International Journal of Analysis and Applications Volume 17, Number 4 (2019), 548-558 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-17-2019-548 FRACTIONAL INTEGRAL INEQUALITIES OF GRÜSS TYPE VIA GENERALIZED MITTAG-LEFFLER FUNCTION G. FARID1, A. U. REHMAN1, VISHNU NARAYAN MISHRA2,3,∗ AND S. MEHMOOD4 1COMSATS University Islamabad, Attock Campus, Kamra Road, Atoock 43600, Pakistan 2Department of Mathematics, Indira Gandhi National Tribal University, Lalpur, Amarkantak 484 887, Madhya Pradesh, India 3L. 1627 Awadh Puri Colony, Beniganj, Phase-III, Opposite Industrial Training Institute (I.T.I.), Ayodhya-224 001, Uttar Pradesh, India 4GBPS Sherani, Hazro Attock, Pakistan ∗Corresponding author: vishnunarayanmishra@gmail.com Abstract. We use generalized fractional integral operator containing the generalized Mittag-Leffler function to establish some new integral inequalities of Grüss type. A cluster of fractional integral inequalities have been identified by setting particular values to parameters involved in the Mittag-Leffler special function. Presented results contain several fractional integral inequalities which reflects their importance. Received 2018-12-10; accepted 2019-01-16; published 2019-07-01. 2010 Mathematics Subject Classification. 26A33, 26D10, 33E12. Key words and phrases. Grüss inequality; Generalized fractional integral operator; Mittag-Leffler function. c©2019 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 548 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-17-2019-548 Int. J. Anal. Appl. 17 (4) (2019) 549 1. Introduction In 1935, Grüss [5] proved the following inequality 1 a2 −a1 ∫ a2 a1 f1(t)f2(t)dt− ( 1 a2 −a1 ∫ a2 a1 f1(t)dt )( 1 a2 −a1 ∫ a2 a1 f2(t)dt ) (1.1) ≤ (M −m)(N −n) 4 , where f and g are two integrable functions on [a,b] and satisfying the following conditions m ≤ f1(x) ≤ M, n ≤ f2(x) ≤ N m,M,n,N ∈ R, x ∈ [a,b]. In the literature inequality (1.1) is well known as the Grüss inequality. Inequality (1.1) remains in the focus of researchers especially working in the field of mathematical analysis. A lot of authors are working on (1.1) and have produced important results for different kinds of functions. In recent years, many important and fascinating Grüss type inequalities have been established (see for example [7, 9, 15]). Our interest in this paper is to give some generalized fractional integral inequalities of Grüss type by use of generalized fractional integral operators due to the Mittag-Leffler function. In the following we define an extended generalized Mittag-Leffler function E γ,δ,k,c µ,α,l (t; p) as fallows: Definition 1.1. [3] Let µ,α,l,γ,c ∈ C, <(µ),<(α),<(l) > 0, <(c) > <(γ) > 0 with p ≥ 0, δ > 0 and 0 < k ≤ δ + <(µ). Then the extended generalized Mittag-Leffler function Eγ,δ,k,cµ,α,l (t; p) is defined by E γ,δ,k,c µ,α,l (t; p) = ∞∑ n=0 βp(γ + nk,c−γ) β(γ,c−γ) (c)nk Γ(µn + α) tn (l)nδ , (1.2) Here (c)nk denotes the generalized Pochhammer symbol (c)nk = Γ(c + nk) Γ(c) , Bp is an extension of the beta function Bp(x,y) = ∫ 1 0 tx−1(1 − t)y−1e− p t(1−t) dt (<(x),<(y),<(p) > 0) . The corresponding generalized fractional integral operator � γ,δ,k,c µ,α,l,ω,af is defined as fallows: Definition 1.2. [3] Let ω,µ,α,l,γ,c ∈ C, <(µ),<(α),<(l) > 0, <(c) > <(γ) > 0 with p ≥ 0, δ > 0 and 0 < k ≤ δ + <(µ). Let f ∈ L1[a,b] and x ∈ [a,b]. Then the generalized fractional integral operator � γ,δ,k,c µ,α,l,ω,af is defined by: ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) = ∫ x a (x− t)α−1Eγ,δ,k,cµ,α,l (ω(x− t) µ; p)f(t)dt. (1.3) Int. J. Anal. Appl. 17 (4) (2019) 550 From generalized fractional integral operator we have ( � γ,δ,k,c µ,α,l,ω,a1 ) (x; p) = ∫ x a (x− t)α−1Eγ,δ,k,cµ,α,l (w(x− t) µ; p)dt = ∫ x a (x− t)α−1 ∞∑ n=0 Bp(γ + nk,c−γ) B(γ,c−γ) (c)nk Γ(µn + α) wn(x− t)µn (l)nδ dt = ∞∑ n=0 Bp(γ + nk,c−γ) B(γ,c−γ) (c)nk Γ(µn + α) wn (l)nδ ∫ x a (x− t)µn+α−1dt = (x−a)α ∞∑ n=0 Bp(γ + nk,c−γ) B(γ,c−γ) (c)nk Γ(µn + α) wn (l)nδ (x−a)µn 1 µn + α . Hence ( � γ,δ,k,c µ,α,l,ω,a1 ) (x; p) = (x−a)α Eγ,δ,k,cµ,α+1,l(w(x−a) µ; p). We use the following notation in our results Cα(x; p) = ( � γ,δ,k,c µ,α,l,ω,a1 ) (x; p). (1.4) Integral operators are very useful in solving integral as well as differential equations. Several types of integral operators have been studied by the mathematicians (see for example [1, 2, 4, 6, 8, 11–14]). In this paper at first some generalized fractional integral inequalities and their particular cases are es- tablished. Then a generalized fractional Korkine’s identity is proved. At the end Grüss fractional integral inequality via generalized fractional integral operator have been obtained. The presented inequality contained several versions of Grüss inequality in fractional calculus. 2. Main results First we prove the following fractional inequality. Theorem 2.1. Let f,ψ1,ψ2 ∈ L1[a,b] such that ψ1(x) ≤ f(x) ≤ ψ2(x) ∀x ∈ [a,b]. (2.1) Then for extended generalized fractional integral operator (1.3) we have the following inequality: ( � γ,δ,k,c µ,α,l,ω,aψ2 ) (x; p) ( � γ,δ,k,c µ,β,l,ω,af ) (x; p) + ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ( � γ,δ,k,c µ,β,l,ω,aψ1 ) (x; p) (2.2) ≥ ( � γ,δ,k,c µ,α,l,ω,aψ2 ) (x; p) ( � γ,δ,k,c µ,β,l,ω,aψ1 ) (x; p) + ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ( � γ,δ,k,c µ,β,l,ω,af ) (x; p) Proof. From (2.1) we have (ψ2(u) −f(u)) (f(v) −ψ1(v)) ≥ 0 ∀u,v ∈ [a,b]. (2.3) Int. J. Anal. Appl. 17 (4) (2019) 551 This gives the following inequality: ψ2(u)f(v) + ψ1(v)f(u) ≥ ψ1(v)ψ2(u) + f(u)f(v). (2.4) Multiplying (2.4) by (x − u)α−1Eγ,δ,k,cµ,α,l (ω(x − u) µ; p) on both sides and integrating with respect to u over [a,x], the following inequality is obtained:∫ x a (x−u)α−1Eγ,δ,k,cµ,α,l (ω(x−u) µ; p)ψ2(u)f(v)du (2.5) + ∫ x a (x−u)α−1Eγ,δ,k,cµ,α,l (ω(x−u) µ; p)ψ1(v)f(u)du ≥ ∫ x a (x−u)α−1Eγ,δ,k,cµ,α,l (ω(x−u) µ; p)ψ1(v)ψ2(u)du + ∫ x a (x−u)α−1Eγ,δ,k,cµ,α,l (ω(x−u) µ; p)f(u)f(v)du. Using the Definition 1.2 we get f(v) ( � γ,δ,k,c µ,α,l,ω,aψ2 ) (x; p) + ψ1(v) ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) (2.6) ≥ ψ1(v) ( � γ,δ,k,c µ,α,l,ω,aψ2(u) ) (x; p) + f(v) ( � γ,δ,k,c µ,α,l,ω,af ) (x; p). Now multiplying (2.6) by (x − v)β−1Eγ,δ,k,cµ,β,l (ω(x − v) µ; p) on both sides and integrating with respect to v over [a,x], the following inequality is obtained: ( � γ,δ,k,c µ,β,l,ω,aψ2 ) (x; p) ∫ x a (x−v)β−1Eγ,δ,k,cµ,β,l (ω(x−v) µ; p)f(v)dv (2.7) + ( � γ,δ,k,c µ,β,l,ω,af ) (x; p) ∫ x a (x−v)β−1Eγ,δ,k,cµ,β,l (ω(x−v) µ; p)ψ1(v)dv ≥ ( � γ,δ,k,c µ,β,l,ω,aψ2 ) (x; p) ∫ x a (x−v)β−1Eγ,δ,k,cµ,β,l (ω(x−v) µ; p)ψ1(v)dv + ( � γ,δ,k,c µ,β,l,ω,af ) (x; p) ∫ x a (x−v)β−1Eγ,δ,k,cµ,β,l (ω(x−v) µ; p)f(v)dv. By using the Definition 1.2 and then after simple calculation we get the required inequality (2.2). � A particular case is given as follows. Corollary 2.1. Let f ∈ L1[a,b] and m1,m2 be two real numbers such that m1 ≤ f(x) ≤ m2 ∀x ∈ [a,b]. Then we have m2Cα(x; p) ( � γ,δ,k,c µ,β,l,ω,af ) (x; p) + m1 ( � γ,δ,k,c µ,α,l,ω,af ) (x; p)Cβ(x; p) ≥ m1m2Cα(x; p)Cβ(x; p) + ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ( � γ,δ,k,c µ,β,l,ω,af ) (x; p). (2.8) Int. J. Anal. Appl. 17 (4) (2019) 552 Proof. Proof follows on the same lines as the proof of Theorem 2.1 just use ψ1(x) = m1 and ψ2(x) = m2 as constant functions. � Some more inequalities are given in the next result. Theorem 2.2. Let f,ψ1,ψ2 ∈ L1[a,b] such that (2.1) holds. Also let g ∈ L1[a,b] and there exist φ1 and φ2 such that φ1(x) ≤ g(x) ≤ φ2(x) ∀x ∈ [a,b]. (2.9) Then for extended generalized fractional integral (1.3) we have the following inequalities: (i) ( � γ,δ,k,c µ,β,l,ω,aφ1 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) + ( � γ,δ,k,c µ,β,l,ω,aψ2 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) (2.10) ≥ ( � γ,δ,k,c µ,β,l,ω,aφ1 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,aψ2 ) (x; p) + ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ( � γ,δ,k,c µ,β,l,ω,ag ) (x; p), (ii) ( � γ,δ,k,c µ,β,l,ω,aψ1 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) + ( � γ,δ,k,c µ,β,l,ω,aφ2 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ≥ ( � γ,δ,k,c µ,β,l,ω,aψ1 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,aφ2 ) (x; p) + ( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) ( � γ,δ,k,c µ,β,l,ω,af ) (x; p), (iii) ( � γ,δ,k,c µ,β,l,ω,aψ2 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,aφ2 ) (x; p) + ( � γ,δ,k,c µ,β,l,ω,af ) (x; p) ( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) ≥ ( � γ,δ,k,c µ,β,l,ω,aψ2 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) + ( � γ,δ,k,c µ,α,l,ω,aφ2 ) (x; p) ( � γ,δ,k,c µ,β,l,ω,af ) (x; p), (iv) ( � γ,δ,k,c µ,β,l,ω,aψ1 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,aφ1 ) (x; p) + ( � γ,δ,k,c µ,β,l,ω,af ) (x; p) ( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) ≥ ( � γ,δ,k,c µ,β,l,ω,aψ1 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) + ( � γ,δ,k,c µ,α,l,ω,aφ1 ) (x; p) ( � γ,δ,k,c µ,β,l,ω,af ) (x; p), Proof. (i) From (2.1) and (2.9) we have (ψ2(u) −f(u))(g(v) −φ1(v)) ≥ 0, (2.11) that gives ψ2(u)g(v) + φ1(v)f(u) ≥ ψ1(v)ψ2(u) + f(u)g(v). (2.12) Multiplying (2.12) by (x − u)α−1Eγ,δ,k,cµ,α,l (ω(x − u) µ; p)(x − v)β−1Eγ,δ,k,cµ,β,l (ω(x − v) µ; p) on both sides and integrating with respect to u and v over [a,x] then by using Definition 1.2 we get (i). To prove (ii) − (iv), we use the following inequalities instead of (2.11) respectively (ii) (φ2(u) −g(u))(f(v) −ψ1(v)) ≥ 0, (iii) (ψ2(u) −f(u))(g(v) −φ2(v)) ≤ 0, (iv) (ψ1(u) −f(u))(g(v) −φ1(v)) ≤ 0, then on the same lines as done to obtain (i) one can get inequalities (ii) − (iv). � Special cases are stated as follows. Int. J. Anal. Appl. 17 (4) (2019) 553 Corollary 2.2. Let f,g ∈ L1[a,b]. Also let m1,m2,n1 and n2 be real constants such that m1 ≤ f(x) ≤ m2, n1 ≤ g(x) ≤ n2, ∀x ∈ [a,b]. Then we have (i) n1Cβ(x; p) ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) + m2Cα(x; p) ( � γ,δ,k,c µ,β,l,ω,ag ) (x; p) ≥ n1m2Cβ(x; p)Cα(x; p) + ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ( � γ,δ,k,c µ,β,l,ω,ag ) (x; p), (ii) m1Cβ(x; p) ( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) + n2Cα(x; p) ( � γ,δ,k,c µ,β,l,ω,af ) (x; p) ≥ m1n2Cβ(x; p)Cα(x; p) + ( � γ,δ,k,c µ,β,l,ω,af ) (x; p) ( � γ,δ,k,c µ,α,l,ω,ag ) (x; p), (iii) m2Cα(x; p) ( � γ,δ,k,c µ,β,l,ω,ag ) (x; p) + n2Cβ(x; p) ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ≥ m2n2Cβ(x; p)Cα(x; p) + ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ( � γ,δ,k,c µ,β,l,ω,ag ) (x; p), (iv) m1Cα(x; p) ( � γ,δ,k,c µ,β,l,ω,ag ) (x; p) + n1Cβ(x; p) ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ≥ m1n1Cβ(x; p)Cα(x; p) + ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ( � γ,δ,k,c µ,β,l,ω,ag ) (x; p). Proof. Proof follows on the same lines as the proof of Theorem 2.1 just use ψ1(x) = m1,ψ2(x) = m2,φ1(x) = n1 and φ2(x) = n2 as constant functions. � Next we give the Korkine’s identity which is used in the next result. Theorem 2.3. Let f,ψ1,ψ2 ∈ L1[a,b] such that (2.1) holds. Then for extended generalized fractional integral (1.3) we have the following equality: Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,af 2 ) (x; p) − [( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ]2 (2.13) = [( � γ,δ,k,c µ,α,l,ω,aψ2 ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ] × [( � γ,δ,k,c µ,α,l,ω,af ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,aψ1 ) (x; p) ] −Cα(x; p) [( � γ,δ,k,c µ,α,l,ω,aψ2 ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ] × [( � γ,δ,k,c µ,α,l,ω,af ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,aψ1 ) (x; p) ] + Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,aψ1f ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,aψ1 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) + Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,aψ2f ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,aψ2 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) −Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,aψ1ψ2 ) (x; p) + ( � γ,δ,k,c µ,α,l,ω,aψ1 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,aψ2 ) (x; p). Int. J. Anal. Appl. 17 (4) (2019) 554 Proof. For any u,v ∈ [a,b] we have (ψ2(v) −f(v))(f(u) −ψ1(u)) + (ψ2(u) −f(u))(f(v) −ψ1(v)) (2.14) − (ψ2(u) −f(u))(f(u) −ψ1(u)) − (ψ2(v) −f(v))(f(v) −ψ1(v)) = f2(u) + f2(v) − 2f(u)f(v) + ψ2(v)f(u) + ψ1(u)f(v) −ψ1(u)ψ2(v) + ψ2(u)f(v) + ψ1(v)f(u) −ψ1(v)ψ2(v) −ψ2(u)f(u) + ψ1(u)ψ2(u) −ψ1(u)f(u) −ψ2(v)f(v) + ψ1(v)ψ2(v) −ψ1(v)f(v). Now multiplying (2.14) by (x − u)α−1Eγ,δ,k,cµ,α,l (ω(x − u) µ; p)(x − v)α−1Eγ,δ,k,cµ,α,l (ω(x − v) µ; p) on both sides and integrating with respect to u and v over [a,x] then by using Definition 1.2 we get the required identity (2.13). � The last result is the generalized fractional Grüss inequality. Theorem 2.4. Let f and g be a two functions such that f,g ∈ L1[a,b]. Also let ψ1,ψ2,φ1 and φ2 be four integrable functions satisfying (2.1) and (2.9). Then for extended generalized fractional integral (1.3) we have the following inequality: ∣∣∣Cα(x; p) (�γ,δ,k,cµ,α,l,ω,afg) (x; p) −(�γ,δ,k,cµ,α,l,ω,af) (x; p) (�γ,δ,k,cµ,α,l,ω,ag) (x; p)∣∣∣ (2.15) ≤ √ G(f,ψ1,ψ2)G(g,φ1,φ2), where G(u,v,w) = [( � γ,δ,k,c µ,α,l,ω,aw ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,au ) (x; p) ] × [( � γ,δ,k,c µ,α,l,ω,au ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,av ) (x; p) ] + Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,avu ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,av ) (x; p) ( � γ,δ,k,c µ,α,l,ω,au ) (x; p) + Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,awu ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,aw ) (x; p) ( � γ,δ,k,c µ,α,l,ω,au ) (x; p) −Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,avw ) (x; p) + ( � γ,δ,k,c µ,α,l,ω,av ) (x; p) ( � γ,δ,k,c µ,α,l,ω,aw ) (x; p). Proof. Since f and g are two integrable functions we have [f(u) −f(v)] [g(u) −g(v)] (2.16) = f(u)g(u) + f(v)g(v) −f(u)g(v) −f(v)g(u). Int. J. Anal. Appl. 17 (4) (2019) 555 Multiplying (2.16) by 1 2 (x−u)α−1Eγ,δ,k,cµ,α,l (ω(x−u) µ; p)(x−v)α−1Eγ,δ,k,cµ,α,l (ω(x−v) µ; p) and integrating with respect to u and v over [a,x], the following inequality is obtained:( 1 2 ∫ x a ∫ x a (x−u)α−1Eγ,δ,k,cµ,α,l (ω(x−u) µ; p)(x−v)α−1Eγ,δ,k,cµ,α,l (ω(x−v) µ; p) (2.17) × [f(u) −f(v)] [g(u) −g(v)] dudv) = ( � γ,δ,k,c µ,α,l,ω,afg ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ( � γ,δ,k,c µ,α,l,ω,ag ) (x; p). Now by using Cauchy-Schwarz inequality we have( 1 2 ∫ x a ∫ x a (x−u)α−1Eγ,δ,k,cµ,α,l (ω(x−u) µ; p)(x−v)α−1Eγ,δ,k,cµ,α,l (ω(x−v) µ; p) (2.18) × [f(u) −f(v)] [g(u) −g(v)] dudv)2 ≤ 1 2 ∫ x a ∫ x a (x−u)α−1Eγ,δ,k,cµ,α,l (ω(x−u) µ; p)(x−v)α−1Eγ,δ,k,cµ,α,l (ω(x−v) µ; p) × [f(u) −f(v)]2 dudv × 1 2 ∫ x a ∫ x a (x−u)α−1Eγ,δ,k,cµ,α,l (ω(x−u) µ; p)(x−v)α−1Eγ,δ,k,cµ,α,l (ω(x−v) µ; p) × [g(u) −g(v)]2 dudv. From (2.18) one can have 1 2 ∫ x a ∫ x a (x−u)α−1Eγ,δ,k,cµ,α,l (ω(x−u) µ; p)(x−v)α−1Eγ,δ,k,cµ,α,l (ω(x−v) µ; p) (2.19) × [f(u) −f(v)]2 dudv = Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,af 2 ) (x; p) − [( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ]2 . Similarly, 1 2 ∫ x a ∫ x a (x−u)α−1Eγ,δ,k,cµ,α,l (ω(x−u) µ; p)(x−v)α−1Eγ,δ,k,cµ,α,l (ω(x−v) µ; p) (2.20) × [g(u) −g(v)]2 dudv = Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,ag 2 ) (x; p) − [( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) ]2 . Using (2.19) and (2.20) in (2.18) we have( 1 2 ∫ x a ∫ x a (x−u)α−1Eγ,δ,k,cµ,α,l (ω(x−u) µ; p)(x−v)α−1Eγ,δ,k,cµ,α,l (ω(x−v) µ; p) (2.21) × [f(u) −f(v)] [g(u) −g(v)] dudv)2 ≤ Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,af 2 ) (x; p) − [( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ]2 ×Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,ag 2 ) (x; p) − [( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) ]2 . Int. J. Anal. Appl. 17 (4) (2019) 556 Now combining (2.17) and (2.21) we have (( � γ,δ,k,c µ,α,l,ω,afg ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) )2 (2.22) ≤ Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,af 2 ) (x; p) − [( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ]2 ×Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,ag 2 ) (x; p) − [( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) ]2 . Since (ψ2(x) −f(x))(f(x) −ψ1(x)) ≥ 0 and (φ2(x) −g(x))(g(x) −φ1(x)) ≥ 0, therefore Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,a(ψ2 −f)(f −ψ1) ) (x; p) ≥ 0, and Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,a(φ2 −g)(g −φ1) ) (x; p) ≥ 0. By Theorem 2.3 we have Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,af 2 ) (x; p) − [( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ]2 (2.23) ≤ [( � γ,δ,k,c µ,α,l,ω,aψ2 ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) ] × [( � γ,δ,k,c µ,α,l,ω,af ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,aψ1 ) (x; p) ] + Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,aψ1f ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,aψ1 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) + Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,aψ2f ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,aψ2 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,af ) (x; p) −Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,aψ1ψ2 ) (x; p) + ( � γ,δ,k,c µ,α,l,ω,aψ1 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,aψ2 ) (x; p). = G(f,ψ1,ψ2). Int. J. Anal. Appl. 17 (4) (2019) 557 Similarly Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,ag 2 ) (x; p) − [( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) ]2 (2.24) ≤ [( � γ,δ,k,c µ,α,l,ω,aφ2 ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) ] [( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,aφ1 ) (x; p) ] + Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,aφ1f ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,aφ1 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) + Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,aφ2g ) (x; p) − ( � γ,δ,k,c µ,α,l,ω,aφ2 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,ag ) (x; p) −Cα(x; p) ( � γ,δ,k,c µ,α,l,ω,aφ1φ2 ) (x; p) + ( � γ,δ,k,c µ,α,l,ω,aφ1 ) (x; p) ( � γ,δ,k,c µ,α,l,ω,aφ2 ) (x; p). = G(g,φ1,φ2). Combining (2.23),(2.24) with (2.22), we get the required inequality (2.15). � Concluding remarks Since the extended generalized fractional integral operator contains itself several known fractional integral operators for particular values of involved parameters. 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