EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 5997 ISSN 1307-5543 – ejpam.com Published by New York Business Global New Extension of Inequalities through Extended Version of Fractional Operators for s-Convexity with Applications Miguel Vivas-Cortez1, Rana Safdar Ali2,∗, Naila Talib 2, Imen Kebaili3, Imed Boukhris3, Gauhar Rahman4 1 Pontificia Universidad Católica del Ecuador, Faculty of Exact, Natural and Environmen- tal Sciences, FRACTAL Laboratory (Fractional Research in Analysis, Convexity and Their Applications Laboratory, Ecuador 2 Department of Mathematics and Statistics, The University of Lahore, Lahore, Pakistan 3 Department of Physics, Faculty of Science, King Khalid University, P.O. Box 960, Abha, Saudi Arabia 4 Department of Mathematics and Statistics, Hazara University Mansehra 21300, Pakistan Abstract. Fractional integral inequalities play a significant role in both pure and applied math- ematics, contributing to the advancement and extension of various mathematical techniques. An accurate formulation of such inequalities is essential to establish the existence and uniqueness of fractional methods. Additionally, convexity theory serves as a fundamental component in the study of fractional integral inequalities due to its defining characteristics and properties. Moreover, there is a strong interconnection between convexity and symmetric theories, allowing results from one to be effectively applied to the other. This correlation has become particularly evident in recent decades, further enhancing their importance in mathematical research. This article investigate two innovative approaches of differentiable functions to modify Hermite-Hadamard inequalities and their refinements by implementation of generalized fractional operators through the s-convex functions. The study aims to extend and refine existing inequalities with a fractional operator that has extended the Bessel-Maitland functions as a kernel, providing a more generalized framework. By incorporating these special functions, the results encompass and improve numerous classical in- equalities found in the literature, offering deeper insights and broader applicability in mathematical analysis. 2020 Mathematics Subject Classifications: 9B62, 33C10, 26A33 Key Words and Phrases: Convex function, Extended Bessel-Maitland function; Fractional operators ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.5997 Email addresses: mjvivas@puce.edu.ec M. V. Cortez ), rsafdar0@gmail.com (R. S. Ali),∗, 20nailatalib@gmail.com N. Talib), eqabaeli@kku.edu.sa (I. Kebaili), eakhdar@kku.edu.sa I. Boukhris), gauhar55uom@gmail (G. Rahman) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 2 of 23 1. Introduction Mathematical inequalities are now widely recognized as one of the most valuable and applicable branches of mathematics. Their versatility and effectiveness have been demon- strated in various scientific and engineering disciplines, with significant applications in information theory, economics, finance, and engineering. The theory of inequalities plays a fundamental role in nearly all areas of pure and applied mathematics. The study of convex functions has been instrumental in advancing inequality theory, tracing back to the pioneering work of Jensen in 1905–1906. Many modern analytical inequalities stem directly from the properties of convex functions. In addition, convex functions and vari- ous forms of convexity serve as powerful tools to derive numerous important and practical inequalities. Due to its expanding range of applications, the study of inequalities remains one of the most actively researched fields in mathematical analysis. The broad spectrum of applications of fractional calculus [1] in domains such as fluid dynamics, mathematical biology, and mathematical physics has made it an essential field of ongoing research [2–5]. In order to validate various solutions in theoretical and prac- tical contexts, numerous researchers have generated fractional integral inequalities using fractional operators [6, 7]. Recent research has explored fractional integral inequalities in extensive detail, examining their various manifestations and possible uses. These studies have tremendously broadened the discipline and provided new tools and insights. A notable advancement in this field is the formulation of integral expressions that include special functions. Fractional integral operators that utilize particular kernel func- tions are essential in many fields of study [8, 9]. The Bessel-Maitland function, first proposed by Daniel Bernoulli, is related to the linear differential equation. The fractional calculus, with its extensive applications, extends its scope. Mathematical analysis and fractional theory have greatly benefited from its extensions. More research and develop- ment in the topic is being motivated by quick developments in fractional calculus, which have brought attention to the need for creative transformations and generalized fractional operators. Convex analysis plays a crucial role in variational analysis, as it encompasses a gener- alized differentiation theory applicable to mathematical models that do not require differ- entiability assumptions. Convex optimization is well known to be just one of many areas in which convex analysis has demonstrated its significance. The convexity of a problem allows for the development of efficient numerical algorithms to solve convex optimization problems, even in the presence of non-differentiable data, while also enabling an in-depth study of the qualitative properties of optimal solutions. The impact of convex analysis and optimization continues to expand across various mathematical disciplines and practi- cal applications, including estimation, control systems, communications, networks, signal processing, data analysis, electrical circuit design, finance, statistics, economics, and math- ematical modeling.Inequalities have applications as tools in various fields of mathematics, such as differential and integral equations, and these ideas serve as a basis for their anal- ysis. One of the most well-known inequalities is the Hermite-Hadamard inequality, which was first introduced by Charles Hermite and Jacques Hadamard and describes how con- M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 3 of 23 vex functions behave and is used extensively in mathematical modeling, optimization, and numerical analysis. In recent years the researchers have extended the classical convexity into h-convexity, h-Godunova-Levin convexity, s-convexity and (η1, η2) convex functions etc. For s-convex functions, the integral identities linked to Hermite-Hadamard inequality were first presented by Barsam et al. [10]. Sattarzadeh and Barsam [11] discovered the Hermite-Hadamard type problems with fractional integrals for functions that are uniformly convex. It is meant to explore the HermiteHadamard type inequalities involving fractional integrals due to the numerous applications of fractional calculus [12–16] and Hermite-Hadamard type inequalities [17– 21]. The inequalities of fractional integral by convex functions with respect to increasing functions was obtained by Mohammed [22]. Abdeljawad et al. [23] proposed new Simpson- type inequalities for (s, m)-convex functions . Some inequalities for s-convex functions with fractional integrals were presented by I̧ scan [24]. Trapezoid type inequalities for s- convex functions involving generalized fractional operators established by Usta et al. [25] . Butt et al. [26] introduced integral identity; by using that identity, new inequalities were obtained via a general form of fractional integral operators. Agarwal et al. [27] proposed Hermite-Hadamard type inequalities for generalized k-fractional integrals. In our present work we use the s-convex function for the class of first odder deriva- tives and second order derivatives and modified the Hermite-Hadamard (H-H) integral inequalities by utilizing the Bessel-Maitland function as its kernel. 2. Preliminaries In this section, we discuss the basic definitions which will help us to understand our main work. Definition 1. [28, 29] Let ϕ : R → R be real valued function, is said to be convex if the following inequalities holds: ϕ(℘α+ (1− ℘)ω) ≤ ℘ϕ(α) + (1− ℘)ϕ(ω), (1) where ℘ ∈ [0, 1] and ∀ ω, α ∈ R. Definition 2. [30] Let ϕ : R → R be real valued function, is said to be s-convex function, if the following relation holds: ϕ(℘α+ (1− ℘)ω) ≤ ℘sϕ(α) + (1− ℘)sϕ(ω), (2) where α, ω ∈ R and ℘ ∈ (0, 1) and s ∈ (0, 1]. Definition 3. [29, 31–33] The Hermite-Hadamard type inequality for convex function ϕ : R → R, is defined as follows; ϕ ( α+ ω 2 ) ≤ 1 ω − α ∫ ω α ϕ(℘)d℘ ≤ ϕ(α) + ϕ(ω) 2 . (3) where α, ω ∈ R, α < ω. M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 4 of 23 Definition 4. [34] The gamma function in integral type is defined for ℜ(t) > 0, as follows Γ(t) = ∫ ∞ 0 xt−1e−xdx. (4) Definition 5. [34] The Pochammer’s symbol is defined as follows: (ð)η = { 1, for η = 0, ð ̸= 0 ð(ð+ 1) · · · (ð+ η − 1), for η ≥ 1, } (5) For η ∈ N and ð ∈ C Here, is some relations of gamma functions. (℘)n = Γ(℘+ n) Γ(℘) , (℘)kn = Γ(℘+ kn) Γ(℘) . where Γ being the gamma notation. Definition 6. [35] The beta function is defined for ℜ(m) > 0 and ℜ(n) > 0 as follows: B(g, h) = ∫ 1 0 ηg−1(1− η)h−1dη, = Γ(g)Γ(h) Γ(g + h) . (6) Definition 7. [36] The extended form of beta function is define for ℜ(g) > 0, ℜ(h) > 0, ℜ(p) > 0 as follows Bp(g, h) = ∫ 1 0 zg−1(1− z)h−1exp ( −p z(1− z) ) dz. (7) Taking the value of p = 1, the extended beta function becomes the classical beta function. Definition 8. [37] The Bessel-Maitland function is defined for ϕ, ψ, υ, χ,ϖ ∈ C and ℜ(ϕ) > 0,ℜ(ξ) > 0,ℜ(υ) > 0,ℜ(χ) > 0,ℜ(ϖ) > 0, ρ, η,m ≥ 0 and m, η > ℜ(ϕ) + η ; Jψ,υ,χ,ϖϕ,ρ,m,η (y) = ∞∑ p=0 (υ)ρp(ϖ)ηp(−y)p Γ(ϕp+ ψ + 1)(χ)mp . (8) Definition 9. [38] The extended version of Bessel-Maitland is defined for µ, ξ, ζ, ς, c, κ1 ∈ C,ℜ(µ) > 0,ℜ(ξ) > 0,ℜ(ζ) > 0,ℜ(ς) > 0,ℜ(κ1) > 0, ρ,m, η ≥ 0 and m, ρ > ℜ(µ) + η, as follows Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) = ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n. (9) M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 5 of 23 Definition 10. [39] The left and right sided of extended version of Bessel-Maitland func- tion is defined for the same assumption of definition (9), as follows:( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,p+ f ) (x, r) = ∫ x p (x− t)ξJµ,ρ,m,ηξ,ζ,ς,κ1 (κ(x− t)µ; r)f(t)dt, (10)( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,q− f ) (x, r) = ∫ q x (t− x)ξJµ,ρ,m,ηξ,ζ,ς,κ1 (κ(t− x)µ; r)f(t)dt. (11) Remark 1. If we replace r = 0, κ = 0, and ξ = ξ− 1 in the definition (10), we obtain the left-and right-sided Riemann fractional operators. 3. Applications of differentiable function with the refinements of Hermite-Hadamard (H −H) inequalities Here, we discuss the refinements of Hermite-Hadamard type fractional inequalities with the differentiable functions by implementations of extended version of fractional operators for s-convexity. Lemma 1. Let ϕ : I ⊆ R → R be a differentiable mapping on Io and ω, τ ∈ Io with ω < τ . If ϕ ′ ∈ L[ω, τ ], and µ, ξ, ζ, ς, c, κ1 ∈ C,ℜ(µ) > 0,ℜ(ξ) > 0,ℜ(ζ) > 0,ℜ(ς) > 0,ℜ(κ1) > 0, ρ,m, η ≥ 0 and m, ρ > ℜ(µ) + η, then following integral equality holds; ϕ(α)Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ 1 α− ω + 1 α− τ ] − (ξ ′ + µn)[ 1 (α− ω)ξ ′+1 ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α+ϕ ) (ω, κ1) + 1 (α− τ)ξ ′+1 ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,τ− ϕ ) (ω, κ) ] = ∫ 1 0 ℘ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)ϕ ′ (℘α+ (1− ℘)ωd℘+ ∫ 1 0 ℘ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)ϕ ′ (℘α+ (1− ℘)τd℘. Proof. Consider the integral I1 = ∫ 1 0 ℘ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)ϕ ′ (℘α+ (1− ℘)ωd℘ = ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∫ 1 0 ℘ξ ′ +µnϕ ′ (℘α+ (1− ℘)ωd℘ = ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n [ ∫ 1 0 ℘ξ ′ +µn α− ω ϕ(℘α+ (1− ℘)ω ∣∣1 0 − ∫ 1 0 ξ ′ + µn α− ω ℘ξ ′ +µn−1ϕ(℘α+ (1− ℘)ωd℘ ] = ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n [ ϕ(α) α− ω − ξ ′ + µn α− ω ∫ 1 0 ℘ξ ′ +µn−1ϕ(℘α+ (1− ℘)ωd℘ ] .(12) M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 6 of 23 Using substitution ℘α+ (1− ℘)ω = x in equation (12), we get I1 = ϕ(α) α− ω Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)− ξ ′ + µn (α− ω)ξ ′+1 ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α+ϕ ) (ω, κ). (13) Now, consider the integral I2, we have I2 = ∫ 1 0 ℘ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)ϕ ′ (℘α+ (1− ℘)τd℘. Proceeding same as I1 and we get the result, I2 = ϕ(α) α− τ Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)− ξ ′ + µn (α− τ)ξ ′+1 ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,τ− ϕ ) (ω, κ). (14) By combining equation (13) and (14) we get the required result. Corollary 1. If we replace p = 0, κ = 0, and ξ = ξ − 1 in the Lemma (1), we have a result [19]. Theorem 1. Let ϕ : [ω, τ ] → R be a positive function with 0 ≤ ω < α < τ and ϕ ∈ L[ω, τ ]. If ϕ is s-convex function on [ω, τ ], and µ, ξ, ζ, ς, c, κ1 ∈ C,ℜ(µ) > 0,ℜ(ξ) > 0,ℜ(ζ) > 0,ℜ(ς) > 0,ℜ(κ1) > 0, ρ,m, η ≥ 0 and m, ρ > ℜ(µ) + η, then we get the following inequality ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α+ϕ ) (ω, κ) (α− ω)ξ′+µn+1 + ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,τ− ϕ ) (ω, κ) (α− τ)ξ′+µn+1 ≤ ϕ(α)Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ β(ξ ′ + µn+ s+ 1, 1) + β(ξ ′ + s, 1) ] + [ ϕ(ω) + ϕ(τ) ] Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)β(ξ ′ + 1, s+ 1). Proof. From definition of s-convexity of ϕ, ϕ(℘α+ (1− ℘)ω) + ϕ(℘α+ (1− ℘)τ) ≤ ℘sϕ(α) + (1− ℘)sϕ(ω) + ℘sϕ(α) + (1− ℘)sϕ(τ) Multiplying both sides with ℘ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p) and integrate w.r.t ℘ on [0, 1]∫ 1 0 ℘ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)ϕ(℘α+ (1− ℘)ω) + ∫ 1 0 ℘ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)ϕ(℘α+ (1− ℘)τ) ≤ ∫ 1 0 ℘ξ ′ +sEµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)ϕ(α) + ∫ 1 0 ℘ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)(1− ℘)sϕ(ω) +∫ 1 0 ℘ξ ′ +sEµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)ϕ(α) + ∫ 1 0 ℘ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)(1− ℘)sϕ(τ) (15) Now, solving the following integral∫ 1 0 ℘ξ ′ +sEµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)ϕ(α) = ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n [ ϕ(α) ∫ 1 0 ℘ξ ′ +µn+sd℘ ] M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 7 of 23 = ϕ(α) ξ′ + µn+ s+ 1 Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) = ϕ(α)Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)β(ξ ′ + µn+ s+ 1, 1). (16) Similarly, we resolve another fractional integral∫ 1 0 ℘ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)(1− ℘)sϕ(ω) = ϕ(ω)Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)β(ξ ′ + µn+ 1, s+ 1) (17) ∫ 1 0 ℘ξ ′ +sEµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)ϕ(α) = ϕ(α)Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)β(ξ ′ + µn+ s, 1) (18) ∫ 1 0 ℘ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)(1− ℘)sϕ(τ) = ϕ(τ)Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)β(ξ ′ + µn+ 1, s+ 1) (19) Putting the values (16), (17), (18) and (19) in equation (15), we have∫ 1 0 ℘ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)ϕ(℘α+ (1− ℘)ω) + ∫ 1 0 ℘ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)ϕ(℘α+ (1− ℘)τ) ≤ ϕ(α)Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ β(ξ ′ + µn+ s+ 1, 1) + β(ξ ′ + µn+ s, 1) ] +[ ϕ(ω) + ϕ(τ) ] Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)β(ξ ′ + µn+ 1, s+ 1). (20) By substitution ℘α + (1 − ℘)ω = x, and ℘α + (1 − ℘)τ = ℓ in left side of equation (20), and then simplify, we have the required result. Corollary 2. If we replace p = 0, κ = 0, and ξ = ξ − 1 in the Theorem (1), we have a result [19]. Theorem 2. Let ϕ : I ⊆ R → R be a differentiable mapping on Io and ω, τ ∈ Io with ω < α < τ such that ϕ ′ ∈ L[ω, τ ]. If |ϕ′ | is s-convex function on [ω, τ ], and µ, ξ, ζ, ς, c, κ1 ∈ C,ℜ(µ) > 0,ℜ(ξ) > 0,ℜ(ζ) > 0,ℜ(ς) > 0,ℜ(κ1) > 0, ρ,m, η ≥ 0 and m, ρ > ℜ(µ) + η, then we have the following integral inequality in result∣∣∣∣ϕ(α)Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ 1 α−ω + 1 α−τ ] − (ξ ′ + µn) [ 1 (α−ω)ξ ′+1 ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α+ϕ ) (ω, κ) + 1 (α−τ)ξ ′+1 ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,τ− ϕ ) (ω, κ) ]∣∣∣∣ ≤ 2 ∣∣ϕ′ (α) ∣∣Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)β(ξ ′ + µn+ s+ 1, 1) + [∣∣ϕ′ (ω) ∣∣+ ∣∣ϕ′ (τ) ∣∣]Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)β(ξ ′ + µn+ 1, s+ 1) Proof. Using Lemma 1∣∣∣∣ϕ(α)Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ 1 α−ω + 1 α−τ ] − (ξ ′ + µn) [ 1 (α−ω)ξ ′+1 ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α+ϕ ) (ω, κ1) + 1 (α−τ)ξ ′+1 ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,τ− ϕ ) (ω, κ1) ]∣∣∣∣ = ∫ 1 0 ℘ ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p) ∣∣ϕ′ (℘α+ (1− ℘)ω ∣∣d℘+ ∫ 1 0 ℘ ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p) ∣∣ϕ′ (℘α+ (1− ℘)τ ∣∣d℘ M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 8 of 23 Utilizing the definition of s-convex function on |ϕ′ |, we have ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣[∣∣ϕ′ (α) ∣∣ ∫ 1 0 ℘ξ ′ +µn+sd℘+ ∣∣ϕ′ (ω) ∣∣ ∫ 1 0 ℘ξ ′ +µn(1− ℘)sd℘ ] + ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣[∣∣ϕ′ (α) ∣∣ ∫ 1 0 ℘ξ ′ +µn+sd℘+ ∣∣ϕ′ (τ) ∣∣ ∫ 1 0 ℘ξ ′ +µn(1− ℘)sd℘ ] ≤ ∣∣ϕ′ (α) ∣∣Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)β(ξ ′ + µn+ s+ 1, 1) + ∣∣ϕ′ (ω) ∣∣Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)β(ξ ′ + µn+ 1, s+ 1) +∣∣ϕ′ (α) ∣∣Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)β(ξ ′ + µn+ s, 1) + ∣∣ϕ′ (τ) ∣∣Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)β(ξ ′ + µn+ 1, s+ 1) ≤ 2 ∣∣ϕ′ (α) ∣∣Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)β(ξ ′ + µn+ s+ 1, 1) + [∣∣ϕ′ (ω) ∣∣+ ∣∣ϕ′ (τ) ∣∣]Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)β(ξ ′ + µn+ 1, s+ 1) The proof is completed. Corollary 3. If we replace p = 0, κ = 0, and ξ = ξ − 1 in the Theorem (2), we have a result [19]. Theorem 3. Let ϕ : I ⊆ R → R be a differentiable mapping on Io and ω, τ ∈ Io with ω < α < τ such that ϕ ′ ∈ L[ω, τ ]. If |ϕ′ |q(q > 0) is s-convex function on [ω, τ ], and µ, ξ, ζ, ς, c, κ1 ∈ C,ℜ(µ) > 0,ℜ(ξ) > 0,ℜ(ζ) > 0,ℜ(ς) > 0,ℜ(κ1) > 0, ρ,m, η ≥ 0 and m, ρ > ℜ(µ) + η, then we have the following integral inequality in result;∣∣∣∣ϕ(α)Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ 1 α− ω + 1 α− τ ] − (ξ ′ + µn) [ 1 (α− ω)ξ ′+1 ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α+ϕ ) (ω, κ) + 1 (α− τ)ξ ′+1 ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,τ− ϕ ) (ω, κ) ]∣∣∣∣ ≤ Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) ( 1 ξ′ + µn+ 1 )1− 1 q [( ∣∣ϕ′ (α) ∣∣q ξ′ + µn+ s+ 1 + ∣∣ϕ′ (ω) ∣∣qβ(ξ′ + µn+ s+ 1, 1) ) 1 q +( ∣∣ϕ′ (α) ∣∣q ξ′ + µn+ s+ 1 + ∣∣ϕ′ (τ) ∣∣qβ(ξ′ + µn+ s+ 1, 1) ) 1 q ] . Proof. By using Lemma 1 ℵ := ∣∣∣∣ϕ(α)Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ 1 α−ω + 1 α−τ ] − (ξ ′ + µn) [ 1 (α−ω)ξ ′+1 ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α+ϕ ) (ω, κ) + 1 (α−τ)ξ ′+1 ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,τ− ϕ ) (ω, κ) ]∣∣∣∣ ≤ ∫ 1 0 ℘ ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p) ∣∣ϕ′ (℘α+ (1− ℘)ω ∣∣d℘+ ∫ 1 0 ℘ ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p) ∣∣ϕ′ (℘α+ (1− ℘)τ ∣∣d℘ M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 9 of 23 Applying Power mean inequality; ℵ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣[(∫ 1 0 ℘ξ ′ +µnd℘ )1− 1 q (∫ 1 0 ℘ξ ′ +µn ∣∣ϕ′ (℘α+ (1− ℘)ω ∣∣qd℘) 1 q ] + ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣[(∫ 1 0 ℘ξ ′ +µnd℘ )1− 1 q (∫ 1 0 ℘ξ ′ +µn ∣∣ϕ′ (℘α+ (1− ℘)τ ∣∣qd℘) 1 q ] As ∣∣ϕ′∣∣q is s-convex function; ℵ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣[( 1 ξ′ + µn+ 1 )1− 1 q (∣∣ϕ′ (α) ∣∣q ∫ 1 0 ℘ξ ′ +µn+sd℘+ ∣∣ϕ′ (ω) ∣∣q ∫ 1 0 ℘ξ ′ +µn(1− ℘)sd℘ ) 1 q ] + ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣[( 1 ξ′ + µn+ 1 )1− 1 q (∣∣ϕ′ (α) ∣∣q ∫ 1 0 ℘ξ ′ +µn+sd℘ + ∣∣ϕ′ (τ) ∣∣q ∫ 1 0 ℘ξ ′ +µn(1− ℘)sd℘ ) 1 q ] . After simplification, we have ℵ ≤ Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [( 1 ξ′ + µn+ 1 )1− 1 q ( ∣∣ϕ′ (α) ∣∣q ξ′ + µn+ s+ 1 + ∣∣ϕ′ (ω) ∣∣qβ(ξ′ + µn+ s+ 1, 1) ) 1 q ] +Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [( 1 ξ′ + µn+ 1 )1− 1 q ( ∣∣ϕ′ (α) ∣∣q ξ′ + µn+ s+ 1 + ∣∣ϕ′ (τ) ∣∣qβ(ξ′ + µn+ s+ 1, 1) ) 1 q ] . Now, the final result is obtained as follows ℵ ≤ Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) ( 1 ξ′ + µn+ 1 )1− 1 q [( ∣∣ϕ′ (α) ∣∣q ξ′ + µn+ s+ 1 + ∣∣ϕ′ (ω) ∣∣qβ(ξ′ + µn+ s+ 1, 1) ) 1 q + ( ∣∣ϕ′ (α) ∣∣q ξ′ + µn+ s+ 1 + ∣∣ϕ′ (τ) ∣∣qβ(ξ′ + µn+ s+ 1, 1) ) 1 q ] . Corollary 4. If we replace p = 0, κ = 0, and ξ = ξ − 1 in Theorem (3), we have a result [19]. Theorem 4. Let ϕ : I ⊆ R → R be a differentiable mapping on Io and ω, τ ∈ Io with ω < α < τ such that ϕ ′ ∈ L[ω, τ ]. If |ϕ′ |q is s-convex function on [ω, τ ], and µ, ξ, ζ, ς, c, κ1 ∈ C,ℜ(µ) > 0,ℜ(ξ) > 0,ℜ(ζ) > 0,ℜ(ς) > 0,ℜ(κ1) > 0, ρ,m, η ≥ 0 and m, ρ > ℜ(µ) + η, then following fractional integral inequality holds; M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 10 of 23 ∣∣∣∣ϕ(α)Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ 1 α− ω + 1 α− τ ] − (ξ ′ + µn) [ 1 (α− ω)ξ ′+1 ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α+ϕ ) (ω, κ) + 1 (α− τ)ξ ′+1 ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,τ− ϕ ) (ω, κ) ]∣∣∣∣ ≤ Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) ( 1 pξ′ + pµn+ 1 ) 1 p [(∣∣ϕ′ (α) ∣∣q 1 s+ 1 + ∣∣ϕ′ (ω) ∣∣q 1 s+ 1 ) 1 q + (∣∣ϕ′ (α) ∣∣q 1 s+ 1 + ∣∣ϕ′ (τ) ∣∣q 1 s+ 1 ) 1 q ] . Proof. According to Lemma 1 ℵ = ∣∣∣∣ϕ(α)Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ 1 α− ω + 1 α− τ ] − (ξ ′ + µn) [ 1 (α− ω)ξ ′+1 ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α+ϕ ) (ω, κ) + 1 (α− τ)ξ ′+1 ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,τ− ϕ ) (ω, κ) ]∣∣∣∣ ≤ ∫ 1 0 ℘ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p) ∣∣ϕ′ (℘α+ (1− ℘)ω ∣∣d℘+ ∫ 1 0 ℘ξ ′ Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p) ∣∣ϕ′ (℘α+ (1− ℘)τ ∣∣d℘. Applying Holder’s inequality; ℵ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣[(∫ 1 0 ℘pξ ′ +pµnd℘ ) 1 p (∫ 1 0 ∣∣ϕ′ (℘α+ (1− ℘)ω ∣∣qd℘) 1 q ] + ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣[(∫ 1 0 ℘pξ ′ +pµnd℘ ) 1 q (∫ 1 0 ∣∣ϕ′ (℘α+ (1− ℘)τ ∣∣qd℘) 1 q ] . As ∣∣ϕ′∣∣q is s-convex function; ℵ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣[( 1 pξ′ + pµn+ 1 ) 1 p (∣∣ϕ′ (α) ∣∣q ∫ 1 0 ℘sd℘ + ∣∣ϕ′ (ω) ∣∣q ∫ 1 0 (1− ℘)sd℘ ) 1 q ] + ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣[( 1 pξ′ + pµn+ 1 ) 1 p (∣∣ϕ′ (α) ∣∣q ∫ 1 0 ℘sd℘ + ∣∣ϕ′ (τ) ∣∣q ∫ 1 0 (1− ℘)sd℘ ) 1 q ] . ℵ ≤ Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) ( 1 pξ′ + pµn+ 1 ) 1 p [∣∣ϕ′ (α) ∣∣q 1 s+ 1 + ∣∣ϕ′ (ω) ∣∣q 1 s+ 1 ] 1 q + M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 11 of 23 Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) ( 1 pξ′ + pµn+ 1 ) 1 p [∣∣ϕ′ (α) ∣∣q 1 s+ 1 + ∣∣ϕ′ (τ) ∣∣q 1 s+ 1 ] 1 q The proof is completed. Corollary 5. If we replace p = 0, κ = 0, and ξ = ξ − 1 in Theorem (4), we have an inequality [19]. 4. Behavior of Hermite-Hadamard type fractional integral inequalities for the class of twice differentiable function In this section, we develop the lemma for twice differentiable s-convex function with extended Bessel-Maitland function as a kernel which is helpful to prove our main results. Lemma 2. Let ϕ : I ⊆ R → R be twice differentiable mapping on Io and let, if ω, τ ∈ Io with ω < α < τ such that ϕ ′′ ∈ L[ω, τ ], and µ, ξ, ζ, ς, c, κ1 ∈ C,ℜ(µ) > 0,ℜ(ξ) > 0,ℜ(ζ) > 0,ℜ(ς) > 0,ℜ(κ1) > 0, ρ,m, η ≥ 0 and m, ρ > ℜ(µ) + η, then we get the following result; ( ξ ′) Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ϕ(α) + ϕ(ω) 2 ] + [ (µn+ 1)(µn) 2(α− ω)µn − (ξ ′ + µn+ 1)(ξ ′ + µn) 2(α− ω)ξ ′+µn ] ×[( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,ω+ϕ ) (α, κ) + ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α−ϕ ) (ω, κ) ] = (α− ω)2 2 ∫ 1 0 ℘(1− ℘ξ ′ )Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p) [ ϕ ′′( ℘ω + (1− ℘)α+ ϕ ′′( (1− ℘)ω + ℘α )] d℘. Proof. Consider the integral I = ∫ 1 0 ℘(1− ℘ξ ′ )Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p) [ ϕ ′′( ℘ω + (1− ℘)α+ ϕ ′′( (1− ℘)ω + ℘α )] d℘ I = ∫ 1 0 ℘(1− ℘ξ ′ )Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)ϕ ′′( ℘ω + (1− ℘)α ) d℘ + ∫ 1 0 ℘(1− ℘ξ ′ )Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)ϕ ′′( (1− ℘)ω + ℘α ) d℘ (21) I = I1 + I2. Consider the integral I1 I1 = ∫ 1 0 ℘(1− ℘ξ ′ )Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p)ϕ ′′( ℘ω + (1− ℘)α ) d℘ = ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∫ 1 0 (℘− ℘ξ ′ +1)℘µnϕ ′′( ℘ω + (1− ℘)α ) d℘ M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 12 of 23 = ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∫ 1 0 (℘µn+1 − ℘ξ ′ +µn+1)ϕ ′′( ℘ω + (1− ℘)α ) d℘ = ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n [ (℘µn+1 − ℘ξ ′ +µn+1) ϕ ′( ℘ω + (1− ℘)α ) ω − α ∣∣∣∣1 0 − ∫ 1 0 ( (µn+ 1)℘µn − (ξ ′ + µn+ 1)℘ξ ′ +µn ) ϕ ′( ℘ω + (1− ℘)α ) ω − α d℘ ] . After simplification, we have I1 = ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n [ (µn+ 1) α− ω ∫ 1 0 ℘µnϕ ′( ℘ω + (1− ℘)α ) d℘ ] − ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n [ (ξ ′ + µn+ 1) α− ω ∫ 1 0 ℘ξ ′ +µnϕ ′( ℘ω + (1− ℘)α ) d℘ ] (22) I1 = I3 − I4 Taking I3 from above equation I3 = ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n [ (µn+ 1) α− ω ∫ 1 0 ℘µnϕ ′( ℘ω + (1− ℘)α ) d℘ ] = ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n [ (µn+ 1) α− ω { ℘µn ϕ ( ℘ω + (1− ℘)α ) ω − α ∣∣∣∣1 0 − ∫ 1 0 (µn)℘µn−1ϕ ( ℘ω + (1− ℘)α ) ω − α d℘ }] . I3 = ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n [ (µn+ 1) α− ω { − ϕ(ω) α− ω + ∫ 1 0 (µn)℘µn−1ϕ ( ℘ω + (1− ℘)α ) α− ω d℘ }] = Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ (µn+ 1) (α− ω)2 ( − ϕ(ω) )] + ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n [ (µn+ 1)(µn) (α− ω)2 ×∫ 1 0 ℘µn−1ϕ ( ℘ω + (1− ℘)α ) d℘ ] Use substitution ℘ω+(1−℘)α = ℓ in above equation and after solving, we get the result; I3 = Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ (µn+ 1) (α− ω)2 ( − ϕ(ω) )] + (µn+ 1)(µn) (α− ω)2+µn ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,ω+ϕ ) (α, κ1) (23) M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 13 of 23 Now, solving I4 from equation (22) I4 = ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n [ (ξ ′ + µn+ 1) α− ω ∫ 1 0 ℘ξ ′ +µnϕ ′( ℘ω + (1− ℘)α ) d℘ ] = ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n [ (ξ ′ + µn+ 1) α− ω { ℘ξ ′ +µnϕ ( ℘ω + (1− ℘)α ) ω − α ∣∣∣∣1 0 − ∫ 1 0 (ξ ′ + µn)℘ξ ′ +µn−1ϕ ( ℘ω + (1− ℘)α ) ω − α d℘ }] = ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n [ (ξ ′ + µn+ 1) α− ω { − ϕ(ω) α− ω + ∫ 1 0 (ξ ′ + µn)℘ξ ′ +µn−1ϕ ( ℘ω + (1− ℘)α ) α− ω d℘ }] = Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ (ξ ′ + µn+ 1) (α− ω)2 ( − ϕ(ω) )] + ∞∑ n=0 βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ×[ (ξ ′ + µn+ 1)(ξ ′ + µn) (α− ω)2 ∫ 1 0 ℘ξ ′ +µn−1ϕ ( ℘ω + (1− ℘)α ) d℘ ] Use substitution ℘ω+(1−℘)α = ℓ in above equation and after solving, we get the result; I4 = Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ (ξ ′ + µn+ 1) (α− ω)2 ( − ϕ(ω) )] + (ξ ′ + µn+ 1)(ξ ′ + µn) (α− ω)2+ξ ′+µn ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,ω+ϕ ) (α, κ1)(24) Using equation (23), (24) in (22), we have I1 = ( ξ ′ (α− ω)2 ) Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)[ϕ(ω)] + ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,ω+ϕ ) (α, κ1) [ (µn+ 1)(µn) (α− ω)2+µn − (ξ ′ + µn+ 1)(ξ ′ + µn) (α− ω)2+ξ ′+µn ] .(25) Similarly, we solve I2 and get the result; I2 = ( ξ ′ (α− ω)2 ) Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p)[ϕ(α)] + ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α−ϕ ) (ω, κ) [ (µn+ 1)(µn) (α− ω)2+µn − (ξ ′ + µn+ 1)(ξ ′ + µn) (α− ω)2+ξ ′+µn ] (26) Combining equations (25) and (26), then simplify, we have I = ( ξ ′) Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ϕ(α) + ϕ(ω) 2 ] + [ (µn+ 1)(µn) 2(α− ω)µn − (ξ ′ + µn+ 1)(ξ ′ + µn) 2(α− ω)ξ ′+µn ] ×[( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,ω+ϕ ) (α, κ) + ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α−ϕ ) (ω, κ) ] . Hence the required result is proved. M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 14 of 23 Corollary 6. If we replace p = 0, κ = 0, and ξ = ξ − 1 in the Lamma (2), we have a result [19]. Theorem 5. Let ϕ : I ⊆ R → R be twice differentiable mapping on Io and ω, τ ∈ Io with ω < α < τ such that ϕ ′′ ∈ L[ω, τ ]. If |ϕ′′ | is s-convex function on I, and µ, ξ, ζ, ς, c, κ1 ∈ C,ℜ(µ) > 0,ℜ(ξ) > 0,ℜ(ζ) > 0,ℜ(ς) > 0,ℜ(κ1) > 0, ρ,m, η ≥ 0 and m, ρ > ℜ(µ) + η, then following fractional integral inequality holds;∣∣∣∣(ξ′)Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ϕ(α) + ϕ(ω) 2 ] + [ (µn+ 1)(µn) 2(α− ω)µn − (ξ ′ + µn+ 1)(ξ ′ + µn) 2(α− ω)ξ ′+µn ] ×[( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,ω+ϕ ) (α, κ) + ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α−ϕ ) (ω, κ) ]∣∣∣∣ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 [( |ϕ′′ (ω) + |ϕ′′ (α)| )[ β(µn+ 2 + s, ξ ′ + 1) +β(µn+ 2, ξ ′ + s+ 1) ]] . Proof. By using Lemma 2 ℶ := ∣∣∣∣(ξ′)Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ϕ(α) + ϕ(ω) 2 ] + [ (µn+ 1)(µn) 2(α− ω)µn − (ξ ′ + µn+ 1)(ξ ′ + µn) 2(α− ω)ξ ′ +µn ] ×[( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,ω+ϕ ) (α, κ) + ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α−ϕ ) (ω, κ) ]∣∣∣∣ ≤ (α− ω)2 2 ∫ 1 0 ℘(1− ℘ξ ′ )Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p) ∣∣[ϕ′′( ℘ω + (1− ℘)α+ ϕ ′′( (1− ℘)ω + ℘α )]∣∣d℘ ℵ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 ×[ ∫ 1 0 ℘µn+1(1− ℘ξ ′ ) ∣∣[ϕ′′( ℘ω + (1− ℘)α+ ϕ ′′( (1− ℘)ω + ℘α )]∣∣]d℘ ℶ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 ×[ ∫ 1 0 ℘µn+1(1− ℘ξ ′ ) ∣∣[ϕ′′( ℘ω + (1− ℘)α) ]∣∣d℘+ ∫ 1 0 ℘µn+1(1− ℘ξ ′ ) ∣∣[ϕ′′( (1− ℘)ω + ℘α )]∣∣]d℘ As ∣∣ϕ′′∣∣ is s-convex function; ℶ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 × M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 15 of 23[ ∫ 1 0 ℘µn+1(1− ℘ξ ′ ) ∣∣[℘s|ϕ′′ (ω)|+ (1− ℘)s|ϕ′′ (α)| ] d℘+∫ 1 0 ℘µn+1(1− ℘ξ ′ ) ∣∣[(1− ℘)s|ϕ′′ (ω)|+ ℘s|ϕ′′ (α)| ]] d℘ Since ℘ξ ′ ≥ ℘, ξ ′ ∈ (0, 1] and ℘ ∈ [0, 1], we have −℘ξ ′ ≤ ℘⇒ 1− ℘ξ ′ ≤ 1− ℘ ≤ (1− ℘)ξ ′ ℶ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 [ |ϕ′′ (ω)| ∫ 1 0 ℘µn+1+s(1− ℘)ξ ′ d℘+ |ϕ′′ (α)| ∫ 1 0 ℘µn+1(1− ℘)ξ ′ +sd℘+ |ϕ′′ (ω)| ∫ 1 0 ℘µn+1(1− ℘)ξ ′ +sd℘+ |ϕ′′ (α)| ∫ 1 0 ℘µn+1+s(1− ℘)ξ ′ d℘ ]] ℶ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 [ |ϕ′′ (ω)|β(µn+ 2 + s, ξ ′ + 1) + |ϕ′′ (α)|β(µn+ 2, ξ ′ + s+ 1) + |ϕ′′ (ω)β(µn+ 2, ξ ′ + s+ 1) + |ϕ′′ (α)|β(µn+ 2 + s, ξ ′ + 1) ] ℶ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 [( |ϕ′′ (ω) + |ϕ′′ (α)| )[ β(µn+ 2 + s, ξ ′ + 1) +β(µn+ 2, ξ ′ + s+ 1) ]] Corollary 7. If we replace p = 0, κ = 0, and ξ = ξ − 1 in Theorem (5), we have a result [19]. Theorem 6. Let ϕ : I ⊆ R → R be twice differentiable mapping on Io and ω, τ ∈ Io with ω < α < τ such that ϕ ′′ ∈ L[ω, τ ]. If |ϕ′′ |q(q > 1) is s-convex function on I, and µ, ξ, ζ, ς, c, κ1 ∈ C,ℜ(µ) > 0,ℜ(ξ) > 0,ℜ(ζ) > 0,ℜ(ς) > 0,ℜ(κ1) > 0, ρ,m, η ≥ 0 and m, ρ > ℜ(µ) + η, then we have the following fractional inequality ∣∣∣∣(ξ′)Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ϕ(α) + ϕ(ω) 2 ] + [ (µn+ 1)(µn) 2(α− ω)µn − (ξ ′ + µn+ 1)(ξ ′ + µn) 2(α− ω)ξ ′+µn ] ×[( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,ω+ϕ ) (α, κ) + ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α−ϕ ) (ω, κ) ]∣∣∣∣ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 × [( β(µnp+ p+ 1, ξ ′ p+ 1) ) 1 p ( |ϕ′′ (ω)|q + |ϕ′′ (α)|q s+ 1 ) 1 q ] . (27) M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 16 of 23 Proof. By using Lemma 2 ℶ = ∣∣∣∣(ξ′)Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ϕ(α) + ϕ(ω) 2 ] + [ (µn+ 1)(µn) 2(α− ω)µn − (ξ ′ + µn+ 1)(ξ ′ + µn) 2(α− ω)ξ ′+µn ] ×[( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,ω+ϕ ) (α, κ) + ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α−ϕ ) (ω, κ) ]∣∣∣∣ ≤ (α− ω)2 2 ∫ 1 0 ℘(1− ℘ξ ′ )Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p) ∣∣[ϕ′′( ℘ω + (1− ℘)α+ ϕ ′′( (1− ℘)ω + ℘α )]∣∣d℘ ℶ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 ×[ ∫ 1 0 ℘µn+1(1− ℘ξ ′ ) ∣∣[ϕ′′( ℘ω + (1− ℘)α+ ϕ ′′( (1− ℘)ω + ℘α )]∣∣]d℘ ℶ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 ×[ ∫ 1 0 ℘µn+1(1− ℘ξ ′ ) ∣∣[ϕ′′( ℘ω + (1− ℘)α) ]∣∣d℘+ ∫ 1 0 ℘µn+1(1− ℘ξ ′ ) ∣∣[ϕ′′( (1− ℘)ω + ℘α )]∣∣]d℘. Using Holder inequality, ℶ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 × [(∫ 1 0 ℘µnp+p(1− ℘ξ ′ )p ) 1 p (∫ 1 0 ∣∣[ϕ′′( ℘ω + (1− ℘)α) ]∣∣q) 1 q d℘+(∫ 1 0 ℘µnp+p(1− ℘ξ ′ )p ) 1 p (∫ 1 0 ∣∣[ϕ′′( (1− ℘)ω + ℘α )]∣∣q) 1 q d℘ ] d℘ (28) Since, ℘ξ ′ ≥ ℘, ξ ′ ∈ (0, 1] and ℘ ∈ [0, 1], we have −℘ξ ′ ≤ ℘⇒ 1− ℘ξ ′ ≤ 1− ℘ ≤ (1− ℘)ξ ′ ℶ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 × [(∫ 1 0 ℘µnp+p(1− ℘)ξ ′ p ) 1 p {(∫ 1 0 ∣∣[ϕ′′( ℘ω + (1− ℘)α) ]∣∣q) 1 q d℘+(∫ 1 0 ∣∣[ϕ′′( (1− ℘)ω + ℘α )]∣∣q) 1 q d℘ }] (29) As ∣∣ϕ′′∣∣q is s-convex function;∫ 1 0 ∣∣[ϕ′′( ℘ω + (1− ℘)α) ]∣∣qd℘ ≤ |ϕ′′ (ω)|q ∫ 1 0 ℘sd℘+ |ϕ′′ (α)|q ∫ 1 0 (1− ℘)sd℘ M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 17 of 23 = |ϕ′′ (ω)|q + |ϕ′′ (α)|q s+ 1 (30) ∫ 1 0 ∣∣[ϕ′′( (1− ℘)ω + ℘α )]∣∣qd℘ ≤ ϕ ′′ (ω)|q ∫ 1 0 (1− ℘)sd℘+ |ϕ′′ (α)|q ∫ 1 0 ℘sd℘ = |ϕ′′ (ω)|q + |ϕ′′ (α)|q s+ 1 (31) β(µnp+ p+ 1, ξ ′ p+ 1) = ∫ 1 0 ℘µnp+p(1− ℘)ξ ′ pd℘ (32) Substitute equations (30), (31),(32) in equation (29), then we get ℶ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 × [( β(µnp+ p+ 1, ξ ′ p+ 1) ) 1 p {( |ϕ′′ (ω)|q + |ϕ′′ (α)|q s+ 1 ) 1 q + ( |ϕ′′ (ω)|q + |ϕ′′ (α)|q s+ 1 ) 1 q }] The proof is completed. Corollary 8. If we replace p = 0, κ = 0, and ξ = ξ − 1 in Theorem (6), we have a result [19]. Theorem 7. Let ϕ : I ⊆ R → R be twice differentiable mapping on Io and ω, τ ∈ Io with ω < α < τ such that ϕ ′′ ∈ L[ω, τ ]. If |ϕ′′ |q(q ≥ 1) is s-convex function on I, and µ, ξ, ζ, ς, c, κ1 ∈ C,ℜ(µ) > 0,ℜ(ξ) > 0,ℜ(ζ) > 0,ℜ(ς) > 0,ℜ(κ1) > 0, ρ,m, η ≥ 0 and m, ρ > ℜ(µ) + η, then following integral inequality holds as a result; ∣∣∣∣(ξ′)Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ϕ(α) + ϕ(ω) 2 ] + [ (µn+ 1)(µn) 2(α− ω)µn − (ξ ′ + µn+ 1)(ξ ′ + µn) 2(α− ω)ξ ′+µn ] ×[( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,ω+ϕ ) (α, κ) + ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α−ϕ ) (ω, κ) ]∣∣∣∣ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 [( ξ ′ (µn+ 2)(ξ′ + µn+ 2) )1− 1 q × {( |ϕ′′ (ω)|qβ(µn+ s+ 2, ξ ′ + 1) + |ϕ′′ |q(α)β(µn+ 2, ξ ′ + s+ 1) ) 1 q +( |ϕ′′ (ω)|qβ(µn+ 2, ξ ′ + s+ 1) + |ϕ′′ |qβ(µn+ s+ 2, ξ ′ + 1) ) 1 q }] M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 18 of 23 Proof. By using Lemma 2, we have ℶ = ∣∣∣∣(ξ′)Jµ,ρ,m,η,cξ,ζ,ς,κ1 (κ, p) [ϕ(α) + ϕ(ω) 2 ] + [ (µn+ 1)(µn) 2(α− ω)µn − (ξ ′ + µn+ 1)(ξ ′ + µn) 2(α− ω)ξ ′+µn ] ×[( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,ω+ϕ ) (α, κ) + ( Eµ,ρ,m,η,c ξ,ζ,ς,κ1,α−ϕ ) (ω, κ) ]∣∣∣∣ ≤ (α− ω)2 2 ∫ 1 0 ℘(1− ℘ξ ′ )Eµ,ρ,m,η,cξ,ζ,ς,κ1 (κ℘µ; p) ∣∣[ϕ′′( ℘ω + (1− ℘)α+ ϕ ′′( (1− ℘)ω + ℘α )]∣∣d℘. ℶ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 ×[ ∫ 1 0 ℘µn+1(1− ℘ξ ′ ) ∣∣[ϕ′′( ℘ω + (1− ℘)α) ]∣∣d℘+ ∫ 1 0 ℘µn+1(1− ℘ξ ′ ) ∣∣[ϕ′′( (1− ℘)ω + ℘α )]∣∣]d℘. Using Power mean inequality, ℶ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 × [(∫ 1 0 ℘µn+1(1− ℘ξ ′ ) )1− 1 q (∫ 1 0 ℘µn+1(1− ℘ξ ′ ) ∣∣[ϕ′′( ℘ω + (1− ℘)α) ]∣∣q) 1 q d℘+(∫ 1 0 ℘µn+1(1− ℘ξ ′ ) )1− 1 q (∫ 1 0 ℘µn+1(1− ℘ξ ′ ) ∣∣[ϕ′′( (1− ℘)ω + ℘α )]∣∣q) 1 q d℘ ] (33) Using Modulus property, we have ℶ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 × [(∫ 1 0 ℘µn+1(1− ℘ξ ′ ) )1− 1 q {(∫ 1 0 ℘µn+1(1− ℘ξ ′ ) ∣∣[ϕ′′( ℘ω + (1− ℘)α) ]∣∣q) 1 q d℘+(∫ 1 0 ℘µn+1(1− ℘ξ ′ ) ∣∣[ϕ′′( (1− ℘)ω + ℘α )]∣∣q) 1 q d℘ }] (34) Now, consider the integral∫ 1 0 ℘µn+1(1− ℘ξ ′ )d℘ = ∫ 1 0 (℘µn+1 − ℘ξ ′ +µn+1)d℘ = ξ ′ (µn+ 2)(ξ′ + µn+ 2) (35) Putting the value (35) in equation (34), we have ℶ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 [( ξ ′ (µn+ 2)(ξ′ + µn+ 2) )1− 1 q × M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 19 of 23{(∫ 1 0 ℘µn+1(1− ℘ξ ′ )[|ϕ′′ (ω)|q℘s + |ϕ′′ (α)|q(1− ℘)s]d℘ ) 1 q +(∫ 1 0 ℘µn+1(1− ℘ξ ′ )[|ϕ′′ (ω)|q(1− ℘)s + |ϕ′′ (α)|q℘s]d℘ ) 1 q }] (36) Since ℘ξ ′ ≥ ℘, ξ ′ ∈ (0, 1] and ℘ ∈ [0, 1], we have −℘ξ ′ ≤ ℘⇒ 1− ℘ξ ′ ≤ 1− ℘ ≤ (1− ℘)ξ ′ Simplify; ∫ 1 0 ℘µn+1(1− ℘)ξ ′ [|ϕ′′ (ω)|q℘s + |ϕ′′ (α)|q(1− ℘)s]d℘ = |ϕ′′ (ω)|q ∫ 1 0 ℘µn+s+1(1− ℘)ξ ′ d℘+ |ϕ′′ (α)|q ∫ 1 0 ℘µn+1(1− ℘)ξ ′ +s = |ϕ′′ (ω)|qβ(µn+ s+ 2, ξ ′ + 1) + |ϕ′′ |q(α)β(µn+ 2, ξ ′ + s+ 1). (37) Consider the integral∫ 1 0 ℘µn+1(1− ℘)ξ ′ [|ϕ′′ (ω)|q(1− ℘)s + |ϕ′′ (α)|q℘s]d℘ = |ϕ′′ (ω)|q ∫ 1 0 ℘µn+1(1− ℘)ξ ′ +sd℘+ |ϕ′′ (α)|q ∫ 1 0 ℘µn+s+1(1− ℘)ξ ′ d℘ = |ϕ′′ (ω)|qβ(µn+ 2, ξ ′ + s+ 1) + |ϕ′′ |qβ(µn+ s+ 2, ξ ′ + 1) (38) Use equations (37) and (38) in (36) then we get; ℶ ≤ ∞∑ n=0 ∣∣∣∣ βp(ζ + ρn, c− ζ)(cρn)(κ1)ηn β(ζ, c− ζ)Γ(µn+ ξ + 1)(ς)mn (−κ)n ∣∣∣∣(α− ω)2 2 [( ξ ′ (µn+ 2)(ξ′ + µn+ 2) )1− 1 q × {( |ϕ′′ (ω)|qβ(µn+ s+ 2, ξ ′ + 1) + |ϕ′′ |q(α)β(µn+ 2, ξ ′ + s+ 1) ) 1 q +( |ϕ′′ (ω)|qβ(µn+ 2, ξ ′ + s+ 1) + |ϕ′′ |qβ(µn+ s+ 2, ξ ′ + 1) ) 1 q }] The proof is completed. Corollary 9. If we replace p = 0, κ = 0, and ξ = ξ − 1 in Theorem (7), we have a result [19]. 5. Conclusion This study introduced two innovative approaches to proving Hermite-Hadamard type inequalities using the extended Bessel-Maitland function as a kernel within the frame- work of s-convex functions. The first approach used differentiable functions and their M. Vivas-Cortez et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5997 20 of 23 first derivatives to derive a key identity that serves as the basis for establishing these inequalities. Taking a more comprehensive perspective, the second approach employs the integral of a second derivative as an alternative, yet equally effective, way to establish the fundamental identity. Beyond proving these inequalities, we explore various appli- cations, particularly in relation to different types of means. Furthermore, this approach can be extended to other classes of convex functions, which enhances its broader mathe- matical significance. Ultimately, this study deepens the understanding of convexity-based inequalities and paves the way for further research in mathematical analysis. Acknowledgements The authors extend their appreciation to the Deanship of Research and Graduate Studies at King Khalid University, Saudi Arabia for funding this work through Large Groups Project under grant number R.G.P2/76/46. 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