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Simpson’s type integral inequalities for ĸ-fractional integrals and their applications

  • In this paper, some new inequalities of Simpson's type are set up for the classes of functions whose derivatives of absolute are preinvex by means of κ-fractional integrals. Additionally, by extraordinary choices of n and κ, we give some diminished outcomes. Meanwhile, we also provide the inequalities for F-divergence measures and in probabilistic versions.

    Citation: Saima Rashid, Ahmet Ocak Akdemir, Fahd Jarad, Muhammad Aslam Noor, Khalida Inayat Noor. Simpson’s type integral inequalities for ĸ-fractional integrals and their applications[J]. AIMS Mathematics, 2019, 4(4): 1087-1100. doi: 10.3934/math.2019.4.1087

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  • In this paper, some new inequalities of Simpson's type are set up for the classes of functions whose derivatives of absolute are preinvex by means of κ-fractional integrals. Additionally, by extraordinary choices of n and κ, we give some diminished outcomes. Meanwhile, we also provide the inequalities for F-divergence measures and in probabilistic versions.


    The following inequality is named the Simpson's integral inequality:

    |16[φ(ε)+4φ(ε+ζ2)+φ(ζ)]1ζεζεφ(t)dt|φ(4)(ζε)4, (1.1)

    where φ:[ε,ζ]R is a four-order differentiable mapping on (ε,ζ) and φ(4)=supt(ε,ζ)|φ(4)(t)|<.

    Considering the Simpson-type inequality via mappings of different classes, many results involving the ordinary integrals can be found in [1,3,4,5,18,19] and the references therein.

    Definition 1.1. A set ΩRn is named as invex set with respect to the mapping η:Ω×ΩRn, if

    ε+η(ζ,ε)Ω

    holds, for all ε,ζΩ and θ[0,1].

    Definition 1.2. A mapping φ:ΩR is called preinvex respecting η, if

    φ(ε+η(ζ,ε))(1θ)φ(ε)+θφ(ζ),

    holds for all ε,ζΩ and θ[0,1].

    The preinvex function is an important substantive generalization of the convex function. For the properties, applications, integral inequalities and other aspects of preinvex functions, see [5,7,8,9,10,11,12,13,14] and the references therein.

    Definition 1.3. Let s(0,1) and a mapping φ:ΩR is called s-preinvex respecting η, if

    φ(ε+η(ζ,ε))(1θs)φ(ε)+θsφ(ζ),

    holds for all ε,ζΩ and θ[0,1].

    Fractional calculus has recently been the focus of mathematics and many related sciences. In addition to defining new integral-derivative operators and their properties, many researchers have achieved important results in applied mathematics, engineering and statistics. It is appropriate for the reader to review articles [2,15,16,17] for some recent studies. We end this section by reciting a well known κ-fractional integral operators in the literature.

    Definition 1.4. ([6]) Let φL[ε,ζ], the κ-fractional integrals Jα,κε+ and Jα,κζ of order α>0 are defined by

    Jα,κε+=1κΓκ(α)xε(xθ)ακ1φ(θ)dθ,(0α<x<ζ) (1.2)

    and

    Jα,κζ+=1κΓκ(α)ζx(θx)ακ1φ(θ)dθ,(0α<x<ζ), (1.3)

    respectively, where κ>0, and Γκ is the κ-gamma function defined as Γκ(x):=0θx1eθκκdθ,(x)>0, with the properties Γκ(x+κ)=xΓκ(x) and Γκ(κ)=1.

    This paper aims to obtain estimation type results of Simpson-type inequality related to κ- fractional integral operators. Next, we derive the results with the boundedness of the derivative and with a Lipschitzian condition for the derivative of the considered mapping to derive integral inequalities with new bounds. Application of our results to random variables are also provided.

    Throughout of this work, let ΩR be an open subset respecting η:Ω×ΩR{0} and ε,ζΩ with ε<ζ.

    To prove our results, we obtain a new integral identity as following:

    Lemma 2.1. Let φ:Ω=[ε+η(ζ,ε)]R be a differentiable function on (ε+η(ζ,ε)) with η(ζ,ε)>0. If φL[ε,ε+η(ζ,ε)],n0 and α,κ>0, then the following identity holds:

    Ψ(ε,ζ;n,α,κ):=η(ζ,ε)2(n+1)[10(2(1θ)ακθακ3)φ(ε+1θn+1η(ζ,ε))dθ+10(θακ2(1θ)ακ3)]φ(ε+n+θn+1η(ζ,ε))dθ], (2.1)

    where

    Ψ(ε,ζ;n,α,κ):=16[φ(ε)+φ(ε+η(ζ,ε))+2φ(ε+1n+1η(ζ,ε))+2φ(ε+nn+1η(ζ,ε))Γκ(α+κ)(n+1)ακ6(η(ζ,ε))ακ[Jα,κ(ε)+φ(a+1n+1η(ζ,ε))]+Jα,κ(ε+η(ζ,ε))φ(ε+nn+1η(ζ,ε))]Γκ(α+κ)(n+1)ακ3(η(ζ,ε))ακ[Jα,κ(ε+nn+1η(ζ,ε))+φ(ε+η(ζ,ε))+Jα,κ(ε+1n+1η(ζ,ε))+φ(ε)]].

    Proof. Integration by parts, one can have

    I1=10(2(1θ)ακθακ3)φ(ε+1θn+1η(ζ,ε))dθ=n+13η(ζ,ε)φ(ε)+2(n+1)3η(ζ,ε)φ(ε+1n+1η(ζ,ε))α(n+1)ακ+13κ(η(ζ,ε))ακ+1ε+1n+1η(ζ,ε)ε(ε+1n+1η(ζ,ε)x)ακ1φ(x)dx2α(n+1)ακ+13κ(η(ζ,ε))ακε+1n+1η(ζ,ε)ε(xε)ακ1φ(x)dx

    and

    I2=10(θακ2(1θ)ακ3)]φ(ε+n+θn+1η(ζ,ε))dθ=n+13(η(ζ,ε))φ(ε+η(ζ,ε))+2(n+1)3η(ζ,ε)φ(ε+nn+1η(ζ,ε))α(n+1)ακ+13κ(η(ζ,ε))ακ+1ε+η(ζ,ε)ε+nn+1η(ζ,ε)(x(ε+nn+1η(ζ,ε)))ακ1φ(x)dxα(n+1)ακ+13κ(η(ζ,ε))ακ+1ε+η(ζ,ε)ε+nn+1η(ζ,ε)((ε+η(ζ,ε))x)ακ1φ(x)dx.

    By adding I1 and I2 and multiplying the both sides η(ζ,ε)2(n+1), we can write

    I1+I2=16[φ(ε)+φ(ε+η(ζ,ε))+2φ(ε+1n+1η(ζ,ε))+2φ(ε+nn+1η(ζ,ε))]α(n+1)ακ6κ(η(ζ,ε))ακ[ε+1n+1η(ζ,ε)ε(ε+1n+1η(ζ,ε))ακ1φ(x)dx+ε+η(ζ,ε)ε+nn+1η(ζ,ε)(x(ε+nn+1η(ζ,ε)))ακ1φ(x)dx]α(n+1)ακ3κ(η(ζ,ε))ακ[ε+nn+1η(ζ,ε)ε(ε+η(ζ,ε)x)ακ1φ(x)dx+ε+1n+1η(ζ,ε)ε(xε)ακ1φ(x)dx].

    Using the fact that

    1κΓκ(α)ε+1n+1η(ζ,ε)ε(xε)ακ1φ(x)dx=Jα,κ(ε+1n+1η(ζ,ε))φ(ε),1κΓκ(α)ε+η(ζ,ε)ε+nn+1η(ζ,ε)(ε+η(ζ,ε)x)ακ1φ(x)dx=Jα,κ(ε+nn+1η(ζ,ε))+φ(ε+η(ζ,ε)),1κΓκ(α)ε+1n+1η(ζ,ε)ε(ε+1n+1η(ζ,ε)x)ακ1φ(x)dx=Jα,κ(ε)+φ(ε+1n+1η(ζ,ε)),1κΓκ(α)ε+η(ζ,ε)ε+nn+1η(ζ,ε)(x(ε+1n+1η(ζ,ε)))ακ1φ(x)dx=Jα,κ(ε+η(ζ,ε))φ(ε+nn+1η(ζ,ε)),

    we get the result.

    Remark 2.1. If we choose η(ζ,ε)=ζε with κ=1, then under the assumption of Lemma 2.1 one has Lemma 2.1 in [19].

    Theorem 2.2. Let φ:Ω=[ε,ε+η(ζ,ε)]R be a differentiable function on Ω. If φL[ε,ε+η(ζ,ε)]and |φ|λ for λ>1 with μ1+λ1=1 is s-preinvex function, then the following inequality holds for fractional integrals with α,κ>0, then the following integral inequality holds:

    |Ψ(ε,ζ;n,α,κ)|η(ζ,ε)2(n+1)(10|2(1θ)ακθακ3|μdθ)1μ×[[(11(n+1)s(s+1)|φ(ε)|λ+1(n+1)s(s+1)|φ(ζ)|λ)]1λ+[1(n+1)s(s+1)|φ(ε)|λ+(21s1(n+1)s(s+1))|φ(ζ)|λ]1λ]. (2.2)

    Proof. From the integral identity given in Lemma 2.1, the H¨older integral inequality and the s-preinvexity of |φ(x)|λ, we have

    |Ψ(ε,ζ;n,α,κ)|η(ζ,ε)2(n+1)[10|(2(1θ)ακθακ3)||φ(ε+1θn+1η(ζ,ε))|dθ+|10(θακ2(1θ)ακ3)||φ(ε+n+θn+1η(ζ,ε))|dθ]η(ζ,ε)2(n+1)[(10|(2(1θ)ακθακ3)|μdθ)1μ×[10((1(1θn+1)s)|φ(ε)|λ+(1θn+1)s|φ(ζ)|λ)dθ]1λ+(10|(θακ2(1θ)ακ3)|μdθ)1μ[10((1(n+θn+1)s)|φ(ε)|λ+(n+θn+1)s|φ(ζ)|λ)dθ]1λ]. (2.3)

    Using the fact that 1χs(1χ)s21sχs for χ[0,1] with s(0,1], we have

    10(1(n+θn+1)s|φ(ε)|λ+(n+θn+1)|φ(ζ)|λ)dθ10[(1θn+1)|φ(ε)|λ+(21s(1θn+1)s)|φ(ζ)|λ]dθ=1(n+1)s(s+1)|φ(ε)|λ+(21s1(n+1)s(s+1))|φ(ζ)|λ (2.4)

    and

    10[(1(1θn+1)s)|φ(ε)|λ+(1θn+1)s|φ(ζ)|λ]dθ=(11(n+1)s(s+1)|φ(ε)|λ+1(n+1)s(s+1)|φ(ζ)|λ). (2.5)

    Remark 2.2. If we choose η(ζ,ε)=ζε with κ=1, then under the assumption of Theorem 2.2, one has Theorem 2.3 in [19].

    Corollary 2.1. If we choose α=κ=1, and n=s=1, then under the assumption of Theoremm 2.2 we have

    |16[φ(ε)+4φ(2ε+η(ζ,ε)2)+φ(ε+η(ζ,ε))]1η(ζ,ε)ε+η(ζ,ε)εφ(x)dx|η(ζ,ε)4[1+2μ+13μ+1(μ+1)]1μ[(34)1λ+(14)1λ][|φ(ε)|+|φ(ζ)|].

    Corollary 2.2. If we choose η(ζ,ε)=ζε with α=κ=1, and n=s=1, then under the assumption of Theoremm 2.2 we have

    |16[φ(ε)+4φ(ε+ζ2)+φ(ζ)]1ζεζεφ(x)dx|ζε4[1+2μ+13μ+1(μ+1)]1μ[(34)1λ+(14)1λ][|φ(ε)|+|φ(ζ)|].

    Proof. The proof of the last inequality is obtained by using the fact that

    ni=1(ωi+νi)jni=1(ωi)j+ni=1(νi)j

    for 0j1,ω1,ω2,ω3,...,ωn0;ν1,ν2,ν3,...,νn0.

    Theorem 2.3. Let φ:Ω=[ε,ε+η(ζ,ε)]R be a differentiable function on Ω. If φL[ε,ε+η(ζ,ε)]and |φ| is s-preinvex function, then the following inequality holds for fractional integrals with α,κ>0, then the following integral inequality holds:

    |Ψ(ε,ζ;n,α,κ)|η(ζ,ε)2(n+1){(Δ1+Δ3)|φ(ε)|+(Δ1+Δ2)|φ(ζ)|}, (2.6)

    where

    Δ1=24(12κα2κα+1)ακ+s+13(n+1)s(ακ+s+1)+13(n+1)s[β(ακ+1,s+1)2β(2κα2κα+1;ακ+1,s+1)],
    Δ2=21s{32(2κα2ακ+1)ακ+14(12κα2ακ+1)ακ+1}3(ακ+1)24(12κα2κα+1)ακ+s+13(n+1)s(ακ+s+1)+13(n+1)s[2β(2κα2ακ+1;ακ+1,s+1)β(ακ+1,s+1)]

    and

    Δ3={32(2κα2ακ+1)ακ+14(12κα2ακ+1)ακ+1}3(ακ+1)24(12κα2κα+1)ακ+s+13(n+1)s(ακ+s+1)+13(n+1)s[2β(2κα2ακ+1;ακ+1,s+1)β(ακ+1,s+1)].

    Proof. From Lemma 2.1 and using the s-preinvexity of |φ(x)|, we have

    |Ψ(ε,ζ;n,α,κ)|η(ζ,ε)2(n+1)[10|2(1θ)ακθακ3|[(1(1θn+1)s)|φ(ε)|+(1θn+1)s|φ(ζ)|]dθ+10|θακ2(1θ)ακ3|[(1(n+θn+1)s)|φ(ε)|+(n+θn+1)s|φ(ζ)|]dθ] (2.7)

    From (2.7), we have

    10|θακ2(1θ)ακ3|[(1(n+θn+1)s)|φ(ε)|+(n+θn+1)s|φ(ζ)|]dθ|φ(ε)|10|θακ2(1θ)ακ3|(1θn+1)sdθ+|φ(ζ)|10|θακ2(1θ)ακ3|(21s1θn+1)sdθ, (2.8)

    By taking into account

    Δ1=10|θακ2(1θ)ακ3|(1θn+1)sdθ=24(12κα2κα+1)ακ+s+13(n+1)s(ακ+s+1)+13(n+1)s[β(ακ+1,s+1)2β(2κα2κα+1;ακ+1,s+1)], (2.9)
    Δ2=10|θακ2(1θ)ακ3|(21s1θn+1)sdθ=21s{32(2κα2ακ+1)ακ+14(12κα2ακ+1)ακ+1}3(ακ+1)24(12κα2κα+1)ακ+s+13(n+1)s(ακ+s+1)+13(n+1)s[2β(2κα2ακ+1;ακ+1,s+1)β(ακ+1,s+1)] (2.10)

    and

    Δ3=10|θακ2(1θ)ακ3|(11θn+1)sdθ={32(2κα2ακ+1)ακ+14(12κα2ακ+1)ακ+1}3(ακ+1)24(12κα2κα+1)ακ+s+13(n+1)s(ακ+s+1)+13(n+1)s[2β(2κα2ακ+1;ακ+1,s+1)β(ακ+1,s+1)], (2.11)

    Theorem 2.4. Let φ:Ω=[ε,ε+η(ζ,ε)]R be a differentiable function on Ω. If φL[ε,ε+η(ζ,ε)]and |φ| is s-preinvex function, then the following inequality holds for fractional integrals with α,κ>0, then the following integral inequality holds:

    |Ψ(ε,ζ;n,α,κ)|η(ζ,ε)2(n+1)(Δ0)11λ{(Δ1|φ(ε)|λ+Δ3|φ(ζ)|λ)1λ+(Δ1|φ(ε)|λ+Δ2|φ(ζ)|λ)1λ}, (2.12)

    where

    Δ0={32(2κα2ακ+1)ακ+14(12κα2ακ+1)ακ+1}3(ακ+1).

    Proof. By using Lemma 2.1 and the H¨older's integral inequality for λ>1 we have

    |Ψ(ε,ζ;n,α,κ)|η(ζ,ε)2(n+1)[(10|2(1θ)ακθακ3|dθ)11λ×[10|2(1θ)ακθακ3||φ(ε+1θn+1η(ζ,ε))|λdθ)1λ+10|θακ2(1θ)ακ3||φ(ε+n+θn+1η(ζ,ε))|λdθ)1λ]

    Using s-preinvexity of |φ|, we get

    |Ψ(ε,ζ;n,α,κ)|η(ζ,ε)2(n+1)[(10|2(1θ)ακθακ3|dθ)11λ×[10|2(1θ)ακθακ3|[((1(1θn+1)s)|φ(ε)|λ+(1θn+1)λ|φ(ζ)|λ)dθ]1λ+10|θακ2(1θ)ακ3|[((1(n+θn+1)s)|φ(ε)|λ+(n+θn+1)λ|φ(ζ)|λ)dθ]1λ,] (2.13)

    using the fact that

    Δ0=10|2(1θ)ακθακ3|dθ={32(2κα2ακ+1)ακ+14(12κα2ακ+1)ακ+1}3(ακ+1). (2.14)

    A combination of (2.9)–(2.11) with (2.14) into (2.13) gives the desired inequality (2.12). The proof is completed.

    Corollary 2.3. If we take n=s=1 and α=κ=1, then under the assumption of Theorem 2.4, we have

    |16(φ(ε)+4φ(2ε+η(ζ,ε)2)+φ(ε+η(ζ,ε)))1η(ζ,ε)ε+η(ζ,ε)εφ(x)dx|η(ζ,ε)4(518)11λ[(61324)1λ+(29324)1λ][|φ(ε)|+|φ(ζ)|].

    Corollary 2.4. If we take η(ζ,ε)=ζε with n=s=1 and α=κ=1, then under the assumption of Theorem 2.4, we have

    |16(φ(ε)+4φ(ε+ζ2)+φ(ζ))1ζεζεφ(x)dx|ζε4(518)11λ[(61324)1λ+(29324)1λ][|φ(ε)|+|φ(ζ)|].

    Remark 2.3. If we take η(ζ,ε)=ζε with λ=1 and κ=s=1, then Theorem 2.4 reduces to Theorem 2.2 in [19].

    Remark 2.4. If we take η(ζ,ε)=ζε with λ=1, κ=s=1, and α=n=1 then Theorem 2.4 reduces to Corollary 1 in [18].

    For deriving further results, we deal with the boundedness and the Lipschitzian condition of φ.

    Theorem 2.5. Let φ:Ω=[ε,ε+η(ζ,ε)]R be a differentiable function on Ω. If |φ| is s-preinvex function, there exists constants m<M satisfying that <mφ(x)M< for all x[ε,ε+η(ζ,ε)] and α,κ>0, and n0, then the following integral inequality holds:

    |Ψ(ε,ζ;n,α,κ)|(Mm)η(ζ,ε)2(n+1)Δ0, (2.15)

    where Δ0 given in (2.14).

    Proof. From Lemma 2.1, we have

    Ψ(ε,ζ;n,α,κ)|=η(ζ,ε)2(n+1)[10(2(1θ)ακθακ3)(φ(ε+1θn+1η(ζ,ε))m+M2)dθ+10(θακ2(1θ)ακ3)(φ(ε+n+θn+1η(ζ,ε))m+M2)dθ] (2.16)

    Using the fact that mm+M2φ(ε+1θn+1η(ζ,ε))m+M2Mm+M2, one has

    |φ(ε+1θn+1η(ζ,ε))m+M2|Mm2,

    similarly,

    |φ(ε+n+θn+1η(ζ,ε))m+M2|Mm2,

    Inequality (2.16) implies that

    |Ψ(ε,ζ;n,α,κ)|(Mm)η(ζ,ε)2(n+1)10|2(1θ)ακθακ3|dθ=(Mm)η(ζ,ε)2(n+1)Δ0.

    The proof is completed.

    Corollary 2.5. If we take η(ζ,ε)=ζε with α=κ=1, then under the assumption of Theorem 2.5, we have

    |Ψ(ε,ζ;n,α,κ)|7(Mm)(ζε)36(n+1).

    Theorem 2.6. Let φ:Ω=[ε,ε+η(ζ,ε)]R be a differentiable function on Ω. If φ satisfies Lipchitz conditions on [ε,ε+η(ζ,ε)] for certain L>0, with α>0,κ>0,n0, then the following inequality holds:

    |Ψ(ε,ζ;n,α,κ)|L(η(ζ,ε))22(n+1)2[(n1)Δ04(2ακ2ακ+1)ακ+223(ακ)+2+83β(2ακ2ακ+1;ακ+1,2)43β(ακ+1,2)], (2.17)

    where Δ0 given in (2.14).

    Proof. From Lemma 2.1, we have

    Ψ(ε,ζ;n,α,κ)|=η(ζ,ε)2(n+1)[10(2(1θ)ακθακ3)[φ(ε+1θn+1η(ζ,ε))φ(ε+n+θn+1η(ζ,ε))]dθ.

    Since φ is Lipschitz condition on [ε,ε+η(ζ,ε)], for certain L>0, we have

    |φ(ε+1θn+1η(ζ,ε))φ(ε+n+θn+1η(ζ,ε))|Lη(ζ,ε)(n1+2tn+1)

    Thus

    |Ψ(ε,ζ;n,α,κ)|η(ζ,ε)2(n+1)[10|2(1θ)ακθακ3||φ(ε+1θn+1η(ζ,ε))φ(ε+n+θn+1η(ζ,ε)|]dθL(η(ζ,ε))22(n+1)10|2(1θ)ακθακ3|(n1+2tn+1)dθ=L(η(ζ,ε))22(n+1)2[(n1)Δ04(2ακ2ακ+1)ακ+223(ακ)+2+83β(2ακ2ακ+1;ακ+1,2)43β(ακ+1,2)].

    The proof is completed.

    Corollary 2.6. If we take η(ζ,ε)=ζε with α=κ=1, then under the assumption of Theorem 2.6, we have

    |Ψ(ε,ζ;n,α,κ)|L(ζε)22(n+1)2[7n1118+22135+83β(23;2,2)].

    Let the set ψ and the δ-finite measure ϱ be given, and let the set of all probability densities on ϱ to be defined on Λ:={z|z:ψR,z(x)>0,ψz(x)dϱ(x)=1}.

    Let F:(0,)R be given mapping and consider DF(z,w) be defined by

    DF(z,w):=ψz(x)F(z(x)w(x))dϱ(x),z,wΛ. (3.1)

    If F is convex, then (3.1) is called as the Csiszar F-divergence.

    Consider the following Hermite-Hadamard divergence

    DFHH(z,w):=ψw(x)z(x)1F(t)dtw(x)z(x)1dϱ(x),w,zΛ, (3.2)

    where F is convex on (0,) with F(1)=0. Note that DFHH(z,w)0 with the equality holds if and only if w=z.

    Proposition 3.1. Suppose all assumptions of Corollary 2.2 hold for (0,) and F(1)=0, if w,zΛ, then the following inequality holds:

    |16[DF(w,z)+4ψz(x)F(z(x)+w(x)2w(x)dϱ(x))]DFHH(w,z)|(1+2λ+1)1λ4(3λ+1(1+λ))1λ[(34)1λ+(14)1λ]×[|F(1)|ψ|z(x)w(x)|dϱ(x)+ψ|z(x)w(x)||Fz(x)w(x)|dϱ(x)]. (3.3)

    Proof. Let Θ1={xψ:z(x)=w(x)}. Θ2={xψ:z(x)<w(x)} and Θ3={xψ:z(x)>w(x)} and Obviously, if xΘ1, then equality holds in (3.3).

    Now if xΘ2, then using Corollary 2.2 for ε=z(x)w(x) and ζ=1, multiplying both sides of the obtained results by w(x) and then integrating over Θ2, we obtain

    |16[4ψ2w(x)F(w(x)+z(x)2w(x))dϱ(x)+ψ2w(x)F(z(x)w(x))dϱ(x)]ψ2w(x)z(x)w(x)1F(t)dtz(x)w(x)1dϱ(x)|(1+2λ+1)1λ4(3λ+1(1+λ))1λ[(34)1λ+(14)1λ]×[|F(1)|ψ2|w(x)z(x)|dϱ(x)+ψ2|w(x)z(x)||Fz(x)w(x)|dϱ(x)]. (3.4)

    Similarly if xΘ3, then using Corollary 2.2 for ε=1 and ζ=z(x)w(x), multiplying both sides of the obtained results by w(x) and then integrating over Θ3, we obtain

    |16[4ψ3w(x)F(w(x)+z(x)2w(x))dϱ(x)+ψ3w(x)F(z(x)w(x))dϱ(x)]ψ3w(x)z(x)w(x)1F(t)dtz(x)w(x)1dϱ(x)|(1+2λ+1)1λ4(3λ+1(1+λ))1λ[(34)1λ+(14)1λ]×[|F(1)|ψ3|z(x)w(x)|dϱ(x)+ψ3|z(x)w(x)||Fz(x)w(x)|dϱ(x)]. (3.5)

    Adding inequalities (3.4) and (3.5) and then utilizing triangular inequality, we get the result.

    Proposition 3.2. Suppose all assumptions of Corollary 2.4 holds for (0,) and f(1)=0, if w,zΛ, then the following inequality holds:

    |16[DF(w,z)+4ψz(x)F(z(x)+w(x)2w(x)dϱ(x))]DFHH(w,z)|(518)1λ[(61324)1λ+(29324)1λ]×[|F(1)|ψ|z(x)w(x)|4dϱ(x)+ψ|z(x)w(x)|4|Fz(x)w(x)|dϱ(x)]. (3.6)

    Proof. The proof is similar as one has done for Corollary 2.2.

    Let G:[ε,ε+η(ζ,ε)][0,1] be the probability density function of a continuous random variable Y with the cumulative distribution function of G

    F(y)=P(Yy)=xεG(t)dt.

    As we know E(y)=ε+η(ζ,ε)εtdF(t)=(ε+η(ζ,ε))ε+η(ζ,ε)εF(t)

    Proposition 3.3. By Corollary 2.1, we get the inequality

    |16[4P(Y2ε+η(ζ,ε)2)+1]1η(ζ,ε)(ε+η(ζ,ε)E(y))|η(ζ,ε)4[1+2μ+13μ+1(μ+1)]1μ[(34)1λ+(14)1λ][|G(ε)|+|G(ζ)|].

    Proposition 3.4. By Corollary 2.3, we get the inequality

    |16[4P(Y2ε+η(ζ,ε)2)+1]1η(ζ,ε)(ε+η(ζ,ε)E(y))|η(ζ,ε)4(518)11λ[(61324)1λ+(29324)1λ][|G(ε)|+|G(ζ)|].

    Same applications can be found for Corollary 2.2 and Corollary 2.4, respectively. We leave it to the interested readers.

    In this paper, we have established several Simpson's type inequalities via κ-fractional integrals in terms of preinvex functions. We also have obtained the inequalities applied to F-divergence measures and application for probability density functions. These results can be viewed as refinement and significant improvements of the previously known for [18,19] and preinvex functions. Applications can be provided in terms of the obtained results to special means. The ideas and techniques of this paper may be attracted to interested readers.

    The authors declare that there is no conflicts of interest in this paper.



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