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Research article

New developments in fractional integral inequalities via convexity with applications

  • Received: 09 December 2022 Revised: 15 March 2023 Accepted: 20 March 2023 Published: 04 May 2023
  • MSC : 26A33, 26D07, 26D10, 26D15

  • The main objective of this article is to build up a new integral equality related to Riemann Liouville fractional (RLF) operator. Based on this integral equality, we show numerous new inequalities for differentiable convex as well as concave functions which are similar to celebrated Hermite-Hadamard and Simpson's integral inequalities. The present outcomes of this paper are a unification and generalization of the comparable results in the literature on Hermite-Hadamard and Simpson's integral inequalities. Furthermore as applications in numerical analysis, we find some means, q-digamma function and modified Bessel function type inequalities.

    Citation: Maimoona Karim, Aliya Fahmi, Shahid Qaisar, Zafar Ullah, Ather Qayyum. New developments in fractional integral inequalities via convexity with applications[J]. AIMS Mathematics, 2023, 8(7): 15950-15968. doi: 10.3934/math.2023814

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  • The main objective of this article is to build up a new integral equality related to Riemann Liouville fractional (RLF) operator. Based on this integral equality, we show numerous new inequalities for differentiable convex as well as concave functions which are similar to celebrated Hermite-Hadamard and Simpson's integral inequalities. The present outcomes of this paper are a unification and generalization of the comparable results in the literature on Hermite-Hadamard and Simpson's integral inequalities. Furthermore as applications in numerical analysis, we find some means, q-digamma function and modified Bessel function type inequalities.



    An important inequality for classical convex functions which has been extensively studied in recent decades is the Hermite-Hadamard's inequality, which was obtained by Hermite and Hadamard independently. This inequality gives lower and upper estimates for the integral average of any convex function formed on a compact interval encompassing the domain midpoint and endpoints. To, more precise, In [1] Let f:IRR be a convex function on the interval I of real numbers and α1,α2I with α1<α2. Then

    f(α1+α22)1α2α1α2α1f(λ)dλ f(α1)+f(α2)2.

    In the field of analysis, numerous mathematicians have observed the significance of the double inequality and have also used it in various useful applications. Moreover, it has been extended to various structures utilizing the classical convex function. Fractional calculus has applications in a variety of engineering and science domains, including electromagnetic, photoelasticity, fluid mechanics, electrochemistry, biological population models, optics, and signal processing. Due to its vast variety of applications, many mathematicians employed fractional calculus concepts and studied in various areas, one of which is integral inequalities for different classes of functions. For example, some authors, [2,3,4,5] obtained the inequalities for Riemann-Liouville fractional integrals and AB-fractional integral operator. Dragomir et al. [6], proved some Hermite-Hadamard type inequalities for Riemann-Liouville fractional integrals. In [7], Dragomir used the generalized form of Riemann-Liouville fractional integrals and proved some new Ostrowski type inequalities for bounded functions. In [8], he used the generalized form of Riemann-Liouville fractional integrals and proved some new trapezoid type inequalities for bounded functions. He gave trapezoid and ostrowski type inequalities using the fractional integrals. Iqbal et al. [9] presented some fractional midpoint type inequalities for convex functions. In recent years much attention has been devoted to the theory of convex sets and theory of convex functions by generalizing and extending these concepts in different dimensions using innovative techniques. Here, we recall the important definitions related to convex function and left-right Riemann-Liouville fractional integrals.

    Definition 1.1. [1] Let I be an interval in R. A function f:IRR is said to be convex on [α1,α2], with α1<α2, where α1,α2I, if,

    f(λα1+(1λ)α2)λf(α1)+(1λ)f(α2),   λ[0,1].

    Definition 1.2. [10] For fL[α1,α2]. The left-sided and right-sided Riemann-Liouville fractional integrals of order κR+ that are defined by

    (Jκ(α1)+f)(x)=1Γ(κ)xα1(xt)κ1f(t)dt,   ( 0α1<xα2),

    and

    (Jκ(α2)f)(x)=1Γ(κ)α2x(tx)κ1f(t)dt,   ( 0α1<x<α2).

    respectively, where Γ(.) is Gamma function and its definition is Γ(κ)=0euuκ1du. It is to be noted that J0(α1)+f(x)=Jκ(α2)_f(x)=f(x).

    If we put κ=1, the fractional integral becomes the classical integral. The recent results and the properties concerning this operator can be found [11,12].

    An inequality which is notable as Simpson's inequality in [1]:

    Theorem 1.1. Suppose f:[α1,α2] R is four times continuously differentiable function on (α1,α2) and f(4)=supθ(α1,α2)|f(4)(θ)|<, then the following inequality holds:

    |[16f(α1)+23f(α1+α22)+16f(α2)]1α2α1α2α1f(θ)dθ|(α2α1)42880f(4). (1.1)

    The accompanying ongoing improvements for Riemann-Liouville fractional integral on double and Simpson's inequalities are demonstrated by Hwang et al. (see [13]).

    Theorem 1.2. Let f:[α1,α2]R be a differentiable function on (α1,α2) and 0<κ1. If |f| is convex function on [α1,α2], then the following inequality holds:

    |f(α1)+f(α2)2Γ(κ+1)2(α2α1)κ[Jκ(α2)f(α1)+Jκ(α1)+f(α2)]|(α2α1)(2κ1)2κ+1(κ+1)[|f(α1)|+|f(α2)|]. (1.2)

    Proposition 1.1. Suppose that all the assumptions of Theorem 2, are satisfied. If we choose κ=1, we have trapezoid inequality:

    |f(α1)+f(α2)21α2α1α2α1f(x)dx|α2α18[|f(α1)|+|f(α2)|], (1.3)

    which is obtained by Dragomir in [14].

    Theorem 1.3. Let f:[α1,α2]R be a differentiable function on (α1,α2) and 0<κ1. If |f| is convex function on [α1,α2], then the following inequality holds:

    |Γ(κ+1)2(α2α1)κ[Jκ(α2)f(α1)+Jκ(α1)+f(α2)]f(α1+α22)|α2α14(κ+1)(2κ1(κ1)+12κ1)[|f(α1)|+|f(α2)|]. (1.4)

    Proposition 1.2. Suppose that all the assumptions of Theorem 3, are satisfied. If we choose κ=1, we have midpoint inequality:

    |1α2α1α2α1f(x)dxf(α1+α22)|α2α18[|f(α1)|+|f(α2)|], (1.5)

    which is obtained by Kirmaci in [15].

    Theorem 1.4. Let f:[α1,α2]R be a differentiable function on (α1,α2) and 0<κ1. If |f| is convex on [α1,α2], then the following inequality holds:

    |Γ(κ+1)2(α2α1)κ[Jκ(α2)f(α1)+Jκ(α1)+f(α2)][5κ16κf(α1+α22)+6κ5κ+16κf(α1)+f(α2)2]|[1κ+1(2κ+12κ+15κ+1+16κ+1)+(5κ112×6κ)](α2α1)[|f(α1)|+|f(α2)|]. (1.6)

    Proposition 1.3. Suppose that all the assumptions of Theorem 1.4, are satisfied. If we choose κ=1, we have Simpson's inequality:

    |16[f(α1)+4f(α1+α22)+f(α2)]1α2α1α2α1f(x)dx|5(α2α1)72[|f(α1)|+|f(α2)|], (1.7)

    which is obtained by Sarikaya in [16].

    In [17], Lian et al. presented fractional integrals inequalities for concave function as follows,

    Theorem 1.5. Let f:[α1,α2]R be a differentiable function on (α1,α2) and 0<κ1. If |f| is convex function on [α1,α2], then the following inequality holds:

    |f(α1+α22)+Γ(κ+1)(α2α1)[(2α1α2)κJκ(α1+α22)+f(α1)(2α1α2)κJκ(α1+α22)f(α2)]|α2α14(κ+1)[|f((κ+3)α1+(κ+1)α22(κ+2))|+|f((κ+1)α1+(κ+3)α22(κ+2))|]. (1.8)

    Proposition 1.4. Suppose that all the assumptions of Theorem 1.5, are satisfied. If we choose κ=1, we have midpoint inequality:

    |1α2α1α2α1f(x)dxf(α1+α22)|α2α18[|f(α1+2α23)|+|f(2α1+α23)|]. (1.9)

    The current study is organized in two sections. The first section is related to the introductory body, where ideas and the hypotheses that provides the foundation for the advancement of the work has been discussed. While the second section has been divided into three sub-sections which shows the outcomes acquired for each of the inequalities under investigation. Moreover, the purpose of this paper is to study Hermite-Hadamard and Simpson's-like integral inequalities for convex functions as well as concave functions by applying the fractional concept. We also discuss the relation of our results with comparable results existing in the literature. Furthermore as applications, we find some means, q-digamma function and modified Bessel function type inequalities. We expect that the study initiated in this paper may inspire new research in this area.

    Here, we prove an important new Lemma for Riemann-Livouille fractional integrals, which plays a key role to prove our main results as follows:

    Lemma 2.1. Let f:[α1,α2]R be a differentiable function on (α1,α2) with α1<α2. If fL[α1,α2] with 0 < κ1, λ[0,1], and ρ,ϑ[0,1], then the following equality holds:

    2κ2+ρ+ϑ2κ(f(α1)+f(α2)2)+2ρϑ2κf(α1+α22)Γ(κ+1)2(α2α1)κ[Jκ(α2)f(α1)+Jκ(α1)+f(α2)]=α2α12κ+2(Q1+Q2+Q3+Q4),

    where

    Q1=10[(1λ)κρ]f(λα1+(1λ)α1+α22)dλ,  Q2=10[ϑ(1λ)κ]f(λα2+(1λ)α1+α22)dλ,Q3=10[(2λ)κ+ρ2]f(λα1+α22+(1λ)α2)dλ,Q4=10[2ϑ(2λ)κ]f(λα1+α22+(1λ)α1)dλ.

    Proof. Integrating by parts successively, in order to compute each integral, one obtain

    Q1=10[(1λ)κρ]f(λα1+(1λ)α1+α22)dλ=2[(1λ)κρ]f(λα1+(1λ)α1+α22)dλα1α2|10+2κα1α210(1λ)κ1f(λα1+(1λ)α1+α22)dλ=2ρα2α1f(α1)+2(1ρ)α2α1f(α1+α22)κ(2κ+1)(α2α1)κ+1α1+α22α1(xα1)κ1f(x)dx. (2.1)

    Simple calculations analogously

    Q2=10[ϑ(1λ)κ]f(λα2+(1λ)α1+α22)dλ=2[ϑ(1λ)κ]f(λα2+(1λ)α1+α22)dλα2α1|10+2κα1α210(1λ)κ1f(λα2+(1λ)α1+α22)dλ=2ϑα2α1f(α2)+2(1ϑ)α2α1f(α1+α22)κ(2κ+1)(α2α1)κ+1α2α1+α22(α2x)κ1f(x)dx. (2.2)
    Q3=10[(2λ)κ+ρ2]f(λα1+α22+(1λ)α2)dλ=2[(2λ)κ+ρ2]f(λα1+α22+(1λ)α2)dλα1α2|10+2κα1α210(2λ)κ1f(λα1+α22+(1λ)α2)dλ=2(2κ2+ρ)α2α1f(α2)+2(1ρ)α2α1f(α1+α22)κ(2κ+1)(α2α1)κ+1α2α1+α22(xα1)κ1f(x)dx. (2.3)
    Q4=10[2ϑ(2λ)κ]f(λα1+α22+(1λ)α1)dλ=2[2ϑ(2λ)κ]f(λα1+α22+(1λ)α1)dλα2α1|10+2κα1α210(2λ)κ1f(λα1+α22+(1λ)α1)dλ=2(2κ2+ϑ)α2α1f(α1)+2(1ϑ)α2α1f(α1+α22)κ(2κ+1)(α2α1)κ+1α1+α22α1(α2x)κ1f(x)dx. (2.4)

    Hence, by adding (2.1)–(2.4), and multiplying the resultant one by α2α12κ+2, we obtain the resultant equality.

    Theorem 2.1. Let f:[α1,α2]R be a differentiable function on (α1,α2) with α1<α2. If fL1[α1,α2] with 0 <κ1 and ρ,ϑ[0,1], and |f| is a convex on [α1,α2], then the following inequality holds:

    |2κ2+ρ+ϑ2κ(f(α1)+f(α2)2)+2ρϑ2κf(α1+α22)Γ(κ+1)2(α2α1)κ[Jκ(α2)f(α1)+Jκ(α1)+f(α2)]|(α2α1)(U1+U2+U3+U4+U5+U6+U7+U8)2κ+2( |f(α1)|+|f(α2)| ), (2.5)

    where,

    U1=10|[(1λ)κρ]|λdλ=[12(1(1ρ1κ))κ+2(κ+1)(κ+2)2(1ρ1κ)(1(1ρ1κ))κ+1(κ+1)+ρ2(12((1ρ1κ))2)].
    U2=10|[(1λ)κρ]|(1λ)dλ=[2(ρ1κ)κ+1(1ρ1κ)(κ+1)+12(ρ1κ)κ+1(κ+1)12(ρ1κ)κ+2(κ+1)(κ+2)+ρ(12(1ρ1κ))ρ2(12(1ρ1κ)2)].
    U3=10|[ϑ(1λ)κ]|λdλ=[12(1(1ϑ1κ))κ+2(κ+1)(κ+2)2(1ϑ1κ)(1(1ϑ1κ))κ+1(κ+1)+ϑ2(12((1ϑ1κ))2)].U4=10|[ϑ(1λ)κ]|(1λ)dλ=[2(ϑ1κ)κ+1(1ϑ1κ)(κ+1)+12(ϑ1κ)κ+1(κ+1)12(ϑ1κ)κ+2(κ+1)(κ+2)+ϑ(12(1ϑ1α))ϑ2(12(1ϑ1α)2)].
    U5=10|[(2λ)κ+ρ2]|λdλ=1+2κ+22(2(2(2ρ))1κ)κ+2κ2+3κ+2+12(2(2ρ))1κ((2(2(2ρ))1κ)κ+1)(κ+1)+(2ρ)(12((2(2ρ))1κ)2).
    U6=10|[(2λ)κ+ρ2]|(1λ)dλ=12ρ(4(κ(ρ3)+ρ2)(2ρ)κ(κ+1)(κ+2)2ρ2+8ρ9)+κ(4(2ρ)κ+2κ+1+κ+3)+1(κ+1)(κ+2).
    U7=10|[2ϑ(2λ)κ]|λdλ=1+2κ+22(2(2(2ϑ))1κ)κ+2κ2+3κ+2+12(2(2ϑ))1κ((2(2(2ϑ))1κ)κ+1)(κ+1)+(2ϑ)(12((2(2ϑ))1κ)2).U8=10|[2ϑ(2λ)κ]|(1λ)dλ=12ϑ(4(κ(ϑ3)+ϑ2)(2ϑ)κ(κ+1)(κ+2)2ϑ2+8ϑ9)+κ(4(2ϑ)κ+2κ+1+κ+3)+1(κ+1)(κ+2).

    Proof. From Lemma 2.1 and convexity, it follows that,

    |2κ2+ρ+ϑ2κ(f(α1)+f(α2)2)+2ρϑ2κf(α1+α22)Γ(κ+1)2(α2α1)κ[Jκ(α2)f(α1)+Jκ(α1)+f(α2)]|=|α2α12κ+210[(1λ)κρ]f(λα1+(1λ)α1+α22)dλ|+|α2α12κ+210[ϑ(1λ)κ]f(λα2+(1λ)α1+α22)dλ|+|α2α12κ+210[(2λ)κ+ρ2]f(λα1+α22+(1λ)α2)dλ|+|α2α12κ+210[2ϑ(2λ)κ]f(λα1+α22+(1λ)α1)dλ|
    α2α12κ+210|[(1λ)κρ]||f(λα1+(1λ)α1+α22)|dλ+α2α12κ+210|[ϑ(1λ)κ]||f(λα2+(1λ)α1+α22)|dλ+α2α12κ+210|[(2λ)κ+ρ2]||f(λα1+α22+(1λ)α2)|dλ+α2α12κ+210|[2ϑ(2λ)κ]||f(λα1+α22+(1λ)α1)|dλ
    α2α12κ+210|[(1λ)κρ]|[λ|f(α1)|+(1λ)|f(α1+α22)|]dλ+α2α12κ+210|[ϑ(1λ)κ]|[λ|f(α2)|+(1λ)|f(α1+α22)|]dλ+α2α12κ+210|[(2λ)κ+ρ2]|[λ|f(α1+α22)|+(1λ)|f(α2)|]dλ+α2α12κ+210|[2ϑ(2λ)κ]|[λ|f(α1+α22)|+(1λ)|f(α1)|]dλ|2κ2+ρ+ϑ2κ(f(α1)+f(α2)2)+2ρϑ2κf(α1+α22)Γ(κ+1)2(α2α1)κ[Jκ(α2)f(α1)+Jκ(α1)+f(α2)]|(α2α1)(U1+U2+U3+U4+U5+U6+U7+U8)2κ+2( |f(α1)|+|f(α2)| ).

    The proof is completed.

    Example 2.1. Let [α1,α2]=[0,1] and define the function f:[0,1]R as f(λ)=λ3+3. Let us consider the right-hand side of the inequality (2.5) as follows:

    (α2α1)(U1+U2+U3+U4+U5+U6+U7+U8)2κ+2=2(ρκ+1+κ(2ρ23ρ+2)2(2ρ)κ+ρ(2ρ)κ+2κ+1+2ρ2)(κ+1)+2(3ρϑκ+1+2(κ+1)ϑ22(2ϑ)κ+ϑ(3κ+(2ϑ)κ3)+4)(κ+1)

    From the definitions of fractional integrals, the equalities

    |2κ2+ρ+ϑ2κ(f(0)+f(1)2)+2ρϑ2κf(12)Γ(κ+1)2(α2α1)κ[Jκ(α2)f(α1)+Jκ(α1)+f(α2)]|=2ρϑ2κ(338)+2κ2+ρ+ϑ2κ(92)12(33κ2+99κ+728κ2+24κ+16)

    are valid. Finally, we have the following inequality:

    2ρϑ2κ(338)12(33κ2+99κ+728κ2+24κ+16)12κ+2[2(κ28.2κ+2κ+2+4κ+5)(κ+1)(κ+2)+2(4.2κ+2κ+1+κ+3)(κ+1)(κ+2)+2(κ+1)+2(κ+1)(κ+2)]. (2.6)

    As one can see in Figure 1, (2.6) in Example 1 shows the correctness of this inequality for all values of κ(0,1] and special choices of ρ,ϑ. The Figure 1 represents the Graphical description of inequality (2.6) and their difference.

    Figure 1.  Graphical description of inequality (2.6).

    Remark 2.1. If we choose ρ=ϑ=1, κ=1, in Theorem 2.1, then inequality (2.5) reduces to inequality (1.3).

    Remark 2.2. If we choose ρ=ϑ=0, κ=1, in Theorem 2.1, then inequality (2.5) reduces to inequality (1.5).

    Remark 2.3. If we choose ρ=ϑ=13, κ=1, in Theorem 2.1, then inequality (2.5) reduces to inequality (1.7).

    Theorem 2.2. Let f:[α1,α2]R be a differentiable function on (α1,α2) with α1<α2 with 0 < κ1, λ[0,1], and ρ,ϑ[0,1]. If fL[α1,α2] and mapping |f|q with q1, is convex on [α1,α2], then the following inequality holds:

    |2κ2+ρ+ϑ2κ(f(α1)+f(α2)2)+2ρϑ2κf(α1+α22)Γ(κ+1)2(α2α1)κ[Jκ(α2)f(α1)+Jκ(α1)+f(α2)]|α2α12κ+2 ×[{(U9)11/q(U1|f(α1)|q+U2|f(α1+α22)|q)1/q+(U10)11/q(U3|f(α1)|q+U4|f(α1+α22)|q)1/q}+{(U11)11/q(U5|f(α1+α22)|q+U6|f(α2)|q)1/q+(U12)11/q(U7|f(α1+α22)|q+U8|f(α2)|q)1/q}], (2.7)

    where,

    U9=10|[(1λ)κρ]|dλ=12(1η1)κ+1κ+1+(12η1)ρ,   η1=1ρ1κ.U10=10|[ϑ(1λ)κ]|dλ=1+2(1η2)κ+1κ+1(12η2)ϑ,   η2=1ϑ1κ.
    U11=10|[(2λ)κ(2ρ)]|dλ=12κ+1+2(2η3)κ+1κ+1(2κρ)(12η2), η3=2(2κρ)1κ.
    U12=10|[2ϑ(2λ)κ]|dλ=1+2κ+12(2η4)κ+1κ+1+(2κϑ)(12η4), η4=2(2κϑ)1κ.

    Proof. Utilizing the Lemma 2.1 and Power-mean inequality, we obtain

    |2κ2+ρ+ϑ2κ(f(α1)+f(α2)2)+2ρϑ2κf(α1+α22)Γ(κ+1)2(α2α1)κ[Jκ(α2)f(α1)+Jκ(α1)+f(α2)]|α2α12κ+2(10|(1λ)κρ|dλ)11/q×(10|(1λ)κρ||f(λα1+(1λ)α1+α22)|qdλ)1/q+α2α12κ+2(10[ϑ(1λ)κ]dλ)11/q×(10[ϑ(1λ)κ]|f(λα2+(1λ)α1+α22)|qdλ)1/q+α2α12κ+2(10|(2λ)κ+ρ2|dλ)11/q×(10|(2λ)κ+ρ2||f(λα1+α22+(1λ)α2)|qdλ)1/q+α2α12κ+2(10[2ϑ(2λ)κ]dλ)11/q×(10[(2λ)κ+ρ2]|f(λα1+α22+(1λ)α1)|qdλ)1/q
    α2α12κ+2(10|(1λ)κρ|dλ)11/q×(10|(1λ)κρ|(λ|f(α1)|q+(1λ)|f(α1+α22)|q)dλ)1/q+α2α12κ+2(10|ϑ(1λ)κ|dλ)11/q×(10|ϑ(1λ)κ|(λ|f(α2)|q+(1λ)|f(α1+α22)|q)dλ)1/q
    +α2α12κ+2(10|(2λ)κ+ρ2|dλ)11/q×(10|(2λ)κ+ρ2|(λ|f(α1+α22)|q+(1λ)|f(α2)|q)dλ)1/q+α2α12κ+2(10[|2ϑ(2λ)κ|]dλ)11/q×(10|(2λ)κ+ρ2|(λ|f(α1+α22)|q+(1λ)|f(α1)|q)dλ)1/q.
    |2κ2+ρ+ϑ2κ(f(α1)+f(α2)2)+2ρϑ2κf(α1+α22)Γ(κ+1)2(α2α1)κ[Jκ(α2)f(α1)+Jκ(α1)+f(α2)]|α2α12κ+2 ×[{(U9)11/q(U1|f(α1)|q+U2|f(α1+α22)|q)1/q+(U10)11/q(U3|f(α1)|q+U4|f(α1+α22)|q)1/q}+{(U11)11/q(U5|f(α1+α22)|q+U6|f(α2)|q)1/q+(U12)11/q(U7|f(α1+α22)|q+U8|f(α2)|q)1/q}]. (2.8)

    The proof is completed.

    Now we discuss the particular inequalities which generalize inequalities in classical sense.

    Corollary 2.1. Suppose that all the assumptions of Theorem 2.2, are satisfied. If we choose ρ=ϑ=1, κ=1, the following inequality holds:

    |f(α1)+f(α2)21α2α1α2α1f(u)du|α2α18[(23|f(α1)|q+13|f(α1+α22)|q)1q+(23|f(α2)|q+13|f(α1+α22)|q)1q].

    Corollary 2.2. Suppose that all the assumptions of Theorem 2.2, are satisfied. If we choose ρ=ϑ=0, κ=1, the following inequality holds:

    |f(α1+α22)1α2α1α2α1f(u)du|α2α18[(13|f(α1)|q+23|f(α1+α22)|q)1q+(13|f(α2)|q+23|f(α1+α22)|q)1q].

    Corollary 2.3. Suppose that all the assumptions of Theorem 2.2, are satisfied. If we choose ρ=ϑ=13, κ=1, the following inequality holds:

    |13{2f(α1+α22)+f(α1)+f(α2)2}1α2α1α2α1f(u)du|5(α2α1)72×[(881|f(α1)|q+29162|f(α1+α22)|q)1q+(881|f(α2)|q+29162|f(α1+α22)|q)1q].

    Now we obtain some estimates of Simpson's and Hermite-Hadamard-inequalities for concavity.

    Theorem 2.3. Let f:[α1,α2]R be a differentiable mapping on (α1,α2) with α1<α2,0 < κ1, and ρ,ϑ[0,1], q1. If |f|q is concave on [α1,α2], then the following inequality holds:

    |2κ2+ρ+ϑ2κ(f(α1)+f(α2)2)+2ρϑ2κf(α1+α22)Γ(κ+1)2(α2α1)κ[Jκ(α2)f(α1)+Jκ(α1)+f(α2)]|α2α12κ+2[{U9×|f({U1×(α1)+U2×(α1+α22)U9})|+U10×|f({U3×(α2)+U4×(α1+α22)U10})|}+U11×|f({(U5×(α1+α22)+U6×(α2)U11})|+U12×|f({(U7×(α1+α22)+U8×(α1)U12})|]. (2.9)

    Proof. Using the concavity of |f|q and the power-mean inequality, we know that for λ[0,1],

    |f(λα1+(1λ)α2)|q>λ|f(α1)|q+(1λ)|f(α2)|q(λ|f(α1)|+(1λ)|f(α2)|)q

    Hence

    |f(λα1+(1λ)α2)|λ|f(α1)|+(1λ)|f(α2)|.

    By the concavity and Jensen integral inequality, we have

    |2κ2+ρ+ϑ2κ(f(α1)+f(α2)2)+2ρϑ2κf(α1+α22)Γ(κ+1)2(α2α1)κ[Jκ(α2)f(α1)+Jκ(α1)+f(α2)]|α2α12κ+2(10|[(1λ)κρ]|dλ)|f(10[(1λ)κρ]|(λα1+(1λ)α1+α22)|dλ10|[(1λ)κρ]|dλ)|+α2α12κ+2(10|[ϑ(1λ)κ]|dλ)|f(10[ϑ(1λ)κ]|(λα2+(1λ)α1+α22)|dλ10|[ϑ(1λ)κ]|dλ)|+α2α12κ+2(10|[(2λ)κ+ρ2]|dλ)|f(10[(2λ)κ+ρ2]|(λα1+α22+(1λ)α2)|dλ10|[(2λ)κ+ρ2]|dλ)|+α2α12κ+2(10|[2ϑ(2λ)κ]|dλ)|f(10[2ϑ(2λ)κ]|(λα1+α22+(1λ)α1)|dλ10|[2ϑ(2λ)κ]|dλ)|α2α12κ+2(U9)|f(U1(α1)+U2(α1+α22)U9)|+α2α12κ+2(U10)|f(U3(α2)+U4(α1+α22)U9)|+α2α12κ+2(U11)|f((U5(α1+α22)+U6(α2)U6)|+α2α12κ+2(U12)|f((U7(α1+α22)+U8(α1)U6)|
    |2κ2+ρ+ϑ2κ(f(α1)+f(α2)2)+2ρϑ2κf(α1+α22)Γ(κ+1)2κ+1(α2α1)κ[Jκ(α2)f(α1)+Jκ(α1)+f(α2)]|α2α12κ+2[{U9×|f({U1×(α1)+U2×(α1+α22)U9})|+U10×|f({U3×(α2)+U4×(α1+α22)U10})|}+U11×|f({(U5×(α1+α22)+U6×(α2)U11})|+U12×|f({(U7×(α1+α22)+U8×(α1)U12})|],

    which completes the proof.

    As a special case of Theorem 2.3, we obtain the following result,

    Corollary 2.4. Suppose that all the assumptions of Theorem 2.3, are satisfied. If we choose ρ=ϑ=13, κ=1, we have Simpson's inequality:

    |13{2f(α1+α22)+f(α1)+f(α2)2}1α2α1α2α1f(u)du|5(α2α1)72×[|f(16α1+29α245)|+|f(29α1+16α245)|]. (2.10)

    Remark 2.4. Our inequality (2.10) is an improvement of Alomari inequality as obtained in [18].

    Let consider the following special means for α1α2.

    The arithmetic mean:

    A(α1,α2)=α1+α22, α1,α2R.

    The logarithmic-mean:

    L(α1,α2)=α2α1ln|α2|ln|α1|,|α1||α2|, α1,α20, α1,α2R.

    The generalized logarithmic-mean:

    Lr(α1,α2)=[(α2)r+1(α1)r+1(r+1)(α2α1)]1r;rR{1,0}, α1,α2>0.

    Proposition 2.1. Suppose rR{1,0} and α1,α2R such that 0<α1<α2 with q1, then the following inequality holds:

    |A(α1,α2)Lrr(α1,α2)|r(α2α1)8[(23|α1|q+13|(α1+α22)r1|q)1q+(23|α2|q+13|(α1+α22)r1|q)1q].

    Proof. The assertion follows from Corollary 1 for the function f(x)=xr and r as specified above.

    Proposition 2.2. Suppose q1 and α1,α2R, such that 0<α1<α2, then the following inequality holds:

    |A1(α1,α2)L1(α1,α2)|(α2α1)8[(13|α1|q+23|A2(α1,α2)|q)1q+(13|α2|q+23|A2(α1,α2)|q)1q].

    Proof. The assertion follows from Corollary 2.2 for the function f(x)=1x.

    Suppose 0<q<1, the q-digamma function φq, is the q-analogue of the digamma function φ defined by (see [19,20]).

    φq=ln(1q)+ lnqk=0qk+x1qk+x    =ln(1q)+ lnqk=0qkx1qkx

    For q>1 and x>0, q-digamma function φqdefined by

    φq=ln(q1)+ lnq[x12k=0q(k+x)1q(k+x)]    =ln(q1)+ lnq[x12k=0qkx1qkx]

    Proposition 2.3. Suppose α1,α2 be real numbers such that 0<α1<α2, with q1, then the following inequality holds:

    |A(φq(α1),φq(α2))1α2α1α2α1φq(u)du|(α2α1)8[(23|φq(α1)|q+13|φq(α1+α22)|q)1q+(23|φq(α2)|q+13|φq(α1+α22)|q)1q].

    Proof. The assertion can be obtained immediately by using Corollary 2.1 to f(ε)=φq(ε) and ε>0, f(ε)=φq(ε) is convex on (0,+).

    Proposition 2.4. Suppose α1,α2 be real numbers such that 0<α1<α2, with q1, then the following inequality holds:

    |13{2φq(α1+α22)+A(φq(α1),φq(α2))}1α2α1α2α1φq(u)du|5(α2α1)72×[(881|φq(α1)|q+29162|φq(α1+α22)|q)1q+(881|φq(α2)|q+29162|φq(α1+α22)|q)1q].

    Proof. The assertion can be obtained immediately by using Corollary 2.3 to f(ε)=φq(ε) and ε>0, f(ε)=φq(ε) is convex on (0,+).

    Recall the first kind of modified Bessel function Iρ, which has the series representation ([19], p.77).

    Iρ(x)=m0(x2)ρ+2mm!Γ(ρ+m+1),

    where xR and ρ>1, while the second kind modified Bessel function Kρ ([19], p.78) is usually defined as

    Kρ(x)=π2Iρ(x)Iρ(x)sinρπ.

    Here, we consider the function Ωρ(x):R[1,) defined by

    Ωρ(x)=2ρΓ(ρ+1)xρIρ(x),

    where Γ is the Gamma function.

    Proposition 2.5. Suppose ρ>1 and 0<α1<α2. Then

    |Ωρ(α1+α22)1α2α1α2α1Ωρ(ε)dε|(α2α1)16(ρ+1)[(13α1|Ωρ+1(α1)|q+α1+α23|Ωρ+1(α1+α22)|q)1q+(13α2|Ωρ+1(α2)|q+α1+α23|Ωρ+1(α1+α22)|q)1q].

    Specifically, if ρ=12, then

    |cosh(α1+α22)1α2α1α2α1cosh(ε)dε|(α2α1)8[(13α1|sinh(α1)α1|q+α1+α23|sinh(α1+α22)α1+α22|q)1q+(13α2|sinh(α2)α2|q+α1+α23|sinh(α1+α22)α1+α22|q)1q].

    Proof. The assertion can be obtained immediately by using Corollary 2.2 to f(ε)=Ωρ(ε), ε>0, and  Ωρ(ε)=ερ+1Ωρ+1(ε). Now taking into account the relations Ω12(ε)=cosh(ε) and Ω12(ε)=sinh(ε)ε.

    In this article, we have established an integral identity via Riemann-Livouille fractional integral. Based on this identity, we present several midpoint, trapezoid and Simpson's-type inequalities whose absolute values are convex and concave. It is also shown that several results are given by special cases of the main results. We deduce that the findings proved in this work are naturally universal, contribute to the theory of inequalities. Finally, we have presented some applications to special means, q-digamma and modifies Bessel functions with respect to our deduced results. In future studies, researchers can obtain generalized versions of our results by utilizing other kinds of convex function classes or different types of generalized fractional integral operators.

    We would like to thank you for following the instructions above very closely in advance. It will definitely save us lot of time and expedite the process of your paper's publication.

    There is no conflict of interest among the authors.



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