{p} dt \right)^{\frac{1}{p}} \\ & \quad \times \left[ \left( \frac{\left| f'''(b) \right|^{q} + 3\left| f''(a) \right|^{q}}{8} \right)^{\frac{1}{q}} + \left( \frac{3\left| f''(b) \right|^{q} + \left| f''(a) \right|^{q}}{8} \right)^{\frac{1}{q}} \right]. \end{split}$$

This finishes the proof of Theorem 3.

**Remark 5.** *If we choose ϕ*(*t*) = *t in Theorem 3, then we obtain*

$$\begin{aligned} &\left|\frac{1}{6}\left[f(a)+4f\left(\frac{a+b}{2}\right)+f(b)\right]-\frac{1}{(b-a)}\int\_{a}^{b}f(\mathbf{x})d\mathbf{x}\right|\\ &\leq \quad \frac{(b-a)^{2}}{6}\left(\int\_{0}^{\frac{1}{2}}|f^{p}|1-3t|^{p}dt\right)^{\frac{1}{p}}\\ &\times\left[\left(\frac{|f''(b)|^{q}+3|f''(a)|^{q}}{8}\right)^{\frac{1}{q}}+\left(\frac{3|f''(b)|^{q}+|f''(a)|^{q}}{8}\right)^{\frac{1}{q}}\right].\end{aligned}$$

*which is given by Budak et al. in [36].*

**Remark 6.** *Let us consider ϕ*(*t*) = *<sup>t</sup> α* <sup>Γ</sup>(*α*), *α* > 0 *in Theorem 3, then the Simpson-type inequality for Riemann–Liouville fractional integrals*

$$\begin{aligned} &\left|\frac{1}{6}\left[f(a)+4f\left(\frac{a+b}{2}\right)+f(b)\right]-\frac{2^{a-1}\Gamma(a+1)}{(b-a)^{a}}\left[f\_{\left(\frac{a+b}{2}\right)+}^{\mathfrak{a}}f(b)+f\_{\left(\frac{a+b}{2}\right)-}^{\mathfrak{a}}f(a)\right]\right|\\ &\leq \quad \frac{(b-a)^{2}}{6}\mathcal{Y}(a,p)\left[\left(\frac{|f''(b)|^{q}+3|f''(a)|^{q}}{8}\right)^{\frac{1}{q}}+\left(\frac{3|f''(b)|^{q}+|f''(a)|^{q}}{8}\right)^{\frac{1}{q}}\right] \end{aligned}$$

*is valid. Here,* <sup>1</sup> *<sup>p</sup>* <sup>+</sup> <sup>1</sup> *<sup>q</sup>* = 1 *and*

$$\Upsilon(\mathfrak{a}, p) = \left( \int\_0^{\frac{1}{2}} t^p \left| 1 - \frac{3 \cdot 2^{\alpha}}{\alpha + 1} t^{\alpha} \right|^p dt \right)^{\frac{1}{p}}.$$

*which is given by Budak et al. in [36].*

**Corollary 3.** *If we choose ϕ*(*t*) = *<sup>t</sup> α k <sup>k</sup>*Γ*<sup>k</sup>* (*α*), *<sup>α</sup>*, *<sup>k</sup>* <sup>&</sup>gt; <sup>0</sup> *in Theorem 3, then we have the following Simpson-type inequality for k-Riemann–Liouville fractional integrals*

$$\begin{split} & \left| \frac{1}{6} \left[ f(a) + 4f\left(\frac{a+b}{2}\right) + f(b) \right] - \frac{2\frac{a-k}{k}\Gamma(a+k)}{(b-a)^{\frac{a}{k}}} \left[ l^{\frac{a}{\left(\frac{a+b}{2}\right)+k}}\_{\left(\frac{a+b}{2}\right)+k} f(b) + l^{\frac{a}{\left(\frac{a+b}{2}\right)-k}}\_{\left(\frac{a+b}{2}\right)} f(a) \right] \right| \\ & \leq \quad \frac{(b-a)^{2}}{6} \mathcal{Y}(a,p,k) \left[ \left( \frac{|f''(b)|^{q} + 3|f''(a)|^{q}}{8} \right)^{\frac{1}{q}} + \left( \frac{3|f''(b)|^{q} + |f''(a)|^{q}}{8} \right)^{\frac{1}{q}} \right]. \end{split}$$

*Here,* <sup>1</sup> *<sup>p</sup>* <sup>+</sup> <sup>1</sup> *<sup>q</sup>* = 1 *and*

$$\Upsilon(\mathfrak{a}, p, k) = \left( \int\_0^{\frac{1}{2}} t^p \left| 1 - \frac{3k \cdot 2^{\frac{\mathfrak{a}}{k}}}{\mathfrak{a} + k} t^{\frac{\mathfrak{b}}{k}} \right|^p dt \right)^{\frac{1}{p}}.$$

**Proof.** For *ϕ*(*τ*) = *<sup>τ</sup> α k <sup>k</sup>*Γ*<sup>k</sup>* (*α*), the proof can be seen easily by the equalities (3)–(5).

**Theorem 4.** *Assume that the assumptions of Lemma 2 hold. If the mapping* | *f* | *q , q* ≥ 1 *is convex on* [*a*, *b*]*, then we have the following Simpson-type inequality for generalized fractional integrals*

$$\begin{split} & \left| \frac{1}{6} \left[ f(a) + 4f\left(\frac{a+b}{2}\right) + f(b) \right] - \frac{1}{2T\left(\frac{1}{2}\right)} \left[ \left(\frac{a+b}{2}\right) + \ell\_{\theta}f(b) + \left(\frac{a+b}{2}\right) - l\_{\theta}f(a) \right] \right| \\ & \leq \quad \frac{(b-a)^{2}}{6} \left( \int\_{0}^{\frac{1}{2}} \left| t - \frac{3A(t)}{T(1/2)} \right| dt \right)^{1-\frac{1}{6}} \\ & \quad \times \left\{ \left[ \left(\int\_{0}^{\frac{1}{2}} t \left| t - \frac{3A(t)}{T(1/2)} \right| dt \right) \left| f''(b) \right|^{q} + \left(\int\_{0}^{\frac{1}{2}} (1-t) \left| t - \frac{3A(t)}{T(1/2)} \right| dt \right) \left| f''(a) \right|^{\frac{1}{6}} \right. \\ & \left. + \left[ \left(\int\_{0}^{\frac{1}{2}} (1-t) \left| t - \frac{3A(t)}{T(1/2)} \right| dt \right) \left| f''(b) \right|^{q} + \left(\int\_{0}^{\frac{1}{2}} \left| t - \frac{3A(t)}{T(1/2)} \right| dt \right) \left| f''(a) \right|^{\frac{1}{6}} \right] \right\}. \end{split}$$

**Proof.** By applying the power-mean inequality in (6), we obtain

$$\begin{split} & \left| \frac{1}{6} \left[ f(a) + 4f\left(\frac{a+b}{2}\right) + f(b) \right] - \frac{1}{2T\left(\frac{1}{2}\right)} \left[ \left(\frac{a+b}{2}\right) + I\_q f(b) + \left(\frac{a+b}{2}\right) - I\_q f(a) \right] \right| \\ & \leq \quad \frac{(b-a)^2}{6} \left[ \left( \int\_0^{\frac{1}{2}} \left| t - \frac{3A(t)}{T(1/2)} \right| dt \right)^{1 - \frac{1}{q}} \\ & \quad \times \left( \sum\_{0}^{\frac{1}{2}} \left| t - \frac{3A(t)}{T(1/2)} \right| \left| f''(tb + (1-t)a) \right|^q dt \right)^{\frac{1}{q}} \\ & \quad + \left( \int\_0^{\frac{1}{2}} \left| 1 - t - \frac{3A(1-t)}{T(1/2)} \right| dt \right)^{1 - \frac{1}{q}} \\ & \quad \times \left( \int\_{\frac{1}{2}}^1 \left| 1 - t - \frac{3A(1-t)}{T(1/2)} \right| \left| f''(tb + (1-t)a) \right|^q dt \right)^{\frac{1}{q}} \end{split}$$

Since | *f* | *<sup>q</sup>* is convex, we obtain

$$\int\_0^{\frac{1}{2}} \left| t - \frac{3A(t)}{T(1/2)} \right| \left| f''(tb + (1-t)a) \right|^q dt \tag{10}$$

$$\leq \quad \int\_0^{\frac{1}{2}} \left| t - \frac{3A(t)}{T(1/2)} \right| \left[ t \left| f''(b) \right|^q + (1-t) \left| f''(a) \right|^q \right] dt$$

$$= \quad \left| f''(b) \right|^q \int\_0^{\frac{1}{2}} t \left| t - \frac{3A(t)}{T(1/2)} \right| dt + \left| f''(a) \right|^q \int\_0^{\frac{1}{2}} (1-t) \left| t - \frac{3A(t)}{T(1/2)} \right| dt$$

and similarly

$$\begin{split} \int\_{\frac{1}{2}}^{\frac{1}{2}} \left| 1 - t - \frac{3A(1-t)}{T(1/2)} \right| \left| f''(tb + (1-t)a) \right|^q dt \\ \leq \quad \left| f'''(b) \right|^q \int\_{\frac{1}{2}}^{\frac{1}{2}} t \left| 1 - t - \frac{3A(1-t)}{T(1/2)} \right| dt + \left| f''(a) \right|^q \int\_{\frac{1}{2}}^{1} (1-t) \left| 1 - t - \frac{3A(1-t)}{T(1/2)} \right| dt \\ = \quad \left| f''(b) \right|^q \int\_{0}^{\frac{1}{2}} (1-t) \left| t - \frac{3A(t)}{T(1/2)} \right| dt + \left| f''(a) \right|^q \int\_{0}^{t} t \left| t - \frac{3A(t)}{T(1/2)} \right| dt. \end{split} \tag{11}$$

If we substitute the inequalities (10) and (11) in (9), then we obtain the desired result.

**Remark 7.** *Consider ϕ*(*t*) = *t in Theorem 4, then Theorem 4 reduces to [32] (Theorem 2.5).*

**Remark 8.** *If we take ϕ*(*t*) = *<sup>t</sup> α* <sup>Γ</sup>(*α*), *α* > 0 *in Theorem 4, then we obtain the following Simpson-type inequality for Riemann–Liouville fractional integrals*

$$\begin{split} & \quad \left| \frac{1}{6} \Big[ f(a) + 4f\left(\frac{a+b}{2}\right) + f(b) \Big] - \frac{2^{a-1}\Gamma(a+1)}{(b-a)^a} \Big[ f^{\mu}\_{\left(\frac{a+b}{2}\right)} + f(b) + f^{\mu}\_{\left(\frac{a+b}{2}\right)} \cdot f(a) \Big] \right| \\ & \leq \quad \frac{(b-a)^{2}}{6} (\Theta(a))^{1-\frac{1}{q}} \Big[ \Big( \Xi(a) |f''(b)|^{q} + \Omega(a) |f''(a)|^{q} \Big)^{\frac{1}{q}} + \Big( \Omega(a) |f''(b)|^{q} + \Xi(a) |f''(a)|^{q} \Big)^{\frac{1}{q}} \Big]. \end{split}$$

*Here,* Θ(*α*) *is defined as in (7) and*

$$\begin{aligned} \Xi(\alpha) &= \begin{array}{c} 1 \\ \frac{\alpha}{4(\alpha+3)} \left[ \frac{\alpha}{3} \left( \frac{\alpha+1}{3} \right)^{\frac{3}{4}} + \frac{3}{2(\alpha+1)} \right] - \frac{1}{24}, \\ \Omega(\alpha) &= \begin{array}{c} \Theta(\alpha) - \Xi(\alpha) \end{array} \\ &= \frac{1}{4(\alpha+2)} \left( \alpha \left( \frac{\alpha+1}{3} \right)^{\frac{2}{4}} + \frac{3}{\alpha+1} \right) \\ &- \frac{1}{4(\alpha+3)} \left[ \frac{\alpha}{3} \left( \frac{\alpha+1}{3} \right)^{\frac{3}{4}} + \frac{3}{2(\alpha+1)} \right] - \frac{1}{12}, \end{array} \end{aligned}$$

*which is given by Budak et al. in [36].*

**Corollary 4.** *Let us consider ϕ*(*t*) = *<sup>t</sup> α k <sup>k</sup>*Γ*<sup>k</sup>* (*α*), *<sup>α</sup>*, *<sup>k</sup>* <sup>&</sup>gt; <sup>0</sup> *in Theorem 4, then the following Simpsontype inequality for k-Riemann–Liouville fractional integrals holds:*

$$\begin{split} & \left| \frac{1}{6} \Big[ f(a) + 4f\left(\frac{a+b}{2}\right) + f(b) \Big] - \frac{2^{\frac{a-b}{k}} \Gamma(a+k)}{(b-a)^{\frac{\pi}{k}}} \Big[ f^{\left(\frac{a+b}{2}\right) +,k}\_{\left(\frac{a+b}{2}\right) +,k} f(b) + f^{\left(\frac{a+b}{2}\right) -,k}\_{\left(\frac{a+b}{2}\right) -,k} f(a) \Big] \right| \\ & \leq \ \ \frac{(b-a)^{2}}{6} (\Theta(a,k))^{1-\frac{1}{\theta}} \Big[ \Big( \Xi(a,k) |f''(b)|^{q} + \Omega(a,k) |f''(a)|^{q} \Big)^{\frac{1}{\theta}} + \Big( \Omega(a,k) |f^{\prime\prime}(b)|^{q} + \Xi(a,k) |f^{\prime\prime}(a)|^{q} \Big)^{\frac{1}{\theta}} .\end{split}$$

*where* Θ(*α*, *k*) *is defined as in (8) and*

$$\begin{split} \Xi(\alpha,k) &=& \frac{k}{4(\alpha+3k)} \left[ \frac{\alpha}{3k} \left( \frac{\alpha+k}{3k} \right)^{\frac{3k}{\alpha}} + \frac{3k}{2(\alpha+k)} \right] - \frac{1}{24}, \\ \Omega(\alpha,k) &=& \Theta(\alpha,k) - \Xi(\alpha,k) \\ &=& \frac{k}{4(\alpha+2k)} \left( \frac{\alpha}{k} \left( \frac{\alpha+k}{3k} \right)^{\frac{3k}{\alpha}} + \frac{3k}{\alpha+k} \right) \\ &- \frac{k}{4(\alpha+3k)} \left[ \frac{\alpha}{3k} \left( \frac{\alpha+k}{3k} \right)^{\frac{3k}{\alpha}} + \frac{3k}{2(\alpha+k)} \right] - \frac{1}{12}. \end{split}$$

**Proof.** Let *ϕ*(*τ*) = *<sup>τ</sup> α k <sup>k</sup>*Γ*<sup>k</sup>* (*α*). By the equalities (3)–(5), we have

$$\begin{aligned} &\int\_0^{\frac{1}{2}} \tau \left| \tau - \frac{3\Lambda(\tau)}{\Lambda(1/2)} \right| d\tau \\ &= \quad \int\_0^{\frac{1}{2}} \tau \left| \tau - \frac{3k \cdot 2^{\frac{a}{\frac{a}{\pi}}} \tau^{\frac{a}{\frac{a}{\pi}}} \right| d\tau \\ &= \quad \frac{k}{4(a+3k)} \left[ \frac{a}{3k} \left( \frac{a+k}{3k} \right)^{\frac{3k}{a}} + \frac{3k}{2(a+k)} \right] - \frac{1}{24} \end{aligned}$$

and

$$\begin{aligned} &\int\_0^{\frac{1}{2}} (1-\tau) \left| \tau - \frac{3\Lambda(\tau)}{\Lambda(1/2)} \right| d\tau \\ &= \quad \int\_0^{\frac{1}{2}} (1-\tau) \left| \tau - \frac{3k \cdot 2^{\frac{k}{2}}}{\alpha+k} \tau^{\frac{k}{k}} \right| d\tau \\ &= \quad \frac{k}{4(\alpha+2k)} \left( \frac{\alpha}{k} \left( \frac{\alpha+k}{3k} \right)^{\frac{2k}{\alpha}} + \frac{3k}{\alpha+k} \right) \\ &\quad - \frac{k}{4(\alpha+3k)} \left[ \frac{\alpha}{3k} \left( \frac{\alpha+k}{3k} \right)^{\frac{3k}{\alpha}} + \frac{3k}{2(\alpha+k)} \right] - \frac{1}{12} . \end{aligned}$$

## **3. Conclusions**

For twice-differentiable functions, we have developed a generalized fractional version of the Simpson-type inequality in this paper. After that, we explained how our findings generalize a number of inequalities found in previous research. For *k*-Riemann–Liouville fractional integrals, we additionally provided novel Simpson-type inequalities. The findings of this study can be utilized in symmetry. The results for the case of symmetric convex functions can be obtained in future studies. In future studies, researchers can obtain generalized versions of our results by utilizing other kinds of convex function classes or different types of generalized fractional integral operators.

**Author Contributions:** All authors contributed equally in the preparation of the present work. Theorems and corollaries: M.A.A., H.K., J.T., S.A., H.B. and F.H.; review of the articles and books cited: M.A.A., H.K., J.T., S.A., H.B. and F.H.; formal analysis: M.A.A., H.K., J.T., S.A., H.B. and F.H.; writing—original draft preparation and writing—review and editing: M.A.A., H.K., J.T., S.A., H.B. and F.H. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research was funded by King Mongkut's University of Technology, North Bangkok. Contract no. KMUTNB-62-KNOW-29.

**Institutional Review Board Statement:** Not applicable.

**Informed Consent Statement:** Not applicable.

**Data Availability Statement:** Not applicable.

**Acknowledgments:** We thank the referees for their valuable comments.

**Conflicts of Interest:** The authors declare no conflict of interest.

## **References**

