Refinement of Fractional Weddle's-Type Inequalities via Tempered Fractional Integrals

Refinement of Fractional Weddle’s-Type Inequalities via Tempered Fractional Integrals Asra Ahmad1, Hüseyin Budak2, Bilal Ahmad3, Asia Shehzadi4, Wali Haider5 1 Department of Mathematics and Statistics,University of Agriculture Faisalabad, Pakistan 2Department of Mathematics, Faculty of Science and Arts, Kocaeli University, Kocaeli 41001, Türkiye 3School of Mathematics and Computational Science, Xiangtan University, Xiangtan, 411105, China 4 School of Mathematics and Statistics, Central South University, Changsha 410083, China 5School of Mathematics and Statistics, Shaanxi Normal University, Xi’an, Shaanxi, 710119, China Corresponding author: haiderw[email protected] ORCID IDs: First Author: 0009-0008-7864-3635 Second Author: 0000-0001-8843-955X Third Author: 0009-0002-8603-0421 Fourth Author: 0009-0005-1101-5536 Fifth Author: 0009-0001-7065-2755 DOI : 10.5281/zenodo.18011928 Abstract Fractional calculus generalizes classical differentiation and integration of arbitrary order. It offers practical analytical tools for describing complex phenomena exhibiting memory and nonlocal behavior. This investigation introduces unique inequalities for convex functions employing tempered fractional integrals, focusing on Weddle’s type inequalities. A significant equality is formulated within the framework of tempered fractional integrals. Based on this identity, we develop various Weddle-type inequalities for differentiable convex mappings connected with tempered fractional integrals. These mappings extend the applicability of fractional calculus by extending classical notions and presenting further insight into the behavior of convex functions. We attain distinctive types of Weddle’s type inequalities by involving key mathematical tools such as Holder’s and power mean inequality, enhancing the implementation of these inequalities in diverse mathematical domains. Keywords: Weddle’s formula type inequality, convex function, tempered fractional integrals M6-1 KOSC-2025 Proceedings 1 Introduction and Motivation By adding derivatives and integrals of fractional/non-integer orders, fractional calculus (FC), which was originally presented in 1695 through the correspondence between Leibniz and the Marquis de l’Hospital, expands on classical calculus. Later, mathematicians like Euler, Liouville, and Riemann developed the theory, and during the 19th century, it was acknowledged as a crucial tool in both theoretical and applied mathematics. FC is unique because of its non-local nature, which makes it particularly helpful for explaining memory and inherited effects in materials and systems. FC has become an increasingly important tool for representing non-standard occurrences in various fields, such as geometry, dynamics, and fractional differential equations [ 23 , 12 , 10 ]. These days, it is used in viscoelasticity, bioengineering, control theory, and acoustics [ 1 ]. Foundational works like those by Samko, Kilbas, and Marichev (1993) and Podlubny (1999), which allow the study of the complex and unpredictable ways that natural systems function, have solidified its place in science and engineering. The well-known Riemann-Liouville (RL) fractional integrals are listed as follows: Definition 1.1 (See [ 15 , 11 ]).Presume F ∈L1 [Θ , ρ ] , the RL fractional integral of order α > 0, expressed as: Jα Θ+F(κ) = 1 Γ(α)Zκ Θ (κ−κ)α−1F(κ)dκ, κ>Θ and Jα ρ−F(κ) = 1 Γ(α)Zρ κ (κ−κ)α−1F(κ)dκ, κ< ρ. Definition 1.2. The Gamma function, as described in [ 7 ], is described using the subsequent integral expression. Γ(χ) = Z∞ 0 e−κκχ−1dκ, Re(χ)>0. See references [ 21 , 2 , 8 ] for a more thorough explanation of RL fractional integrals. However, fractional integral inequalities are crucial for practical mathematics and differential equation analysis. A key concept in mathematics since ancient times is mathematical inequality, which compares two values or expressions to show their relationships or relative magnitudes. Convex functions are closely related to the theory of inequality; several famous inequalities, such as Hermite–Hadamard, Simpson, Newton, Euler-Maclaurin, and Boole’s type inequalities, have been developed for convex functions. Convexity serves as a reliable tool for studying inequalities, laying the groundwork for using geometric and analytical techniques to verify these conclusions.Convex function analysis and inequality analysis are closely related, with each impacting the development of the other. Please refer to [ 13 , 3 , 25 ] and the sources listed therein for more details on convexity and Newton-type inequalities associated with convex differentiable functions. The tempered fractional integral was initially studied in this subject by Buschman [ 4 ]. Following this, further comprehensive contributions were made by Liu et al. [ 16 ] and Meerschaert et al. [ 20 ]. A prominent subfield of fractional calculus called tempered fractional calculus M6-2 2nd Kocaeli Science Congress, November 19-21, 2025 finds a special intersection between weighted fractional calculus and fractional calculus with analytic kernels [ 9 ]. It can be considered an extension of fractional calculus due to its wide application and importance. Its significance is demonstrated by its rediscovery under many titles, such as significant fractional calculus [ 9 , 5 ] and generalized proportional fractional calculus [ 14 ]. The next section reviews the essential preparations that form the basis of our key conclusions. Definition 1.3 (See [ 22 ]).For real values α > 0and κ, λ ≥ 0, the λ -incomplete gamma function is identified as: Υλ(α, κ) := Zκ 0 κα−1e−λκdκ. λ= 1 corresponds to the incomplete gamma function [6]: Υ(α, κ) := Zκ 0 κα−1e−κdκ. Where, 0<α<∞and λ≥0. Remark 1.1 (See [22]).For the real values α > 0and κ, λ ≥0, we gain I. Υλ(ρ−Θ)(α, 1) = R1 0κα−1e−λ(ρ−Θ)κdκ =1 (ρ−Θ)αΥλ(α, ρ −Θ). II. R1 0Υλ(ρ−Θ)(α, κ)dκ=Υλ(α,ρ−Θ) (ρ−Θ)α−Υλ(α+1,ρ−Θ) (ρ−Θ)α+1 . Definition 1.4 (See [ 17 , 21 ]).Tempered fractional integral operators of order α > 0and λ≥ 0 are presented as: Z(α,λ) Θ+ F(κ) = 1 Γ(α)Zκ Θ (κ−κ)α−1e−λ(κ−κ)F(κ)dκ, κ∈[Θ, ρ] and Z(α,λ) ρ−F(κ) = 1 Γ(α)Zρ κ (κ−κ)α−1e−λ(κ−κ)F(κ)dκ, κ∈[Θ, ρ], respectively for F∈L1[Θ, ρ]. The RL fractional integral equations in Definition 1.1 are obviously comparable if we assume λ= 0 in Definition 1.4.. Using tempered fractional integral operators, Gul and Yalcin [ 11 ] developed Minkowski and Hermite–Hadamard integral inequalities. They displayed a number of integral inequality instances for tempered fractional integrals from earlier research. For more details, interested readers are referred to [23,24,26]. Inspired by previous research, we employed tempered fractional integrals to identify some of Boole’s formula-type inequalities in the class of functions whose derivatives are convex functions. This paper examines important mathematical tools such as Holder’s and power mean inequality, enhancing the implementation of these inequalities in diverse mathematical domains. The most important advantage of these inequalities is that, by setting λ = 0, these inequalities turned into 2nd Kocaeli Science Congress, November 19-21, 2025 M6-3 KOSC-2025 Proceedings Riemann–Liouville fractional Boole’s type inequalities. With α = 1, the modified inequalities transform into classical Boole’s type inequalities. The study is broken up into three pieces, the first of which is an introduction and preliminary material that includes basic concepts of fractional calculus and a synopsis of important research in the area. Section 1.1 will establish some results regarding Boole’s type inequality for differentiable convex functions. We also assess Boole’s formula for new mathematical tools such as holder’s inequality and power mean. We summarize our results and potential study opportunities in the last part. 1.1 Main Contributions Lemma 1.1. If F: [Θ , ρ ] →R be a function that is absolutely continuous on (Θ , ρ )such that F′∈L1[Θ, ρ].Then, the following equality is valid: 1 20 F(Θ) + 5F5Θ+ρ 6+F2Θ+ρ 3+ 6FΘ+ρ 2+FΘ+2ρ 3+ 5FΘ+5ρ 6+F(ρ) −Γ(α) 2⋎λα, ρ−Θ 2"Z(α,λ) (Θ+ρ 2)+F(ρ)+Z(α,λ) (Θ+ρ 2)−F(Θ)#=(ρ−Θ)α+1 2α+2 ⋎λα, ρ−Θ 2 ×"Z1 3 0⋎λ(ρ−Θ 2)(α, ξ)−1 10 ⋎λ(ρ−Θ 2)(α, 1)F′ξ 2ρ+2−ξ 2Θ−F′ξ 2Θ + 2−ξ 2ρdξ +Z2 3 1 3⋎λ(ρ−Θ 2)(α, ξ)−3 5⋎λ(ρ−Θ 2)(α, 1)F′ξ 2ρ+2−ξ 2Θ−F′ξ 2Θ + 2−ξ 2ρdξ +Z1 2 3⋎λ(ρ−Θ 2)(α, ξ)−7 10 ⋎λ(ρ−Θ 2)(α, 1)F′ξ 2ρ+2−ξ 2Θ−F′ξ 2Θ + 2−ξ 2ρdξ#. (1) Proof. By acknowledging the essential rules of integration, it is enough to state that I1=Z1 3 0⋎λ(ρ−Θ 2)(α, ξ)−1 10 ⋎λ(ρ−Θ 2)(α, 1)F′ξ 2ρ+2−ξ 2Θdξ =2 ρ−Θ⋎λ(ρ−Θ 2)(α, ξ)−1 10 ⋎λ(ρ−Θ 2)(α, 1)Fξ 2ρ+2−ξ 2Θ 1 3 0 −2 ρ−ΘZ1 3 0 ξα−1e−λ(ρ−Θ 2)ξFξ 2ρ+2−ξ 2Θdξ =2 ρ−Θ⋎λ(ρ−Θ 2)α, 1 3−1 10 ⋎λ(ρ−Θ 2)(α, 1)F5Θ+ρ 6+ ⋎λ(ρ−Θ 2)(α, 1) 5(ρ−Θ) F(Θ) −2 ρ−ΘZ1 3 0 ξα−1e−λ(ρ−Θ 2)ξFξ 2ρ+2−ξ 2Θdξ, (2) I2=2 ρ−Θ⋎λ(ρ−Θ 2)α, 2 3−3 5⋎λ(ρ−Θ 2)(α, 1)F2Θ+ρ 3 −⋎λ(ρ−Θ 2)α, 1 3−3 5⋎λ(ρ−Θ 2)(α, 1)F5Θ+ρ 6 M6-4 2nd Kocaeli Science Congress, November 19-21, 2025 1.1 Main Contributions −2 ρ−ΘZ2 3 1 3 ξα−1e−λ(ρ−Θ 2)ξFξ 2ρ+2−ξ 2Θdξ. (3) In the same way, we find I3=2 ρ−Θ⋎λ(ρ−Θ 2)(α, 1) −7 10 ⋎λ(ρ−Θ 2)(α, 1)FΘ+ρ 2 −⋎λ(ρ−Θ 2)α, 2 3−7 10 ⋎λ(ρ−Θ 2)(α, 1)F2Θ+ρ 3 −2 ρ−ΘZ1 2 3 ξα−1e−λ(ρ−Θ 2)ξFξ 2ρ+2−ξ 2Θdξ. (4) Consequently, we arrive at the subsequent equality by merging (2)-(4) I1+I2+I3=2 10(ρ−Θ) ⋎λ(ρ−Θ 2)(α, 1)F(Θ) + 5 ⋎λ(ρ−Θ 2)(α, 1)F5Θ+ρ 6 +⋎λ(ρ−Θ 2)(α, 1)F2Θ+ρ 3+3⋎λ(ρ−Θ 2)(α, 1)FΘ+ρ 2 −2 ρ−ΘZ1 0 ξα−1e−λ(ρ−Θ 2)ξFξ 2ρ+2−ξ 2Θdξ.(5) Similarly I4+I5+I6=−2 10(ρ−Θ) ⋎λ(ρ−Θ 2)(α, 1)F(ρ) + ⋎λ(ρ−Θ 2)(α, 1)FΘ+2ρ 3 +5 ⋎λ(ρ−Θ 2)(α, 1)FΘ+5ρ 6+3⋎λ(ρ−Θ 2)(α, 1)FΘ+ρ 2 +2 ρ−ΘZ1 0 ξα−1e−λ(ρ−Θ 2)ξFξ 2Θ + 2−ξ 2ρdξ.(6) By subtracting equalities (5) and (6) , then we make the substitutions ϖ = ξ 2ρ + 2−ξ 2 Θand ϖ=ξ 2Θ + 2−ξ 2ρfor ξ∈[0,1], it becomes [I1+I2+I3−(I4+I5+I6)] = ⋎λ(ρ−Θ 2)(α, 1) 5(ρ−Θ) F(Θ) + 5F5Θ+ρ 6+F2Θ+ρ 3+ 6FΘ+ρ 2+FΘ+2ρ 3 +5FΘ+5ρ 6+F(ρ)−2α+1Γ(α) (ρ−Θ)α+1 "Z(α,λ) (Θ+ρ 2)+F(ρ) + Z(α,λ) (Θ+ρ 2)−F(Θ)#.(7) Multiplying both sides of (7) by ρ−Θ 4⋎ λ(ρ−Θ 2)(α,1) , the equality (1) is obtained. Theorem 1.1. If all the conditions in Lemma 1.1 are accomplished and | F ′| is convex on [Θ , ρ ], then one can prove the following inequality:  1 20 F(Θ) + 5F5Θ+ρ 6+F2Θ+ρ 3+ 6FΘ+ρ 2+FΘ+2ρ 3+ 5FΘ+5ρ 6+F(ρ) −Γ(α) 2⋎λα, ρ−Θ 2"Z(α,λ) (Θ+ρ 2)+F(ρ) + Z(α,λ) (Θ+ρ 2)−F(Θ)# 2nd Kocaeli Science Congress, November 19-21, 2025 M6-5 KOSC-2025 Proceedings ≤(ρ−Θ)α+1 2α+2 ⋎λα, ρ−Θ 2(∆1(α, λ) + ∆2(α, λ) + ∆3(α, λ)) F′(Θ)+F′(ρ),(8) where ∆1(α, λ) = Z1 3 0 ⋎λ(ρ−Θ 2)(α, ξ)−1 10 ⋎λ(ρ−Θ 2)(α, 1) dξ, ∆2(α, λ) = Z2 3 1 3 ⋎λ(ρ−Θ 2)(α, ξ)−3 5⋎λ(ρ−Θ 2)(α, 1) dξ, ∆3(α, λ) = Z1 2 3 ⋎λ(ρ−Θ 2)(α, ξ)−7 10 ⋎λ(ρ−Θ 2)(α, 1) dξ. Proof. By referencing Lemma 1.1 and leveraging the convexity of |F′|, we acquire  1 20 F(Θ) + 5F5Θ+ρ 6+F2Θ+ρ 3+ 6FΘ+ρ 2+FΘ+2ρ 3+ 5FΘ+5ρ 6+F(ρ) −Γ(α) 2⋎λα, ρ−Θ 2"Z(α,λ) (Θ+ρ 2)+F(ρ) + Z(α,λ) (Θ+ρ 2)−F(Θ)# ≤(ρ−Θ)α+1 2α+2 ⋎λα, ρ−Θ 2"Z1 3 0 ⋎λ(ρ−Θ 2)(α, ξ)−1 10 ⋎λ(ρ−Θ 2)(α, 1) × F′ξ 2ρ+2−ξ 2Θ−F′ξ 2Θ + 2−ξ 2ρ dξ +Z2 3 1 3 ⋎λ(ρ−Θ 2)(α, ξ)−3 5⋎λ(ρ−Θ 2)(α, 1) F′ξ 2ρ+2−ξ 2Θ−F′ξ 2Θ + 2−ξ 2ρ dξ# +Z1 2 3 ⋎λ(ρ−Θ 2)(α, ξ)−7 10 ⋎λ(ρ−Θ 2)(α, 1) F′ξ 2ρ+2−ξ 2Θ−F′ξ 2Θ + 2−ξ 2ρ dξ#(9) ≤(ρ−Θ)α+1 2α+2 ⋎λα, ρ−Θ 2"Z1 3 0 ⋎λ(ρ−Θ 2)(α, ξ)−1 10 ⋎λ(ρ−Θ 2)(α, 1) ×ξ 2|F′(ρ)|+2−ξ 2|F(Θ)|+ξ 2|F′(Θ)|+2−ξ 2|F(ρ)|dξ +Z2 3 1 3 ⋎λ(ρ−Θ 2)(α, ξ)−3 5⋎λ(ρ−Θ 2)(α, 1) ×ξ 2|F′(ρ)|+2−ξ 2|F(Θ)|+ξ 2|F′(Θ)|+2−ξ 2|F(ρ)|dξ +Z1 2 3 ⋎λ(ρ−Θ 2)(α, ξ)−7 10 ⋎λ(ρ−Θ 2)(α, 1) ×ξ 2|F′(ρ)|+2−ξ 2|F(Θ)|+ξ 2|F′(Θ)|+2−ξ 2|F(ρ)|dξ =(ρ−Θ)α+1 2α+2 ⋎λα, ρ−Θ 2(∆1(α, λ) + ∆2(α, λ) + ∆3(α, λ)) F′(Θ)+F′(ρ). We have reached the conclusion of the proof of Theorem 1.1. M6-6 2nd Kocaeli Science Congress, November 19-21, 2025 1.1 Main Contributions Remark 1.2. If λ= 0 in Theorem 1.1, we conclude  1 20 F(Θ) + 5F5Θ+ρ 6+F2Θ+ρ 3+ 6FΘ+ρ 2+FΘ+2ρ 3+ 5FΘ+5ρ 6+F(ρ) −2α−1Γ(α+ 1) (ρ−Θ)αJα (Θ+ρ 2)+F(ρ)+Jα (Θ+ρ 2)−F(Θ) ≤α(ρ−Θ) 4(∆1(α, 0) + ∆2(α, 0) + ∆3(α, 0)) F′(Θ)+F′(ρ), which is derived in [18]. Remark 1.3. If we assume λ= 0 and α= 1 in Theorem 1.1, then we gain  1 20 F(Θ) + 5F5Θ+ρ 6+F2Θ+ρ 3+ 6FΘ+ρ 2+FΘ+2ρ 3+ 5FΘ+5ρ 6+F(ρ) −1 ρ−ΘZρ Θ F(ϖ)dϖ ≤13(ρ−Θ) 450 |F′(Θ)|+|F′(ρ)|, which is proved in [27]. Theorem 1.2. If all the conditions in Lemma 1.1 are accomplished and | F ′|q, q > 1is convex on [Θ, ρ], then the subsequent inequality is valid:  1 20 F(Θ) + 5F5Θ+ρ 6+F2Θ+ρ 3+ 6FΘ+ρ 2+FΘ+2ρ 3+ 5FΘ+5ρ 6+F(ρ) −Γ(α) 2⋎λα, ρ−Θ 2"Z(α,λ) (Θ+ρ 2)+F(ρ) + Z(α,λ) (Θ+ρ 2)−F(Θ)# ≤(ρ−Θ)α+1 2α+2 ⋎λα, ρ−Θ 2 γ1(α, p)  |F′(ρ)|q+ 7|F′(Θ)|q 16  1 q+|F′(Θ)|q+ 7|F′(ρ)|q 16  1 q   +γ2(α, p)  3|F′(ρ)|q+ 5|F′(Θ)|q 16  1 q+3|F′(Θ)|q+ 5|F′(ρ)|q 16  1 q   ,(10) where p−1+q−1= 1,, γp 1(α, λ) = Z1 3 0 ⋎λ(ρ−Θ 2)(α, ξ)−1 10 ⋎λ(ρ−Θ 2)(α, 1) p dξ!1 p , γp 2(α, λ) = Z2 3 1 3 ⋎λ(ρ−Θ 2)(α, ξ)−3 5⋎λ(ρ−Θ 2)(α, 1) p dξ!1 p , γp 3(α, λ) = Z1 2 3 ⋎λ(ρ−Θ 2)(α, ξ)−7 10 ⋎λ(ρ−Θ 2)(α, 1) p dξ!1 p . Proof. By applying Hölder’s inequality (9), it becomes  1 20 F(Θ) + 5F5Θ+ρ 6+F2Θ+ρ 3+ 6FΘ+ρ 2+FΘ+2ρ 3+ 5FΘ+5ρ 6+F(ρ) 2nd Kocaeli Science Congress, November 19-21, 2025 M6-7 KOSC-2025 Proceedings −Γ(α) 2⋎λα, ρ−Θ 2"Z(α,λ) (Θ+ρ 2)+F(ρ)+Z(α,λ) (Θ+ρ 2)−F(Θ)# ≤(ρ−Θ)α+1 2α+2 ⋎λα, ρ−Θ 2  Z1 3 0 ⋎λ(ρ−Θ 2)(α, ξ)−1 10 ⋎λ(ρ−Θ 2)(α, 1) p dξ!1 p ×   Z1 3 0 F′ξ 2ρ+2−ξ 2Θ q dξ!1 q + Z1 3 0 F′ξ 2Θ + 2−ξ 2ρ q dξ!1 q   + Z1 3 1 3 ⋎λ(ρ−Θ 2)(α, ξ)−3 5⋎λ(ρ−Θ 2)(α, 1) p dξ!1 p ×   Z2 3 1 3 F′ξ 2ρ+2−ξ 2Θ q dξ!1 q + Z2 3 1 3 F′ξ 2Θ + 2−ξ 2ρ q dξ!1 q   + Z1 2 3 ⋎λ(ρ−Θ 2)(α, ξ)−7 10 ⋎λ(ρ−Θ 2)(α, 1) p dξ!1 p ×   Z1 2 3 F′ξ 2ρ+2−ξ 2Θ q dξ!1 q + Z1 2 3 F′ξ 2Θ + 2−ξ 2ρ q dξ!1 q   . By exploiting the convexity |F′|q, we conclude that  1 20 F(Θ) + 5F5Θ+ρ 6+F2Θ+ρ 3+ 6FΘ+ρ 2+FΘ+2ρ 3+ 5FΘ+5ρ 6+F(ρ) −Γ(α) 2⋎λα, ρ−Θ 2"Z(α,λ) (Θ+ρ 2)+F(ρ) + Z(α,λ) (Θ+ρ 2)−F(Θ)# ≤(ρ−Θ)α+1 2α+2 ⋎λα, ρ−Θ 2 × γp 1(α, λ)   Z1 3 0 F′ξ 2ρ+2−ξ 2Θ q dξ!1 q + Z1 3 0 F′ξ 2Θ + 2−ξ 2ρ q dξ!1 q   +γp 2(α, λ)   Z2 3 1 3 F′ξ 2ρ+2−ξ 2Θ q dξ!1 q + Z2 3 1 3 F′ξ 2Θ + 2−ξ 2ρ q dξ!1 q    +γp 3(α, λ)   Z1 2 3 F′ξ 2ρ+2−ξ 2Θ q dξ!1 q + Z1 2 3 F′ξ 2Θ + 2−ξ 2ρ q dξ!1 q    ≤(ρ−Θ)α+1 2α+2 ⋎λα, ρ−Θ 2 γp 1(α, λ)   Z1 3 0ξ 2|F′(ρ)|q+2−ξ 2|F′(Θ)|qdξ!1 q + Z1 3 0ξ 2|F′(Θ)|q+2−ξ 2|F′(ρ)|qdξ!1 q   +γp 2(α, λ)   Z2 3 1 3ξ 2|F′(ρ)|q+2−ξ 2|F′(Θ)|qdξ!1 q + Z2 3 1 3ξ 2|F′(Θ)|q+2−ξ 2|F′(ρ)|qdξ!1 q   +γp 3(α, λ)   Z1 2 3ξ 2|F′(ρ)|q+2−ξ 2|F′(Θ)|qdξ!1 q + Z1 2 3ξ 2|F′(Θ)|q+2−ξ 2|F′(ρ)|qdξ!1 q    M6-8 2nd Kocaeli Science Congress, November 19-21, 2025 1.1 Main Contributions ≤(ρ−Θ)α+1 2α+2 ⋎λα, ρ−Θ 2 γp 1(α, λ)  |F′(ρ)|q+ 11|F′(Θ)|q 36  1 q+|F′(Θ)|q+ 11|F′(ρ)|q 36  1 q   +γp 2(α, λ)  |F′(ρ)|q+ 3|F′(Θ)|q 12  1 q+|F′(Θ)|q+ 3|F′(ρ)|q 12  1 q   +γp 3(α, λ)  5|F′(ρ)|q+ 7|F′(Θ)|q 36  1 q+5|F′(Θ)|q+ 7|F′(ρ)|q 36  1 q   . We have successfully concluded the proof for Theorem 1.2. Remark 1.4. If λ= 0 in Theorem 1.2, we conclude  1 20 F(Θ) + 5F5Θ+ρ 6+F2Θ+ρ 3+ 6FΘ+ρ 2+FΘ+2ρ 3+ 5FΘ+5ρ 6+F(ρ) −2α−1Γ(α+ 1) (ρ−Θ)αJα (Θ+ρ 2)+F(ρ)+Jα (Θ+ρ 2)−F(Θ) ≤α(ρ−Θ) 4 γp 1(α, 0)   |F′(ρ)|q+ 11|F′(Θ)|q 36  1 q+|F′(Θ)|q+ 11|F′(ρ)|q 36  1 q   +γp 2(α, 0)   |F′(ρ)|q+ 3|F′(Θ)|q 12  1 q+|F′(Θ)|q+ 3|F′(ρ)|q 12  1 q   +γp 3(α, 0)   5|F′(ρ)|q+ 7|F′(Θ)|q 36  1 q+5|F′(Θ)|q+ 7|F′(ρ)|q 36  1 q   , which is derived by Mateeen et al. in [18, Theorem 6]. Remark 1.5. If we assume λ= 0 and α= 1 in Theorem 1.2, then we gain  1 20 F(Θ) + 5F5Θ+ρ 6+F2Θ+ρ 3+ 6FΘ+ρ 2+FΘ+2ρ 3+ 5FΘ+5ρ 6+F(ρ) −1 ρ−ΘZρ Θ F(ϖ)dϖ ≤    2  3−p−17p20−2p−13p+1 20 7p+ 7 20p p+ 1   1 p ×   |Ψ′(υ)|q+ 11|Ψ′(κ)|q 72 !1 q + 11|Ψ′(υ)|q+|Ψ′(κ)|q 72 !1 q   + 21−1 p  15−2p−115 2p+ 2p+215p p+ 1   1 p ×   |Ψ′(υ)|q+ 3|Ψ′(κ)|q 24 !1 q + 3|Ψ′(υ)|q+|Ψ′(κ)|q 24 !1 q   +2×3−1/p   20−2p−120 3p+ 3p+220p p+ 1   1 p 2nd Kocaeli Science Congress, November 19-21, 2025 M6-9