EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 6089 ISSN 1307-5543 – ejpam.com Published by New York Business Global Generalized Fractional Integral Extensions of Hermite-Hadamard Inequalities Saima Naheed1,∗, Adeeba Rafi1, Gauhar Rahman2, Irshad Ayoob3, Nabil Mlaiki3 1 Department of Mathematics, University of Sargodha P.O. Box 40100, Sargodha, Pakistan 2 Department of Mathematics and Statistics, Hazara University, Mansehra 21300, Pakistan 3 Department of Mathematics and Sciences, Prince Sultan University, Riyadh 11586, Saudi Arabia Abstract. In this article, we provide a number of Hermite-Hadamard type fractional integral inequalities for the Atangana-Baleanu and Prabhakar fractional operators, using extended gener- alized Mittag-Leffler functions as their kernel. Significant findings are provided for the integral inequalities involving fractional integrals of the type (ℑ1+,ℑ2−) and (ℑ1+ℑ2 2 ). By employing cer- tain functions to create visual graphs with matching numerical entries that depict the inequalities, we show the veracity of our findings. 2020 Mathematics Subject Classifications: 33E12, 26A33, 26D15 Key Words and Phrases: Atangana-Baleanu fractional calculus, Fractional calculus, Hermite- Hadamard inequality, Generalized fractional integral operators, Mittag-Leffler function 1. Introduction The expansion of differentiation and integration to non-integer orders is the work of fractional calculus, a subfield of mathematics, making it possible to represent complex pro- cesses more intricately [1, 2]. Fractional integral operators are utilized as mathematical tools in the study of fractional calculus because they are integrals of a certain order, which is not limited to integer values but can be any real or complex number. In the solution of fractional differential equations, fractional integral operators are crucial because they generalize several classical operators, including the integral and derivative [3], and they aid in describing processes that are not fully represented by traditional calculus, which is limited to integer orders. Fractional calculus uses definitions such as Riemann-Liouville to extend these operations to arbitrary real or complex numbers [4, 5]. Moreover, fractional ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.6089 Email addresses: saima.naheed@uos.edu.pk (S. Naheed), osa436rj@gmail.com (A. Rafi), gauhar55uom@gmail.com (G. Rehman), iayoub@psu.edu.sa (I. Ayoob), nmlaiki@psu.edu.sa (N. Mlaiki) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 2 of 34 calculus intersects with concepts like convex functions [6], leading to advancements in op- timization and engineering [7, 8]. Overall, fractional calculus offers a robust framework for understanding and analyzing complex phenomena. Giving complex functions boundaries and approximations is an essential part that inequal- ities play in providing important insights into their behavior. Integral inequalities are a key area in mathematical analysis, crucial for studying integral equations and differential equations [9, 10]. They play a crucial role in fractional differential equations, where frac- tional integral inequalities help to establish the uniqueness of solutions and provide bounds for fractional boundary value problems. Inequalities involving fractional derivatives are especially valuable in determining solutions for Cauchy problems as well as their upper limits [11, 12]. The goal of expanding the theory of integral inequalities through the use of fractional integral operators to generalize classical inequalities has been spurred by this significance [13], which improves theoretical comprehension and real-world applications [14, 15]. Sajid et al. have discussed some new Grüss type inequalities associated with generalized fractional derivatives in [16]. Special functions are closely related to fractional calculus in many ways [9, 17], like the Mittag-Leffler function, which extends the concepts of fractional operators [18, 19] and plays a crucial role in fractional calculus [20, 21]. Named after Gösta Mittag-Leffler, this function is essential for solving fractional differential equations. Usually accomplished by adding more parameters to its definition, the extended generalized Mittag-Leffler function is a further expanded form of the standard Mittag-Leffler function that provides more flexibility in modeling complex phenomena [22]. Two prominent models in this area that incorporate Mittag-Leffler functions are the Atangana-Baleanu [23] and Prabhakar models [24, 25]. These models advance fractional calculus by offering refined tools for describing systems with memory, non-singular and non-local effects [26, 27], making them valuable in various scientific and engineering fields [28, 29] while tackling practical issues in a variety of fields [30, 31]. The modified (k, s) fractional integral operator involving k-Mittag-Leffler function along with its properties is discussed in [32]. In this research, the Hermite-Hadamard (H−H) inequality [33] by applying generalized fractional integral operators through the extended generalized Mittag-Leffler function will be studied. The goal is to derive new inequalities that not only generalize but also en- hance the classical (H−H) inequality [34], utilizing fractional calculus and special func- tion methodologies. As we continue our work, it is crucial to remember important definitions to ensure clar- ity and consistency. These concepts help to manage complex activities and initiatives by providing a strong foundation. Definition 1. [7] A function Υ : I → ℜ is known as a convex function if it satisfies the following inequality Υ(cℑ1 + (1− c)ℑ2) ≤ cΥ(ℑ1) + (1− c)Υ(ℑ2), where c ∈ [0, 1] and ℑ1,ℑ2 ∈ I. S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 3 of 34 Definition 2. [4] Let Υ be a function in L1 on the interval [ℑ1,ℑ2]. The G-th order left and right sided Reimann-Liouville integrals for any c ∈ [ℑ1,ℑ2], applied to Υ(c) are defined as, provided that ℜ(G) > 0 R−LIGℑ1+Υ(c) = 1 Γ(G) ∫ c ℑ1 (c− ψ)G−1Υ(ψ)dψ, (1) and R−LIGℑ2−Υ(c) = 1 Γ(G) ∫ ℑ2 c (ψ − c)G−1Υ(ψ)dψ. (2) Definition 3. [23] Let Υ be a function in L1 on the interval [ℑ1,ℑ2]. The G-th order left and right sided Atangana-Baleanu integrals for any c ∈ [ℑ1,ℑ2], applied to Υ(c) and for 1 > G > 0, written as A−BIGℑ1+Υ(c) = G B(G) (R−LIGℑ1+Υ(c) ) + (1−G) B(G) Υ(c), (3) and A−BIGℑ2−Υ(c) = G B(G) (R−LIGℑ2−Υ(c) ) + (1−G) B(G) Υ(c), (4) where B(G) is a normalization function that is both real and positive, having properties B(0) = B(1) = 1. Definition 4. [24, 25] Given a function Υ that belongs to L1 on the interval [ℑ1,ℑ2] and for any c ∈ [ℑ1,ℑ2], then the left and right sided Prabhakar fractional integral operators applied to Υ(c) with ℜ(α∗) > 0, ℜ(β∗) > 0 and γ, ♭ ∈ C, given as PIα ∗,β∗,γ,♭ ℑ1+ Υ(c) = ∫ c ℑ1 (c− ψ)β ∗−1Eγα∗,β∗ ( ♭(c− ψ)α ∗ ) Υ(ψ)dψ, (5) and PIα ∗,β∗,γ,♭ ℑ2− Υ(c) = ∫ ℑ2 c (ψ − c)β ∗−1Eγα∗,β∗ ( ♭(ψ − c)α ∗ ) Υ(ψ)dψ, (6) where Eγα∗,β∗(z) symbolizes the three parameters Mittag-Leffler function. Definition 5. [29] For any function Υ ∈ L1 on the interval [ℑ1,ℑ2] ⊂ R and c ∈ [ℑ1,ℑ2], then the infinite series formula for left and right Prabhakar integrals applied to Υ(c), stated as PIℵ,℘,γ,♭ℑ1+ Υ(c) = ∞∑ ð=0 Γ(γ + ð)♭ð Γ(γ)ð! R−LI (ℵð+℘) ℑ1+ Υ(c), (7) and PIℵ,℘,γ,♭ℑ2− Υ(c) = ∞∑ ð=0 Γ(γ + ð)♭ð Γ(γ)ð! R−LI (ℵð+℘) ℑ2− Υ(c). (8) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 4 of 34 Definition 6. [18] The Mittag-Leffler function with one parameter can be defined as Eℵ(z) = ∞∑ ð=0 zð Γ(ℵð+ 1) , (z ∈ C,ℜ(ℵ) > 0). The first generalization of Mittag-Leffler function for two parameters, is given as Eℵ,℘(z) = ∞∑ ð=0 zð Γ(ℵð+ ℘) , (z,ℵ, ℘ ∈ C,ℜ(ℵ) > 0). Prabhakar defined the Mittag-Leffler function of three parameters [25] as Eδℵ,℘(z) = ∞∑ ð=0 (δ)ð Γ(ℵð+ ℘) zð ð! , (z,ℵ, ℘, δ ∈ C,ℜ(ℵ) > 0). Definition 7. [22] Let ℵ, ℘, τ, δ, c ∈ C, with ℜ(ℵ),ℜ(℘),ℜ(τ) > 0 and ℜ(b) > ℜ(δ) > 0 with 0 ≤ g, 1 > 0 and l + ℜ(ℵ) ≥ s > 0. Then the extended generalized Mittag-Leffler function Eδ,b,s,lℵ,℘,τ (z; g) is defined by Eδ,b,s,lℵ,℘,τ (z; g) = ∞∑ ð=0 Bg(δ + ðs, b− δ) B(δ, b− δ) (b)ðs Γ(ℵð+ ℘) zð (τ)ðl , (9) where (b)ðs = Γ(b+ðs) Γ(b) , is the generalized Pochhammer symbol and Bg(i, j) = ∫ 1 0 t i−1 (1− t)j−1e − g t(1−t)dt with ℜ(i),ℜ(j),ℜ(g) > 0, is an extended beta function. Definition 8. [22] Let ♭,ℵ, ℘, τ, δ, b ∈ C with ℜ(ℵ),ℜ(℘),ℜ(τ) > 0 and ℜ(b) > ℜ(δ) > 0 and let g ≥ 0, l > 0 and l+ℜ(ℵ) ≥ s > 0. For a function Υ ∈ L1[ℑ1,ℑ2] and c ∈ [ℑ1,ℑ2], the left and right sided generalized fractional integral operators ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ, ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ given as ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(c; g) = ∫ c ℑ1 (c− ψ)℘−1Eδ,b,s,lℵ,℘,τ ( ♭(c− ψ)ℵ; g ) Υ(ψ)dψ, (10) and ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ(c; g) = ∫ ℑ2 c (ψ − c)℘−1Eδ,b,s,lℵ,℘,τ ( ♭(ψ − c)ℵ; g ) Υ(ψ)dψ. (11) To derive the main results, we relied on theorems and lemmas provided in references [35], [36] and [37]. Theorem 1. For an L1 continuous convex function Υ : [ℑ1,ℑ2] → ℜ, with ℑ2 > ℑ1, the standard (H−H) inequality is stated as Υ ( ℑ1 + ℑ2 2 ) ≤ 1 ℑ2 −ℑ1 ∫ ℑ2 ℑ1 Υ(c)dc ≤ ( Υ(ℑ1) + Υ(ℑ2) 2 ) . S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 5 of 34 Theorem 2. For Υ : [ℑ1,ℑ2] → ℜ be a positive convex function with 0 ≤ ℑ1 < ℑ2 and Υ ∈ L1[ℑ1,ℑ2], then we have Υ ( ℑ1 + ℑ2 2 ) ≤ Γ(α∗ + 1) 2(ℑ2 −ℑ1)α ∗ ( R−LIα ∗ ℑ1+Υ(ℑ2) + R−LIα ∗ ℑ2−Υ(ℑ1) ) ≤ ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Theorem 3. If Υ : [ℑ1,ℑ2] → ℜ is L1 and convex, and α∗ ∈ (0, 1), then we have the following inequality Υ ( ℑ1 + ℑ2 2 ) ≤ B(α∗)Γ(α∗) 2 ((ℑ2 −ℑ1)α ∗ + (1− α∗)Γ(α∗)) × ( A−BIα ∗ ℑ1+Υ(ℑ2) + A−BIα ∗ ℑ2−Υ(ℑ1) ) ≤ ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Theorem 4. For Υ : [ℑ1,ℑ2] → ℜ be a positive function with 0 ≤ ℑ1 < ℑ2 and Υ ∈ L1[ℑ1,ℑ2] be convex function, then we have Υ ( ℑ1 + ℑ2 2 ) ≤ 2α ∗−1Γ(α∗ + 1) (ℑ2 −ℑ1)α ∗ ( R−LIα ∗( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LIα ∗( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) ≤ ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Lemma 1. For Υ : [ℑ1,ℑ2] → ℜ in L1 and have a differentiable mapping on (ℑ1,ℑ2) with ℑ1 < ℑ2 and if Υ′ ∈ L1[ℑ1,ℑ2] with α ∗ > 0 then the following equality for fractional integrals holds( Υ(ℑ1) + Υ(ℑ2) 2 ) − Γ(α∗ + 1) 2(ℑ2 −ℑ1)α ∗ ( R−LIα ∗ ℑ1+Υ(ℑ2) + R−LIα ∗ ℑ2−Υ(ℑ1) ) = ℑ2 −ℑ1 2 ∫ 1 0 ( (1− t)α ∗ − tα ∗ ) Υ′(tℑ1 + (1− t)ℑ2)dt. Lemma 2. For Υ : [ℑ1,ℑ2] → ℜ in L1 be a differentiable mapping on (ℑ1,ℑ2) with ℑ1 < ℑ2 and if Υ′ ∈ L1[ℑ1,ℑ2] with α ∗ > 0 then the following equality holds 2α ∗−1Γ(α∗ + 1) (ℑ2 −ℑ1)α ∗ ( R−LIα ∗( ℑ1+ℑ2 2 ) + Υ(c) + R−LIα ∗( ℑ1+ℑ2 2 ) − Υ(y) ) −Υ ( ℑ1 + ℑ2 2 ) = ℑ2 −ℑ1 4 ∫ 1 0 tα ∗ ( Υ′ ( t 2 ℑ1 + 2− t 2 ℑ2 ) −Υ′ ( 2− t 2 ℑ1 + t 2 ℑ2 )) dt. S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 6 of 34 2. Generalized Fractional Integral Operators and Hermite-Hadamard Inequality in the Fractional Framework In this section, we examine inequalities involving fractional integral of the type (ℑ1+,ℑ2−) and (ℑ1+ℑ2 2 ). We also present some examples and their graphical representations to con- firm our results. 2.1. Inequalities Involving Fractional Integral of the Type (ℑ1+,ℑ2−) Proposition 1. If ♭,ℵ, ℘, τ, δ, b ∈ C with ℜ(ℵ),ℜ(℘),ℜ(τ) > 0 and ℜ(b) > ℜ(δ) > 0 and let g ≥ 0, l > 0 and 0 < s ≤ l+ℜ(ℵ). For a function Υ ∈ L1[ℑ1,ℑ2] and c ∈ [ℑ1,ℑ2], then the addition of left and right sided generalized fractional integral operators, ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ, ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ are defined by ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(ℑ2; g) + ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ(ℑ1; g) = ∞∑ ð=0 að ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) , where að = Bg(δ+ðs,b−δ) B(δ,b−δ) (b)ðs♭ ð (τ)ðl . Proof. Using left sided generalized fractional integral operator (10), we get ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(c; g) = ∫ c ℑ1 (c− ψ)℘−1Eδ,b,s,lℵ,℘,τ ( ♭(c− ψ)ℵ; g ) Υ(ψ)dψ, using extended generalized Mittag-Leffler function (9) in the above expression, we get ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(c; g) = ∫ c ℑ1 (c− ψ)℘−1 × ∞∑ ð=0 ( Bg(δ + ðs, b− δ) B(δ, b− δ) (b)ðs Γ(ℵð+ ℘) ♭ð(y − ψ)ℵð (τ)ðl ) Υ(ψ)dψ, after rearranging, the above equation can be expressed as ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(c; g) = ∞∑ ð=0 ( Bg(δ + ðs, b− δ) B(δ, b− δ) (b)ðs♭ ð (τ)ðl ) × ( 1 Γ(ℵð+ ℘) ∫ c ℑ1 (c− ψ)℘−1+ℵðΥ(ψ)dψ ) , Using left sided Reimann-Liouville integral (1) in above equation, we acquire ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(c; g) = ∞∑ ð=0 ( Bg(δ + ðs, b− δ) B(δ, b− δ) (b)ðs♭ ð (τ)ðl ) R−LI (ℵð+℘) ℑ1+ Υ(c). (12) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 7 of 34 Similarly for right sided generalized fractional integral operator (11), we have ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ(ℑ1; g) = ∞∑ ð=0 ( Bg(δ + ðs, b− δ) B(δ, b− δ) (b)ðs♭ ð (τ)ðl ) R−LI (ℵð+℘) ℑ2− Υ(c). (13) Adding (12) and (13) ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(ℑ2; g) + ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ(ℑ1; g) = ∞∑ ð=0 ( Bg(δ + ðs, b− δ) B(δ, b− δ) (b)ðs♭ ð (τ)ðl )( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) = ∞∑ ð=0 að ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) . Here, the integral transform gives (ℵð+℘)th order left and right sided Reimann-Liouville fractional integrals of Υ(c), provided that ℜ(ℵð+ ℘) > 0. Theorem 5. Let Υ : [ℑ1,ℑ2] → ℜ be a convex function with Υ ∈ L1[ℑ1,ℑ2] and ♭,ℵ, ℘, τ, δ, b ∈ C such that ℜ(ℵ),ℜ(℘),ℜ(τ) > 0, ℜ(b) > ℜ(δ) > 0. Let g ≥ 0, l > 0 and 0 < s ≤ l + ℜ(ℵ), then for (ℵð+ ℘) > 0, we can write ∞∑ ð=0 aðvðΥ ( ℑ1 + ℑ2 2 ) ≤ ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(ℑ2; g) + ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ(ℑ1; g) ≤ ∞∑ ð=0 aðvð ( Υ(ℑ1) + Υ(ℑ2) 2 ) , where að = Bg(δ+ðs,b−δ) B(δ,b−δ) (b)ðs♭ ð (τ)ðl and vð = 2(ℑ2−ℑ1)(ℵð+℘) Γ(ℵð+℘+1) . Proof. Replacing α∗ by (ℵð+ ℘) in Theorem 2, we get Υ ( ℑ1 + ℑ2 2 ) ≤ Γ(ℵð+ ℘+ 1) 2(ℑ2 −ℑ1)(ℵð+℘) ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) ≤ ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Multiplying the above inequality with 2(ℑ2−ℑ1)(ℵð+℘) Γ(ℵ,weobtainð+℘+1) , we obtain 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 ) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 8 of 34 ≤ ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) ≤ 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Again the above inequality is multiplied with að to obtain að 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 ) ≤ að ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) ≤ að 2(ℑ2 −ℑ1) (ℵð+℘) Γ(ℵð+ ℘+ 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Summing over all ð ∞∑ ð=0 að 2(ℑ2 −ℑ1) (ℵð+℘) Γ(ℵð+ ℘+ 1) Υ ( ℑ1 + ℑ2 2 ) ≤ ∞∑ ð=0 að ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) ≤ ∞∑ ð=0 að 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Using Proposition 1, we get ∞∑ ð=0 aðvðΥ ( ℑ1 + ℑ2 2 ) ≤ ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(ℑ2; g) + ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ(ℑ1; g) ≤ ∞∑ ð=0 aðvð ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Hence the result is proved. Proposition 2. If Υ : [ℑ1,ℑ2] → ℜ is L1 (convex) and (ℵð + ℘) ∈ (0, 1), we have (H−H) inequality for Atangana-Baleanu fractional integrals, where, ♭,ℵ, ℘, τ, δ, b ∈ C with ℜ(ℵ),ℜ(℘),ℜ(τ) > 0 and ℜ(b) > ℜ(δ) > 0 and let g ≥ 0, l > 0 and 0 < s ≤ l +ℜ(ℵ) ∞∑ ð=0 aðuðΥ ( ℑ1 + ℑ2 2 ) ≤ ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(ℑ2; g) + ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ(ℑ1; g) ≤ ∞∑ ð=0 aðuð ( Υ(ℑ1) + Υ(ℑ2) 2 ) , S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 9 of 34 where að = Bg(δ+ðs,b−δ) B(δ,b−δ) (b)ðs♭ ð (τ)ðl and uð = 2((ℑ2−ℑ1)ℵð+℘+(1−ℵð−℘)Γ(ℵð+℘)) Γ(ℵð+℘+1) − 2(1−ℵð−℘) (ℵð+℘) . Proof. Replacing α∗ by (ℵð+ ℘) in Theorem 3, we obtain Υ ( ℑ1 + ℑ2 2 ) ≤ B(ℵð+ ℘)Γ(ℵð+ ℘) 2 ( (ℑ2 −ℑ1)(ℵð+℘) + (1− ℵð− ℘)Γ(ℵð+ ℘) ) × ( A−BI (ℵð+℘) ℑ1+ Υ(ℑ2) + A−BI (ℵð+℘) ℑ2− Υ(ℑ1) ) ≤ ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Multiplying the above inequality with 2((ℑ2−ℑ1)(ℵð+℘)+(1−ℵð−℘),wehaveΓ(ℵð+℘)) B(ℵð+℘)Γ(ℵð+℘) , we get 2 ( (ℑ2 −ℑ1) (ℵð+℘) + (1− ℵð− ℘)Γ(ℵð+ ℘) ) B(ℵð+ ℘)Γ(ℵð+ ℘) Υ ( ℑ1 + ℑ2 2 ) ≤ ( A−BI (ℵð+℘) ℑ1+ Υ(ℑ2) + A−BI (ℵð+℘) ℑ2− Υ(ℑ1) ) ≤ 2 ( (ℑ2 −ℑ1) (ℵð+℘) + (1− ℵð− ℘)Γ(ℵð+ ℘) ) B(ℵð+ ℘)Γ(ℵð+ ℘) ( Υ(ℑ1) + Υ(ℑ2) 2 ) . (14) Adding left and right sided Atangana-Baleanu integrals (3) and (4), we acquire A−BIα ∗ ℑ1+Υ(ℑ2) + A−BIα ∗ ℑ2−Υ(ℑ1) = α∗ B(ℑ1) ( R−LIα ∗ ℑ1+Υ(ℑ2) + R−LIα ∗ ℑ2−Υ(ℑ1) ) + 1− α∗ B(ℑ1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Using α∗ = (ℵð+ ℘) in the above expression and then put the results in (14), we get 2 ( (ℑ2 −ℑ1) (ℵð+℘) + (1− ℵð− ℘)Γ(ℵð+ ℘) ) B(ℵð+ ℘)Γ(ℵð+ ℘) Υ ( ℑ1 + ℑ2 2 ) ≤ (ℵð+ ℘) B(ℵð+ ℘) ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) + (1− ℵð− ℘) B(ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)) ≤ 2 ( (ℑ2 −ℑ1) (ℵð+℘) + (1− ℵð− ℘)Γ(ℵð+ ℘) ) B(ℵð+ ℘)Γ(ℵð+ ℘) ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Subtracting 1−ℵð−℘ B(ℵð+℘)(Υ(ℑ1)+Υ(ℑ2)) from the above inequality and then multiplying with B(ℵð+℘) (ℵð+℘) , we obtain 2 ( (ℑ2 −ℑ1) (ℵð+℘) + (1− ℵð− ℘)Γ(ℵð+ ℘) ) Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 ) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 10 of 34 − 1− ℵð− ℘ (ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)) ≤ ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) ≤ 2 ( (ℑ2 −ℑ1) (ℵð+℘) + (1− ℵð− ℘)Γ(ℵð+ ℘) ) Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) − 1− ℵð− ℘ (ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)) . Again we multiply the above inequality with að to obtain að ( 2 ( (ℑ2 −ℑ1) (ℵð+℘) + (1− ℵð− ℘)Γ(ℵð+ ℘) ) Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 )) − að ( 1− ℵð− ℘ (ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)) ) ≤ að ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) ≤ að ( 2 ( (ℑ2 −ℑ1) (ℵð+℘) + (1− ℵð− ℘)Γ(ℵð+ ℘) ) Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 )) − að ( 1− ℵð− ℘ (ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)) ) . By convexity of Υ we have, Υ ( ℑ1+ℑ2 2 ) ≤ ( Υ(ℑ1)+Υ(ℑ2) 2 ) að ( 2 ( (ℑ2 −ℑ1) (ℵð+℘) + (1− ℵð− ℘)Γ(ℵð+ ℘) ) Γ(ℵð℘+ 1) − 2(1− ℵð− ℘) (ℵð+ ℘) ) ×Υ ( ℑ1 + ℑ2 2 ) ≤ að ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) ≤ að ( 2 ( (ℑ2 −ℑ1) (ℵð+℘) + (1− ℵð− ℘)Γ(ℵð+ ℘) ) Γ((ℵð+ ℘) + 1) − 2(1− ℵð− ℘) ℵð℘ ) × ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Summing over all ð ∞∑ ð=0 að ( 2 ( (ℑ2 −ℑ1) (ℵð+℘) + (1− ℵð− ℘)Γ(ℵð+ ℘) ) Γ((ℵð+ ℘) + 1) − 2(1− ℵð− ℘) ℵð℘ ) ×Υ ( ℑ1 + ℑ2 2 ) ≤ ∞∑ ð=0 að ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) ≤ ∞∑ ð=0 að ( 2 ( (ℑ2 −ℑ1) ℵð+℘ + (1− ℵð− ℘)Γ(ℵð+ ℘) ) Γ((ℵð+ ℘) + 1) − 2(1− ℵð− ℘) (ℵð+ ℘) ) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 11 of 34 × ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Using Proposition 1, we get ∞∑ ð=0 aðuðΥ ( ℑ1 + ℑ2 2 ) ≤ ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(ℑ2; g) + ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ(ℑ1; g) ≤ ∞∑ ð=0 aðuð ( Υ(ℑ1) + Υ(ℑ2) 2 ) . It is our required result. Proposition 3. For a function Υ ∈ L1[ℑ1,ℑ2] and c ∈ [ℑ1,ℑ2], the addition of the left and right sided Prabhakar and the left and right sided generalized fractional integral operators applied to Υ(c) are defined by the following integral transforms, where ℜ(ℵð + ℘) > 0 also ♭, γ,ℵ, ℘, τ, δ, b ∈ C with ℜ(♭),ℜ(γ),ℜ(ℵ),ℜ(℘),ℜ(τ) > 0 and ℜ(b) > ℜ(δ) > 0 and let g ≥ 0, l > 0 and 0 < s ≤ l + ℜ(ℵ) PIℵ,℘,γ,♭ℑ1+ Υ(c) + ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(c; g) + PIℵ,℘,γ,♭ℑ2− Υ(c) + ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ(c; g) = ∞∑ ð=0 hð ( R−LI (ℵð+℘) ℑ1+ Υ(c) + R−LI (ℵð+℘) ℑ2− Υ(c) ) , where hð = Bg(δ+ðs,b−δ) B(δ,b−δ) (b)ðs♭ ð (τ)ðl + Γ(γ+ð)♭ð Γ(γ)ð! . Proof. Adding infinite series formula for left Prabhakar integral (7) and left sided generalized fractional integral operator (10), we get PIℵ,℘,γ,♭ℑ1+ Υ(c) + ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(c; g) = ∞∑ ð=0 ( Bg(δ + ðs, b− δ) B(δ, b− δ) (b)ðs♭ ð (τ)ðl + ∞∑ ð=0 Γ(γ + ð)♭ð Γ(γ)ð! ) R−LI (ℵð+℘) ℑ1+ Υ(c). (15) Similarly adding infinite series formula for right Prabhakar integral (8) and right sided generalized fractional integral operator (11), we have PIℵ,℘,γ,♭ℑ2− Υ(c) + ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ(c; g) = ∞∑ ð=0 ( Bg(δ + ðs, b− δ) B(δ, b− δ) (b)ðs♭ ð (τ)ðl + ∞∑ ð=0 Γ(γ + ð)♭ð Γ(γ)ð! ) R−LI (ℵð+℘) ℑ2− Υ(c). (16) Finally, we add (15) and (16) to get following required result PIℵ,℘,γ,♭ℑ1+ Υ(c) + ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(c; g) + PIℵ,℘,γ,♭ℑ2− Υ(c) + ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ(c; g) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 12 of 34 = ( ∞∑ ð=0 Bg(δ + ðs, b− δ) B(δ, b− δ) (b)ðs♭ ð (τ)ðl + ∞∑ ð=0 Γ(γ + ð)♭ð Γ(γ)ð! ) × (R−LI (ℵð+℘) ℑ1+ Υ(c) + R−LI (ℵð+℘) ℑ2− Υ(c)) = ∞∑ ð=0 hð ( R−LI (ℵð+℘) ℑ1+ Υ(c) + R−LI (ℵð+℘) ℑ2− Υ(c) ) . Theorem 6. If Υ : [ℑ1,ℑ2] → ℜ is L1 and convex and the parameters, ℜ(ℵð + ℘) > 0 also ℜ(♭),ℜ(γ),ℜ(ℵ),ℜ(℘),ℜ(τ) > 0 and ℜ(b) > ℜ(δ) > 0 and let g ≥ 0, l > 0 and 0 < s ≤ l + ℜ(ℵ), then we have the following (H−H) inequality for Prabhakar fractional integrals and generalized fractional integral operators ∞∑ ð=0 hðvðΥ ( ℑ1 + ℑ2 2 ) ≤ PIℵ,℘,γ,♭ℑ1+ Υ(c) + ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(c; g) + PIℵ,℘,γ,♭ℑ2− Υ(c) + ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ(c; g) ≤ ∞∑ ð=0 hðvð ( Υ(ℑ1) + Υ(ℑ2) 2 ) , (17) where hð = Bg(δ+ðs,b−δ) B(δ,b−δ) (b)ðs♭ ð (τ)ðl + Γ(γ+ð)♭ð Γ(γ)ð! and vð = 2(ℑ2−ℑ1)(ℵð+℘) Γ((ℵð+℘)+1) . Proof. Replacing α∗ by (ℵð+ ℘) in Theorem 2, we obtain Υ ( ℑ1 + ℑ2 2 ) ≤ Γ((ℵð+ ℘) + 1) 2(ℑ2 −ℑ1)(ℵð+℘) ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) ≤ Υ(ℑ1) + Υ(ℑ1) 2 . Multiplying the above inequality with 2(ℑ2−ℑ1)(ℵð+℘) Γ((ℵð+℘)+1) , we get 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 ) ≤ ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) ≤ 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Again we multiply the above inequality with hð to obtain hð 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 ) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 13 of 34 ≤ hð ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) ≤ hð 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Summing over all ð ∞∑ ð=0 hð 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 ) ≤ ∞∑ ð=0 hð ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) ≤ ∞∑ ð=0 hð 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Using Proposition 3 in the above expression, we get ∞∑ ð=0 hðvðΥ ( ℑ1 + ℑ2 2 ) ≤ PIℵ,℘,γ,♭ℑ1+ Υ(c) + ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(c; g) + PIℵ,℘,γ,♭ℑ2− Υ(c) + ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ(c; g) ≤ ∞∑ ð=0 hðvð ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Hence the required result is proved. Example 1. We verify the result of Theorem 6 for convex function Υ(c) = c2 on the interval [0, 1]. Using substitution t = ψ c in left and right sided Reimann-Liouville integrals (1) and (2), we get R−LI (ℵð+℘) ℑ1+ ℑ2 2 = Γ(3) Γ((ℵð+ ℘) + 3) , (18) R−LI (ℵð+℘) ℑ2− ℑ1 2 = 0. (19) Using (18) in infinite series formula for left Prabhakar integrals (7) and (19) in infinite series formula for right Prabhakar integrals (8), we have PIℵ,℘,γ,♭ℑ1+ ℑ2 2 = ∞∑ ð=0 Γ(γ + ð)♭ð Γ(γ)ð! × Γ(3) Γ((ℵð+ ℘) + 3) , (20) PIℵ,℘,γ,♭ℑ2− ℑ1 2 = ∞∑ ð=0 Γ(γ + ð)♭ð Γ(γ)ð! × (0). (21) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 14 of 34 Also substitute (18) in left sided generalized fractional integral operator (10) and (19) in right sided generalized fractional integral operator (11), we acquire ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ ( ℑ2 2; g ) = ∞∑ ð=0 að × Γ(3) Γ((ℵð+ ℘) + 3) , (22) ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ ( ℑ1 2; g ) = ∞∑ ð=0 að × (0). (23) Substituting these expressions (20), (21), (22) and (23) in the inequality (17) and after some simplification, we get ∞∑ ð=0 hð Γ((ℵð+ ℘) + 1) 2 ≤ ∞∑ ð=0 ( Γ(γ + ð)♭ð Γ(γ)ð! × Γ(3) Γ((ℵð+ ℘) + 3) ) + ∞∑ ð=0 ( að × Γ(3) Γ((ℵð+ ℘) + 3) ) ≤ ∞∑ ð=0 hð Γ((ℵð+ ℘) + 1) . Figure 1: The 2D graph exhibiting the inequality (17) for ð = 1. 2.2. Inequalities Involving Fractional Integral of the Type (ℑ1+ℑ2 2 ) Theorem 7. If Υ : [ℑ1,ℑ2] → ℜ is L1 and convex and the parameters, ℜ(ℵð + ℘) > 0 also ℜ(♭),ℜ(γ),ℜ(ℵ),ℜ(℘),ℜ(τ) > 0 and ℜ(b) > ℜ(δ) > 0 and let g ≥ 0, l > 0 and S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 15 of 34 Figure 2: The 3D graph exhibiting the inequality (17) for convex function Υ(c) = c2 on the interval [0, 1] and for ð = 1. 0 < s ≤ l+ℜ(ℵ), then we have a distinct fractional development of the (H−H) inequality ∞∑ ð=0 aðoðΥ ( ℑ1 + ℑ2 2 ) ≤ ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) +,ℵ,℘,τ Υ(ℑ2; g) + ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) −,ℵ,℘,τ Υ(ℑ1; g) ≤ ∞∑ ð=0 aðoð ( Υ(ℑ1) + Υ(ℑ2) 2 ) , where að = Bg(δ+ðs,b−δ) B(δ,b−δ) (b)ðs♭ ð (τ)ðl and oð = (ℑ2−ℑ1)(ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+℘)+1) . Proof. Replacing α∗ by (ℵð+ ℘) in Theorem 4, we get Υ ( ℑ1 + ℑ2 2 ) ≤ 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) (ℑ2 −ℑ1)(ℵð+℘) ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) ≤ ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Multiply the above expression with (ℑ2−ℑ1)(ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+℘)+1) to get (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 ) ≤ ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) ≤ (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) . S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 16 of 34 Again the above inequality is multiplied with að and then summing over all ð ∞∑ ð=0 að (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 ) ≤ ∞∑ ð=0 að ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) ≤ ∞∑ ð=0 að (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Using Proposition 1 from the middle of interval [ℑ1,ℑ2] ∞∑ ð=0 aðoðΥ ( ℑ1 + ℑ2 2 ) ≤ ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) +,ℵ,℘,τ Υ(ℑ2; g) + ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) −,ℵ,℘,τ Υ(ℑ1; g) ≤ ∞∑ ð=0 aðoð ( Υ(ℑ1) + Υ(ℑ2) 2 ) . This proves the desired result. Proposition 4. If Υ : [ℑ1,ℑ2] → ℜ is L1 and convex and the parameters, ℜ(ℵð+℘) > 0 also ℜ(♭),ℜ(γ),ℜ(ℵ),ℜ(℘),ℜ(τ) > 0 and ℜ(b) > ℜ(δ) > 0 and let g ≥ 0, l > 0 and 0 < s ≤ l + ℜ(ℵ), then the (H−H) inequality for Atangana-Baleanu fractional integrals becomes ∞∑ ð=0 aðsðΥ ( ℑ1 + ℑ2 2 ) ≤ ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) +,ℵ,℘,τ Υ(ℑ2; g) + ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) −,ℵ,℘,τ Υ(ℑ1; g) ≤ ∞∑ ð=0 aðsð ( Υ(ℑ1) + Υ(ℑ2) 2 ) , where að = Bg(δ+ðs,b−δ) B(δ,b−δ) (b)ðs♭ ð (τ)ðl and sð = ((ℑ2−ℑ1)(ℵð+℘)+2(ℵð+℘)(1−ℵð−℘)Γ(ℵð+℘)) 2(ℵð+℘)−1Γ((ℵð+℘)+1) − 2(1−ℵð−℘) (ℵð+℘) . Proof. Replacing α∗ by (ℵð+ ℘) in Theorem 3, we have Υ ( ℑ1 + ℑ2 2 ) ≤ 2(ℵð+℘)−1B(ℵð+ ℘)Γ(ℵð+ ℘)( (ℑ2 −ℑ1)(ℵð+℘) + 2(ℵð+℘)(1− ℵð− ℘)Γ(ℵð+ ℘) ) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 17 of 34 × ( A−BI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + A−BI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) ≤ ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Multiplying the above inequality with ((ℑ2−ℑ1)(ℵð+℘)+2(ℵð+℘)(1−ℵð−℘)Γ(ℵð+℘)) 2(ℵð+℘)−1B(ℵð+℘)Γ(ℵð+℘) , we get( (ℑ2 −ℑ1) (ℵð+℘) + 2(ℵð+℘)(1− ℵð− ℘)Γ(ℵð+ ℘) ) 2(ℵð+℘)−1B(ℵð+ ℘)Γ(ℵð+ ℘) Υ ( ℑ1 + ℑ2 2 ) ≤ ( A−BI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + A−BI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) ≤ ( (ℑ2 −ℑ1) (ℵð+℘) + 2(ℵð+℘)(1− ℵð− ℘)Γ(ℵð+ ℘) ) 2(ℵð+℘)−1B(ℵð+ ℘)Γ(ℵð+ ℘) × ( Υ(ℑ1) + Υ(ℑ2) 2 ) . (24) Adding left and right sided Atangana-Baleanu integrals (3) and (4) from the middle of interval [ℑ1,ℑ2], we have( A−BIα ∗( ℑ1+ℑ2 2 ) + Υ(ℑ2) + A−BIα ∗( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) = α∗ B(α∗) ( R−LIα ∗( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LIα ∗( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) + 1− α∗ B(α∗) (Υ(ℑ1) + Υ(ℑ2)). Using α∗ = (ℵð+ ℘) in the above equation and then put the results in (24), we get ( (ℑ2 −ℑ1) (ℵð+℘) + 2(ℵð+℘)(1− ℵð− ℘)Γ(ℵð+ ℘) ) 2(ℵð+℘)−1B(ℵð+ ℘)Γ(ℵð+ ℘) Υ ( ℑ1 + ℑ2 2 ) ≤ (ℵð+ ℘) B(ℵð+ ℘) ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + (ℑ2) + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) + 1− ℵð− ℘ B(ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)) ≤ ( (ℑ2 −ℑ1) (ℵð+℘) + 2(ℵð+℘)(1− ℵð− ℘)Γ(ℵð+ ℘) ) 2(ℵð+℘)−1B(ℵð+ ℘)Γ(ℵð+ ℘) Υ(ℑ1) + (ℑ2) 2 . Subtracting 1−ℵð−℘ B(ℵð+℘) [Υ(ℑ1)+Υ(ℑ2)] from the above expression and then multiplying with B(ℵð+℘) (ℵð+℘) , we obtain( (ℑ2 −ℑ1) (ℵð+℘) + 2(ℵð+℘)(1− ℵð− ℘)Γ(ℵð+ ℘) ) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 ) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 18 of 34 − 1− ℵð− ℘ (ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)) ≤ ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LIℵð+℘( ℑ1+ℑ2 2 ) − (ℑ1) ) ≤ ( (ℑ2 −ℑ1) (ℵð+℘) + 2(ℵð+℘)(1− ℵð− ℘)Γ(ℵð+ ℘) ) 2ℵ̂ð+℘−1Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) − 1− ℵð− ℘ (ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)). Again multiplying the above inequality with að að (( (ℑ2 −ℑ1) (ℵð+℘) + 2(ℵð+℘)(1− ℵð− ℘)Γ(ℵð+ ℘) ) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 )) − að ( 2(1− ℵð− ℘) (ℵð+ ℘) ( Υ(ℑ1) + Υ(ℑ2) 2 )) ≤ að ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) ≤ að (( (ℑ2 −ℑ1) (ℵð+℘) + 2(ℵð+℘)(1− ℵð− ℘)Γ(ℵð+ ℘) ) 2(ℵð+℘)−1Γ(ℵð+ ℘+ 1) ( Υ(ℑ1) + Υ(ℑ2) 2 )) − ( 2(1− ℵð− ℘) (ℵð+ ℘) ( Υ(ℑ1) + Υ(ℑ2) 2 )) . By convexity of Υ we have, Υ ( ℑ1+ℑ2 2 ) ≤ ( Υ(ℑ1)+Υ(ℑ2) 2 ) að (( (ℑ2 −ℑ1) (ℵð+℘) + 2(ℵð+℘)(1− ℵð− ℘)Γ(ℵð+ ℘) ) 2ℵð+℘−1Γ((ℵð+ ℘) + 1) − 2(1− ℵð− ℘) (ℵð+ ℘) ) ×Υ ( ℑ1 + ℑ2 2 ) ≤ að ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + + R−LIℵð+℘( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) ≤ að (( (ℑ2 −ℑ1) (ℵð+℘) + 2(ℵð+℘)(1− ℵð− ℘)Γ(ℵð+ ℘) ) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) − 2(1− ℵð− ℘) (ℵð+ ℘) ) × ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Summing over all ð ∞∑ ð=0 að (( (ℑ2 −ℑ1) (ℵð+℘) + 2(ℵð+℘)(1− ℵð− ℘)Γ(ℵð+ ℘) ) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) − 2(1− ℵð− ℘) (ℵð+ ℘) ) ×Υ ( ℑ1 + ℑ2 2 ) ≤ ∞∑ ð=0 að ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 19 of 34 ≤ ∞∑ ð=0 að (( (ℑ2 −ℑ1) (ℵð+℘) + 2(ℵð+℘)(1− ℵð− ℘)Γ(ℵð+ ℘) ) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) − 2(1− ℵð− ℘) (ℵð+ ℘) ) × ( Υ(ℑ1) + Υ(ℑ2) 2 ) , Using Proposition 1 from the middle of interval [ℑ1,ℑ2], we acquire ∞∑ ð=0 aðsðΥ ( ℑ1 + ℑ2 2 ) ≤ ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) +,ℵ,℘,τ Υ(ℑ2; g) + ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) −,ℵ,℘,τ Υ(ℑ1; g) ≤ ∞∑ ð=0 aðsð ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Hence the result is established. Theorem 8. If Υ : [ℑ1,ℑ2] → ℜ is L1 convex and the parameters, ℜ(ℵð + ℘) > 0 also ℜ(♭),ℜ(γ),ℜ(ℵ),ℜ(℘),ℜ(τ) > 0 and ℜ(b) > ℜ(δ) > 0 and let g ≥ 0, l > 0 and 0 < s ≤ l + ℜ(ℵ), then we have the following (H−H) inequality for Prabhakar fractional integrals and generalized fractional integral operators ∞∑ ð=0 hðoðf ( ℑ1 + ℑ2 2 ) ≤ PIℵ,℘,γ,♭( ℑ1+ℑ2 2 ) + Υ(c) + ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) +,ℵ,℘,τ Υ(c; g) + PIℵ,℘,γ,♭( ℑ1+ℑ2 2 ) − Υ(c) + ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) −,ℵ,℘,τ Υ(c; g) ≤ ∞∑ ð=0 hðoð ( Υ(ℑ1) + Υ(ℑ2) 2 ) , (25) where hð = Bg(δ+ðs,b−δ) B(δ,b−δ) (b)ðs♭ ð (τ)ðl + Γ(γ+ð)♭ð Γ(γ)ð! and oð = (ℑ2−ℑ1)(ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+℘)+1) . Proof. Replacing α∗ by (ℵð+ ℘) in Theorem 4, we obtain Υ ( ℑ1 + ℑ2 2 ) ≤ 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) (ℑ2 −ℑ1)(ℵð+℘) × ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(c) + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(c) ) ≤ ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Multiplying the above inequality with (ℑ2−ℑ1)(ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+℘)+1) , we obtain (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 ) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 20 of 34 ≤ ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(c) + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(c) ) ≤ (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Again we multiply the above expression with hð, then it becomes hð (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 ) ≤ hð ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(c) + R−LIℵi+℘( ℑ1+ℑ2 2 ) − Υ(c) ) ≤ hð (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) . Summing over all ð ∞∑ ð=0 hð (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 ) ≤ ∞∑ ð=0 hð ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(c) + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(c) ) ≤ ∞∑ ð=0 hð (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) . (26) From the middle of interval [ℑ1,ℑ2], Proposition 3 becomes PIℵ,℘,γ,♭( ℑ1+ℑ2 2 ) + Υ(c) + ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) +,ℵ,℘,τ Υ(c; g) + PIℵ,℘,γ,♭( ℑ1+ℑ2 2 ) − Υ(c) + ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) −,ℵ,℘,τ Υ(c; g) = ∞∑ ð=0 hð ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(c) + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(c) ) . (27) Using (27) in (26), we obtain ∞∑ ð=0 hðoðΥ ( ℑ1 + ℑ2 2 ) ≤ PIℵ,℘,γ,♭( ℑ1+ℑ2 2 ) + Υ(c) + ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) +,ℵ,℘,τ Υ(c; g) + PIℵ,℘,γ,♭( ℑ1+ℑ2 2 ) − Υ(c) + ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) −,ℵ,℘,τ Υ(c; g) ≤ ∞∑ ð=0 hðoð ( Υ(ℑ1) + Υ(ℑ2) 2 ) . That is our required result. S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 21 of 34 Example 2. We verify the result of Theorem 8 for convex function Υ(c) = c4n, n ∈ N on the interval [−1, 1]. Using substitution t = ψ c in left and right sided Reimann-Liouville integrals (1) and (2), we get R−LI (ℵð+℘) 0+ (1)4n = (4n)! Γ(ℵð+ ℘+ 4n+ 1) , (28) R−LI (ℵð+℘) 0+ (−1)4n = (4n)! Γ(ℵð+ ℘+ 4n+ 1) . (29) Use (28) in infinite series formula for left Prabhakar integral (7) and (29) in infinite series formula for right Prabhakar integral (8), we have PIℵ,℘,γ,♭0+ (1)4n = ∞∑ ð=0 Γ(γ + ð)♭ð Γ(γ)ð! (4n)! Γ(ℵð+ ℘+ 4n+ 1) , (30) PIℵ,℘,γ,♭0− (−1)4n = ∞∑ ð=0 Γ(γ + ð)♭ð Γ(γ)ð! (4n! Γ(ℵð+ ℘+ 4n+ 1) . (31) Also substitute (28) in left sided generalized fractional integral operator (10) and (29) in right sided generalized fractional integral operator (11), we acquire ε♭,δ,b,s,l 0+,ℵ,℘,τΥ ( (1)4n; g ) = ∞∑ ð=0 að (4n)! Γ(ℵð+ ℘+ 4n+ 1) , (32) ε♭,δ,b,s,l 0−,ℵ,℘,τΥ ( (−1)4n; g ) = ∞∑ ð=0 að (4n)! Γ(ℵð+ ℘+ 4n+ 1) . (33) Substituting these expressions (30), (31), (32) and (33) in the inequality (25) and after some simplification, we get 04ð ≤ ∞∑ ð=0 ( Γ(γ + ð)♭ð Γ(γ)ð! + að ) (4n)! Γ(ℵð+ ℘+ 4n+ 1) ≤ ∞∑ ð=0 hð Γ(ℵð+ ℘+ 1) . 3. Applications of Key Results in Terms of Means (H−H) inequality are often connected to additional integral inequalities, such as trapezoid-type (utilizing the interval’s endpoints ℑ1 and ℑ2) and midpoint-type (utilizing the midpoint (ℑ1+ℑ2 2 ) of the interval). Many researchers have contributed to establishing these inequalities [15, 38]. In this section, we employed an equality of trapezoid type and an inequality of midpoint type for the (H−H) integrals. S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 22 of 34 Figure 3: The 2D graph exhibiting the inequality (25) for ð = 1. Figure 4: The 3D graph exhibiting the inequality (25) for convex function Υ(c) = (c)4ð on the interval [−1, 1] and for ð = 1. 3.1. An Equality of Trapezoid Type for the Hermite-Hadamard Integrals Lemma 3. Let Υ : [ℑ1,ℑ2] → ℜ is an L1 function and (ℵð+ ℘) ∈ (0, 1) also a dif- ferentiable function on (ℑ1,ℑ2) with ℑ1 < ℑ2, and assume Υ′ ∈ L1[ℑ1,ℑ2]. Consider ♭,ℵ, ℘, τ, δ, b ∈ C with ℜ(ℵ),ℜ(℘),ℜ(τ) > 0 and ℜ(b) > ℜ(δ) > 0. Moreover, let g ≥ 0, l > 0 and 0 < s ≤ l + ℜ(ℵ) with (ℵð+ ℘) > 0, then we have ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(ℑ2; g) + ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ(ℑ1; g) = ∞∑ ð=0 aðvð ( Υ(ℑ1) + Υ(ℑ2) 2 ) − ∞∑ ð=0 aðvð × ( ℑ2 −ℑ1 2 ∫ 1 0 ( (1− t)(ℵð+℘) − t(ℵð+℘) ) Υ′(tℑ1 + (1− t)ℑ2)dt ) , where að = Bg(δ+ðs,b−δ) B(δ,b−δ) (b)ðs♭ ð (τ)ðl and vð = 2(ℑ2−ℑ1)(ℵð+℘) Γ((ℵð+℘)+1) . Proof. Replacing α∗ by (ℵð+ ℘) in Lemma 1, we get( Υ(ℑ1) + Υ(ℑ2) 2 ) − Γ((ℵð+ ℘) + 1) 2(ℑ2 −ℑ1)(ℵð+℘) ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) = ℑ2 −ℑ1 2 ∫ 1 0 ( (1− t)(ℵð+℘) − t(ℵð+℘) ) Υ′(tℑ1 + (1− t)ℑ2)dt. S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 23 of 34 Multiplying the above equation with 2(ℑ2−ℑ1)(ℵð+℘) Γ((ℵð+℘)+1) , we get 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) − ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) = 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) × ( ℑ2 −ℑ1 2 ∫ 1 0 ( (1− t)(ℵð+℘) − tℵð+℘ ) Υ′(tℑ1 + (1− t)ℑ2)dt ) . Again the above expression is multiplied with að to get að 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) − að ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) = að 2(ℑ2 −ℑ1) (ℵð+℘) Γ(ℵð+ ℘+ 1) × ( ℑ2 −ℑ1 2 ∫ 1 0 ( (1− t)(ℵð+℘) − t(ℵð+℘) ) Υ′(tℑ1 + (1− t)ℑ2)dt ) . Summing over all ð ∞∑ ð=0 að 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) − ∞∑ ð=0 að ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) = ∞∑ ð=0 að 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) × ( ℑ2 −ℑ1 2 ∫ 1 0 ( (1− t)(ℵð+℘) − t(ℵð+℘) ) Υ′(tℑ1 + (1− t)ℑ2)dt ) . Using Proposition 1 ∞∑ ð=0 að 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) − ( ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(ℑ2; g) + ( ε♭,δ,b,s,lℑ2−,ℵ,℘,τf ) (ℑ1; g) ) = ∞∑ ð=0 að 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) × ( ℑ2 −ℑ1 2 ∫ 1 0 ( (1− t)(ℵð+℘) − t(ℵð+℘) ) Υ′(tℑ1 + (1− t)ℑ2)dt ) . After rearranging the above expression, we get ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ(ℑ2; g) + ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ(ℑ1; g) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 24 of 34 = ∞∑ ð=0 aðvð ( Υ(ℑ1) + Υ(ℑ2) 2 ) − ∞∑ ð=0 aðvð × ( ℑ2 −ℑ1 2 ∫ 1 0 ( (1− t)(ℵð+℘) − t(ℵð+℘) ) Υ′(tℑ1 + (1− t)ℑ2)dt ) . Thus the proof is completed. Theorem 9. Let Υ : [ℑ1,ℑ2] → ℜ is an L1 function and (ℵð+ ℘) ∈ (0, 1) also a dif- ferentiable function on (ℑ1,ℑ2) with ℑ1 < ℑ2 and assume Υ′ ∈ L1[ℑ1,ℑ2]. Consider ♭,ℵ, ℘, τ, δ, b ∈ C with ℜ(ℵ),ℜ(℘),ℜ(τ) > 0 and ℜ(b) > ℜ(δ) > 0. Moreover, let g ≥ 0, l > 0 and 0 < s ≤ l + ℜ(ℵ) with (ℵð+ ℘) > 0 ∞∑ ð=0 að B(ℵð+ ℘) (ℵð+ ℘) ( A−BI (ℵð+℘) ℑ1+ Υ(ℑ2) + A−BI (ℵð+℘) ℑ2− Υ(ℑ1) ) − ∞∑ ð=0 að 1− (ℵð+ ℘) (ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)) = ∞∑ ð=0 aðvð ( Υ(ℑ1) + Υ(ℑ2) 2 ) − ∞∑ ð=0 aðvð × ( ℑ2 −ℑ1 2 ∫ 1 0 ( (1− t)(ℵð+℘) − t(ℵð+℘) ) Υ′(tℑ1 + (1− t)ℑ2)dt ) , where að = Bg(δ+ðs,b−δ) B(δ,b−δ) (b)ðs♭ ð (τ)ðl and vð = 2(ℑ2−ℑ1)(ℵð+℘) Γ((ℵð+℘)+1) . Proof. Adding left and right sided Atangana-Baleanu integrals (3) and (4), we have A−BIα ∗ ℑ1+Υ(ℑ2) + A−BIα ∗ ℑ2−Υ(ℑ1) = α∗ B(α∗) ( R−LIα ∗ ℑ1+Υ(ℑ2) + R−LIα ∗ ℑ2−Υ(ℑ1) ) + 1− α∗ B(α∗) (Υ(ℑ1) + Υ(ℑ2). Replacing α∗ by (ℵð+ ℘) and then rearrange the above equation, we obtain B(ℵð+ ℘) (ℵð+ ℘) ( A−BI (ℵð+℘) ℑ1+ Υ(ℑ2) + A−BI (ℵð+℘) ℑ2− Υ(ℑ1) ) − 1− ℵð− ℘ (ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)) = ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) . Multiplying the above expression with að, we obtain að B(ℵð+ ℘) (ℵð+ ℘) ( A−BI (ℵð+℘) ℑ1+ Υ(ℑ2) + A−BI (ℵð+℘) ℑ2− Υ(ℑ1) ) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 25 of 34 − að 1− ℵð− ℘ (ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)) = að ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) . Summing over all ð ∞∑ ð=0 að B(ℵð+ ℘) (ℵð+ ℘) ( A−BI (ℵð+℘) ℑ1+ Υ(ℑ2) + A−BI (ℵð+℘) ℑ2− Υ(ℑ1) ) − ∞∑ ð=0 að 1− ℵð− ℘ (ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)) = ∞∑ ð=0 að ( R−LI (ℵð+℘) ℑ1+ Υ(ℑ2) + R−LI (ℵð+℘) ℑ2− Υ(ℑ1) ) . Using Proposition 1, we obtain ∞∑ ð=0 að B(ℵð+ ℘) (ℵð+ ℘) ( A−BI (ℵð+℘) ℑ1+ Υ(ℑ2) + A−BI (ℵð+℘) ℑ2− Υ(ℑ1) ) − ∞∑ ð=0 að 1− ℵð− ℘ (ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)) = (( ε♭,δ,b,s,lℑ1+,ℵ,℘,τΥ ) (ℑ2; g) + ( ε♭,δ,b,s,lℑ2−,ℵ,℘,τΥ ) (ℑ1; g) ) . Comparing with Lemma 3, we get ∞∑ ð=0 að B(ℵð+ ℘) (ℵð+ ℘) ( A−BI (ℵð+℘) ℑ1+ Υ(ℑ2) + A−BI (ℵð+℘) ℑ2− Υ(ℑ1) ) − ∞∑ ð=0 að 1− ℵð− ℘ (ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)) = ∞∑ ð=0 að 2(ℑ2 −ℑ1) (ℵð+℘) Γ(ℵð+ ℘+ 1) ( Υ(ℑ1) + Υ(ℑ2) 2 ) − ∞∑ ð=0 að 2(ℑ2 −ℑ1) (ℵð+℘) Γ((ℵð+ ℘) + 1) × ( ℑ2 −ℑ1 2 ∫ 1 0 ( (1− t)(ℵð+℘) − t(ℵð+℘) ) Υ′(tℑ1 + (1− t)ℑ2)dt ) . This completes desired result. S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 26 of 34 3.2. An Inequality of Midpoint Type for the Hermite-Hadamard Integrals Lemma 4. Let Υ : [ℑ1,ℑ2] → ℜ is an L1 function and (ℵð+ ℘) ∈ (0, 1) also a dif- ferentiable function on (ℑ1,ℑ2) with ℑ1 < ℑ2 and assume Υ′ ∈ L1[ℑ1,ℑ2]. Consider ♭,ℵ, ℘, τ, δ, b ∈ C with ℜ(ℵ),ℜ(℘),ℜ(τ) > 0 and ℜ(b) > ℜ(δ) > 0. Moreover, let g ≥ 0, l > 0 and 0 < s ≤ l + ℜ(ℵ). Under these conditions, the following equality for fractional integrals holds ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) +,ℵ,℘,τ Υ(ℑ2; g) + ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) −,ℵ,℘,τ Υ(ℑ1; g) = ∞∑ ð=0 aðoð × ( ℑ2 −ℑ1 4 ∫ 1 0 t(ℵð+℘) ( Υ′ ( t 2 ℑ1 + 2− t 2 ℑ2 ) −Υ′ ( 2− t 2 ℑ1 + t 2 ℑ2 )) dt ) + ∞∑ ð=0 aðoðΥ ( ℑ1 + ℑ2 2 ) , where að = Bg(δ+ðs,b−δ) B(δ,b−δ) (b)ðs♭ ð (τ)ðl and oð = (ℑ2−ℑ1)(ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+℘)+1) with (ℵð+ ℘) > 0. Proof. Replacing α∗ by (ℵð+ ℘) in Lemma 2, we get 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) (ℑ2 −ℑ1)(ℵð+℘) ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) −Υ ( ℑ1 + ℑ2 2 ) = ℑ2 −ℑ1 4 ∫ 1 0 t(ℵð+℘) × ( Υ′ ( t 2 ℑ1 + 2− t 2 ℑ2 ) −Υ′ ( 2− t 2 ℑ1 + t 2 ℑ2 )) dt. Multiplying the above expression with (ℑ2−ℑ1)(ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+℘)+1) , we get( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) = (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) × ( ℑ2 −ℑ1 4 ∫ 1 0 t(ℵð+℘) ( Υ′ ( t 2 ℑ1 + 2− t 2 ℑ2 ) −Υ′ ( 2− t 2 ℑ1 + t 2 ℑ2 )) dt ) + (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 ) . Again we multiply the above equation with að að ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) = að (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 27 of 34 × ( ℑ2 −ℑ1 4 ∫ 1 0 t(ℵð+℘) ( Υ′ ( t 2 ℑ1 + 2− t 2 ℑ2 ) −Υ′ ( 2− t 2 ℑ1 + t 2 ℑ2 )) dt ) + að (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 ) . Summing over all ð ∞∑ ð=0 að ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) = ∞∑ ð=0 að (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) × ( ℑ2 −ℑ1 4 ∫ 1 0 t(ℵð+℘) ( Υ′ ( t 2 ℑ1 + 2− t 2 ℑ2 ) −Υ′ ( 2− t 2 ℑ1 + t 2 ℑ2 )) dt ) + ∞∑ ð=0 að (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) Υ ( ℑ1 + ℑ2 2 ) . Using Proposition 1 from the middle of interval [ℑ1,ℑ2], we obtain ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) +,ℵ,℘,τ Υ(ℑ2; g) + ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) −,ℵ,℘,τ Υ(ℑ1; g) = ∞∑ ð=0 aðoð × ( ℑ2 −ℑ1 4 ∫ 1 0 t(ℵð+℘) ( Υ′ ( t 2 ℑ1 + 2− t 2 ℑ2 ) − f ′ ( 2− t 2 ℑ1 + t 2 ℑ2 )) dt ) + ∞∑ ð=0 aðoðΥ ( ℑ1 + ℑ2 2 ) . This is our required result. Theorem 10. Let Υ : [ℑ1,ℑ2] → ℜ is an L1 function and (ℵð+ ℘) ∈ (0, 1) also a differentiable function on (ℑ1,ℑ2) with ℑ1 < ℑ2 and assume Υ′ ∈ L1[ℑ1,ℑ2]. Consider ♭,ℵ, ℘, τ, δ, b ∈ C with ℜ(ℵ),ℜ(℘),ℜ(τ) > 0 and ℜ(b) > ℜ(δ) > 0. Moreover, let g ≥ 0, l > 0 and 0 < s ≤ l + ℜ(ℵ)( ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) +,ℵ,℘,τ Υ ) (ℑ2; g) + ( ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) −,ℵ,℘,τ Υ ) (ℑ1; g) − ∞∑ ð=0 að (oð − 1)Υ ( ℑ1 + ℑ2 2 ) ≥ ∞∑ ð=0 aðoð S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 28 of 34 × ( ℑ2 −ℑ1 4 ∫ 1 0 t(ℵð+℘) { Υ′ ( t 2 ℑ1 + 2− t 2 ℑ2 ) −Υ′ ( 2− t 2 ℑ1 + t 2 ℑ2 )} dt ) + ∞∑ ð=0 aðΥ ( ℑ1 + ℑ2 2 ) . where að = Bg(δ+ðs,b−δ) B(δ,b−δ) (b)ðs♭ ð (τ)ðl and oð = (ℑ2−ℑ1)(ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+℘)+1) with (ℵð+ ℘) > 0. Proof. Adding left and right sided Atangana-Baleanu integrals (3) and (4) from the middle of interval [ℑ1,ℑ2], we have( A−BIα ∗( ℑ1+ℑ2 2 ) + Υ(ℑ2) + A−BIα ∗( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) = α∗ B(α∗) ( R−LIα ∗( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LIα ∗( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) + 1− α∗ B(α∗) (Υ(ℑ1) + Υ(ℑ2)). Using α∗ = (ℵð+ ℘) in the above expression, we get ℵð+ ℘ B(ℵð+ ℘) ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) = ( A−BI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + A−BI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) − 1− ℵð− ℘ B(ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)). Multiplying the above equality with B(ℵð+℘) (ℵð+℘) and then subtracting Υ ( ℑ1+ℑ2 2 ) , we obtain( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LIℵð+℘( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) −Υ ( ℑ1 + ℑ2 2 ) = B(ℵð+ ℘) (ℵð+ ℘) ( A−BI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + A−BIℵð+℘( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) − 2(1− ℵð− ℘) (ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)) 2 −Υ ( ℑ1 + ℑ2 2 ) . Again we multiply the above equality with að to get að ( R−LIℵð+℘( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) − aðΥ ( ℑ1 + ℑ2 2 ) = að B(ℵð+ ℘) (ℵð+ ℘) ( A−BI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + A−BIℵð+℘( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 29 of 34 − að 2(1− ℵð− ℘) (ℵð+ ℘) (Υ(ℑ1) + Υ(ℑ2)) 2 − aðΥ ( ℑ1 + ℑ2 2 ) . Summing over all ð ∞∑ ð=0 að ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) − ∞∑ ð=0 aðΥ ( ℑ1 + ℑ2 2 ) = ∞∑ ð=0 að B(ℵð+ ℘) (ℵð+ ℘) ( A−BI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + A−BI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) − ∞∑ ð=0 að 2(1− ℵð− ℘) ℵð+ ℘ (Υ(ℑ1) + Υ(ℑ2)) 2 − ∞∑ ð=0 aðΥ ( ℑ1 + ℑ2 2 ) . (34) By convexity of Υ we have, Υ ( ℑ1+ℑ2 2 ) ≤ ( Υ(ℑ1)+Υ(ℑ2) 2 ) with positive multiplier að ∞∑ ð=0 að ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + + R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) − ∞∑ ð=0 aðΥ ( ℑ1 + ℑ2 2 ) ≤ ∞∑ ð=0 að B(ℵð+ ℘) (ℵð+ ℘) ( A−BI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + A−BI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) − ∞∑ ð=0 að ( 2(1− ℵð− ℘) (ℵð+ ℘) + 1 )( Υ(ℑ1) + Υ(ℑ2) 2 ) . Using Proposition 1 from the middle of interval [ℑ1,ℑ2], we acquire ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) +,ℵ,℘,τ Υ(ℑ2; g) + ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) −,ℵ,℘,τ Υ(ℑ1; g)− ∞∑ ð=0 aðΥ ( ℑ1 + ℑ2 2 ) ≤ ∞∑ ð=0 að B(ℵð+ ℘) (ℵð+ ℘) ( A−BI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + A−BI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) − ∞∑ ð=0 að ( 2(1− ℵð− ℘) (ℵð+ ℘) + 1 )( Υ(ℑ1) + Υ(ℑ2) 2 ) . Using Lemma 4 in the above inequality, we get ∞∑ ð=0 að (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) × ( ℑ2 −ℑ1 4 ∫ 1 0 t(ℵð+℘) { Υ′ ( t 2 ℑ1 + 2− t 2 ℑ2 ) −Υ′ ( 2− t 2 ℑ1 + t 2 ℑ2 )} dt ) + ∞∑ ð=0 að ( (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) − 1 ) Υ ( ℑ1 + ℑ2 2 ) S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 30 of 34 ≤ ∞∑ ð=0 að B(ℵð+ ℘) (ℵð+ ℘) ( A−BI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + A−BI (ℵð+℘)( ℑ1+ℑ2 2 ) − Υ(ℑ1) ) − ∞∑ ð=0 að ( 2(1− ℵð− ℘) (ℵð+ ℘) + 1 )( Υ(ℑ1) + Υ(ℑ2) 2 ) . Using (34) in the above expression ∞∑ ð=0 að (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) × ( ℑ2 −ℑ1 4 ∫ 1 0 t(ℵð+℘) { Υ′ ( t 2 ℑ1 + 2− t 2 b ) −Υ′ ( 2− t 2 ℑ1 + t 2 ℑ2 )} dt ) + ∞∑ ð=0 að ( (ℑ2 −ℑ1) (ℵð+℘) 2(ℵð+℘)−1Γ((ℵð+ ℘) + 1) − 1 ) Υ ( ℑ1 + ℑ2 2 ) ≤ ∞∑ ð=0 að ( R−LI (ℵð+℘)( ℑ1+ℑ2 2 ) + Υ(ℑ2) + R−LI (ℵð+℘)( ℑ̇1+ℑ2 2 ) − Υ(ℑ1) ) − ∞∑ ð=0 aðΥ ( ℑ1 + ℑ2 2 ) . Using Proposition 1 from the middle of interval [ℑ1,ℑ2], we have( ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) +,ℵ,℘,τ Υ ) (ℑ2; g) + ( ε♭,δ,b,s,l( ℑ1+ℑ2 2 ) −,ℵ,℘,τ Υ ) (ℑ1; g) − ∞∑ ð=0 að (oð − 1)Υ ( ℑ1 + ℑ2 2 ) ≥ ∞∑ ð=0 aðoð × ( ℑ2 −ℑ1 4 ∫ 1 0 t(ℵð+℘) { Υ′ ( t 2 ℑ1 + 2− t 2 ℑ2 ) −Υ′ ( 2− t 2 ℑ1 + t 2 ℑ2 )} dt ) + ∞∑ ð=0 aðΥ ( ℑ1 + ℑ2 2 ) . Hence the required result is obtained. 4. Conclusion Hermite-Hadamard inequalities are essential to many areas of mathematics, such as calculus, real analysis, and convex functions with practical applications in diverse areas S. Naheed et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6089 31 of 34 such as physics, economics, optimization and engineering. In this article, we look into fractional integral inequalities to Atangana-Baleanu and Prabhakar fractional calculus op- erators. Using extended generalized Mittag-Leffler functions as their kernel, we present several Hermite-Hadamard type fractional integral inequalities for the Atangana-Baleanu and Prabhakar fractional operators. For the integral inequalities involving fractional in- tegral of the kind (ℑ1+,ℑ2−) and (ℑ1+ℑ2 2 ), important results are given. We demonstrate the validity of our results by using certain functions to generate visual graphs that illus- trate the inequalities with corresponding numerical entries. This article seeks to provide more precise bounds and enhance the theoretical foundation for further studies to broaden the classical (H−H) inequality with generalized fractional integral operators. The latest inequalities will assist to boost comprehension in fractional calculus and convex analysis, with inference through several domains. Acknowledgements The authors I. Ayoob and N. 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