EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 5816 ISSN 1307-5543 – ejpam.com Published by New York Business Global Advancements in Ostrowski type fractional Integral Inequalities via Applications of Jensen’s and Young’s Inequalities Gauhar Rahman1,∗, Muhammad Samraiz2, Çetin Yıldız3,∗, Maryam Ali Alghafli4, Nabil Mlaiki4 1 Department of Mathematics and Statistics, Hazara University, Mansehra 21300, Pakistan 2 Department of Mathematics, University of Sargodha P.O. Box 40100, Sargodha, Pakistan 3 Deparment of Mathematics, K.K. Education Faculty, Atatürk University, 25240 Erzurum, Turkey 4 Department of Mathematics and Sciences, Prince Sultan University, Riyadh, 11586, Saudi Arabia Abstract. Fractional integral operators and convexity have a close link due to their fascinating properties in the mathematical sciences. In this paper, we first establish an integral identity involv- ing the generalized Hattaf-fractional integral operators. By using the Jensen integral inequality, Young’s inequality, power-mean inequality, and Hölder inequality, we then apply this identity to provide some new generalizations of Ostrowski type inequality for the convexity of |ℵ|. Further- more, we deduce several special cases from the main results. The results of this novel investigation should lead to new discoveries in the area of fractional calculus and inequalities. 2020 Mathematics Subject Classifications: 26A33, 26A51, 26D10 Key Words and Phrases: Young inequality, convex function, power mean inequality, fractional operators 1. Introduction Fractional integrals and derivatives have attracted a lot of attention from researchers nowadays. In many cases, fractional derivatives and fractional integrals provide more accurate representations of the frameworks than ordinary derivatives and integrals. The fractional calculus is currently widely employed in many scientific domains due to its nu- merous applications. The interest of researchers in fractional derivative and fractional ∗Corresponding author. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.5816 Email addresses: gauhar55uom@gmail.com, drgauhar.rahman@hu.edu.pk (G. Rahman), muhammad.samraiz@uos.edu.pk;msamraizuos@gmail.com (M. Samraiz), cetin@atauni.edu.tr (Ç. Yıldız), mghafli@psu.edu.sa (M. A. Alghafli), nmlaiki@psu.edu.sa; nmlaiki2012@gmail.com (N. Mlaiki) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) G. Rahman et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5816 2 of 18 integration has grown recently due to their wide applications in diverse domain, for exam- ple (see, [1–5]). Dumitru and Arran [6] have given a new formula for fractional derivatives and integrals using the Mittag-Leffler kernel. While more theoretical ideas about fractional operators with Mittag-Leffler kernels (Atangana-Baleanu operators) and the higher-order case have been discussed in [7–9], the generalization to the generalized Mittag-Leffler ker- nels to gain a semigroup property has recently been developed in [10, 11]. In the beginning, many scientists working in different areas of theory of inequalities em- ployed fractional calculus as an essential tool, for example, [12–16]. Shuang and Qi [17] proved a number of Hermite-Hadamard-type inequalities and examined specific methods for a class of s-convex functions. Mehrez and Agarwal [18] proved new integral inequali- ties and looked at specific cases of their discoveries with application to special means by employing the conventional Hermite-Hadamard inequalities. Park et al. [19] researched and used new generalized inequalities to stability analysis. By utilizing the local fractional approach, fractional integral inequalities were generated by Sarikaya et al. [20], expanding upon the findings found in the classical literature. In [21], Set et al. presented integral inequalities for differentiable convex functions via Atangana-Baleanu fractional integral operators. The various researchers examined a few noteworthy integral inequalities using various fractional methods. We refer the readers to the research conducted by [22–25]. 2. Preliminaries It is clear that the convex function is essential for the study of mathematical inequal- ities since it has several applications in the fields of pure and practical mathematics, mechanics, probability and statistics theory, economics, engineering, and optimization theory. Recently, there have been several mathematicians working on convexity’s theories, variations, augmentations, generalizations, and refinements. For example, in a number of scientific and mathematical domains, it is a useful tool for presenting a variety of chal- lenges and demonstrating awareness [26–29]. The convexity property can be used to generalize a number of well-known inequalities, such as the Opial type inequality, the Hermite–Hadamard inequality, the Ostrowski inequality, the Simpson inequality, the Bullen type inequality, and many more. The Ostrowski type inequality is one of the most extensively studied conclusions involving several kinds of convexities. Definition 1. [30] Let ℵ : I ⊆ R → R be a differentiable function ℵ ∈ L1[r, s] with r < s ∈ I. If |ℵ′(t)| ≤ K, for t ∈ [r, s], then the Ostrowski type integral inequality is given by | ℵ(t)− 1 s− r ∫ s r ℵ(t)dt |≤ K(s− r) [ 1 4 + ( t− r+s 2 ) (r + s)2 ] , where 1 4 is the least possible value. Mathematicians and scholars have been studying this inequality with significant at- tention and effort in recent years. This inequality was studied in 1997 by Dragomir and G. Rahman et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5816 3 of 18 Wang [31, 32] with relation to the lower and upper bounds of the first derivative. It was investigated by Barnett et al. and Cerone et al. [33, 34] that this inequality involving twice differentiable convex functions involved. Definition 2. [35] A function ℵ : [r, s] ⊆ R → R is said to be convex if ℵ (ρu+ (1− ρ)v) ≤ ρℵ(u) + (1− ρ)ℵ(v), for all u, v ∈ [r, s] and ρ ∈ [0, 1]. Convex functions are a concept that is frequently utilized in inequality theory. The Hermite-Hadamard inequality, which derives upper and lower bounds from averages of the mean value of a convex function, is as follows: Definition 3. Given a convex mapping ℵ : I ⊆ R → R, let r < s be on the interval I of R. Then the Hermite-Hadamard inequality is defined by ℵ ( r + s 2 ) ≤ 1 s− r ∫ s r ℵ(x)dx ≤ ℵ(r) + ℵ(s) 2 . Definition 4. [36] The ABC-fractional derivative is defined by ABCDκ r,ξℵ(ξ) = M(κ) 1− κ ∫ ξ r ℵ′(η)Eκ ( −κ(ξ − η)κ 1− κ ) dη, where 0 < κ < 1, ℵ′ ∈ L1(r, T ] and M(κ) is normalization function which satisfies the condition M(0) = M(1) = 1. Definition 5. [37, 38] The AB-fractional operator for ℵ ∈ L1(r, T ] and 0 < κ < 1 is defined by ABIκ r,ξℵ(ξ) = 1− κ M(κ) ℵ(ξ) + κ M(κ)Γ(κ) ∫ ξ r (ξ − η)κ−1 ℵ(η)dη. (1) Definition 6. [39] The Hattaf fractional derivative is defined by Dκ,σ,δ r,ξ,ωℵ(ξ) = M(κ) 1− κ 1 ω(ξ) ∫ ξ r Eσ ( −κ(ξ − η)δ 1− κ ) d dt (ωℵ)(η)dη, where 0 < κ < 1, ℵ′ ∈ L1(r, T ] ω ∈ C1(a, b), ω, ω′ > 0 on [a, b]. Definition 7. [39] The left sided Hattaf fractional operator for ℵ ∈ L1(r, T ] and 0 < κ < 1 is defined by Iκ,σ,ω r,ξ ℵ(ξ) = 1− κ M(κ) ℵ(ξ) + κ M(κ)Γ(σ)ω(ξ) ∫ ξ r (ξ − η)σ−1 ω(η)ℵ(η)dη. (2) G. Rahman et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5816 4 of 18 Definition 8. The right sided Hattaf fractional operator for ℵ ∈ L1(r, s) and 0 < κ < 1 is defined by Iκ,σ,ω s,ξ ℵ(ξ) = 1− κ M(κ) ℵ(ξ) + κ M(κ)Γ(σ)ω(ξ) ∫ s ξ (η − ξ)σ−1 ω(η)ℵ(η)dη. (3) Remark 1. i. If we consider σ = κ in (2) and (3), then we get AB-operator defined in (1). ii. If we consider σ = κ and ω = 1 in (2) and (3), then we get AB-operator defined in (1). Definition 9. The left sided Hattaf fractional operator for ω = 1, ℵ ∈ L1(r, T ] and 0 < κ < 1 is defined by Iκ,σ r,ξ ℵ(ξ) = 1− κ M(κ) ℵ(ξ) + κ M(κ)Γ(σ ∫ ξ r (ξ − η)σ−1 ℵ(η)dη. (4) Definition 10. The right sided Hattaf fractional operator for ω = 1, ℵ ∈ L1(r, s) and 0 < κ < 1 is defined by Iκ,σ s,ξ ℵ(ξ) = 1− κ M(κ) ℵ(ξ) + κ M(κ)Γ(σ) ∫ s ξ (η − ξ)σ−1 ℵ(η)dη. (5) This paper aims to derive an integral identity by incorporating the Hattaf fractional integral operators (4) and (5) and uses them to prove the refinement of Ostrowski type integral inequalities for differentiable convex functions. 3. Main Result In this section, we first prove the following fractional integral identity which will be used in our main findings. Lemma 1. Assume that ℵ : [r, s] → R represents a differentiable function on (r, s), where ℵ′ ∈ L1[r, s] and r < s. Next, for modified Hattaf-fractional integral operators, we have the identity given below:[ (t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ] = (t− r)σ+1 s− r ∫ 1 0 ρσℵ′(ρt+ (1− ρ)r)dρ− (s− t)σ+1 s− r ∫ 1 0 ρϱℵ′(ρt+ (1− ρ)s)dρ, where κ ∈ (0, 1], t ∈ [r, s]. G. Rahman et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5816 5 of 18 Proof. For simplicity, let us consider I = (t− r)σ+1 s− r ∫ 1 0 ρσℵ′(ρt+ (1− ρ)r)dρ− (s− t)σ+1 s− r ∫ 1 0 ρσℵ′(ρt+ (1− ρ)s)dρ = (t− r)σ+1 s− r I1 − (s− t)σ+1 s− r I2, (6) where I1 = ∫ 1 0 ρσℵ′(ρt+ (1− ρ)r)dρ = ρσℵ(ρt+ (1− ρ)r) t− r |10 + σ t− r ∫ 1 0 ρσ−1ℵ(ρt+ (1− ρ)r)dρ By substituting ρ = u−r t−r in the integral part, we get I1 = ℵ(t) t− r + σ (t− r)σ+1 ∫ t r (u− r)σ−1ℵ (u) du. (7) Similarly, one can get I2 = − ℵ(t) s− t + σ (s− t)σ+1 ∫ s t (s− u)σ−1ℵ (u) du. (8) By substituting (7) and (8) in (6) and then after a simple computation, we get the desired Lemma 1. Remark 2. Applying Lemma 1 for σ = κ, we get the Lemma 1 given in [40]. Theorem 1. Assume that ℵ : [r, s] → R be a differentiable function on (r, s), with ℵ′ ∈ L1[r, s] and r < s. Then, the following inequality holds for Hattaf-fractional integral operators (4) and (5) if |ℵ′| is a convex function∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r { ℵ′ (t) σ + 2 + ℵ′ (r) (σ + 1)(σ + 2) } + (s− t)σ+1 s− r { ℵ′ (t) σ + 2 + ℵ′ (s) (σ + 1)(σ + 2) } , where t ∈ [r, s], κ ∈ (0, 1]. Proof. By using Lemma 1, we have∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] G. Rahman et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5816 6 of 18 −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r ∫ 1 0 ρσ|ℵ′(ρt+ (1− ρ)r)|dρ− (s− t)σ+1 s− r ∫ 1 0 ρσ|ℵ′(ρt+ (1− ρ)s)|dρ. By applying the convexity of |ℵ′|, we get∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r ∫ 1 0 ρσ[ρ ∣∣ℵ′(t) ∣∣+ (1− ρ) ∣∣ℵ′(r) ∣∣]dρ + (s− t)σ+1 s− r ∫ 1 0 ρσ[ρ ∣∣ℵ′(s) ∣∣+ (1− ρ) ∣∣ℵ′(t) ∣∣]dρ. After solving the above integrals, we get∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r { 1 (σ + 2) |ℵ′(t)|+ 1 (σ + 1)(σ + 2) |ℵ′(r)| } + (s− t)σ+1 s− r { 1 (σ + 2) |ℵ′(t)|+ 1 (σ + 1)(σ + 2) |ℵ′(s)| } , which complete the desired proof. Corollary 1. Applying Theorem 1 for |ℵ′| ≤ K where K > 0 , we get the following inequality ∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤ K s− r ( 1 σ + 2 + 1 (σ + 1)(σ + 2) ){ (t− r)σ+1 + (s− t)σ+1 } . Corollary 2. Applying Corollary 1 for t = r+s 2 , we get the following inequality ∣∣∣(s− r)σ−1 2σ−1 ℵ ( r + s 2 ) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] G. Rahman et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5816 7 of 18 −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r, r+s 2 ℵ(r) + Iκ,σ s, r+s 2 ℵ(s) ]∣∣∣ ≤K ( 1 σ + 1 ) (s− r)σ 2σ . Remark 3. Applying Theorem 1 for σ = κ, we get Theorem 1 proved by Ahmad et al. [40]. Remark 4. Applying Corollary 2 for σ = κ, we get Corollary 2 proved earlier by Ahmad et al. [40]. Theorem 2. Let ℵ : [r, s] → R be a differentiable function on (r, s) with r < s and ℵ′ ∈ L1[r, s]. For Hattaf-fractional integral operators, we have the following inequality if |ℵ′|q is a convex function.∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r ( 1 σp+ 1 ) 1 p [ | ℵ′(t) |q + | ℵ′(r) |q 2 ] 1 q + (s− t)σ+1 s− r ( 1 σp+ 1 ) 1 p [ | ℵ′(s) |q + | ℵ′(t) |q 2 ] 1 q , where 1 p + 1 q = 1, t ∈ [r, s], κ ∈ (0, 1] and M(κ) > 0. Proof. By applying Lemma 1, we have∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− t ∫ 1 0 ρσ | ℵ′(ρt+ (1− ρ)r) | dρ + (s− t)σ+1 s− t ∫ 1 0 ρσ | ℵ′(ρt+ (1− ρ)s) | dρ. By employing Hölder inequality, we obtain∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] (9) −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ G. Rahman et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5816 8 of 18 ≤(t− r)σ+1 s− t [(∫ 1 0 ρϱpdρ ) 1 p (∫ 1 0 | ℵ′(ρt+ (1− ρ)r) |q dρ ) 1 q ] + (s− t)ϱ+1 s− t [(∫ 1 0 ρσpdρ ) 1 p (∫ 1 0 | ℵ′(ρt+ (1− ρ)s) |q dρ ) 1 q ] . (10) By employing the convexity of | ℵ′ |q, we have∫ 1 0 | ℵ′(ρt+ (1− ρ)r) |q dρ ≤ ∫ 1 0 [ρ | ℵ′(t) |q +(1− ρ) | ℵ′(r) |q]dρ = | ℵ′(t) |q + | ℵ′(r) |q 2 (11) and ∫ 1 0 | ℵ′(ρt+ (1− ρ)s) |q dρ ≤ ∫ 1 0 [ρ | ℵ′(t) |q +(1− ρ) | ℵ′(s) |q]dρ = | ℵ′(t) |q + | ℵ′(s) |q 2 . (12) Substituting (11) and (12) in (9) and then solving the integrals, we get the required inequality. Corollary 3. Applying Theorem 2 for |ℵ′| ≤ K where K > 0 , we get the following inequality ∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤ K s− r ( 1 σp+ 1 ) 1 p { (t− r)σ+1 + (s− t)σ+1 } . Corollary 4. Applying Corollary 3 for t = r+s 2 , we get the following inequality∣∣∣(s− r)σ−1 2σ−1 ℵ ( r + s 2 ) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r, r+s 2 ℵ(r) + Iκ,σ s, r+s 2 ℵ(s) ]∣∣∣ ≤K ( 1 σp+ 1 ) 1 p (s− r)σ 2σ . Remark 5. Applying Theorem 2 for σ = κ, we get Theorem 2 proved by Ahmad et al. [40]. G. Rahman et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5816 9 of 18 Remark 6. Applying Corollary 4 for σ = κ, we get Corollary 4 proved earlier by Ahmad et al. [40]. Theorem 3. Let ℵ : [r, s] → R be a differentiable function on (r, s), where ℵ′ ∈ L1[r, s] and r < s. The following inequality holds for Hattaf-fractional integral operators (4) and (5) if |ℵ′|q is a convex function.∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r ( 1 p(σp+ 1) + | ℵ′(t) |q + | ℵ′(r) |q 2q ) + (s− t)σ+1 s− t ( 1 p(σp+ 1) + | ℵ′(s) |q + | ℵ′(t) |q 2q ) , where 1 p + 1 q = 1, t ∈ [r, s], κ ∈ (0, 1] and M(κ) > 0. Proof. By utilizing Lemma 1, we have∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r ∫ 1 0 ρσ | ℵ′(ρt+ (1− ρ)r) | dρ + (s− t)σ+1 s− r ∫ 1 0 ρσ | ℵ′(ρt+ (1− ρ)s) | dρ. By employing Young inequality as uv ≤ up p + vq q in above, we have∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] (13) −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r [ 1 p ∫ 1 0 ρσpdρ+ 1 q ∫ 1 0 | ℵ′(ρt+ (1− ρ)r) |q dρ ] + (s− t)σ+1 s− r [ 1 p ∫ 1 0 ρσpdρ+ 1 q ∫ 1 0 | ℵ′(ρt+ (1− ρ)s) |q dρ ] . (14) Since | ℵ′ |q is convex, so therefore we have∫ 1 0 | ℵ′(ρt+ (1− ρ)r) |q dρ ≤ ∫ 1 0 [ρ | ℵ′(t) |q +(1− ρ) | ℵ′(r) |q]dρ G. Rahman et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5816 10 of 18 = | ℵ′(t) |q + | ℵ′(r) |q 2 (15) and ∫ 1 0 | ℵ′(ρt+ (1− ρ)s) |q dρ ≤ ∫ 1 0 [ρ | ℵ′(t) |q +(1− ρ) | ℵ′(s) |q]dρ = | ℵ′(t) |q + | ℵ′(s) |q 2 . (16) Substituting (15) and (16) in (13) and then by solving the integrals, we get the desired proof. Corollary 5. Applying Theorem 3 for |ℵ′| ≤ K where K > 0 , we get the following inequality ∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤ ( 1 p(σp+ 1) + Kq q ){ (t− r)σ+1 s− r + (s− t)σ+1 s− r } . Corollary 6. Applying Corollary 5 for t = r+s 2 , we get the following inequality ∣∣∣(s− r)σ−1 2σ−1 ℵ ( r + s 2 ) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r, r+s 2 ℵ(r) + Iκ,σ s, r+s 2 ℵ(s) ]∣∣∣ ≤ ( 1 p(σp+ 1) + Kq q ){ (s− r)σ 2σ } . Remark 7. Applying Theorem 3 for σ = κ, we get Theorem 3 proved by Ahmad et al. [40]. Remark 8. Applying Corollary 6 for σ = κ, we get Corollary 6 proved earlier by Ahmad et al. [40]. Theorem 4. Let ℵ : [r, s] → R be a differentiable function on (r, s), where ℵ′ ∈ L1[r, s] and r < s. The following inequality holds for Hattaf-fractional integral operators if |ℵ′|q is a convex function ∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] G. Rahman et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5816 11 of 18 −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r ( 1 σ + 1 )1− 1 q [ | ℵ′(t) |q (σ + 2) + | ℵ′(r) |q (σ + 1)(σ + 2) ] 1 q + (s− t)σ+1 s− r ( 1 σ + 1 )1− 1 q [ | ℵ′(s) |q (σ + 1)(σ + 2) + | ℵ′(t) |q (σ + 2) ] 1 q , where q ≥ 1, t ∈ [r, s], κ ∈ (0, 1] and M(κ) > 0. Proof. By utilizing Lemma 1, we have∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r ∫ 1 0 ρσ | ℵ′(ρt+ (1− ρ)r) | dρ + (s− t)σ+1 s− r ∫ 1 0 ρσ | ℵ′(ρt+ (1− ρ)s) | dρ. By employing power mean inequality, we obtain∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r (∫ 1 0 ρσdρ )1− 1 q (∫ 1 0 ρσ | ℵ′(ρt+ (1− ρ)r) |q dρ ) 1 q + (s− t)σ+1 s− r (∫ 1 0 ρσdρ )1− 1 q (∫ 1 0 ρσ | ℵ′(ρt+ (1− ρ)s) |q dρ ) 1 q . Now, by utilizing the convexity of | ℵ′ |q, we get∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r (∫ 1 0 ρσdρ )1− 1 q (∫ 1 0 ρσ[ρ | ℵ′(t) |q +(1− ρ) | ℵ′(r) |q]dρ ) 1 q + (s− t)σ+1 s− r (∫ 1 0 ρσdρ )1− 1 q (∫ 1 0 ρσ[ρ | ℵ′(t) |q +(1− ρ) | ℵ′(s) |q]dρ ) 1 q G. Rahman et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5816 12 of 18 ≤(t− r)σ+1 s− r ( 1 σ + 1 )1− 1 q [ | ℵ′(t) |q (σ + 2) + | ℵ′(r) |q (σ + 1)(σ + 2) ] 1 q + (s− t)σ+1 s− r ( 1 σ + 1 )1− 1 q [ | ℵ′(t) |q (σ + 2) + | ℵ′(s) |q (σ + 1)(σ + 2) ] 1 q , which complete the proof. Corollary 7. Applying Theorem 4 for |ℵ′| ≤ K where K > 0 , we get the following inequality ∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤ K s− r ( 1 σ + 1 ){ (t− r)σ+1 + (s− t)σ+1 } . Corollary 8. Applying Corollary 7 for t = r+s 2 , we get the following inequality ∣∣∣ [(s− r)σ−1 2σ−2 ] ℵ ( r + s 2 ) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r, r+s 2 ℵ(r) + Iκ,σ s, r+s 2 ℵ(s) ]∣∣∣ ≤ K s− r ( 1 σ + 1 ) (s− r)σ 2σ . Remark 9. Applying Theorem 4 for σ = κ, we get Theorem 4 proved by Ahmad et al. [40]. Remark 10. Applying Corollary 8 for σ = κ, we get Corollary 8 proved earlier by Ahmad et al. [40]. Theorem 5. Let ℵ : [r, s] → R be a differentiable function on (r, s), where ℵ′ ∈ L1[r, s] and r < s. For Hattaf-fractional integral operators (4) and (5), we have the following inequality if |ℵ′| is a concave function∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r ( 1 σ + 1 ) | ℵ′ ( (σ + 1)t+ r σ + 2 ) | +(s− t)σ+1 s− r ( 1 σ + 1 ) | ℵ′ ( (σ + 1)t+ s σ + 2 ) |, where κ ∈ (0, 1] and M(κ) > 0. G. Rahman et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5816 13 of 18 Proof. By utilizing Lemma 1, we have∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r ∫ 1 0 ρσ | ℵ′(ρt+ (1− ρ)r) | dρ + (s− t)σ+1 s− r ∫ 1 0 ρσ | ℵ′(ρt+ (1− ρ)s) | dρ. By employing the Jensen integral inequality, we get∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r (∫ 1 0 ρσdρ ) | ℵ′ (∫ 1 0 ρσ(ρt+ (1− ρ)r)dρ∫ 1 0 ρσdρ ) | + (s− t)σ+1 s− r (∫ 1 0 ρσdρ ) | ℵ′ (∫ 1 0 ρσ(ρt+ (1− ρ)s)dρ∫ 1 0 ρσdρ ) | . After simple calculation of above integrals, we get the required inequality. Corollary 9. Applying Theorem 5 for |ℵ′| ≤ K where K > 0 , we get the following inequality ∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤ K s− r ( 1 σ + 1 ){ (t− r)σ+1 + (s− t)σ+1 } . Remark 11. Applying Theorem 5 for κ = σ, we get Theorem 5 proved by Ahmad et al. [40]. Remark 12. Applying Corollary 9 for κ = σ, we get Corollary 9 proved earlier by Ahmad et al. [40]. Theorem 6. Assume that the function Ψ : [r, s] → R has a continuous derivative on [r, s] and is strictly increasing and positive. Let ℵ : [r, s] → R be a differentiable function on G. Rahman et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5816 14 of 18 (r, s), where ℵ′ ∈ L1[r, s] and r < s. For Hattaf-fractional integral operators, we have the following inequality if |ℵ′|q is a concave function∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r ( 1 σp+ 1 ) 1 p | ℵ′ ( t+ r 2 ) | +(s− t)σ+1 s− r ( 1 σp+ 1 ) 1 p | ℵ′ ( s+ t 2 ) |, where 1 p + 1 q = 1, q > 1, t ∈ [r, s], κ ∈ (0, 1] and M(κ) > 0. Proof. By utilizing Lemma 1, we have∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r ∫ 1 0 ρσ | ℵ′(ρt+ (1− ρ)r) | dρ + (s− t)σ+1 s− r ∫ 1 0 ρσ | ℵ′(ρt+ (1− ρ)s) | dρ. By employing the Hölder integral inequality, we obtain∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r (∫ 1 0 ρσpdρ ) 1 p (∫ 1 0 | ℵ′(ρt+ (1− ρ)r) |q dρ ) 1 q + (s− t)σ+1 s− r (∫ 1 0 ρσpdρ ) 1 p (∫ 1 0 | ℵ′(ρt+ (1− ρ)s) |q dρ ) 1 q . Now, by employing the convexity of | ℵ′ |q and Jensen integral inequality, we obtain∣∣∣ [(t− r)σ + (s− t)σ s− r ] ℵ (t) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] (17) −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r,t ℵ(r) + Iκ,σ s,t ℵ(s) ]∣∣∣ ≤(t− r)σ+1 s− r (∫ 1 0 ρσpdρ ) 1 p (∫ 1 0 | ℵ′(ρt+ (1− ρ)r) |q dρ ) 1 q G. Rahman et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5816 15 of 18 + (s− t)σ+1 s− r (∫ 1 0 ρσpdρ ) 1 p (∫ 1 0 | ℵ′(ρt+ (1− ρ)s) |q dρ ) 1 q . (18) Now, since∫ 1 0 | ℵ′(ρt+ (1− ρ)r) |q dρ ≤ ∫ 1 0 ρ0 | ℵ′(ρt+ (1− ρ)r) |q dρ ≤ (∫ 1 0 ρ0dρ ) | ℵ′ ( 1∫ 1 0 ρ0dρ ∫ 1 0 (ρt+ (1− ρ)r)dρ ) |q ≤ | ℵ′ ( t+ r 2 ) |q . (19) Similarly ∫ 1 0 | ℵ′(ρt+ (1− ρ)s) |q dρ ≤| ℵ′ ( t+ s 2 ) |q . (20) By substituting (20) and (19) in (17) and then solving the integrals, we get the desire assertion. Corollary 10. Applying Theorem 6 for t = r+s 2 , we get the following inequality∣∣∣(s− r)σ−1 2σ−1 ℵ ( r + s 2 ) + σ(1− κ) κ(s− r) Γ(σ) [ℵ (r) + ℵ (s)] −σM(κ)Γ(σ) κ(s− r) [ Iκ,σ r, r+s 2 ℵ(r) + Iκ,σ s, r+s 2 ℵ(s) ]∣∣∣ ≤ ( 1 σp+ 1 ) 1 p 1 s− r { (s− t)σ+1 | ℵ′ ( 3r + s 4 ) | +(s− t)σ+1 | ℵ′ ( r + 3s 4 ) | } . Remark 13. Applying Theorem 6 for σ = κ, we get Theorem 6 proved by Ahmad et al. [40]. Remark 14. Applying Corollary 10 for σ = κ, we get Corollary 10 proved earlier by Ahmad et al. [40]. 4. Concluding Remarks In this paper, we developed Ostrowski-type inequalities for convex functions containing the Hattaf fractional integral operators. The results presented in this paper are original to the best of our knowledge. Given the widespread applications of convex functions in numerous scientific fields, our unique advancements are believed to be applicable to sev- eral special functions, including convexity, interval analysis, quantum calculus, fractional calculus, and coordinates. The inequalities in term of AB operators will be restored if we put κ = σ and classical inequalities if we put κ = σ = 1. One can obtain Grüss type inequalities, Chebyshev type inequalities, Reverse Minkowski’s type inequalities and certain other type inequalities by using Hattaf fractional integral operators. G. Rahman et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5816 16 of 18 Acknowledgements The authors M. A. Alghafli and N. 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