Neural architecture search via standard machine learning methodologies

Giorgia Franchini; Valeria Ruggiero; Federica Porta; Luca Zanni; Giorgia Franchini; Valeria Ruggiero; Federica Porta; Luca Zanni

doi:10.3934/mine.2023012

Mathematics in Engineering

2023, Volume 5, Issue 1: 1-21. doi: 10.3934/mine.2023012

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Neural architecture search via standard machine learning methodologies

1.
Department of Mathematics and Computer Science, University of Ferrara, Via Machiavelli, 30, 44121 Ferrara (FE), Italy
2.
Department of Physics, Informatics and Mathematics, University of Modena and Reggio Emilia, Via Campi 213/B, 41121 Modena (MO), Italy

Received: 24 March 2021 Revised: 19 October 2021 Accepted: 13 January 2022 Published: 11 February 2022

In the context of deep learning, the more expensive computational phase is the full training of the learning methodology. Indeed, its effectiveness depends on the choice of proper values for the so-called hyperparameters, namely the parameters that are not trained during the learning process, and such a selection typically requires an extensive numerical investigation with the execution of a significant number of experimental trials. The aim of the paper is to investigate how to choose the hyperparameters related to both the architecture of a Convolutional Neural Network (CNN), such as the number of filters and the kernel size at each convolutional layer, and the optimisation algorithm employed to train the CNN itself, such as the steplength, the mini-batch size and the potential adoption of variance reduction techniques. The main contribution of the paper consists in introducing an automatic Machine Learning technique to set these hyperparameters in such a way that a measure of the CNN performance can be optimised. In particular, given a set of values for the hyperparameters, we propose a low-cost strategy to predict the performance of the corresponding CNN, based on its behavior after only few steps of the training process. To achieve this goal, we generate a dataset whose input samples are provided by a limited number of hyperparameter configurations together with the corresponding CNN measures of performance obtained with only few steps of the CNN training process, while the label of each input sample is the performance corresponding to a complete training of the CNN. Such dataset is used as training set for a Support Vector Machines for Regression and/or Random Forest techniques to predict the performance of the considered learning methodology, given its performance at the initial iterations of its learning process. Furthermore, by a probabilistic exploration of the hyperparameter space, we are able to find, at a quite low cost, the setting of a CNN hyperparameters which provides the optimal performance. The results of an extensive numerical experimentation, carried out on CNNs, together with the use of our performance predictor with NAS-Bench-101, highlight how the proposed methodology for the hyperparameter setting appears very promising.

Keywords:

Citation: Giorgia Franchini, Valeria Ruggiero, Federica Porta, Luca Zanni. Neural architecture search via standard machine learning methodologies[J]. Mathematics in Engineering, 2023, 5(1): 1-21. doi: 10.3934/mine.2023012

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Abstract

1. Introduction

On different time ranges, fractional calculus has a great impact due to a diversity of applications that have contributed to several fields of technical sciences and engineering ^{[1,2,3,4,5,6,7,8,9,10,11,12]}. One of the principal options behind the popularity of the area is that fractional-order differentiations and integrations are more beneficial tools in expressing real-world matters than the integer-order ones. Various studies in the literature, on distinct fractional operators such as the classical Riemann-Liouville, Caputo, Katugamploa, Hadamard, and Marchaud versions have shown versatility in modeling and control applications across various disciplines. However, such forms of fractional derivatives may not be able to explain the dynamic performance accurately, hence, many authors are found to be sorting out new fractional differentiations and integrations which have a kernel depending upon a function and this makes the range of definition expanded ^[13,14]. Furthermore, models based on these fractional operators provide excellent results to be compared with the integer-order differentiations ^{[15,16,17,18,19,20,21,22,23,24,25,26,27]}.

The derivatives in this calculus seemed complicated and lost some of the basic properties that usual derivatives have such as the product rule and the chain rule. However, the semigroup properties of these operators behave well in some cases. Recently, the authors in ^[28] defined a new well-behaved simple derivative called "conformable fractional derivative" which depends just on the basic limit definition of the derivative. It will define the derivative of higher-order (i.e., order $\delta > 1$ ) and also define the integral of order $0 < \delta\leq1$ only. It will also prove the product rule and the mean value theorem and solve some (conformable) differential equations where the fractional exponential function $e^{\frac{\vartheta^{\delta}}{\delta}}$ plays an important rule. Inequalities and their utilities assume a crucial job in the literature of pure and applied mathematics ^{[29,30,31,32,33,34,35,36,37]}. The assortment of distinct kinds of classical variants and their modifications were built up by using the classical fractional operators.

Convexity and its applications exist in almost every field of mathematics due to impermanence in several areas of science, technology in nonlinear programming and optimization theory. By utilizing the idea of convexity, numerous variants have been derived by researchers, for example, Hardy, Opial, Ostrowski, Jensen and the most distinguished one is the Hermite-Hadamard inequality ^{[38,39,40,41]}.

Let $\mathcal{I}\subset\mathbb{R}$ be an interval and $\mathcal{Q}:\mathcal{I}\rightarrow\mathbb{R}$ be a convex function. Then the double inequality

$\begin{equation} (l_{2}-l_{1})\mathcal{Q}\Big(\frac{l_{1}+l_{2}}{2}\Big)\leq\int\limits_{l_{1}}^{l_{2}}\mathcal{Q}(z)dz\leq(l_{2}-l_{1})\frac{\mathcal{Q}(l_{1})+\mathcal{Q}(l_{2})}{2}, \end{equation}$

(1.1)

holds for all $l_{1}, l_{2}\in\mathcal{I}$ with $l_{1}\neq l_{2}.$ Clearly, if $\mathcal{Q}$ is concave on $\mathcal{I},$ then one has the reverse of inequality (1.1). By taking into account fractional integral operators, several lower and upper bounds for the mean value of a convex function can be obtained by utilizing of inequality (1.1).

Exponentially convex functions have emerged as a significant new class of convex functions, which have potential applications in technology, data science, and statistics. In ^[42], Bernstein introduced the concept of exponentially convex function in covariance formation, then the idea of an exponentially convex function is extended by inserting the condition of $r$ -convexity ^[43]. Following this tendency, Jakšetić and Pečarić introduced various kinds of exponentially convex functions in ^[44] and have contemplated the applications in Euler-Radau expansions and Stolarsky means. Our aim is to utilize the exponential convexity property of the functions as well as the absolute values of their derivatives in order to establish estimates for conformable fractional integral introduced by Abdeljawed ^[45] and Jarad et al. ^[46].

Following the above propensity, we present a novel technique for establishing new generalizations of Hermite-Hadamard inequalities that correlate with exponentially $tgs$ -convex functions and conformable fractional operator techniques in this paper. The main purpose is that our consequences, which are more consistent and efficient, are accelerated via the fractional calculus technique. In addition, our consequences also taking into account the estimates for Hermite-Hadamard inequalities for exponentially $tgs$ -convex functions. We also investigate the applications of the two proposed conformable fractional operator to exponentially $tgs$ -convex functions and fractional calculus. The proposed numerical experiments show that our results are superior to some related results.

2. Preliminaries

Before coming to the main results, we provide some significant definitions, theorems and properties of fractional calculus in order to establish a mathematically sound theory that will serve the purpose of the current article.

Awan et al. ^[47] proposed a new class of functions called exponentially convex functions.

Definition 2.1. (See ^[47]) A positive real-valued function $\mathcal{Q}:\mathcal{K}\subset\mathbb{R}\rightarrow(0, \infty)$ is said to be exponentially convex on $\mathcal{K}$ if the inequality

$\begin{equation} \mathcal{Q}(\vartheta l_{1}+(1-\vartheta)l_{2})\leq\vartheta\frac{\mathcal{Q}(l_{1})}{e^{\alpha l_{1}}} +(1-\vartheta)\frac{\mathcal{Q}(l_{2})}{e^{\alpha l_{2}}}, \end{equation}$

(2.1)

holds for all $l_{1}, l_{2}\in\mathbb{R}, \alpha\in\mathbb{R}$ and $\vartheta\in[0, 1].$

Now, we introduce a novel concept of convex function which is known as the exponentially $tgs$ -convex function.

Definition 2.2. A positive real-valued function $\mathcal{Q}:\mathcal{K}\subset\mathbb{R}\rightarrow(0, \infty)$ is said to be exponentially $tgs$ -convex on $\mathcal{K}$ if the inequality

$\begin{equation} \mathcal{Q}(\vartheta l_{1}+(1-\vartheta)l_{2})\leq\vartheta(1-\vartheta)\Big[\frac{\mathcal{Q}(l_{1})}{e^{\alpha l_{1}}} +\frac{\mathcal{Q}(l_{2})}{e^{\alpha l_{2}}}\Big], \end{equation}$

(2.2)

holds for all $l_{1}, l_{2}\in\mathbb{R}, \alpha\in\mathbb{R}$ and $\vartheta\in[0, 1].$

The conformable fractional integral operator was introduced by Abdeljawad ^[45].

Definition 2.3. (See ^[45]) Let $\rho\in(n, n+1]$ and $\delta = \rho-n.$ Then the left and right-sided conformable fractional integrals of order $\rho > 0$ is defined by

$\begin{equation} \mathcal{J}_{l_{1}^{+}}^{\rho}\mathcal{Q}(z) = \frac{1}{n!}\int\limits_{l_{1}}^{z}(z-\vartheta)^{n}(\vartheta-l_{1})^{\rho-1}\mathcal{Q}(\vartheta)d\vartheta \end{equation}$

(2.3)

and

$\begin{equation} \mathcal{J}_{l_{2}^{-}}^{\rho}\mathcal{Q}(z) = \frac{1}{n!}\int\limits_{z}^{l_{2}}(\vartheta-z)^{n}(l_{2}-\vartheta)^{\rho-1}\mathcal{Q}(\vartheta)d\vartheta. \end{equation}$

(2.4)

Next, we demonstrate the following fractional integral operator introduced by Jarad et al. ^[46].

Definition 2.4. (See ^[46]) Let $\delta\in\mathbb{C}$ and $\Re(\delta) > 0.$ Then the left and right-sided fractional conformable integral operators of order $\rho > 0$ are stated as:

$\begin{equation} \mathcal{J}_{l_{1}^{+}}^{\rho, \delta}\mathcal{Q}(z) = \frac{1}{\Gamma(\delta)}\int\limits_{l_{1}}^{z}\Big(\frac{(z-l_{1})^{\rho}-(\vartheta-l_{1})^{\rho}}{\rho}\Big)^{\delta-1}\frac{\mathcal{Q}(\vartheta)}{(\vartheta-l_{1})^{1-\rho}}d\vartheta \end{equation}$

(2.5)

and

$\begin{equation} \mathcal{J}_{l_{2}^{-}}^{\rho, \delta}\mathcal{Q}(z) = \frac{1}{\Gamma(\delta)}\int\limits_{l_{1}}^{z}\Big(\frac{(l_{2}-z)^{\rho}-(l_{2}-\vartheta)^{\rho}}{\rho}\Big)^{\delta-1}\frac{\mathcal{Q}(\vartheta)}{(l_{2}-\vartheta)^{1-\rho}}d\vartheta. \end{equation}$

(2.6)

Recalling some special functions which are known as beta and incomplete beta function.

$\begin{equation*} \mathbb{B}(l_{1}, l_{2}) = \int\limits_{0}^{1}\vartheta^{l_{1}-1}(1-\vartheta)^{l_{2}-1}d\vartheta, \end{equation*}$

$\begin{equation*} \mathbb{B}_{v}(l_{1}, l_{2}) = \int\limits_{0}^{v}\vartheta^{l_{1}-1}(1-\vartheta)^{l_{2}-1}d\vartheta, \quad \quad v\in[0, 1]. \end{equation*}$

Further, the following relationship holds between classical Beta and incomplete Beta functions:

$\begin{equation*} \mathbb{B}(l_{1}, l_{2}) = \mathbb{B}_{v}(l_{1}, l_{2})+\mathbb{B}_{1-v}(l_{1}, l_{2}), \end{equation*}$

$\begin{equation*} \mathbb{B}_{v}(l_{1}+1, l_{2}) = \frac{l_{1}\mathbb{B}_{v}(l_{1}, l_{2})-(\frac{1}{2})^{l_{1}+l_{2}}}{l_{1}+l_{2}} \end{equation*}$

and

$\begin{equation*} \mathbb{B}_{v}(l_{1}, l_{2}+1) = \frac{l_{2}\mathbb{B}_{v}(l_{1}, l_{2})-(\frac{1}{2})^{l_{1}+l_{2}}}{l_{1}+l_{2}}. \end{equation*}$

3. Hermite-Hadamard type inequality for exponentially tgs-convex functions via conformable fractional integrals

Throughout the article, let $\mathcal{I} = [l_{1}, l_{2}]$ be an interval in real line $\mathbb{R}$ . In this section, we shall demonstrate some integral versions of exponentially tgs-convex functions via conformable fractional integrals.

Theorem 3.1. For $\rho\in(n, n+1])$ with $\rho > 0$ and let $\mathcal{Q}: \mathcal{I}\subset \mathbb{R} \rightarrow \mathbb{R}$ be an exponentially tgs-convex function such that $\mathcal{Q} \in L_{1} ([l_{1}, l_{2}])$ , then the following inequalities hold:

$\begin{equation*} \frac{4 \Gamma (\rho -n)}{\Gamma (\rho +1)} \mathcal{Q}\left(\frac{l_{1} + l_{2}}{2}\right) \end{equation*}$

$\begin{equation*} \leq \frac{1}{(l_{2} - l_{1})^{\rho}} \left[{\mathcal{J}_{l_{1}^{+}}^{\rho}} \frac{\mathcal{Q} (l_{2})}{e^{\alpha l_{2}}} +{\mathcal{J}_{l_{2}^{-}}^{\rho}} \frac{\mathcal{Q} (l_{1})}{e^{\alpha l_{1}}}\right] \end{equation*}$

$\begin{equation} \leq \frac{2(n+1) \Gamma (\rho - n+1)}{\Gamma (\rho+3)} \left(\frac{\mathcal{Q} (l_{1})}{e^{\alpha l_{1}}} + \frac{\mathcal{Q} (l_{2})}{e^{\alpha l_{2}}}\right). \end{equation}$

(3.1)

Proof. By using exponentially tgs-convexity of $\mathcal{Q}$ , we have

$\begin{equation} \mathcal{Q} \bigg(\frac{x+y}{2} \bigg)\leq \frac{1}{4}\left(\frac{\mathcal{Q}(x)}{e^{\alpha x}} + \frac{\mathcal{Q}(y)}{e^{\alpha y}}\right). \end{equation}$

(3.2)

Let $x = \vartheta l_{1}+ (1- \vartheta)l_{2}$ and $y = (1-\vartheta) l_{1}+ \vartheta l_{2}$ , we get

$\begin{equation} 4 \mathcal{Q}\bigg(\frac{l_{1} + l_{2}}{2}\bigg)\leq \frac{\mathcal{Q}(\vartheta l_{1}+ (1-\vartheta)l_{2})}{e^{\alpha \mathcal{Q} (\vartheta l_{1}+ (1-\vartheta)l_{2})}}+ \frac{\mathcal{Q} (\vartheta l_{2}+(1-\vartheta) l_{1})}{e^{\alpha[(1-\vartheta) l_{1}+\vartheta l_{2}]}}. \end{equation}$

(3.3)

If we multiply (3.3) by $\frac{1}{n!} {\vartheta}^{n}(1-\vartheta)^{\rho-n-1}$ with $\vartheta \in (0, 1), \rho > 0$ and then integrating the resulting estimate with respect to $\vartheta$ over $[0, 1]$ , we find

$\begin{equation*} \frac{4}{n!}\mathcal{Q} \bigg(\frac{l_{1} + l_{2}}{2}\bigg)\int_{0}^{1} {\vartheta}^{n} (1- \vartheta)^{\rho-n-1} d\vartheta \end{equation*}$

$\begin{equation*} \leq\frac{1}{n!}\int_{0}^{1}{\vartheta}^{n} (1-\vartheta)^{\rho-n-1}\frac{\mathcal{Q}(\vartheta l_{1}+ (1-\vartheta)l_{2})}{e^{\alpha \mathcal{Q} (\vartheta l_{1} +(1-\vartheta)l_{2})}} d\vartheta \end{equation*}$

$\begin{equation*} +\frac{1}{n!}\int_{0}^{1} {\vartheta}^{n} (1- \vartheta)^{\rho-n-1} \frac{\mathcal{Q}(\vartheta l_{2}+(1-\vartheta) l_{1})}{e^{\alpha[(1-\vartheta) l_{1}+\vartheta l_{2}]}}d\vartheta \end{equation*}$

$\begin{equation} = I_{1} + I_{2} \end{equation}$

(3.4)

By setting $u = \vartheta l_{1}+(1-\vartheta)l_{2}$ , we have

$\begin{equation*} I_{1} = \frac{1}{n!} \int_{0}^{1} {\vartheta}^{n} (1- \vartheta)^{\rho-n-1}\frac{\mathcal{Q} (\vartheta l_{1}+ (1- \vartheta)l_{2})}{e^{\alpha \mathcal{Q} (\vartheta l_{1}+ (1-\vartheta)l_{2})}} d\vartheta \end{equation*}$

$\begin{equation*} = \frac{1}{n!(l_{2}-l_{1})^{\rho}} \int_{l_{1}}^{l_{2}} (l_{2}-1)^{n}(u-l_{1})^{\rho-m-1} \frac{\mathcal{Q}(u)}{e^{\alpha u}} du \end{equation*}$

$\begin{equation} = \frac{1}{(l_{2}-l_{1})^{\rho}} {\mathcal{J}_{l_{1}^{+}}^{\rho}} \frac{\mathcal{Q}(l_{2})}{e^{\alpha l_{2}}}. \end{equation}$

(3.5)

Analogously, by setting $v = \vartheta l_{2}+(1-\vartheta)l_{1}$ , we have

$\begin{equation*} I_{2} = \frac{1}{n!} \int_{0}^{1} {\vartheta}^{n} (1- \vartheta)^{\rho-n-1} \mathcal{Q} (\vartheta l_{2} + (1-\vartheta)l_{1}) d\vartheta \end{equation*}$

$\begin{equation*} = \frac{1}{n!(l_{2}-l_{1})^{\rho}} \int_{l_{1}}^{l_{2}} (v-l_{1})^{n} (l_{2}-v)^{\rho-n-1} \frac{\mathcal{Q}(v)}{e^{\alpha v}} dv \end{equation*}$

$\begin{equation} = \frac{1}{(l_{2}-l_{1})^{\rho}} {{J}_{l_{2}^{-}}^{\rho}} \frac{\mathcal{Q}(l_{1})}{e^{\alpha l_{1}}}. \end{equation}$

(3.6)

Thus by using (3.5) and (3.6) in (3.4), we get the first inequality of (3.1).

Consider

$\begin{equation*} \mathcal{Q} (\vartheta l_{1}+(1-\vartheta)l_{2}) \leq \vartheta (1-\vartheta) \Big(\frac{\mathcal{Q} (l_{1})}{e^{\alpha l_{1}}}+\frac{\mathcal{Q}(l_{2})}{e^{\alpha l_{2}}}\Big) \end{equation*}$

and

$\begin{equation*} \mathcal{Q} (\vartheta l_{2}+(1-\vartheta)l_{1}) \leq \vartheta (1-\vartheta) \Big(\frac{\mathcal{Q} (l_{1})}{e^{\alpha l_{1}}}+\frac{\mathcal{Q}(l_{2})}{e^{\alpha l_{2}}}\Big). \end{equation*}$

By adding

$\begin{equation} \mathcal{Q} (\vartheta l_{1}+(1-\vartheta)l_{2}) + \mathcal{Q} (\vartheta l_{2}+(1-\vartheta)l_{1}) \leq 2 \vartheta (1-\vartheta) \Big(\frac{\mathcal{Q} (l_{1})}{e^{\alpha l_{1}}}+\frac{\mathcal{Q}(l_{2})}{e^{\alpha l_{2}}}\Big). \end{equation}$

(3.7)

If we multiply (3.7) by $\frac{1}{n!} {\vartheta}^{n}(1-\vartheta)^{\rho-n-1}$ with $\vartheta \in (0, 1), \rho > 0$ and then integrating the resulting inequality with respect to $\vartheta$ over $[0, 1]$ , we get

$\begin{equation*} \frac{1}{(l_{2}-l_{1})^{\rho}} \Big[{\mathcal{J}_{l_{1}^{+}}^{\rho}} \frac{\mathcal{Q}(l_{2})}{e^{\alpha l_{2}}} +{\mathcal{J}_{l_{2}^{-}}^{\rho}} \frac{\mathcal{Q} (l_{1})}{e^{\alpha l_{1}}}\Big] \end{equation*}$

$\begin{equation} \leq\frac{2(n+1)\Gamma(\rho-n+1)}{\Gamma(\rho+3)} \Big(\frac{\mathcal{Q}(l_{1})}{e^{\alpha l_{1}}}+\frac{\mathcal{Q}(l_{2})}{e^{\alpha l_{2}}}\Big), \end{equation}$

(3.8)

which is the required result.

Some special cases of above theorem are stated as follows:

Corollary 3.1. Choosing $\alpha = 0,$ then Theorem 3.1 reduces to a new result

$\begin{equation*} \frac{4\Gamma (\rho-n)}{\Gamma(\rho+1)} \mathcal{Q}\bigg(\frac{l_{1} + l_{2}}{2}\bigg) \end{equation*}$

$\begin{equation*} \leq \frac{1}{(l_{2} - l_{1})^{\rho}} \Big[{\mathcal{J}_{l_{1}^{+}}^{\rho}} {\mathcal{Q} (l_{2})} +{\mathcal{J}_{l_{2}^{-}}^{\rho}} {\mathcal{Q} (l_{1})}\Big] \end{equation*}$

$\begin{equation*} \leq \frac{2(n+1) \Gamma (\rho - n+1)}{\Gamma (\rho+3)} \Big({\mathcal{Q} (l_{1})} +{\mathcal{Q} (l_{2})}\Big). \end{equation*}$

Remark 3.1. Choosing $\rho = n+1$ and $\alpha = 0,$ then Theorem 3.1 reduces to Theorem 3.1 in ^[19].

4. Hermite-Hadamard type inequality for differentiable exponentially tgs-convex functions via conformable fractional integrals

Our next result is the following lemma which plays a dominating role in proving our coming results.

Lemma 4.1. For $\rho\in(n, n+1])$ with $\rho > 0$ and let $\mathcal{Q}:\mathcal{I}\subset\mathbb{R}\rightarrow \mathbb{R}$ be differentiable function on $\mathcal{I}^{\circ}$ (interior of $\mathcal{I}$ ) with $l_{1} < l_{2}$ such that ${\mathcal{Q}}^{\prime}\in L_{1}([l_{1}, l_{2}]),$ then the following inequality holds:

$\begin{equation*} \mathbb{B}(n+1, \rho-n) \bigg(\frac{\mathcal{Q}(l_{1})+\mathcal{Q}(l_{2})}{2}\bigg)-\frac{n!}{2(l_{2}-l_{1})^{\rho}}[{\mathcal{J}_{l_{1}^{+}}^{\rho}} \mathcal{Q} (l_{2})+{\mathcal{J}_{l_{2}^{-}}^{\rho}} \mathcal{Q} (l_{1})] \end{equation*}$

$\begin{equation} = \int_{0}^{1} (\mathbb{B}_{1-u}(n+1, \rho-n)-\mathbb{B}_{u}(n+1, \rho-n)) {\mathcal{Q}}^{\prime} (\vartheta l_{1}+(1-\vartheta)l_{2})d\vartheta. \end{equation}$

(4.1)

Proof. It suffices that

$\begin{equation*} \int_{0}^{1} (\mathbb{B}_{1-u}(n+1, \rho-n)-\mathbb{B}_{u}(n+1, \rho-n)) {\mathcal{Q}}^{\prime} (\vartheta l_{1}+(1-\vartheta)l_{2})d\vartheta \end{equation*}$

$\begin{equation*} = \int_{0}^{1}\mathbb{B}_{1-u}(n+1, \rho-n){\mathcal{Q}}^{\prime}(\vartheta l_{1}+(1-\vartheta)l_{2})d\vartheta \end{equation*}$

$\begin{equation*} -\int_{0}^{1}\mathbb{B}_{u}(n+1, \rho-n){\mathcal{Q}}^{\prime}(\vartheta l_{1}+(1-\vartheta)l_{2})d\vartheta \end{equation*}$

$\begin{equation} = S_{1}-S_{2} \end{equation}$

(4.2)

Then by integration by parts, we have

$\begin{equation*} S_{1} = \int_{0}^{1} \mathbb{B}_{1-u}(n+1, \rho-n){\mathcal{Q}}^{\prime}(\vartheta l_{1}+(1-\vartheta)l_{2})d\vartheta \end{equation*}$

$\begin{equation*} = \int_{0}^{1} \bigg(\int_{0}^{1-u} v^{n} (1-v)^{\rho-n-1} dv \bigg) \mathcal{Q}^{\prime} (\vartheta l_{1}+(1-\vartheta)l_{2})d\vartheta \end{equation*}$

$\begin{equation*} = \frac{1}{l_{2}-l_{1}} \mathbb{B} (n+1, \rho-n) \mathcal{Q} (l_{2}) \end{equation*}$

$\begin{equation*} -\frac{1}{l_{2}-l_{1}} \int_{0}^{1}(1-u)^{n} u^{\rho-n-1} \mathcal{Q} (\vartheta l_{1} + (1-\vartheta)l_{2})d\vartheta \end{equation*}$

$\begin{equation*} = \frac{1}{l_{2}-l_{1}} \mathbb{B}(n+1, \rho-n) \mathcal{Q}(l_{2}) \end{equation*}$

$\begin{equation*} - \frac{1}{l_{2}-l_{1}} \int_{l_{2}}^{l_{1}} \bigg(\frac{l_{1}-z}{l_{1}-l_{2}} \bigg)^{n} \bigg(\frac{z-l_{2}}{l_{1}-l_{2}} \bigg)^{\rho-n-1} \frac{\mathcal{Q} (z)}{l_{1}-l_{2}} dz \end{equation*}$

$\begin{equation} = \frac{1}{l_{2}-l_{1}} \mathbb{B}(n+1, \rho-n) \mathcal{Q} (l_{2}) - \frac{n!}{(l_{2}-l_{1})^{\rho+1}} {\mathcal{J}_{l_{2}^{-}}^{\rho}} \mathcal{Q}(l_{1}). \end{equation}$

(4.3)

Analogously

$\begin{equation*} S_{2} = \int_{0}^{1} \mathbb{B}_{u}(n+1, \rho-n){\mathcal{Q}}^{\prime}(\vartheta l_{1}+(1-\vartheta)l_{2})d\vartheta \end{equation*}$

$\begin{equation*} = \int_{0}^{1}\bigg(\int_{0}^{u} v^{m} (1-v)^{\rho-n-1} dv \bigg)\mathcal{Q}^{\prime} (\vartheta l_{1}+(1-\vartheta)l_{2})d\vartheta \end{equation*}$

$\begin{equation*} = -\frac{1}{l_{2}-l_{1}} \mathbb{B} (n+1, \rho-n) \mathcal{Q} (l_{1}) \end{equation*}$

$\begin{equation*} + \frac{1}{l_{2}-l_{1}} \int_{0}^{1}(u)^{n}(1-u)^{\rho-n-1} \mathcal{Q} (\vartheta l_{1} + (1-\vartheta)l_{2})d\vartheta \end{equation*}$

$\begin{equation*} = -\frac{1}{l_{2}-l_{1}} \mathbb{B}(n+1, \rho-n) \mathcal{Q}(l_{1}) \end{equation*}$

$\begin{equation*} + \frac{1}{l_{2}-l_{1}} \int_{l_{2}}^{l_{1}}\bigg(\frac{z-l_{2}}{l_{1}-l_{2}} \bigg)^{n} \bigg( \frac{l_{1}-z}{l_{1}-l_{2}} \bigg)^{\rho-n-1} \frac{\mathcal{Q}(z)}{l_{1}-l_{2}} dz \end{equation*}$

$\begin{equation} = -\frac{1}{l_{2}-l_{1}}\mathbb{B}(n+1, \rho-n) \mathcal{Q} (l_{1}) - \frac{n!}{(l_{2}-l_{1})^{\rho+1}} {\mathcal{J}_{l_{1}^{+}}^{\rho}} \mathcal{Q}(l_{2}). \end{equation}$

(4.4)

By substituting values of $S_{1}$ and $S_{2}$ in (4.2) and then If we multiply by $\frac{l_{2}-l_{1}}{2}$ , we get (4.1).

For the sake of simplicity, we use the following notation:

$\begin{equation*} \Upsilon_{\mathcal{Q}}(\rho;\mathbb{B};n;l_{1}, l_{2}) = \mathbb{B}(n+1, \rho-n) \bigg(\frac{\mathcal{Q}(l_{1})+\mathcal{Q}(l_{2})}{2} \bigg)-\frac{n!}{2(l_{2}-l_{1})^{\rho}}\Big[{\mathcal{J}_{l_{1}^{+}}^{\rho}} \mathcal{Q} (l_{2})+{\mathcal{J}_{l_{2}^{-}}^{\rho}} \mathcal{Q} (l_{1})\Big]. \end{equation*}$

Theorem 4.2. For $\rho\in(n, n+1])$ with $\rho > 0$ and let $\mathcal{Q}: \mathcal{I}\subset \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function on $\mathcal{I}^{\circ}$ with $l_{1} < l_{2}$ such that $\mathcal{Q}^{\prime} \in L_{1} ([l_{1}, l_{2}]).$ If ${\vert\ \mathcal{Q}^{\prime} \vert}^{r}$ , with $r\geq 1$ , is an exponentially tgs-convex function, then the following inequality holds:

$\begin{equation*} \vert\ \Upsilon_{\mathcal{Q}}(\rho;\mathbb{B};n;l_{1}, l_{2}) \vert \leq \frac{l_{2}-l_{1}}{2} \Big(\mathbb{B}(n+1, \rho-n+1) - \mathbb{B}(n+1, \rho-n)+ \mathbb{B}(n+2, \rho-n)\Big)^{1-\frac{1}{r}} \end{equation*}$

$\begin{equation} \times \bigg(\frac{e^{\alpha r l_{2}}\vert\mathcal{Q}^{\prime}(l_{1})\vert^{r}+e^{\alpha r l_{1}}\vert\mathcal{Q}^{\prime}(l_{2})\vert^{r}}{6e^{\alpha r l_{1}}e^{\alpha r l_{2}}} \bigg)^{\frac{1}{r}}. \end{equation}$

(4.5)

Proof. Utilizing exponentially tgs-convex function of $\vert\ \mathcal{Q}^{\prime}\vert^{r},$ Lemma 4.1 and Hölder's inequality, one obtains

$\begin{equation*} \vert\ \Upsilon_{\mathcal{Q}}(\rho;\mathbb{B};n;l_{1}, l_{2}) \vert \end{equation*}$

$\begin{equation*} = \bigg\vert\ \frac{l_{2}-l_{1}}{2} \int_{0}^{1}(\mathbb{B}_{1-u}(n+1, \rho-n) - \mathbb{B}_{u}(n+1, \rho-n)) \mathcal{Q}^{\prime} (\vartheta l_{1}+(1-\vartheta)l_{2})d\vartheta \bigg\vert \end{equation*}$

$\begin{equation*} \leq\frac{l_{2}-l_{1}}{2} \bigg(\int_{0}^{1}( \mathbb{B}_{1-u}(n+1, \rho-n)-\mathbb{B}_{u}(n+1, \rho-n))d\vartheta \bigg)^{1-\frac{1}{r}} \end{equation*}$

$\begin{equation*} \times \bigg(\int_{0}^{1} \vert\ \mathcal{Q}^{\prime} (\vartheta l_{1}+(1-\vartheta)l_{2}) \vert^{r} d\vartheta\bigg)^{\frac{1}{r}} \end{equation*}$

$\begin{equation*} \leq \frac{l_{2}-l_{1}}{2} \Big(\mathbb{B}(n+1, \rho-n+1) - \mathbb{B}(n+1, \rho-n)+ \mathbb{B}(n+2, \rho-n)\Big)^{1-\frac{1}{r}} \end{equation*}$

$\begin{equation*} \times \bigg(\int_{0}^{1} \vartheta(1-\vartheta) \Big(\Big\vert\ \frac{\mathcal{Q}^{\prime} (l_{1})}{e^{\alpha l_{1}}} \Big\vert^{r}+\Big\vert\ \frac{\mathcal{Q}^{\prime} (l_{2})}{e^{\alpha l_{2}}} \Big\vert^{r}\Big) d\vartheta\bigg)^{\frac{1}{r}} \end{equation*}$

$\begin{equation*} \leq \frac{l_{2}-l_{1}}{2} \Big(\mathbb{B}(n+1, \rho-n+1) -\mathbb{B}(n+1, \rho-n)+ \mathbb{B}(n+2, \rho-n)\Big)^{1-\frac{1}{r}} \end{equation*}$

(4.6)

which is the required result.

Theorem 4.3. For $\rho\in(n, n+1]$ with $\rho > 0$ and let $\mathcal{Q}: \mathcal{I}\subset \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function on $\mathcal{I}^{\circ}$ with $l_{1} < l_{2}$ such that $\mathcal{Q}^{\prime} \in L_{1} ([l_{1}, l_{2}]).$ If ${\vert\mathcal{Q}^{\prime} \vert}^{r}$ , with $r, s > 1$ such that $\frac{1}{s}+\frac{1}{r} = 1$ , is exponentially tgs-convex function, then the following inequality holds:

$\begin{equation*} \vert\ \Upsilon_{\mathcal{Q}}(\rho;\mathbb{B};n;l_{1}, l_{2}) \vert \leq \frac{l_{2}-l_{1}}{2}\bigg( 2\int_{0}^{\frac{1}{2}} \bigg(\int_{u}^{1-u} v^{n} (1-v)^{\rho-n-1} dv \bigg)^{s}du \bigg)^{\frac{1}{s}} \end{equation*}$

$\begin{equation} \times\bigg(\frac{e^{\alpha r l_{2}}\vert\mathcal{Q}^{\prime}(l_{1})\vert^{r}+e^{\alpha r l_{1}}\vert\mathcal{Q}^{\prime}(l_{2})\vert^{r}}{6e^{\alpha r l_{1}}e^{\alpha r l_{2}}} \bigg)^{\frac{1}{r}}. \end{equation}$

(4.7)

Proof. Utilizing exponentially tgs-convex function of $\vert\ \mathcal{Q}^{\prime}\vert^{r}$ and well-known Hölder inequality, one obtains

$\begin{equation*} \vert\ \Upsilon_{\mathcal{Q}}(\rho;\mathbb{B};n;l_{1}, l_{2}) \vert \end{equation*}$

$\begin{equation*} = \bigg\vert\ \frac{l_{2}-l_{1}}{2} \int_{0}^{1}( \mathbb{B}_{1-u}(n+1, \rho-n) - \mathbb{B}_{u}(n+1, \rho-n)) \mathcal{Q}^{\prime} (\vartheta l_{1}+(1-\vartheta)l_{2})d\vartheta \bigg\vert \end{equation*}$

$\begin{equation*} \leq \frac{l_{2}-l_{1}}{2} \bigg(\int_{0}^{1} \vert\ \mathbb{B}_{1-u}(n+1, \rho-n) - \mathbb{B}_{n}(n+1, \rho-n) \vert^{s} d\vartheta \bigg)^{\frac{1}{s}} \end{equation*}$

$\begin{equation*} \times \bigg(\int_{0}^{1} \vert\ \mathcal{Q}^{\prime} (\vartheta l_{1}+(1-\vartheta)l_{2}) \vert^{r} d\vartheta\bigg)^{\frac{1}{r}} \end{equation*}$

$\begin{equation*} \leq \frac{l_{2}-l_{1}}{2} \bigg( \int_{0}^{\frac{1}{2}} (\mathbb{B}_{1-u}(n+1, \rho-n) - \mathbb{B}_{u}(n+1, \rho-n))^{s} du \end{equation*}$

$\begin{equation*} +\int_{\frac{1}{2}}^{1} (\mathbb{B}_{u}(n+1, \rho-n)-\mathbb{B}_{1-u}(n+1, \rho-n))^{s} du\bigg)^{\frac{1}{s}} \bigg(\int_{0}^{1} \vartheta(1-\vartheta)\Big (\frac{\vert\ \mathcal{Q}^{\prime} (l_{1}) \vert^{r}}{e^{\alpha r l_{1}}}+\frac{\vert\ \mathcal{Q}^{\prime} (l_{2}) \vert^{q}}{e^{\alpha r l_{2}}}\Big) d\vartheta \bigg)^{\frac{1}{r}} \end{equation*}$

$\begin{equation*} = \frac{l_{2}-l_{1}}{2}\bigg( \int_{0}^{\frac{1}{2}} \bigg(\int_{u}^{1-u} v^{n} (1-v)^{\rho-n-1} dv \bigg)^{s} dv+\int_{\frac{1}{2}}^{1} \bigg(\int_{1-u}^{u} v^{n} (1-v)^{\rho-n-1}dv \bigg)^{s} dv\bigg)^{\frac{1}{s}} \end{equation*}$

$\begin{equation*} \times\bigg(\frac{e^{\alpha r l_{2}}\vert\mathcal{Q}^{\prime}(l_{1})\vert^{r}+e^{\alpha r l_{1}}\vert\mathcal{Q}^{\prime}(l_{2})\vert^{r}}{6e^{\alpha r l_{1}}e^{\alpha r l_{2}}} \bigg)^{\frac{1}{r}} \end{equation*}$

$\begin{equation} = \frac{l_{2}-l_{1}}{2}\bigg( 2\int_{0}^{\frac{1}{2}}\bigg(\int_{u}^{1-u} v^{n} (1-v)^{\rho-n-1} dv \bigg)^{s}du \bigg)^{\frac{1}{s}}\bigg(\frac{e^{\alpha r l_{2}}\vert\mathcal{Q}^{\prime}(l_{1})\vert^{r}+e^{\alpha r l_{1}}\vert\mathcal{Q}^{\prime}(l_{2})\vert^{r}}{6e^{\alpha r l_{1}}e^{\alpha r l_{2}}} \bigg)^{\frac{1}{r}}, \end{equation}$

(4.8)

which is the required result.

5. Hermite-Hadamard inequality within the generalized conformble integral operator

This section is devoted to proving some new generalizations for exponentially tgs-convex functions within the generalized conformable integral operator.

Theorem 5.1. For $\rho > 0$ and let $\mathcal{Q} : [l_{1}, l_{2}] \subset \mathbb{R} \rightarrow \mathbb{R}$ be an exponentially tgs-convex function such that $\mathcal{Q} \in L_{1} [l_{1}, l_{2}],$ then the following inequality holds:

$\begin{equation*} \frac{4}{\delta\rho^{\delta}}\mathcal{Q}\Big(\frac{l_{1}+l_{2}}{2}\Big)\leq \frac{\Gamma(\delta)}{(l_{2}-l_{1})^{\rho\delta}} \left[{\mathcal{J}}^{\rho, \delta}_{l_{1}^{+}} \frac{\mathcal{Q} (l_{2})}{e^{\alpha l_{2}}}+{\mathcal{J}}^{\rho, \delta}_{l_{2}^{-}} \frac{\mathcal{Q} (l_{1})}{e^{\alpha l_{1}}} \right] \end{equation*}$

$\begin{equation} \leq \frac{1}{\rho}\left[\mathbb{B}\Big(\frac{\rho+1}{\rho} , \delta\Big)+ \mathbb{B}\Big(\frac{\rho+2}{\rho} , \delta\Big)\right] \Big(\frac{\mathcal{Q}(l_{1})}{e^{\alpha l_{1}}}+\frac{\mathcal{Q}(l_{2})}{e^{\alpha l_{2}}}\Big). \end{equation}$

(5.1)

Proof. Taking into account (3.3) and conducting product of (3.3) by $(\frac{1- \vartheta^{\rho}}{\rho})^{\delta-1} \vartheta^{\rho-1}$ with $\vartheta\in (0, 1), \rho > 0$ and then integrating the resulting estimate with respect to $\vartheta$ over $[0, 1]$ , we find

$\begin{equation*} 4\mathcal{Q} \bigg(\frac{l_{1}+l_{2}}{2} \bigg)\int_{0}^{1} \bigg(\frac{1-\vartheta^{\rho}}{\rho} \bigg)^{\delta-1} \vartheta^{\rho-1} d\vartheta \end{equation*}$

$\begin{equation*} \leq \int_{0}^{1} \bigg(\frac{1-\vartheta^{\rho}}{\rho} \bigg)^{\delta-1} \vartheta^{\rho -1}\frac{\mathcal{Q} (\vartheta l_{1} + (1-\vartheta)l_{2})}{e^{\alpha (\vartheta l_{1} + (1-\vartheta)l_{2})}}d\vartheta \end{equation*}$

$\begin{equation*} + \int_{0}^{1} \bigg(\frac{1-\vartheta^{\rho}}{\rho} \bigg)^{\delta-1} \vartheta^{\rho -1}\frac{\mathcal{Q} (\vartheta l_{2} + (1-\vartheta)l_{1})}{e^{\alpha (\vartheta l_{2} + (1-\vartheta)l_{1})}}d\vartheta \end{equation*}$

$\begin{equation} = R_{1} + R_{2}. \end{equation}$

(5.2)

By making change of variable $u = \vartheta l_{1}+(1-\vartheta)l_{2}$ , we have

$\begin{equation*} R_{1} = \int_{0}^{1} \bigg(\frac{1-\vartheta^{\rho}}{\rho} \bigg)^{\delta-1} \vartheta^{\rho -1}\frac{\mathcal{Q} (\vartheta l_{1}+(1-\vartheta)l_{2})}{e^{\alpha (\vartheta l_{1} + (1-\vartheta)l_{2})}}d\vartheta \end{equation*}$

$\begin{equation*} = \int_{l_{2}}^{l_{1}} \bigg(\frac{1- \big(\frac{u-l_{2}}{l_{1}-l_{2}} \big)^{\rho}}{\rho} \bigg)^{\delta -1} \bigg(\frac{u-l_{2}}{l_{1}-l_{2}} \bigg)^{\rho-1} \frac{\mathcal{Q}(u)}{e^{\alpha u}} \frac{du}{l_{1}-l_{2}} \end{equation*}$

$\begin{equation*} = \frac{1}{(l_{2}-l_{1})^{\rho \delta}} \int_{l_{1}}^{l_{2}} (\frac{(l_{2}-l_{1})^{\rho}-(l_{2}-u)^{\rho}}{\rho})^{\delta-1} (l_{2}-u)^{\rho-1} \frac{\mathcal{Q}(u)}{e^{\alpha u}} du \end{equation*}$

$\begin{equation} = \frac{\Gamma(\delta}{(l_{2}-l_{1})^{\rho \delta}} {\mathcal{J}}^{\rho, \delta}_{l_{2}^{-}} \frac{\mathcal{Q}(l_{1})}{e^{\alpha l_{1}}}. \end{equation}$

(5.3)

Substituting $v = \vartheta l_{2} +(1-\vartheta)l_{1}$ , we have

$\begin{equation*} R_{2} = \int_{0}^{1} \bigg(\frac{1-\vartheta^{\rho}}{\rho} \bigg)^{\delta-1} \vartheta^{\rho -1}\frac{\mathcal{Q} (\vartheta l_{2} + (1-\vartheta)l_{1})}{e^{\alpha (\vartheta l_{2} + (1-\vartheta)l_{1}) }}d\vartheta \end{equation*}$

$\begin{equation*} = \int_{l_{2}}^{l_{1}} \bigg(\frac{1- \big(\frac{v-l_{1}}{l_{2}-l_{1}} \big)^{\rho}}{\rho} \bigg)^{\delta -1} \bigg(\frac{v-l_{1}}{l_{2}-l_{1}} \bigg)^{\rho-1} \frac{\mathcal{Q}(v)}{e^{\alpha v}} \frac{du}{l_{2}-l_{1}} \end{equation*}$

$\begin{equation*} = \frac{1}{(l_{2}-l_{1})^{\rho \delta}} \int_{l_{1}}^{l_{2}} (\frac{(l_{2}-l_{1})^{\rho}-(v-l_{1})^{\rho}}{\rho})^{\delta-1} (v-l_{1})^{\rho-1} \frac{\mathcal{Q}(v)}{e^{\alpha v}} dv \end{equation*}$

$\begin{equation} = \frac{\Gamma (\delta)}{(l_{2}-l_{1})^{\rho \mathcal{Q}}} {\mathcal{J}}^{\rho, \delta}_{l_{2}^{-}}\frac{\mathcal{Q} (l_{2})}{e^{\alpha l_{2}}}. \end{equation}$

(5.4)

Thus by using (5.2) and (5.3) in (5.4), we get the first inequality of (5.1).

Consider

and

By adding

$\begin{equation} \mathcal{Q} (\vartheta l_{1}+(1-\vartheta)l_{2}) + \mathcal{Q} (\vartheta l_{2}+(1-\vartheta)l_{1}) \leq 2\vartheta (1-\vartheta) \Big(\frac{\mathcal{Q} (l_{1})}{e^{\alpha l_{1}}}+\frac{\mathcal{Q}(l_{2})}{e^{\alpha l_{2}}}\Big). \end{equation}$

(5.5)

If we multiply (5.5) by $(\frac{1-\vartheta^{\rho}}{\rho})^{\delta -1} \vartheta^{\rho-1}$ with $\vartheta\in (0, 1), \rho > 0$ and then integrating the resulting estimate with respect to $\vartheta$ over $[0, 1]$ , we get

$\begin{equation*} \frac{\Gamma(\delta)}{(l_{2}-l_{1})^{\rho \delta}} \big[{\mathcal{J}}^{\rho, \delta}_{l_{1}^{+}} \frac{\mathcal{Q} (l_{2})}{e^{\alpha l_{2}}} +{\mathcal{J}}^{\rho, \delta}_{l_{2}^{-}} \frac{\mathcal{Q} (l_{1})}{e^{\alpha l_{1}}} \big] \end{equation*}$

$\begin{equation} \leq \frac{1}{\rho}\Big[\mathbb{B}\Big(\frac{\rho+1}{\rho} , \delta\Big)+ \mathbb{B}\Big(\frac{\rho+2}{\rho} , \delta\Big)\Big] \Big(\frac{\mathcal{Q} (l_{1})}{e^{\alpha l_{1}}}+\frac{\mathcal{Q}(l_{2})}{e^{\alpha l_{2}}}\Big), \end{equation}$

(5.6)

the desired inequality is the right hand side of (5.1).

Our main results depend on the following identity.

Lemma 5.2. For $\rho > 0$ and let $\mathcal{Q}:\mathcal{I}\subset\mathbb{R}\rightarrow \mathbb{R}$ be a differentiable function on $(l_{1}, l_{2})$ with $l_{1} < l_{2}$ such that ${\mathcal{Q}}^{\prime}\in L_{1}[l_{1}, l_{2}],$ then the following identity holds:

$\begin{equation*} \bigg(\frac{\mathcal{Q}(l_{1})+\mathcal{Q}(l_{2})}{2}\bigg)- \frac{\rho^{\delta} \Gamma(\delta+1)}{2(l_{2}-l_{1})^{\rho\delta}} \big[{\mathcal{J}}^{\rho, \delta}_{l_{1}^{+}} \mathcal{Q} (l_{2}) +{\mathcal{J}}^{\rho, \delta}_{l_{2}^{+}} \mathcal{Q} (l_{1}) \big] \end{equation*}$

$\begin{equation} = \frac{(l_{2}-l_{1}) \rho^{\delta}}{2} \int_{0}^{1} \bigg[ \bigg(\frac{1-\vartheta^{\rho}}{\rho}\bigg)^{\delta}-\bigg(\frac{1-(1-\vartheta)^{\rho}}{\rho}\bigg)^{\delta} \bigg] \mathcal{Q}^{\prime} (\vartheta l_{1}+(1-\vartheta)l_{2}) d\vartheta. \end{equation}$

(5.7)

Proof. It suffices that

$\begin{equation*} \int_{0}^{1} \bigg[\bigg(\frac{1-\vartheta^{\rho}}{\rho}\bigg)^{\delta}-\bigg(\frac{1-(1-\vartheta)^{\rho}}{\rho}\bigg)^{\delta} \bigg] \mathcal{Q}^{\prime} (\vartheta l_{1}+(1-\vartheta)l_{2}) d\vartheta \end{equation*}$

$\begin{equation*} = \int_{0}^{1}\bigg(\frac{1-\vartheta^{\rho}}{\rho}\bigg)^{\delta} \mathcal{Q}^{\prime} (\vartheta l_{1}+(1-\vartheta)l_{2}) d\vartheta -\bigg(\frac{1-(1-\vartheta)^{\rho}}{\rho}\bigg)^{\delta}\mathcal{Q}^{\prime} (\vartheta l_{1}+(1-\vartheta)l_{2}) d\vartheta \end{equation*}$

$\begin{equation} = M_{1}-M_{2}. \end{equation}$

(5.8)

Using integration by parts and making change of variable technique, we have

$\begin{equation*} M_{1} = \int_{0}^{1}\bigg(\frac{1-\vartheta^{\rho}}{\rho}\bigg)^{\delta} \mathcal{Q}^{\prime} (\vartheta l_{1}+(1-\vartheta)l_{2}) d\vartheta \end{equation*}$

$\begin{equation*} = \frac{1}{l_{1}-l_{2}}\bigg(\frac{1-\vartheta^{\rho}}{\rho}\bigg)^{\delta} \mathcal{Q}(\vartheta l_{1}+(1-\vartheta)l_{2}) d\vartheta |_{0}^{1} \end{equation*}$

$\begin{equation*} +\frac{\delta}{l_{1}-l_{2}} \int_{0}^{1}\bigg(\frac{1-\vartheta^{\rho}}{\rho}\bigg)^{\delta-1} \vartheta^{\rho-1}\mathcal{Q} (\vartheta l_{1}+(1-\vartheta)l_{2}) d\vartheta \end{equation*}$

$\begin{equation*} = \frac{\mathcal{Q}(l_{2})}{(l_{2}-l_{1})\rho^{\delta}} - \frac{\delta}{l_{2}-l_{1}} \int_{0}^{1}\bigg(\frac{1-\vartheta^{\rho}}{\rho}\bigg)^{\delta-1} \vartheta^{\rho-1} \mathcal{Q} (\vartheta l_{1}+(1-\vartheta)l_{2}) d\vartheta \end{equation*}$

$\begin{equation*} \label{eq.0.27} = \frac{\mathcal{Q}(l_{2})}{(l_{2}-l_{1})\rho^{\delta}} - \frac{\delta\Gamma(\delta)}{(l_{2}-l_{1})^{\rho\delta+1}} \mathcal{J}_{l_{2}^{-}}^{\rho, \delta}\mathcal{Q}(l_{1}) \end{equation*}$

Analogously

$\begin{equation*} M_{2} = \int_{0}^{1} \bigg(\frac{1-(1-\vartheta)^{\rho}}{\rho}\bigg)^{\delta} \mathcal{Q}^{\prime} (\vartheta l_{1}+(1-\vartheta)l_{2}) d\vartheta \end{equation*}$

$\begin{equation*} = \frac{1}{l_{1}-l_{2}} \bigg(\frac{1-(1-\vartheta)^{\rho}}{\rho}\bigg)^{\delta} \mathcal{Q}(\vartheta l_{1}+(1-\vartheta)l_{2}) |_{0}^{1} \end{equation*}$

$\begin{equation*} -\frac{1}{l_{1}-l_{2}} \int_{0}^{1} \mathcal{Q} \bigg(\frac{1-(1-\vartheta)^{\rho}}{\rho}\bigg)^{\delta-1} (1-\vartheta)^{\rho-1} \mathcal{Q}(\vartheta l_{1}+(1-\vartheta)l_{2}) d\vartheta \end{equation*}$

$\begin{equation*} = \frac{-\mathcal{Q} (l_{1})}{(l_{2}-l_{1})\rho^{\delta}} + \frac{\delta}{l_{2}-l_{1}} \int_{0}^{1} \bigg(\frac{1-(1-\vartheta)^{\rho}}{\rho}\bigg)^{\delta-1} (1-\vartheta)^{\rho-1} \mathcal{Q}(\vartheta l_{1}+(1-\vartheta)l_{2}) d\vartheta \end{equation*}$

$\begin{equation} = \frac{-\mathcal{Q} (l_{1})}{(l_{2}-l_{1})\rho^{\delta}} + \frac{\delta\Gamma(\delta)}{(l_{2}-l_{1})^{\rho\delta+1}} {\mathcal{J}}^{\rho, \delta}_{l_{1}^{+}}\mathcal{Q}(l_{2}). \end{equation}$

(5.9)

By substituting values of $M_{1}$ and $M_{2}$ in (5.8) and then conducting product on both sides by $\frac{(l_{2}-l_{1})\rho^{\delta}}{2}$ , we get the desired result.

Theorem 5.3. For $\rho > 0$ and let $\mathcal{Q}:\mathcal{I}\subset \mathbb{R} \rightarrow\mathbb{R}$ be a differentiable function on $\mathcal{I}^{\circ}$ with $l_{1} < l_{2}$ such that $\mathcal{Q}^{\prime} \in L_{1} ([l_{1}, l_{2}]).$ If ${\vert\ \mathcal{Q}^{\prime} \vert}^{r}$ , with $r\geq 1$ , is an exponentially tgs-convex function, then the following inequality holds

$\begin{equation*} \Big\vert\bigg(\frac{\mathcal{Q}(l_{1})+\mathcal{Q}(l_{2})}{2}\bigg)- \frac{\rho^{\delta}\Gamma(\delta+1)}{2(l_{2}-l_{1})^{\rho\delta}} \big[{\mathcal{J}}^{\rho, \delta}_{l_{1}^{+}} \mathcal{Q} (l_{2}) +{\mathcal{J}}^{\rho, \delta}_{l_{2}^{+}} \mathcal{Q} (l_{1}) \big] \Big \vert \end{equation*}$

$\begin{equation} \leq \frac{(l_{2}-l_{1}) \rho^{\delta}}{2}\bigg(\frac{1}{\rho^{\delta+1}}{\mathbb{B}\Big(\frac{1}{\rho}, \delta+1\Big)}+\frac{1}{\rho^{\delta+2}}\mathbb{B}\Big(\frac{1}{\rho^{2}}, \delta+1\Big)\bigg)^{1-\frac{1}{r}}\bigg(\frac{e^{\alpha r l_{2}}\vert\mathcal{Q}^{\prime}(l_{1})\vert^{r}+e^{\alpha r l_{1}}\vert\mathcal{Q}^{\prime}(l_{2})\vert^{r}}{6e^{\alpha r l_{1}}e^{\alpha r l_{2}}} \bigg)^{\frac{1}{r}}. \end{equation}$

(5.10)

Proof. Using exponentially tgs-convexity of $\vert\ \mathcal{Q}^{\prime}\vert^{r},$ Lemma 5.2, and the well-known Hölder inequality, we have

$\begin{equation*} \Big\vert\bigg(\frac{\mathcal{Q}(l_{1})+\mathcal{Q}(l_{2})}{2}\bigg)-\frac{\rho^{\delta} \Gamma(\delta+1)}{2(l_{2}-l_{1})^{\rho \delta}} \big[{\mathcal{J}}^{\rho, \delta}_{l_{1}^{+}} \mathcal{Q} (l_{2}) +{\mathcal{J}}^{\rho, \delta}_{l_{2}^{+}} \mathcal{Q} (l_{1}) \big]\Big \vert \end{equation*}$

$\begin{equation*} = \bigg\vert\ \frac{(l_{2}-l_{1}) \rho^{\delta}}{2}\int_{0}^{1} \bigg[ \bigg(\frac{1-\vartheta^{\rho}}{\rho}\bigg)^{\delta}-\bigg( \frac{1-(1-\vartheta)^{\rho}}{\rho}\bigg)^{\delta} \bigg] \mathcal{Q}^{\prime} (\vartheta l_{1}+(1-\vartheta)l_{2}) d\vartheta \end{equation*}$

$\begin{equation*} \leq\frac{(l_{2}-l_{1}) \rho^{\delta}}{2} \bigg(\int_{0}^{1}\bigg[\bigg(\frac{1-\vartheta^{\rho}}{\rho}\bigg)^{\delta}-\bigg( \frac{1-(1-\vartheta)^{\rho}}{\rho}\bigg)^{\delta} \bigg] d\vartheta \bigg)^{1-\frac{1}{r}} \end{equation*}$

$\begin{equation*} \times \bigg( \int_{0}^{1} \vert\ \mathcal{Q}^{\prime} (\vartheta l_{1}+(1-\vartheta)l_{2}) \vert^{r} d\vartheta \bigg)^{\frac{1}{r}} \end{equation*}$

$\begin{equation*} \leq \frac{(l_{2}-l_{1})\rho^{\delta}}{2} \bigg(\int_{0}^{1} \bigg(\frac{1-\vartheta^{\rho}}{\rho}\bigg)^{\delta} d\vartheta-\int_{0}^{1} \bigg( \frac{1-(1-\vartheta)^{\rho}}{\rho}\bigg)^{\delta}d\vartheta \bigg)^{1-\frac{1}{r}} \end{equation*}$

$\begin{equation*} \times\bigg(\int_{0}^{1}\vartheta(1-\vartheta)\Big(\frac{\vert\ \mathcal{Q}^{\prime} (l_{1})\vert^{r}}{e^{\alpha r l_{1}}}+\frac{\vert\ \mathcal{Q}^{\prime} (l_{2})\vert^{r}}{e^{\alpha r l_{2}}}\Big)d\vartheta \bigg)^{\frac{1}{r}} \end{equation*}$

$\begin{equation*} \label{eq 0.30} = \frac{(l_{2}-l_{1})\rho^{\delta}}{2}\bigg(\frac{1}{\rho^{\delta+1}}{\mathbb{B}\Big(\frac{1}{\rho}, \delta+1\Big)}+\frac{1}{\rho^{\delta+2}}\mathbb{B}\Big(\frac{1}{\rho^{2}}, \delta+1\Big)\bigg)^{1-\frac{1}{r}}\bigg(\frac{e^{\alpha r l_{2}}\vert\mathcal{Q}^{\prime}(l_{1})\vert^{r}+e^{\alpha r l_{1}}\vert\mathcal{Q}^{\prime}(l_{2})\vert^{r}}{6e^{\alpha r l_{1}}e^{\alpha r l_{2}}} \bigg)^{\frac{1}{r}}, \end{equation*}$

the required result.

6. Applications

Let $l_{1}, l_{2} > 0$ with $l_{1}\neq l_{2}.$ Then the arithmetic mean $\mathcal{A}(l_{1}, l_{2}),$ harmonic mean $\mathcal{H}(l_{1}, l_{2}),$ logarithmic mean $\mathcal{L}(l_{1}, l_{2})$ and $n$ -th generalized logarithmic mean $L_{n}(l_{1}, l_{2})$ are defined by

$\begin{equation*} \mathcal{A}(l_{1}, l_{2}) = \frac{l_{1}+l_{2}}{2}, \end{equation*}$

$\begin{equation*} \mathcal{G}(l_{1}, l_{2}) = \sqrt{l_{1}l_{2}}, \end{equation*}$

$\begin{equation*} \mathcal{L}(l_{1}, l_{2}) = \frac{l_{2}-l_{1}}{\ln l_{2}-\ln l_{1}} \end{equation*}$

and

$\begin{equation*} \mathcal{L}_{n}(l_{1}, l_{2}) = \left[\frac{l_{2}^{n+1}-l_{1}^{n+1}}{(n+1)( l_{2}- l_{1})}\right]^{\frac{1}{n}} \quad (n\neq 0, -1), \end{equation*}$

respectively. Recently, the bivariate means have attracted the attention of many researchers ^{[47,48,49,50,51,52,53,54,55,56,57,58]} due to their are closely related to the special functions.

In this section, we use our obtained results in section 5 to provide several novel inequalities involving the special bivariate means mentioned above.

Proposition 6.1. Let $l_{1}, l_{2} > 0$ with $l_{2} > l_{1}$ . Then

$\begin{equation*} \Big\vert\mathcal{A}(l_{1}^{2}, l_{2}^{2})-\frac{1}{2}\mathcal{L}_{3}^{3}(l_{1}, l_{2})\Big\vert\leq\frac{l_{2}-l_{1}}{(6)^{\frac{1}{r}}e^{\alpha(l_{1}+l_{2})}} \Big[(e^{\alpha l_{2}}l_{1})^{r}+(e^{\alpha l_{1}}l_{2})^{r}\Big]^{\frac{1}{r}}. \end{equation*}$

Proof. Let $\rho = \delta = 1$ and $\mathcal{Q}(z) = z^{2}.$ Then the desired result follows from Theorem 5.3.

Proposition 6.2. Let $l_{1}, l_{2} > 0$ with $l_{2} > l_{1}$ . Then

$\begin{equation*} \Big\vert\mathcal{H}^{-1}(l_{1}^{2}, l_{2}^{2})-\frac{1}{2}\mathcal{L}^{-1}(l_{1}, l_{2})\Big\vert\leq\frac{l_{2}-l_{1}}{2(6)^{\frac{1}{r}}e^{\alpha(l_{1}+l_{2})}} \Big[\frac{(e^{\alpha l_{2}}l_{2}^{2})^{r}+(e^{\alpha l_{1}}l_{1}^{2})^{r}}{(l_{1}l_{2})^{2r}}\Big]^{\frac{1}{r}}. \end{equation*}$

Proof. Let $\rho = \delta = 1$ and $\mathcal{Q}(z) = \frac{1}{z}.$ Then the desired result follows from Theorem 5.3.

Proposition 6.3. Let $l_{1}, l_{2} > 0$ with $l_{2} > l_{1}$ . Then

$\begin{equation*} \Big\vert\mathcal{A}(l_{1}^{n}, l_{2}^{n})-\frac{1}{2}\mathcal{L}^{n}_{n}(l_{1}, l_{2})\Big\vert\leq\frac{(l_{2}-l_{1})\vert n\vert}{2} \Big[\frac{(e^{\alpha l_{2}}l_{1}^{n-1})^{r}+(e^{\alpha l_{1}}l_{2}^{n-1})^{r}}{6e^{\alpha r(l_{1}+l_{2})}}\Big]^{\frac{1}{r}}. \end{equation*}$

Proof. Let $\rho = \delta = 1$ and $\mathcal{Q}(z) = z^{n}.$ Then the desired result follows from Theorem 5.3.

7. Conclusions

In this paper, we proposed a novel technique with two different approaches for deriving several generalizations for an exponentially $tgs$ -convex function that accelerates with a conformable integral operator. We have generalized the Hermite-Hadamard type inequalities for exponentially $tgs$ -convex functions. By choosing different parametric values $\rho$ and $\delta$ , we analyzed the convergence behavior of our proposed methods in form of corollaries. Another aspect is that to show the effectiveness of our novel generalizations, our results have potential applications in fractional integrodifferential and fractional Schrödinger equations. Numerical applications show that our findings are consistent and efficient. Finally, we remark that the framework of the conformable fractional integral operator, it is of interest to further our results to the framework of Riemann-Liouville, Hadamard and Katugampola fractional integral operators. Our ideas and the approach may lead to a lot of follow-up research.

Acknowledgements

The authors would like to thank the anonymous referees for their valuable comments and suggestions, which led to considerable improvement of the article.

The work was supported by the Natural Science Foundation of China (Grant Nos. 61673169, 11971142, 11701176, 11626101, 11601485).

Conflict of interest

The authors declare no conflict of interest.

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