Research article

New Hermite-Hadamard type inequalities for exponentially convex functions and applications

  • Received: 11 April 2020 Accepted: 24 August 2020 Published: 04 September 2020
  • MSC : 26A33, 26A51, 26D10

  • The investigation of the proposed techniques is effective and convenient for solving the integrodifferential and difference equations. The present investigation depends on two highlights; the novel Hermite-Hadamard type inequalities for K-conformable fractional integral operator in terms of a new parameter K>0 and weighted version of Hermite-Hadamard type inequalities for exponentially convex functions in the classical sense. By using an integral identity together with the Hölder-İşcan and improved power-mean inequality we establish several new inequalities for differentiable exponentially convex functions. This generalizes the Hadamard fractional integrals and Riemann-Liouville into a single form. Our contribution expands some innovative studies in this line. Moreover, two suitable examples are presented to demonstrate the novelty of the results established, the first one about the contributions of the modified Bessel functions and the other is about σ-digamma function. Finally, various applications for some special means as arithmetic, geometric and logarithmic are given.

    Citation: Shuang-Shuang Zhou, Saima Rashid, Muhammad Aslam Noor, Khalida Inayat Noor, Farhat Safdar, Yu-Ming Chu. New Hermite-Hadamard type inequalities for exponentially convex functions and applications[J]. AIMS Mathematics, 2020, 5(6): 6874-6901. doi: 10.3934/math.2020441

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  • The investigation of the proposed techniques is effective and convenient for solving the integrodifferential and difference equations. The present investigation depends on two highlights; the novel Hermite-Hadamard type inequalities for K-conformable fractional integral operator in terms of a new parameter K>0 and weighted version of Hermite-Hadamard type inequalities for exponentially convex functions in the classical sense. By using an integral identity together with the Hölder-İşcan and improved power-mean inequality we establish several new inequalities for differentiable exponentially convex functions. This generalizes the Hadamard fractional integrals and Riemann-Liouville into a single form. Our contribution expands some innovative studies in this line. Moreover, two suitable examples are presented to demonstrate the novelty of the results established, the first one about the contributions of the modified Bessel functions and the other is about σ-digamma function. Finally, various applications for some special means as arithmetic, geometric and logarithmic are given.


    Convexity has played a crucial role in the advancement of pure and applied mathematics [1,2,3,4,5,6,7,8,9,10]. Due to its robustness, convex functions and convex sets have been generalized and extended in many mathematical areas [11,12,13,14,15,16,17,18,19,20]. In particular, many inequalities can be found in the literature [21,22,23,24,25,26,27,28,29,30] via convexity theory.

    Integral inequalities [31,32,33,34] have numerous applications in number theory, combinatorics, orthogonal polynomials, hypergeometric functions, quantum theory, linear programming, optimization theory, mechanics and in the theory of relativity. This subject has received considerable attention from researchers [35,36,37,38] and hence it is assumed as an incorporative the subject between mathematics, statistics, economics, and physics [39,40,41].

    To the best of our knowledge, the Hermite-Hadamard inequality is a well-known, paramount and extensively useful inequality in the applied literature [42,43,44,45]. This inequality is of pivotal significance because of other classical inequalities such as the Hardy, Opial, Lynger, Ostrowski, Minkowski, Hölder, Ky-Fan, Beckenbach-Dresher, Levinson, arithmetic-geometric, Young, Olsen and Gagliardo-Nirenberg inequalities, which are closely related to the classical Hermite-Hadamard inequality [46]. It can be stated as follows:

    Let φ:IRR be a convex function and c,dI with c<d. Then

    φ(c+d2)1dcdcφ(z)dzφ(c)+φ(d)2. (1.1)

    In [47], Fejér contemplated the important generalizations that is the weighted generalization of the Hermite-Hadamard inequality.

    Let IR and a function φ:IR be a convex function. Then the inequalities

    φ(c+d2)dcw(z)dzdcφ(z)w(z)dzφ(c)+φ(d)2dcw(z)dz. (1.2)

    hold, where w:IR is non-negative, integrable and symmetric with respect to c+d2. If we choose w(z)=1, then (1.2) reduces to (1.1). Several classical inequalities can be obtained with the help of inequality (1.1) by considering the use of peculiar convex function φ. Moreover, these inequalities for convex functions have a very important role in both applied and pure mathematics.

    In recent years, integral inequalities have been derived via fractional analysis, which has emerged as another interesting technique. Due to advancement in inequalities, the comprehensive investigation of exponentially convex functions as the K-conformable fractional integral in the present paper is new.

    The class of exponentially-convex functions were introduced by Dragomir and Gomm [48]. Bernstein [49] and Antczak [50] introduced these exponentially convex functions implicitly and discuss their role in mathematical programming. The proliferating research on big data analysis and deep learning has recently intensified the interest in information theory involving exponentially-convex functions. The smoothness of exponentially-convex functions is exploited for statistical learning, sequential prediction, and stochastic optimization.

    Now we recall the concept of exponentially convex functions, which is mainly due to M. A. Noor and K. I. Noor [51,52].

    Definition 1.1. (See [51,52]) A real-valued function φ:MRR is said to be an exponentially convex on M if the inequality

    eφ(ξc+(1ξ)d)ξeφ(c)+(1ξ)eφ(d) (1.3)

    holds for all c,dM and ξ[0,1].

    It is well-known that a function φ:I=[c,d]RR is an exponentially convex function if and only if it satisfies the inequality

    eφ(c+d2)1dcdceφ(z)dzeφ(c)+eφ(d)2 (1.4)

    for all c,dI with c<d. Inequality (1.4) provide the upper and lower estimates for the exponential integral, is called the Hermite-Hadamard inequality.

    Recently, the fractional calculus has attracted the consideration of several researchers [53,54]. The impact and inspiration of the fractional calculus in both theoretical, applied science and engineering arose out substantially. Fractional integral operators are sometimes the gateway to physical problems that cannot be expressed by classical integral, sometimes for the solution of problems expressed in fractional order. In recent years, a lot of new operator have been defined. Some of these operators are very close to classical operators in terms of their characteristics and definitions. Various studies in the literature, on distinct fractional operators such as the classical Riemann-Liouville, Caputo, Katugampola, Hadamard, and Marchaud versions have shown versatility in modeling and control applications across various disciplines. However, such forms of fractional derivatives may not be able to explain the dynamic performance accurately, hence many authors are found to be sorting out new fractional differentiations and integrations which have a kernel depending upon a function and this makes the range of definition expanded [55,56].

    Now, we recall the basic definitions and new notations of conformable fractional operators.

    Definition 1.2. ([3]). Let φL1([c,d]). Then the Riemann-Liouville integrals Jδc+φ and Jδdφ of order δ>0 are defined by

    (Jδc+)φ(z)=1Γ(δ)zc(zξ)δ1φ(ξ) dξ(z>c) (1.5)

    and

    (Jδd)φ(z)=1Γ(δ)dz(ξz)δ1φ(ξ) dξ(z<d), (1.6)

    respectively, where Γ() is the Euler Gamma function defined by Γ(δ)=0eξξδ1dξ.

    In [57], Jarad et al. defined a new fractional integral operator that has several special cases among many other features as follows:

    γJδc+φ(z)=1Γ(γ)zc((zc)δ(ξc)δδ)γ1φ(ξ)(ξc)1δdξ (1.7)

    and

    γJδdφ(z)=1Γ(γ)dz((dz)δ(dξ)δδ)γ1φ(ξ)(dξ)1δdξ. (1.8)

    Remark 1.1. It is easy to see the following connections:

    (1) Let c=0 and δ=1. Then (1.7) reduces to the Riemann-Liouville operator that is given in (1.5) and alike the other.

    (2) If we set c=0 and δ0, then the new conformable fractional integral coincides with the generalized fractional integral [58].

    (3) Furthermore, (1.8) reduces to the Riemann-Liouville operator if we set d=0 and δ=1. It also corresponds the Hadamard fractional integral [58] once d=0 and δ0 with the generalized fractional integral.

    The generalized K-conformable fractional integrals are defined by

    γKJδc+φ(z)=1KΓK(γ)zc((zc)δ(ξc)δδ)γK1φ(ξ)(ξc)1δdξ (1.9)

    and

    γKJδdφ(z)=1KΓK(γ)dz((dz)δ(dξ)δδ)γK1φ(ξ)(dξ)1δdξ. (1.10)

    Definition 1.3. Let K>0. Then the K-Gamma function ΓK is defined by

    ΓK (δ)=limyy! Ky(yK)δK1(δ)y,K. (1.11)

    If Re(δ)>0, then the K-Gamma function in integral form is defined by

    ΓK(δ)=0 eξKKξδ1dξ (1.12)

    with δΓK(δ)=ΓK(δ+K), where ΓK() stands for the K-gamma function.

    This paper is aimed at establishing some new integral inequalities for exponentially convexity via K-conformable fractional integrals linked with inequality (1.1). We present some inequalities for the class of mappings whose derivatives in absolute values are exponentially convex. In addition, we obtain some new inequalities linked with (1.2) and exponentially convexity via classical integrals. Moreover, we apply the novel approach of Hölder-İşcan and improved power-mean inequality are better than the Hölder and power-mean inequality. Moreover, we illustrate two examples to show the applicability and supremacy of the proposed technique. As an application, the inequalities for special means are derived.

    In this section, we demonstrate the Hermite-Hadamard type inequalities for exponentially convex function via K-conformable fractional integral operator.

    Theorem 2.1. Let K>0, δ>0, γ>0 and φ:IR be an exponentially convex function such that c,dI with c<d and eφL1([c,d]). Then the following inequality holds for K-conformable fractional integrals:

    eφ(c+d2)δγK2δγK1ΓK(γ+K)(dc)δγK{γKJδ(c+d2)eφ(c)+KΓK(γ)γKJδ(c+d2)+eφ(d)}eφ(c)+eφ(d)2. (2.1)

    Proof. Since φ is exponentially convex on I, for ξ=12, we have

    eφ(z1+z22)eφ(z1)+eφ(z2)2

    for all z1,z2I. Thus, if we choose z1=ξ2c+2ξ2d and z2=2ξ2c+ξ2d, for c,dI and ξ[0,1], we have

    2eφ(c+d2)eφ(2ξ2c+ξ2d)+eφ(ξ2c+2ξ2d). (2.2)

    Moreover, multiplying both sides of (2.2) by (1(1ξ)δδ)γK1(1ξ)δ1 and then making use of integration with respect to ξ over [0,1], we can combine the resulting inequality with the definition of integral operator as follows

    2eφ(c+d2)10(1(1ξ)δδ)γK1(1ξ)δ1dξ10(1(1ξ)δδ)γK1(1ξ)δ1eφ(2ξ2c+ξ2d)dξ+10(1(1ξ)δδ)γK1(1ξ)δ1eφ(ξ2c+2ξ2d)dξ(2dc)δγK{c+d2c((dc2)δ(c+d2u)δδ)γK1(c+d2u)δ1eφ(u)du+dc+d2((dc2)δ(uc+d2)δδ)γK1(uc+d2)δ1eφ(u)du}=(2dc)δγK{KΓK(γ)γKJδ(c+d2)eφ(c)+KΓK(γ)γKJδ(c+d2)+eφ(d)}.

    It is clear to see that

    10(1(1ξ)δδ)γK1(1ξ)δ1dξ=KγδγK.

    Consequently, we get

    2KγδγKeφ(c+d2)(2dc)δγK{KΓK(γ)γKJδ(c+d2)eφ(c)+KΓK(γ)γKJδ(c+d2)+eφ(d)}. (2.3)

    This completes the proof of the first inequality (2.1).

    To prove the second part of the inequality (2.1), by a similar discussion, we will start with the exponentially convexity of φ, then for ξ[0,1], we have

    eφ(2ξ2c+ξ2d)+eφ(ξ2c+2ξ2d)eφ(c)+eφ(d). (2.4)

    By multiplying (2.4) by (1(1ξ)δδ)γK1(1ξ)δ1 and then integrating the above estimate with respect to ξ over [0,1], we obtain

    10(1(1ξ)δδ)γK1(1ξ)δ1eφ(2ξ2c+ξ2d)dξ+10(1(1ξ)δδ)γK1(1ξ)δ1eφ(ξ2c+2ξ2d)dξ[eφ(c)+eφ(d)]10(1(1ξ)δδ)γK1(1ξ)δ1dξ.

    After simplification, we get

    (2dc)δγK{KΓK(γ)γKJδ(c+d2)eφ(c)+KΓK(γ)γKJδ(c+d2)+eφ(d)}K[eφ(c)+eφ(d)]γδγK.

    The proof is completed.

    Throughout this investigation, we use I to denote the interior of the interval IR. In order to establish our main results, we need a lemma which we present in this section.

    Lemma 2.2. Let K>0, δ, γ>0 and φ:IR be a differentiable and exponentially convex function on I such that c,dI with c<d and (eφ)L1([c,d]). Then the following equality holds for K-conformable fractional integrals:

    δγK2δγK1ΓK(γ+K)(dc)δγK[γKJδ(c+d2)eφ(c)+γKJδ(c+d2)+eφ(d)]eφ(c+d2)=δγK(dc)410(1(1ξ)δδ)γK{eφ(2ξ2c+ξ2d)φ(2ξ2c+ξ2d)eφ(ξ2c+2ξ2d)φ(ξ2c+2ξ2d)}dξ. (2.5)

    Proof. Integrating by parts and changing variable of definite integral yield

    10(1(1ξ)δδ)γKeφ(2ξ2c+ξ2d)φ(2ξ2c+ξ2d)dξ=2dc(1(1ξ)δδ)γKeφ(2ξ2c+ξ2d)|102δγKδγK(dc)10(1(1ξ)δ)γK(1ξ)δ1eφ(2ξ2c+ξ2d)dξ=2δγK(dc)eφ(c+d2)2δγKδγKδγK1(dc)δγK+1c+d2c((dc2)δ(c+d2u)δδ)γK1(c+d2u)δ1eφ(u)du=2δγK(dc)eφ(c+d2)2δγK+1ΓK(γ+K)δγK1(dc)δγK+1γKJδ(c+d2)eφ(c). (2.6)

    By similar argument, we have

    10(1(1ξ)δδ)γKeφ(ξ2c+2ξ2d)φ(ξ2c+2ξ2d)dξ=2δγK(dc)eφ(c+d2)+2δγK+1ΓK(γ+K)δγK1(dc)δγK+1γKJδ(c+d2)+eφ(d). (2.7)

    Subtracting these two equations leads to Lemma 2.2.

    Now we are in a position to establish some new integral inequalities of Hermite-Hadamard type for differentiable convex functions. The first main result is Theorem 2.3.

    Theorem 2.3. Let K>0, δ, γ>0 and φ:IR be a differentiable and exponentially convex function on I such that c,dI with c<d and (eφ)L1([c,d]). If |(eφ)|q is convex on I for some fixed p,q>1,q1+p1=1. Then the following inequality holds for K-conformable fractional integrals:

    |δγK2δγK1ΓK(γ+K)(dc)δγK[γKJδ(c+d2)eφ(c)+γKJδ(c+d2)+eφ(d)]eφ(c+d2)|δ1q1(dc)41+1q[B(γK+1,1δ)]11q[{[B(γK+1,3δ)+2B(γK+1,2δ)+B(γK+1,1δ)]|eφ(c)φ(c)|q+[B(γK+1,3δ)2B(γK+1,2δ)+B(γK+1,1δ)]|eφ(d)φ(d)|q+Υ1(c,d)[B(γK+1,1δ)B(γK+1,2δ)]}1q+{[B(γK+1,3δ)2B(γK+1,2δ)+B(γK+1,1δ)]|eφ(c)φ(c)|q+[B(γK+1,3δ)+2B(γK+1,2δ)+B(γK+1,1δ)]|eφ(d)φ(d)|q+Υ1(c,d)[B(γK+1,1δ)B(γK+1,2δ)]}1q], (2.8)

    where

    Υ1(c,d)=|eφ(c)φ(d)|q+|eφ(d)φ(c)|q. (2.9)

    Proof. It follows from Lemma 2.2 and the power-mean inequality that

    |δγK2δγK1ΓK(γ+K)(dc)δγK[γKJδ(c+d2)eφ(c)+γKJδ(c+d2)+eφ(d)]eφ(c+d2)|δγK(dc)4{(10(1(1ξ)δδ)γKdξ)11q(10(1(1ξ)δδ)γK|eφ(2ξ2c+ξ2d)φ(2ξ2c+ξ2d)|qdξ)1q+(10(1(1ξ)δδ)γKdξ)11q(10(1(1ξ)δδ)γK|eφ(ξ2c+2ξ2d)φ(ξ2c+2ξ2d)|qdξ)1q}. (2.10)

    Utilizing the convexity of |(eφ)|q on I, we have

    10(1(1ξ)δδ)γK|eφ(2ξ2c+ξ2d)φ(2ξ2c+ξ2d)|qdξ10(1(1ξ)δδ)γK[2ξ2|eφ(c)|q+ξ2|eφ(d)|q][2ξ2|φ(c)|q+ξ2|φ(d)|q]dξ=10(1(1ξ)δδ)γK[(2ξ2)2|eφ(c)φ(c)|q+(ξ2)2|eφ(d)φ(d)|q+ξ(2ξ)4[|eφ(c)φ(d)|q+|eφ(d)φ(c)|q]]dξ=14δγK{[B(γK+1,3δ)+2B(γK+1,2δ)+B(γK+1,1δ)]|eφ(c)φ(c)|q+[B(γK+1,3δ)2B(γK+1,2δ)+B(γK+1,1δ)]|eφ(d)φ(d)|q+Υ1(c,d)[B(γK+1,1δ)B(γK+1,2δ)]}. (2.11)

    Analogously, we have

    10(1(1ξ)δδ)γK|eφ(ξ2c+2ξ2d)φ(ξ2c+2ξ2d)|qdξ14δγK{[B(γK+1,3δ)2B(γK+1,2δ)+B(γK+1,1δ)]|eφ(c)φ(c)|q+[B(γK+1,3δ)+2B(γK+1,2δ)+B(γK+1,1δ)]|eφ(d)φ(d)|q+Υ1(c,d)[B(γK+1,1δ)B(γK+1,2δ)]}. (2.12)

    Substituting the above two inequalities into the inequality (2.10), we get the required inequality (2.8). This completes the proof.

    Theorem 2.4. Let K>0,δ,γ>0 and φ:IR be a differentiable and exponentially convex function on I such that c,dI with c<d and (eφ)L1([c,d]). If |(eφ)|q is convex on I for some fixed q>1,q1+p1=1. Then the following inequality holds for K-conformable fractional integrals:

    |δγK2δγK1ΓK(γ+K)(dc)δγK[γKJδ(c+d2)eφ(c)+γKJδ(c+d2)+eφ(d)]eφ(c+d2)|δγK(dc)4[1δpδK+1B(pγK+1,1δ)]1p[{7|eφ(c)φ(c)|q+|eφ(d)φ(d)|q+2Υ1(c,d)12}1q+{|eφ(c)φ(c)|q+7|eφ(d)φ(d)|q+2Υ1(c,d)12}1q], (2.13)

    where Υ1(c,d) is given in (2.9).

    Proof. It follows from Lemma 2.2 and the noted Hölder's integral inequality that

    |δγK2δγK1ΓK(γ+K)(dc)δγK[γKJδ(c+d2)eφ(c)+γKJδ(c+d2)+eφ(d)]eφ(c+d2)|δγK(dc)4{(10(1(1ξ)δδ)pγKdξ)1p(10|eφ(2ξ2c+ξ2d)φ(2ξ2c+ξ2d)|qdξ)1q+(10(1(1ξ)δδ)pγKdξ)1p(10|eφ(ξ2c+2ξ2d)φ(ξ2c+2ξ2d)|qdξ)1q}. (2.14)

    Utilizing the convexity of |(eφ)|q on I, we have

    10|eφ(2ξ2c+ξ2d)φ(2ξ2c+ξ2d)|qdξ10[2ξ2|eφ(c)|q+ξ2|eφ(d)|q][2ξ2|φ(c)|q+ξ2|φ(d)|q]dξ=10[(2ξ2)2|eφ(c)φ(c)|q+(ξ2)2|eφ(d)φ(d)|q+ξ(2ξ)4[|eφ(c)φ(d)|q+|eφ(d)φ(c)|q]]dξ=712|eφ(c)φ(c)|q+112|eφ(d)φ(d)|q+212Υ1(c,d). (2.15)

    Analogously, we have

    10|eφ(ξ2c+2ξ2d)φ(ξ2c+2ξ2d)|qdξ112|eφ(c)φ(c)|q+712|eφ(d)φ(d)|q+212Υ1(c,d). (2.16)

    Substituting the above two inequalities into the inequality (2.14), we get the required inequality (2.13). This completes the proof.

    In this section, we will derive the new generalizations by employing the Hölder İşcan [59] and improved power-mean [60] inequalities.

    Theorem 3.1. Let K,δ,γ>0 and φ:IR be a differentiable and exponentially convex function on I such that c,dI with c<d and (eφ)L1([c,d]). If |(eφ)|q is convex on I for some fixed q>1,q1+p1=1. Then the following inequality holds for K-conformable fractional integrals:

    |δγK2δγK1ΓK(γ+K)(dc)δγK[γKJδ(c+d2)eφ(c)+γKJδ(c+d2)+eφ(d)]eφ(c+d2)|δγK(dc)2[(B(1+pγK,2δ)δpγK+1)1p{17|eφ(c)φ(c)|q+|eφ(d)φ(d)|q+3Υ1(c,d)48}1q×[B(1+pγK,1δ)B(1+pγK,2δ)δpγK+1]1p{11|eφ(c)φ(c)|q+3|eφ(d)φ(d)|q+5Υ1(c,d)48}1q+[B(1+pγK,2δ)δpγK+1]1p{|eφ(c)φ(c)|q+17|eφ(d)φ(d)|q+3Υ1(c,d)48}1q×[B(1+pγK,1δ)B(1+pγK,2δ)δpγK+1]1p{3|eφ(c)φ(c)|q+11|eφ(d)φ(d)|q+5Υ1(c,d)48}1q], (3.1)

    whereΥ1(c,d) is given in (2.9).

    Proof. It follows from Lemma 2.2 and the Hölder-İşcan inequality [59] that

    |δγK2δγK1ΓK(γ+K)(dc)δγK[γKJδ(c+d2)eφ(c)+γKJδ(c+d2)+eφ(d)]eφ(c+d2)|δγK(dc)4[{(10(1ξ)(1(1ξ)δδ)pγKdξ)1p(10(1ξ)(|eφ(2ξ2c+ξ2d)φ(2ξ2c+ξ2d)|qdξ)1q}×{(10ξ(1(1ξ)δδ)pγKdξ)1p(10ξ(|eφ(2ξ2c+ξ2d)φ(2ξ2c+ξ2d)|qdξ)1q}+{(10(1ξ)(1(1ξ)δδ)pγKdξ)1p(10(1ξ)(|eφ(ξ2c+2ξ2d)φ(ξ2c+2ξ2d)|qdξ)1q×{(10ξ(1(1ξ)δδ)pγKdξ)1p(10ξ(|eφ(ξ2c+2ξ2d)φ(ξ2c+2ξ2d)|qdξ)1q}]. (3.2)

    Utilizing the convexity of |(eφ)|q on I, we get

    10(1ξ)|eφ(2ξ2c+ξ2d)φ(2ξ2c+ξ2d)|qdξ10(1ξ)[2ξ2|eφ(c)|q+ξ2|eφ(d)|q][2ξ2|φ(c)|q+ξ2|φ(d)|q]dξ=10(1ξ)[(2ξ2)2|eφ(c)φ(c)|q+(ξ2)2|eφ(d)φ(d)|q+ξ(2ξ)4[|eφ(c)φ(d)|q+|eφ(d)φ(c)|q]]dξ=1748|eφ(c)φ(c)|q+148|eφ(d)φ(d)|q+348Υ1(c,d), (3.3)
    10ξ|eφ(2ξ2c+ξ2d)φ(2ξ2c+ξ2d)|qdξ1148|eφ(c)φ(c)|q+348|eφ(d)φ(d)|q+548Υ1(c,d), (3.4)
    10(1ξ)|eφ(ξ2c+2ξ2d)φ(ξ2c+2ξ2d)|qdξ148|eφ(c)φ(c)|q+1748|eφ(d)φ(d)|q+348Υ1(c,d) (3.5)

    and

    10ξ|eφ(ξ2c+2ξ2d)φ(ξ2c+2ξ2d)|qdξ348|eφ(c)φ(c)|q+1148|eφ(d)φ(d)|q+548Υ1(c,d), (3.6)

    where we have used the indeitites

    10(1ξ)(1(1ξ)δδ)pγKdξ=1δpγK+1B(1+pγK,2δ),10ξ(1(1ξ)δδ)pγKdξ=1δpγK+1[B(1+pγK,1δ)B(1+pγK,2δ)]. (3.7)

    Combining (3.2)–(3.7) leads to the required inequality (4.4). This completes the proof.

    Theorem 3.2. Let K,δ,γ>0 and φ:IR be a differentiable and exponentially convex function on I such that c,dI with c<d and (eφ)L1([c,d]). If |(eφ)|q is convex on I for some fixed p,q>1,q1+p1=1. Then the following inequality holds for K-conformable fractional integrals:

    |δγK2δγK1ΓK(γ+K)(dc)δγK[γKJδ(c+d2)eφ(c)+γKJδ(c+d2)+eφ(d)]eφ(c+d2)|22/q(dc)2δ×[(B(1+γK,2δ))11q[{Ω1(δ,γ;K)|eφ(c)φ(c)|q+Ω2(δ,γ;K)|eφ(d)φ(d)|q+Υ1(c,d)Ω3(δ,γ;K)}1q×(B(1+γK,1δ)B(1+γK,2δ))11q×{Ω4(δ,γ;K)|eφ(c)φ(c)|q+Ω5(δ,γ;K)|eφ(d)φ(d)|q+Υ1(c,d)Ω6(δ,γ;K)}1q+(B(1+γK,2δ))11q{Ω2(δ,γ;K)|eφ(c)φ(c)|q+Ω1(δ,γ;K)|eφ(d)φ(d)|q+Υ1(c,d)Ω3(δ,γ;K)}1q×(B(1+γK,1δ)B(1+γK,2δ))11q×{Ω5(δ,γ;K)|eφ(c)φ(c)|q+Ω4(δ,γ;K)|eφ(d)φ(d)|q+Υ1(c,d)Ω6(δ,γ;K)}1q], (3.8)

    where

    Ω1(δ,γ;K)=B(γK+1,4δ)+2B(γK+1,3δ)+B(γK+1,2δ),Ω2(δ,γ;K)=B(γK+1,4δ)2B(γK+1,3δ)+B(γK+1,2δ),Ω3(δ,γ;K)=B(γK+1,2δ)B(γK+1,4δ),Ω4(δ,γ;K)=B(γK+1,1δ)+B(γK+1,2δ)B(γK+1,3δ)B(γK+1,4δ),Ω5(δ,γ;K)=B(γK+1,1δ)3B(γK+1,2δ)+3B(γK+1,3δ)B(γK+1,4δ),Ω6(δ,γ;K)=B(γK+1,4δ)B(γK+1,3δ)B(γK+1,2δ)+B(γK+1,1δ), (3.9)

    and Υ1(c,d) is given in (2.9).

    Proof. Making use of Lemma 2.2 and the improved power-mean inequality [60], we have

    |δγK2δγK1ΓK(γ+K)(dc)δγK[γKJδ(c+d2)eφ(c)+γKJδ(c+d2)+eφ(d)]eφ(c+d2)|δγK(dc)4[{(10(1ξ)(1(1ξ)δδ)γKdξ)11q×(10(1ξ)(1(1ξ)δδ)γK|eφ(2ξ2c+ξ2d)φ(2ξ2c+ξ2d)|qdξ)1q}×{(10ξ(1(1ξ)δδ)γKdξ)11q(10ξ(1(1ξ)δδ)γK|eφ(2ξ2c+ξ2d)φ(2ξ2c+ξ2d)|qdξ)1q}+{(10(1ξ)(1(1ξ)δδ)γKdξ)11q(10(1ξ)(1(1ξ)δδ)γK|eφ(ξ2c+2ξ2d)φ(ξ2c+2ξ2d)|qdξ)1q×{(10ξ(1(1ξ)δδ)γKdξ)11q(10ξ(1(1ξ)δδ)γK|eφ(ξ2c+2ξ2d)φ(ξ2c+2ξ2d)|qdξ)1q}]. (3.10)

    Utilizing the convexity of |(eφ)|q on I, we obtain

    10(1ξ)(1(1ξ)δδ)γK|eφ(2ξ2c+ξ2d)φ(2ξ2c+ξ2d)|qdξ10(1ξ)(1(1ξ)δδ)γK[2ξ2|eφ(c)|q+ξ2|eφ(d)|q][2ξ2|φ(c)|q+ξ2|φ(d)|q]dξ=10(1ξ)(1(1ξ)δδ)γK[(2ξ2)2|eφ(c)φ(c)|q+(ξ2)2|eφ(d)φ(d)|q+ξ(2ξ)4[|eφ(c)φ(d)|q+|eφ(d)φ(c)|q]]dξ=14δγK+1[{B(γK+1,4δ)+2B(γK+1,3δ)+B(γK+1,2δ)}|eφ(c)φ(c)|q+{B(γK+1,4δ)2B(γK+1,3δ)+B(γK+1,2δ)}|eφ(d)φ(d)|q+Υ1(c,d){B(γK+1,2δ)B(γK+1,4δ)}]=14δγK+1[Ω1(δ,γ;K)|eφ(c)φ(c)|q+Ω2(δ,γ;K)|eφ(d)φ(d)|q+Υ1(c,d)Ω3(δ,γ;K)], (3.11)
    10ξ(1(1ξ)δδ)γK|eφ(2ξ2c+ξ2d)φ(2ξ2c+ξ2d)|qdξ14δγK+1[{B(γK+1,1δ)+B(γK+1,2δ)B(γK+1,3δ)B(γK+1,4δ)}|eφ(c)φ(c)|q+{B(γK+1,1δ)3B(γK+1,2δ)+3B(γK+1,3δ)B(γK+1,4δ)}|eφ(d)φ(d)|q+Υ1(c,d){B(γK+1,4δ)B(γK+1,3δ)B(γK+1,2δ)+B(γK+1,1δ)}]=14δγK+1[Ω4(δ,γ;K)|eφ(c)φ(c)|q+Ω5(δ,γ;K)|eφ(d)φ(d)|q+Υ1(c,d)Ω6(δ,γ;K)], (3.12)
    10(1ξ)(1(1ξ)δδ)γK|eφ(ξ2c+2ξ2d)φ(ξ2c+2ξ2d)|qdξ14δγK+1[{B(γK+1,4δ)2B(γK+1,3δ)+B(γK+1,2δ)}|eφ(c)φ(c)|q+{B(γK+1,4δ)+2B(γK+1,3δ)+B(γK+1,2δ)}|eφ(d)φ(d)|q+Υ1(c,d){B(γK+1,2δ)B(γK+1,4δ)}]=14δγK+1[Ω2(δ,γ;K)|eφ(c)φ(c)|q+Ω1(δ,γ;K)|eφ(d)φ(d)|q+Υ1(c,d)Ω3(δ,γ;K)] (3.13)

    and

    10ξ(1(1ξ)δδ)γK|eφ(ξ2c+2ξ2d)φ(ξ2c+2ξ2d)|qdξ14δγK+1[{B(γK+1,1δ)3B(γK+1,2δ)+3B(γK+1,3δ)B(γK+1,4δ)}|eφ(c)φ(c)|q+{B(γK+1,1δ)+B(γK+1,2δ)B(γK+1,3δ)B(γK+1,4δ)}|eφ(d)φ(d)|q+Υ1(c,d){B(γK+1,4δ)B(γK+1,3δ)B(γK+1,2δ)+B(γK+1,1δ)}]=14δγK+1[Ω5(δ,γ;K)|eφ(c)φ(c)|q+Ω4(δ,γ;K)|eφ(d)φ(d)|q+Υ1(c,d)Ω6(δ,γ;K)], (3.14)

    where we have used the facts that

    10(1ξ)(1(1ξ)δδ)γKdξ=1δγK+1B(1+γK,2δ),10ξ(1(1ξ)δδ)γKdξ=1δγK+1[B(1+γK,1δ)B(1+γK,2δ)]. (3.15)

    Combining (3.10)–(3.15) gives the required inequality (3.8). This completes the proof.

    In this section, we present some examples to demonstrate the applications of our proposed results on modified Bessel functions and σ-digamma functions.

    Example 4.1. Let zR and Iρ:R[1,) be the modified Bessel function of the first kind defined by

    Iρ(z)=n0zρ+2n2ρ+2nn!Γ(ρ+n+1).

    The the first order derivative formula of Iρ is given by

    I(z)=z2(ρ+1)Iρ+1(z). (4.1)

    By use of Theorem 2.4 and identity (4.1), we get

    |1dcdceIρ(z)dzeIρ(c+d2)|dc8(ρ+1)(1p+1)1p×[{7c|eIρ(c)Iρ+1(c)|q+d|eIρ(d)Iρ+1(d)|q+2{c|eIρ(d)Iρ+1(c)|q+d|eIρ(c)Iρ+1(d)|q}12}1q+{c|eIρ(c)Iρ+1(c)|q+7d|eIρ(d)Iρ+1(d)|q+2{c|eIρ(d)Iρ+1(c)|q+d|eIρ(c)Iρ+1(d)|q}12}1q],

    where ρ>1,c,dR satisfy the assumptions that δ=γ=1 and 0<c<d,K=1. Specifically, for I1/2(z)=coshz and I1/2(z)=sinhzz, one has

    |1dcdcecosh(z)dzecosh(c+d2)|dc8(ρ+1)(1p+1)1p×[{7|esinh(c)sinh(c)|q+|esinh(d)sinh(d)|q+2{|esinh(d)sinh(c)|q+|esinh(c)sinh(d)|q}12}1q+{|esinh(c)sinh(c)|q+7|esinh(d)sinh(d)|q+2{|esinh(d)sinh(c)|q+|esinh(c)sinh(d)|q}12}1q].

    Example 4.2. Consider the σ-analogue of the digamma function ϕσ given by

    ϕσ(z)=ln(1σ)+lnσκ=0σκ+z1σκ+z=ln(1σ)+lnσκ=0σκz1σzκ. (4.2)

    For σ>1 and z>0, the σ-digamma function ϕσ can be expressed by

    ϕσ(z)=ln(1σ)+lnσ[z12κ=0σ(κ+z)1σ(κ+z)]=ln(1σ)+lnσ[z12κ=0σκz1σκz]. (4.3)

    It follows from limσ1+ϕσ(z)=limσ1ϕσ(z)=ϕ(z) that zϕσ(z) is a completely monotonic function on an interval (0,) for all σ>0, and consequently, zϕσ(z) is convex on the same interval. Let φσ(z)=ϕσ(z) with q>0. Then φσ(z)=ϕσ(z) is completely monotonic on the interval (0,), and from Theorem 3.1 we have

    |eϕσ(c+d2)(eϕσ(d)eϕσ(c)dc)|(dc)2[(1(p+1)(p+2))1p{17|eϕσ(c)ϕσ(c)|q+|eϕσ(d)ϕσ(d)|q+3Υ1(c,d)48}1q×(1p+2)1p{11|eϕσ(c)ϕσ(c)|q+3|eϕσ(d)ˇϕσ(d)|q+5Υ1(c,d)48}1q+(1(p+1)(p+2))1p{|eϕσ(c)ϕσ(c)|q+17|eϕσ(d)ϕσ(d)|q+3Υ1(c,d)48}1q×(1p+1)1p{3|eϕσ(c)ϕσ(c)|q+11|eϕσ(d)ϕσ(d)|q+5Υ1(c,d)48}1q], (4.4)

    for all σ(0,1),γ=1=δ,d>c>0, K=1 and Υ1(c,d) is given in (2.9).

    In order to prove our main results in this section, we need the following lemma.

    Lemma 5.1. Let φ:IR be a differentiable and exponentially convex functions on I such that c,dI with c<d and (eφ)L1([c,d]). Also, let w:[c,d]R be a differentiable mapping such that w is symmetric with respect to (c+d)/2. Then the identity

    1dcdceφ(z)ew(z)dz1dceφ(c+d2)dcew(z)dz=(dc)10ϕ(ξ)eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)dξ (5.1)

    holds for all z[c,d], where

    ϕ(ξ)={ξ0ew(cs+(1s)d)ds,ξ[0,1/2),1ξew(cs+(1s)d)ds,ξ[1/2,1].

    Proof. Note that

    θ=10ϕ(ξ)eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)dξ=120(ξ0ew(cs+(1s)d)ds)eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)dξ+112(1ξew(cs+(1s)d)ds)eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)dξ=θ1+θ2.

    Making use of integration by parts, we get

    θ1=(ξ0ew(cs+(1s)d)ds)eφ(ξc+(1ξ)d)cd|120120ew(cξ+(1ξ)d)eφ(ξc+(1ξ)d)cddξ=(120ew(cs+(1s)d)ds)eφ(c+d2)cd120ew(cξ+(1ξ)d)eφ(ξc+(1ξ)d)cddξ.

    Similarly, one has

    θ2=(112ew(cs+(1s)d)ds)eφ(c+d2)cd112ew(cξ+(1ξ)d)eφ(ξc+(1ξ)d)cddξ.

    Therefore,

    θ=θ1+θ2=(10ew(cs+(1s)d)ds)eφ(c+d2)cd10ew(cξ+(1ξ)d)eφ(ξc+(1ξ)d)cddξ,

    and identity (5.1) can be obtained by the change of variable technique and multiplying the both sided by (dc) in the above formula.

    Theorem 5.2. Let φ:IR be a differentiable and exponentially convex function on I such that c,dI with c<d. Also, let w:[c,d]R be a differentiable mapping and symmetric with respect to (c+d)/2. If |(eφ)| is convex on I. Then the inequality

    |1dcdceφ(z)ew(z)dz1dceφ(c+d2)dcew(z)dz|[16(dc)3dc+d2ew(z)dz]{8[(zc)3(dz)3][|eφ(c)φ(c)|+|eφ(d)φ(d)|2]+Υ(c,d)[(dc)32(dz)2(3dc2z)]} (5.2)

    holds for all z[c,d], where

    Υ(c,d)={|eφ(c)φ(d)|+|eφ(d)φ(c)|}.

    Proof. It follows from Lemma 5.1 and the hypothesis given in Theorem 5.2 that

    |1dcdceφ(z)ew(z)dz1dceφ(c+d2)dcew(z)dz|(dc)[{120(ξ0ew(cs+(1s)d)ds)[eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)]dξ}+{112(1ξew(cs+(1s)d)ds)[eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)]dξ}](dc)[{120(ξ0ew(cs+(1s)d)ds)[ξ2|eφ(c)φ(c)|+(1ξ)2|eφ(d)φ(d)|+ξ(1ξ)Υ(c,d)]dξ}+{112(1ξew(cs+(1s)d)ds)[ξ2|eφ(c)φ(c)|+(1ξ)2|eφ(d)φ(d)|+ξ(1ξ)Υ(c,d)]dξ}]=ρ1+ρ2. (5.3)

    Exchanging the integration order gives

    ρ1=120ξ0ew(cs+(1s)d)[ξ2|eφ(c)φ(c)|+(1ξ)2|eφ(d)φ(d)|+ξ(1ξ)Υ(c,d)]dsdξ=12012sew(cs+(1s)d)[ξ2|eφ(c)φ(c)|+(1ξ)2|eφ(d)φ(d)|+ξ(1ξ)Υ(c,d)]dξds120ew(cs+(1s)d)[{(124s33)|eφ(c)φ(c)|}+{(124+s33)|eφ(d)φ(d)|}+Υ(c,d){s2(32s)6+112}]=dc+d2ew(z)[{(124(dz)33(dc)3)|eφ(c)φ(c)|}+{(124+(zc)33(dc)3)|eφ(d)φ(d)|}+2Υ(c,d){(dc)32(dz)2(d3c+2z)}]. (5.4)

    Similarly, we have

    ρ2=1121ξew(cs+(1s)d)[ξ2|eφ(c)φ(c)|+(1ξ)2|eφ(d)φ(d)|+ξ(1ξ)Υ(c,d)]dsdξc+d2cew(z)[{(124+(zc)33(dc)3)|eφ(c)φ(c)|}+{(124(dz)33(dc)3)|eφ(d)φ(d)|}+2Υ(c,d){(dc)32(dz)2(d3c+2z)}]. (5.5)

    Since w(z) is symmetric with respect to z=c+d2, for w(z)=w(c+dz), we get

    ρ1=ρ2.

    combining (5.3)–(5.5) leads to (5.2). This completes the proof

    Theorem 5.3. Let φ:IR be a differentiable and exponentially convex function on I such that c,dI with c<d. Also, let w:[c,d]R be a differentiable mapping and symmetric with respect to (c+d)/2. If |(eφ)|q is convex on I for some q>1 with p1+q1=1. Then the inequality

    |1dcdceφ(z)ew(z)dz1dceφ(c+d2)dcew(z)dz|(dc)[1(dc)2c+d2depw(z)(zc+d2)dz]1/p[(3|eφ(c)φ(c)|q+11|eφ(d)φ(d)|q+6Υ1(c,d)192)1/q+(11|eφ(c)φ(c)|q+3|eφ(d)φ(d)|q+6Υ1(c,d)192)1q] (5.6)

    holds for all z[c,d], where Υ1(c,d) is given in (2.9).

    Proof. Making use of Lemma 5.1 and changing the integration order, we get

    |1dcdceφ(z)ew(z)dz1dceφ(c+d2)dcew(z)dz|(dc)[{120(ξ0ew(cs+(1s)d)ds)[eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)]dξ}+{112(1ξew(cs+(1s)d)ds)[eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)]dξ}]=(dc)[{12012sew(cs+(1s)d)eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)dξds}+{112s12ew(cs+(1s)d)eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)dξds}].

    By Hölder inequality, we have

    |1dcdceφ(z)ew(z)dz1dceφ(c+d2)dcew(z)dz|(dc){(12012sew(cs+(1s)d)dξds)1/p(12012s|eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)|qdξds)1/q+(112s12ew(cs+(1s)d)dξds)1/p(s1212s|eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)|qdξds)1/q}.

    It follows from the convexity of |(eφ)|q that

    |eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)|q[ξ|eφ(c)|q+(1ξ)|eφ(d)|q][ξ|φ(c)|q+(1ξ)|φ(d)|q][ξ2|eφ(c)φ(c)|q+(1ξ)2|eφ(d)φ(d)|q+ξ(1ξ){|eφ(c)φ(d)|q+|eφ(d)φ(c)|q}][ξ2|eφ(c)φ(c)|q+(1ξ)2|eφ(d)φ(d)|q+ξ(1ξ)Υ1(c,d)]

    for ξ[0,1].

    Therefore, one has

    |1dcdceφ(z)ew(z)dz1dceφ(c+d2)dcew(z)dz|(dc)[(12012sew(cs+(1s)d)dξds)1/p(12012s[ξ2|eφ(c)φ(c)|q+(1ξ)2|eφ(d)φ(d)|q+ξ(1ξ)Υ1(c,d)]dξds)1/q+(112s12ew(cs+(1s)d)dξds)1p(s1212s[ξ2|eφ(c)φ(c)|q+(1ξ)2|eφ(d)φ(d)|q+ξ(1ξ)Υ1(c,d)]dξds)1q]=S1+S2. (5.7)

    Note that

    S1=(12(dc)2dc+d2epw(z)(2zcd)dz)1/p(3|eφ(c)φ(c)|q+11|eφ(d)φ(d)|q+6Υ1(c,d)192)1/q (5.8)

    and

    S2=(12(dc)2c+d2cepw(z)(c+d2z)dz)1/p(11|eφ(c)φ(c)|q+3|eφ(d)φ(d)|q+6Υ1(c,d)192)1/q. (5.9)

    Therefore, inequality (5.3) follows from (5.7)–(5.9). This completes the proof.

    Lemma 5.4. Let φ:IR be a differentiable and exponentially convex function on I such that c,dI with c<d and (eφ)L1([c,d]). Also, let w:[c,d]R be a differentiable mapping such that w is symmetric with respect to (c+d)/2. Then the identity

    1dceφ(c)+eφ(d)2dcew(z)dz1dcdcew(z)eφ(z)dz=dc210μ(ξ)eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)dξ

    holds for all z[c,d], where

    μ(ξ)=1ξew(cs+(1s)d)dsξ0ew(cs+(1s)d)ds.

    Proof. We clearly see that

    U=10μ(ξ)eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)dξ=10(1ξew(cs+(1s)d)ds)eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)dξ=c1+c2.

    It follows from integration by parts that

    c1=(1ξew(cs+(1s)d)ds)eφ(ξc+(1ξ)d)cd|10+10ew(cξ+(1ξ)d)eφ(ξc+(1ξ)d)cddξ=(10ew(cs+(1s)d)ds)eφ(d)cd+10ew(cξ+(1ξ)d)eφ(ξc+(1ξ)d)cddξ.

    Similarly, we have

    c2=(10ew(cs+(1s)d)ds)eφ(c)cd+10ew(cξ+(1ξ)d)eφ(ξc+(1ξ)d)cddξ.

    Thus, we get

    U=c1+c2=eφ(c)+eφ(d)cd(10ew(cs+(1s)d)ds)+210ew(cξ+(1ξ)d)eφ(ξc+(1ξ)d)cddξ.

    Therefore, the desired result can be obtained by use of the change of variable technique and multiplying the both sided by (dc)/2.

    Theorem 5.5. Let φ:IR be a differentiable and exponentially convex function on I such that c,dI with c<d. Also, let w:[c,d]R be a differentiable mapping and symmetric with respect to (c+d)/2. If |(eφ)|q is convex on I for some q>1 with p1+q1=1. Then the inequality

    |1dceφ(c)+eφ(d)2dcew(z)dz1dcdceφ(z)ew(z)dz|12(10(η(z))pdξ)1/p(2(|eφ(c)φ(c)|q+|eφ(d)φ(d)|q)+Υ1(c,d)6)

    holds for all z[c,d], where Υ1(c,d) is given in (2.9) and

    η(z)=|d(dc)ξc+(dc)ξew(z)dz|.

    Proof. From Lemma 5.4 and the hypothesis of Theorem 5.5, we get

    |1dceφ(c)+eφ(d)2dcew(z)dz1dcdceφ(z)ew(z)dz|dc2[10|1ξew(cs+(1s)d)dsξ0ew(cs+(1s)d)ds||eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)|dξ]12[10|d(dc)ξcew(z)dzdd(dc)ξew(z)dz||eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)|dξ]. (5.10)

    Since w(z) is symmetric to z=(c+d)/2, we get

    d(dc)ξcew(z)dzdd(dc)ξew(z)dz=d(dc)ξc+(dc)ξew(z)dz (5.11)

    for ξ[0,1/2], and

    d(dc)ξcew(z)dzdd(dc)ξew(z)dz=c+(dc)ξd(dc)ξew(z)dz (5.12)

    for ξ[1/2,1].

    It follows from (5.10)–(5.12) that

    |1dceφ(c)+eφ(d)2dcew(z)dz1dcdceφ(z)ew(z)dz|1210η(z)|eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)|dξ12{10(η(z))pdξ}1p{10|eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)|qdξ}1q.

    It follows from the convexity of |(eφ)|q that

    |eφ(ξc+(1ξ)d)φ(ξc+(1ξ)d)|q[ξ|eφ(c)|q+(1ξ)|eφ(d)|q][ξ|φ(c)|q+(1ξ)|φ(d)|q][ξ2|eφ(c)φ(c)|q+(1ξ)2|eφ(d)φ(d)|q+ξ(1ξ){|eφ(c)φ(d)|q+|eφ(d)φ(c)|q}][ξ2|eφ(c)φ(c)|q+(1ξ)2|eφ(d)φ(d)|q+ξ(1ξ)Υ1(c,d)].

    Therefore,

    |1dceφ(c)+eφ(d)2dcew(z)dz1dcdceφ(z)ew(z)dz|12[10(η(z))pdξ]1/p{10[ξ2|eφ(c)φ(c)|q+(1ξ)2|eφ(d)φ(d)|q+ξ(1ξ)Υ1(c,d)]dξ}1/q=12[10(η(z))pdξ]1/p[2(|eφ(c)φ(c)|q+|eφ(d)φ(d)|q)+Υ1(c,d)6],

    which completes the proof.

    A real-valued function M:(0,)×(0,)(0,) is said to be a bivariate mean if min{c,d}M(c,d)max{c,d} for all c,d(0,). Recently, the inequalities for the bivariate means have attracted the attention of many researchers.

    Let c,d>0 with cd and nZ{1,0}. Then the classical arithmetic mean A(c,d) and n-th generalized logarithmic mean Ln(c,d) are defined by

    A(c,d)=c+d2,Ln(c,d)=[dn+1cn+1(n+1)(dc)]1/n.

    In this section, we shall establish some inequalities for the arithmetic and generalized logarithmic means by use of our results obtained in section 5.

    Theorem 6.1. Let c,d>0 with d>c and nZ with n2. Then the inequality

    |dcznew(z)dzAn(c,d)dcew(z)dz|(dc)[16(dc)3dc+d2ew(z)dz]×{8n[(zc)3(dz)3][A(c2n1,d2n1)]+Υ1(c,d)[(dc)32(dz)2(3dc2z)]} (6.1)

    holds for all z[c,d], where Υ1(c,d) is given in (2.9).

    Proof. By taking φ(z)=nlogz in Theorem 5.2, we get the desired result.

    Let w=1. Then inequality (6.1) leads to Corollary 6.2 immediately.

    Corollary 6.2. Let c,d>0 with d>c and nZ with n2. Then

    |Lnn(c,d)1dcAn(c,d)|[112(dc)2]{8n[(zc)3(dz)3][A(c2n1,d2n1)]+Υ1(c,d){(dc)32(dz)2(3dc2z)}}.

    Theorem 6.3. Let p,q>1 with p1+q1=1, c,d>0 with d>c and nZ with n2. Then the inequality

    |dcznew(z)dzAn(c,d))dcew(z)dz|(dc)2[1(dc)2dc+d2epw(z)(z(c+d)2)dz]1/p[(nq6A(c(2n1)q,d(2n1)q)+8nqd(2n1)q+6Υ1(c,d)192)1/q+(nq6A(c(2n1)q,d(2n1)q)+8nqc(2n1)q+6Υ1(c,d)192)1/q] (6.2)

    holds for all z[c,d].

    Proof. By taking φ(z)=nlnz in Theorem (5.3), we get the desired result.

    Let w=1. Then (6.2) leads to Corollary 6.4 immediately.

    Corollary 6.4. Let p,q>1 with p1+q1=1, c,d>0 with d>c and nZ with n2. Then

    |Lnn(c,d)1dcAn(c,d)|dc23p[(nq6A(c(2n1)q,d(2n1)q)+8nqd(2n1)q+6Υ1(c,d)192)1/q+(nq6A(c(2n1)q,d(2n1)q)+8nqc(2n1)q+6Υ1(c,d)192)1/q].

    Theorem 6.5. Let p,q>1 with p1+q1=1, c,d>0 with d>c and nZ with n2. Then the inequality

    |1dcA(cn,dn)dcew(z)dz1dcdcznew(z)dz|12(10(η(z))pdξ)1/p(4nqA(c(2n1)q,d(2n1)q)+Υ1(c,d)6) (6.3)

    holds for z[c,d].

    Proof. By taking φ(z)=nlnz in Theorem 5.5, we get the desired result.

    Let w=1. Then (6.3) leads to Corollary 6.6 immediately.

    Corollary 6.6. Let u,v>0 with v>u and nZ with n2. Then the inequality

    |1dc(A(cn,dn))Lnn(c,d)|dc2e[4nqA(c(2n1)q,d(2n1)q)+Υ1(c,d)6]

    holds for all q>1.

    Using the K-conformable fractional integrals, certain inequalities related to the Hermite-Hadamard inequalities for exponentially convex functions are established. The inequalities are parameterized by the parameters δ,γ and K. These inequalities generalize and extend parts of the results for Riemann-Liouville and Hadamard fractional integrals. Also, we have derived the weighted Hermite-Hadamard inequalities for exponentially convex functions in the classical sense. Some applications of the obtained results to special means are also presented. With these contributions, we hope to motivate the interested researcher to further explore this enchanting field of the fractional integral inequalities and exponentially convexity based on these techniques and the ideas developed in the present paper.

    The authors would like to thank the anonymous referees for their valuable comments and suggestions, which led to considerable improvement of the article.

    This work was supported by the National Natural Science Foundation of China (Grant Nos. 11971142, 61673169, 11701176, 11871202) and the Natural Science Foundation of Zhejiang Province (Grant No. LY19A010012).

    The authors declare that they have no competing interests.



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