New Results on Fractional Simpson-type Inequalities for Differentiable Convex Functions

New Results on Fractional Simpson-type Inequalities for Differentiable Convex Functions Cavit Alemdar1, Fatih Hezenci1, Asia Shehzadi2 1 Department of Mathematics, Faculty of Science and Arts, Duzce University, Duzce 81620, Türkiye 2 School of Mathematics and Statistics, Central South University, Changsha 410083, China Corresponding author: ca[email protected] ORCID IDs: First Author: 0009-0003-1054-2022 Second Author: 0000-0003-1008-5856 Third Author: 0009-0005-1101-5536 DOI : 10.5281/zenodo.18017420 Abstract Fractional calculus has many applications in various fields such as physics, chemistry, engineering, and mathematics. The use of arithmetic operations from classical analysis in fractional analysis provides more realistic results in solving many problems. In this paper, we establish an identity involving the Riemann-Liouville fractional integrals for differentiable functions. By using this identity, we derive several Simpson-type inequalities for functions whose derivatives in absolute value are convex. Finally, some new results are presented as special cases of the main theorems. Keywords: Simpson’s inequality, convex functions, conformable fractional calculus. 1 Introduction and Preliminaries Simpson’s rules are a well-known technique for numerical estimations of integrals. Thomas Simpson (1710–1761) developed this method to estimate of definite integrals. Simpson’s quadrature formula (sometimes is called Simpson 1/3rule) is stated as Zb a f(x)dx ≈1 6f(a)+4fa+b 2+f(b).(1) There are many estimations related to Simpson’s quadrature rule, one of them is the following estimation known as Simpson’s inequality: Theorem 1.1. Suppose that f : [ a, b ] →R is a four times continuously differentiable mapping on (a, b)and let ||f(4)|| = supx∈(a,b)|f(4)|<∞.Then we have  1 6f(a)+4fa+b 2+f(b)−1 b−aZb a f(x)dx ≤1 2880 ||f(4)|| (b−a)4.(2) M13-1 KOSC-2025 Proceedings Recently, many researchers have played attention on Simpson’s type inequalities for various classes of functions such as convex functions, s -convex functions, etc., see [ 1 , 2 , 3 ] for more details. Fractional calculus is a new subject in applied mathematics which attracted researches in both pure and applied branches of science and engineering. It was firstly studied as a problem about solving some differential equations containing fractional order derivatives. The answer to such the problem arises a new interesting subject that many mathematicians have been interested in the last and present centuries. To get more details about fractional integral and derivative operators, we suggest the readers to read the works in [ 4 , 5 , 6 , 7 , 8 , 9 ] and the references therein. There are many definitions of fractional integrals such as Riemann–Liouville fractional integral, Hadamard fractional integral and Atangana-Baleanu fractional integral. The Riemann–Liouville integral is one of the most famous fractional integrals defined as follows: Definition 1.1. [ 10 ] Let f∈L1 [ a, b ], a, b ∈R with a < b . The Riemann-Liouville integrals Jβ a+fand Jβ b−fof order β > 0are defined by Jβ a+f(x) = 1 Γ(β)Zx a (x−t)β−1f(t)dt, x > a, Jβ b−f(x) = 1 Γ(β)Zb x (t−x)β−1f(t)dt, x < b, respectively. Here, Γdenotes the Gamma function defined by Γ(β) = Z∞ 0 euuβ−1du. In 2021, X-R. Hai and S-H. Wang [ 11 ] presented Simpson’s type inequalities for differentiable convex functions by using Riemann–Liouville fractional integrals as follows:  1 6f(a)+4fa+b 2+f(b)−2β−1Γ(β+ 1) (b−a)βJβ a+b 2 −f(a)+Jβ a+b 2 +f(b) ≤(b−a)2−β+ 2β·3−1/β 12(β+ 1) [|f′(a)|+|f′(b)|] (3) and  1 6f(a)+4fa+b 2+f(b)−2β−1Γ(β+ 1) (b−a)βJβ a+b 2 −f(a)+Jβ a+b 2 +f(b) ≤b−a 22q+1 q·3 q(β+1)+1 βq 2β+(2−β)31/β 1+β!1−1/q ×  " 4β·31/β + 2(2 −β)32/β 1+β−2β+(4−β)32/β 2(2 + β)!|f′q+2β+(4−β)32/β 2(2 + β)|f′q#1/q +" 4β·31/β + 2(2 −β)32/β 1+β−2β+(4−β)32/β 2(2 + β)!|f′q+2β+(4−β)32/β 2(2 + β)|f′q#1/q   ,(4) for β > 0. M13-2 2nd Kocaeli Science Congress, November 19-21, 2025 Some more Simpson type inequalities please refer to ([12,13,14,15,16]). In 2017, Jarad et al. [ 17 ] introduced the following generalized fractional integral operators. They also provided certain characteristics and relationships between these operators and several other fractional operators in the literature Definition 1.2. [ 17 ] Let β > 0and α∈ (0 , 1]. For f∈L1 [ a, b ], the generalized fractional Riemann-Liouville integrals β +Jα afand β −Jα bfare defined by β +Jα af(x) = 1 Γ(β)Zx a(x−a)α−(t−a)α αβ−1f(t) (t−a)1−αdt, x > a, and β −Jα bf(x) = 1 Γ(β)Zb x(b−x)α−(b−t)α αβ−1f(t) (b−t)1−αdt, x < b, respectively. In the papers [ 7 , 18 ], the author s proved some Hermite-Hadamard type inequalities for generalized fractional integral operators defined above. Motivated by the above studies, we prove a new identity for differentiable functions by using generalized Riemann–Liouville fractional integrals. Then we use such the identity to obtain some new Simpson’s type fractional integral inequalities. We also the validity of newly established inequalities with some examples. 2 Main Results In this section, we first introduce the identity in Lemma 2.1 which is important to get our main results in the next considerations. Lemma 2.1. Let f : [ a, b ] →R be a differentiable mapping on ( a, b )with β > 0and α∈ (0 , 1]. If f′1[a, b], then the following equality for fractional integrals hold: 1 6f(a)+4fa+b 2+f(b)−αβΓ(β+ 1) 22 b−aαβ β −Jα a+b 2 f(a) + β +Jα a+b 2 f(b) =αβ(b−a) 4"Z1 0 1 3αβ−1−(1 −t)α αβ!f′t 2a+2−t 2bdt +Z1 0 1−(1 −t)α αβ −1 3αβ!f′2−t 2a+t 2bdt#.(1) Proof. Consider I1: = Z1 0 1 3αβ−1−(1 −t)α αβ!f′t 2a+2−t 2bdt =−2 b−a 1 3αβ−1−(1 −t)α 2β!ft 2a+2−t 2bdt 1 0 −2β b−aZ1 0 ft 2a+2−t 2b1−(1 −t)α αβ−1 (1 −t)α−1dt 2nd Kocaeli Science Congress, November 19-21, 2025 M13-3 KOSC-2025 Proceedings =−2 b−a−2 3αβfa+b 2−1 3αβf(b) −β 22 b−aαβ+1 Zb a+b 2 b−a 2α−x−a+b 2α α  β−1 x−a+b 2α−1 f(x)dx =2 αβ(b−a)2 3fa+b 2+1 3f(b)−Γ(β+ 1) 2 b−aαβ+1 β +Jα a+b 2 f(b) and I2: = Z1 0 1−(1 −t)α αβ −1 3αβ!f′2−t 2a+t 2bdt =2 b−a 1−(1 −t)α 2β −1 3αβ!f2−t 2a+t 2bdt 1 0 −2β b−aZ1 0 f2−t 2a+t 2b1−(1 −t)α αβ−1 (1 −t)α−1dt =2 b−a2 3αβfa+b 2+1 3αβf(a) −β2 b−aαβ+1 Za+b 2 a b−a 2α−a+b 2−xα α  β−1 a+b 2−xα−1 f(x)dx =2 αβ(b−a)2 3fa+b 2+1 3f(b)−Γ(β+ 1) 2 b−aαβ+1 β −Jα a+b 2 f(b). Multiplying I1+I2by αβ(b−a) 4, we get (1). The proof is completed. Now, we are ready to prove our following main theorems by mainly using the identity in Lemma 2.1. Theorem 2.1. Under the assumptions of Lemma 2.1. If |f′| is convex on [ a, b ], then we have the following inequality for fractional integrals.  1 6f(a)+4fa+b 2+f(b)−αβΓ(β+ 1) 22 b−aαβ β −Jα a+b 2 f(a) + β +Jα a+b 2 f(b) ≤αβ(b−a) 4A1(α, β)[|f′(a)|+|f′(b)|],(2) where A1(α, β) = Z1 0 1 3αβ−1−(1 −t)α αβ dt =1 αβ"2c−1 3−2 αB β+ 1,1 α,1 31/β!+1 αBβ+ 1,1 α#, with c = 1 −1−1 31 β1 α , B(·,·) and B(·,·,·) are the Beta function and the incomplete Beta M13-4 2nd Kocaeli Science Congress, November 19-21, 2025 functions defined by as follows, respectively B(x, y) = Z1 0 ux−1(1 −u)y−1du, B(x, y, r) = Zr 0 ux−1(1 −u)y−1du for x, y > 0and r∈[0,1] . Proof. From Lemma 2.1 and convexity of |f′|, we have  1 6f(a)+4fa+b 2+f(b)−αβΓ(β+ 1) 22 b−aαβ β −Jα a+b 2 f(a) + β +Jα a+b 2 f(b) ≤αβ(b−a) 4"Z1 0 1 3αβ−1−(1 −t)α αβ f′t 2a+2−t 2b dt +Z1 01−(1 −t)α αβ −1 3αβ f′2−t 2a+t 2b dt# ≤αβ(b−a) 4"Z1 0 1 3αβ−1−(1 −t)α αβt 2|f′(a)|+2−t 2|f′(b)|dt +Z1 01−(1 −t)α αβ −1 3αβ2−t 2|f′(a)|+t 2|f′(b)|dt# = Z1 01−(1 −t)α αβ −1 3αβ dt![|f′(a)|+|f′(b)|] =A1(α, β)[|f′(a)|+|f′(b)|].(3) The proof is completed. Remark 2.1. In Theorem 2.1, if α= 1 then we have  1 6f(a)+4fa+b 2+f(b)−Γ(β+ 1) 22 b−aβJβ a+b 2 −f(a)+Jβ a+b 2 +f(b) ≤b−a 4B1(β)[|f′(a)|+|f′(b)|], where B1(β) = Z1 0 1 3−tβ dt = 2 β β+ 11 31 β+1 +1 β+ 1 −1 3, which is the same to (3). Remark 2.2. In Theorem 2.1, if α=β= 1 then we have  1 6f(a)+4fa+b 2+f(b)−1 b−aZb a f(x)dx ≤5(b−a) 72 [|f′(a)|+|f′(b)|], which was proved by M. Z. Sarikaya et al. [16]. 2nd Kocaeli Science Congress, November 19-21, 2025 M13-5 KOSC-2025 Proceedings Theorem 2.2. Under the assumptions of Lemma 2.1. If |f′|q is convex on [ a, b ]where 1 p + 1 q = 1 and p, q > 1, then we have the following inequality for fractional integrals.  1 6f(a)+4fa+b 2+f(b)−αβΓ(β+ 1) 22 b−aαβ β −Jα a+b 2 f(a) + β +Jα a+b 2 f(b) ≤αβ(b−a) 4A 1 p 2(α, β, p)(|f′|q+ 3|f′|q 41/q +3|f′|q+|f′|q 41/q),(4) where A2(α, β, p) = Z1 0 1 3αβ−1−(1 −t)α αβ p dt. Proof. From Lemma 2.1, we have  1 6f(a)+4fa+b 2+f(b)−αβΓ(β+ 1) 22 b−aαβ β −Jα a+b 2 f(a) + β +Jα a+b 2 f(b) ≤αβ(b−a) 4"Z1 0 1 3αβ−1−(1 −t)α αβ f′t 2a+2−t 2b dt +Z1 01−(1 −t)α αβ −1 3αβ f′2−t 2a+t 2b dt#.(5) By Hölder inequality and the convexity of |f′|q, we get Z1 0 1 3αβ−1−(1 −t)α αβ f′t 2a+2−t 2b dt ≤ Z1 0 1 3αβ−1−(1 −t)α αβ p dt!1/p Z1 0 f′t 2a+2−t 2b q dt1/q =A 1 p 2(α, β, p)Z1 0 f′t 2a+2−t 2b q dt1/q ≤A 1 p 2(α, β, p)Z1 0 t 2|f′|q+2−t 2|f′|qdt1/q =A 1 p 2(α, β, p)|f′|q+ 3|f′|q 41/q (6) and Z1 01−(1 −t)α αβ −1 3αβ f′2−t 2a+t 2b dt ≤ Z1 01−(1 −t)α αβ −1 3αβ p dt!1/p Z1 0 f′2−t 2a+t 2b q dt1/q =A 1 p 2(α, β, p)Z1 0 f′2−t 2a+t 2b q dt1/q ≤A 1 p 2(α, β, p)Z1 0 2−t 2|f′|q+t 2|f′|qdt1/q M13-6 2nd Kocaeli Science Congress, November 19-21, 2025 =A 1 p 2(α, β, p)3|f′|q+|f′|q 41/q .(7) Replacing (6) and (7) in (5), we obtain  1 6f(a)+4fa+b 2+f(b)−αβΓ(β+ 1) 22 b−aαβ β −Jα a+b 2 f(a) + β +Jα a+b 2 f(b) ≤αβ(b−a) 4A 1 p 2(α, β, p)(|f′|q+ 3|f′|q 41/q +3|f′|q+|f′|q 41/q). The proof is completed. Remark 2.3. In Theorem 2.2, if α= 1 then we have  1 6f(a)+4fa+b 2+f(b)−Γ(β+ 1) 22 b−aβJβ a+b 2 −f(a)+Jβ a+b 2 +f(b) ≤b−a 4B 1 p 2(β, p)(|f′|q+ 3|f′|q 41/q +3|f′|q+|f′|q 41/q), where B2(β, p) = Z1 0 1 3−tβ p dt. Remark 2.4. In Theorem 2.2, if α=β= 1 then we have  1 6f(a)+4fa+b 2+f(b)−1 b−aZb a f(x)dx ≤b−a 4 2p+1 + 1 3p+1(p+ 1)!1 p(|f′|q+ 3|f′|q 41/q +3|f′|q+|f′|q 41/q). Theorem 2.3. Under the assumptions of Lemma 2.1. If |f′|q is convex on [ a, b ]where q≥ 1, then we have the following inequality for fractional integrals.  1 6f(a)+4fa+b 2+f(b)−αβΓ(β+ 1) 22 b−aαβ β −Jα a+b 2 f(a) + β +Jα a+b 2 f(b) ≤αβ(b−a) 4A1−1 q 1(α, β)nA3(α, β)|f′|q+A4(α, β)|f′|q1/q +A4(α, β)|f′|q+A3(α, β)|f′|q1/qo, (8) where A1is defined in Theorem 2.1 and A3(α, β) = Z1 0 t 2 1 3αβ−1−(1 −t)α αβ dt =1 2αβ"2c2−1 6−2 αB β+ 1,1 α,1 31/β!+2 αB β+ 1,2 α,1 31/β! +1 αBβ+ 1,1 α−1 αBβ+ 1,2 α, 2nd Kocaeli Science Congress, November 19-21, 2025 M13-7 KOSC-2025 Proceedings A4(α, β) = Z1 0 2−t 2 1 3αβ−1−(1 −t)α αβ dt =1 2αβ"−2c2+ 8c−3 6−2 αB β+ 1,1 α,1 31/β!−2 αB β+ 1,2 α,1 31/β! +1 αBβ+ 1,1 α+1 αBβ+ 1,2 α. Proof. By power mean inequality and the convexity of |f′|q, we get Z1 0 1 3αβ−1−(1 −t)α αβ f′t 2a+2−t 2b dt ≤ Z1 0 1 3αβ−1−(1 −t)α αβ dt!1−1/q Z1 0 1 3αβ−1−(1 −t)α αβ f′t 2a+2−t 2b q dt!1/q =A1−1 q 1(α, β) Z1 0 1 3αβ−1−(1 −t)α αβ f′t 2a+2−t 2b q dt!1/q ≤A1−1 q 1(α, β) Z1 0 1 3αβ−1−(1 −t)α αβt 2|f′|q+2−t 2|f′|qdt!1/q =A1−1 q 1(α, β)A3(α, β)|f′|q+A4(α, β)|f′|q1/q (9) and Z1 01−(1 −t)α αβ −1 3αβ f′2−t 2a+t 2b dt ≤ Z1 01−(1 −t)α αβ −1 3αβ dt!1−1/q Z1 01−(1 −t)α αβ −1 3αβ f′2−t 2a+t 2b q dt!1/q =A1−1 q 1(α, β) Z1 01−(1 −t)α αβ −1 3αβ f′2−t 2a+t 2b q dt!1/q ≤A1−1 q 1(α, β)Z1 0 2−t 2|f′|q+t 2|f′|qdt1/q =A1−1 q 1(α, β)A4(α, β)|f′|q+A3(α, β)|f′|q1/q .(10) Replacing (9) and (10) in (5), we obtain  1 6f(a)+4fa+b 2+f(b)−αβΓ(β+ 1) 22 b−aαβ β −Jα a+b 2 f(a) + β +Jα a+b 2 f(b) ≤αβ(b−a) 4A1−1 q 1(α, β)nA3(α, β)|f′|q+A4(α, β)|f′|q1/q +A4(α, β)|f′|q+A3(α, β)|f′|q1/qo. The proof is completed. Remark 2.5. In Theorem 2.3, if α= 1 then we have  1 6f(a)+4fa+b 2+f(b)−Γ(β+ 1) 22 b−aβJβ a+b 2 −f(a)+Jβ a+b 2 +f(b) M13-8 2nd Kocaeli Science Congress, November 19-21, 2025 REFERENCES ≤b−a 4B1−1 q 1(β)nB3(β)|f′|q+B4(β)|f′|q1/q +B4(β)|f′|q+B3(β)|f′|q1/qo, where B1is defined in Remark 2.1 and B3(β) = Z1 0 t 2 1 3−tβ dt =1 2"β β+ 2 1 32 β+1 +1 β+ 2 −1 6# and B4(β) = Z1 0 2−t 2 1 3−tβ dt =1 2"4β β+ 1 1 31 β+1 −β β+ 2 1 32 β+1 +β+ 3 (β+ 1)(β+ 2) −1 2#, which is similar to 4. Remark 2.6. In Theorem 2.3, if α=β= 1 then we have  1 6f(a)+4fa+b 2+f(b)−1 b−aZb a f(x)dx ≤5(b−a) 72 (29|f′|q+ 61|f′|q 90 1/q +61|f′|q+ 29|f′|q 90 1/q). References [1] Dragomir, S. S., Agarwal, R. P., Cerone, P., On Simpson’s inequality and applications. J. Inequal. Appl. 2002,5, 533–579. [2] Alomari, M., Darus, M., Dragomir, S. S., New inequalities of Simpson’s type for s -convex functions with applications. RGMIA Res. Rep. Coll. 2009,4, 12. [3] Zarikaya, M. Z., Set, E., Ozdemir, M. E., On new inequalities of Simpson’s type for convex functions. RGMIA Res. Rep. Coll. 2010,13, 2. [4] Abdeljawad, T., On conformable fractional calculus. J. Comput. Appl. Math. 2015,279, 57-66. [5] Awan, M. U., Noor, M. A., Mihai, M. V., Noor, K. I., Conformable fractional Hermite– Hadamard inequalities via preinvex functions. Tbilisi Math. J. 2017,10 (4), 129-141. [6] Tariboon, J., Ntouyas, S. K., Sudsutad, W., Some new Riemann-Liouville fractional integral inequalities. Int. J. Math. Sci. 2014,2014(6). [7] Set, E., New inequalities of Ostrowski type for mappings whose derivatives are s-convex in the second sense via fractional integrals. Comput. Math. Appl. 2012 63 (7), 1147-1154. [8] Set, E., Akdemir, A. O., Ozdemir, M. E., Simpson type integral inequalities for convex functions via Riemann-Liouville integrals. Filomat 2017,31 (14), 4415-4420. [9] Sarikaya, M. Z., Set, E., Yaldiz, H., Basak, N., Hermite-Hadamard’s inequalities for fractional integrals and related fractional inequalities.Math. Comput. Model 2013,57, 2403-2407. 2nd Kocaeli Science Congress, November 19-21, 2025 M13-9