# Some New Inequalities Using Nonintegral Notion of Variables

Authors: Abha Singh; Abdul Hamid Ganie; Mashael M. Albaidani
Journal: Advances in Mathematical Physics
 (2021)

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## Abstract

The object of this paper is to present an extension of the classical Hadamard fractional integral. We will establish some new results of generalized fractional inequalities.


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## Body

## 1. Introduction

It is important to note that the integral inequalities play a basic role in statistics, mathematics, sciences, and technology (SMST). As in [1–8], it has proven to be of great importance from the past few decades. The formation of fractional calculus has straight impact on the theory utilizing the solution of various spaces in SMST and to prove its efficacy, various statements and applications of fractional derivatives have been constructed. Authors Riemann–Liouville and Grunwald–Letnikov are well known in this filed. Caputo reformulated the classical statement of the Riemann–Liouville fractional derivative for finding solutions of fractional differential equations using initial conditions. The notion of fractional calculus given by Leibniz was studied by Grunwald–Letnikov in a different structure [9–12].Recently, in [13–17] and [18], the development between probability theory and fractional calculus was given, and the results of the classical approach were extended. Also, analysis and observations on this direction and several purposes have been found in the concrete problems which include applied mathematics and fluid mechanics as in [2, 18–26], and many others.As in [27–32], for a function gv∈L1α,β, the Hadamard fractional integral of order κ≥0 is given as follows:
(1)HIακgv=1Γκ∫αvlnvuκ−1guduu,0<α<v≤β,which differs from Riemann-Liouville and Caputo’s definition in the sense that the kernel of integral (1) contains a logarithmic function of an arbitrary exponent.We need the following definition while determining some application, and it is called the Beta function.As in [22], the Beta function, symbolized by βl,m, is given as
(2)βl,m=∫01τl−m1−τm−1dτ=ΓlΓmΓl+m.The basic notion of generalization of special functions using a kind of new parameter fascinated many researchers and mathematicians. More details about fractional integrals can be found in [33–36] and others as cited in the text. Accordingly, our main scenario in this paper is to extend the idea of a new fractional integration with parameter κ≥0 that generalizes Hadamard fractional integrals.


## 2. Main Results

In this section, we shall be dealing with the new generalized type of results to random variables of a continuous type of fractional integral orderκ≥0.Definition 1.
For a functiongv∈L1α,β, the generalized fractional integral of the Hadamard type with order κ∈ℝ+ is given by
(3)HIα,v−κ,λgv=1λΓκ∫αvlnvuκ/λ−1guduu,0<α<v≤β,where Γκ=∫0∞e−uuκ−1du represents the Gamma function as can be seen in [37, 38] and many more.Definition 2.
For a r.v.Z having positive p.d.f. g:α,β⟶ℝ+ (α>0), we define the fractional expectation function of order κ≥0 as
(4)EZ,κ,λv=HIα,v−κ,λvgv=1λΓκ∫αvlnvuκ/λ−1uguduu,where α<v<β.Definition 3.
For a r.v.Z having a positive p.d.f. g:α,β⟶ℝ+ (α>0), the fractional expectation function of Z−EZ of order κ≥0 is given as
(5)EZ−EZ,κ,λv=1λΓκ∫αvlnvuκ/λ−1u−EZguduu,where α<v<β.Definition 4.
For a r.v.Z having a positive p.d.f. g:α,β⟶ℝ+ (α>0), the fractional expectation function of order κ≥0 is given as
(6)EZ,κ,λv=HI1,v−κ,λvgv=1λΓκ∫αβlnvuκ/λ−1u−EZguduu.Definition 5.
IfEZ symbolizes the expected value of the r.v. Z having a positive p.d.f. g:α,β⟶ℝ+ with α>0, then the fractional variance function having order κ of Z is given by
(7)σZ,κ,λ2=HI1,v−κ,λv−EZ2gv=1λΓκ∫αvlnvuκ/λ−1u−EZ2guduu,where α<v<β.Definition 6.
IfEZ symbolizes the expected value of the r.v. Z having a positive p.d.f. g:α,β⟶ℝ+ with α>0, then the fractional variance function having order κ of Z is given by
(8)σZ,κ,λ2=1λΓκ∫αβlnvuκ/λ−1u−EZ2guduu.
Now by choosing different values ofκ and λ, we have the following remarks.Remark 7.
(R1) Choosingλ=1 and κ=1, the classical expectation of r.v. Z will be deduced.
(R2) Choosingλ=1 and κ=1, the classical variance of r.v. Z will be deduced.
(R3) Choosingλ=1, we reach to the definition of [31].Theorem 8.
Let the continuous r.v. beZ with p.d.f. g:α,β⟶ℝ+. Then,
(9)σZ,κ,λ2=EZ2,κ,λ−2EZEZ,κ,λ+EZH2I1,v−κ,λgβ.for all κ≥0.Proof.
By definition, we have(10)σZ,κ,λ2=1λΓκ∫αβlnvuκ/λ−1u−EZ2guduu=1λΓκ∫αβlnvuκ/λ−1u2−2uEZ+EZ2guduu=1λΓκ∫αβlnvuκ/λ−1u2guduu−2EZ1λΓκ∫αβlnvuκ/λ−1uguduu+EZ21λΓκ∫αβlnvuκ/λ−1guduu=EZ2,κ,λ−2EZEZ,κ,λ+EZH2I1,v−κ,λgβ.Theorem 9.
Define a r.v.Z having a p.d.f. g:α,β⟶ℝ+. Then, we have the following inequalities:
(i)
HI1,v−κ,λgvσZ,κ,λ2−EZ−EZ,κν2≤g∞22lnv/uκ/λ/Γκ+1HI1,v−κ,λv2−2HI1,v−κ,λv2,holds if g∈L∞α,β and for all α<ν≤β, κ≥0, and λ≥0.
(ii)
HI1,v−κ,λgvσZ,κ,λ2−EZ−EZ,κν2≤1/2v−α2HI1,v−κ,λgv,holds.Proof.
For the proof of the result, we begin by choosing the functionH for x,y∈α,v,α<v≤β as follows:
(11)Hv,y=H1x−H1yH2x−H2y=H1xH2x−H1xH2y−H1yH2v+H1yH2y,where κ≥0.Now on both sides of (11), we multiply by lnv/xκ/λ−1/xλΓκpx, where the p is a function p:α,β⟶ℝ+, and then integrating the resulting identity from α to v, we see
(12)1λΓκ∫αvlnvxκ/λ−1pxHx,ydxx=HI1,v−κ,λpH1H2v−HyHI1,v−κ,λpH1v‐H1yHI1,v−κ,λpH2v+H1yH2yHI1,v−κ,λpv.Now multiplying (12) by lnv/yκ/λ−1/yλΓκpy for y∈α,v, and then integrating the resulting identity over α,v with respect to y, we see
(13)1λ2Γ2κ∫αv∫αvlnvxκ/λ−1lnvyκ/λ−1pypxHx,ydxxdyy=2HI1,v−κ,λpvHI1,v−κ,λpH1H2v−2HI1,v−κ,λpH2vHI1,v−κ,λpH1v.Puttingpv=gv and H1v=H2v=v−EZ,v∈α,β in (13), we see
(14)1λ2Γ2κ∫αv∫αvlnvxκ/λ−1lnvyκ/λ−1gygxx−y2dxxdyy=2HI1,v−κ,λgvHI1,v−κ,λgvv−EX2−2HI1,v−κ,λgvv−EX2.But we can also write that as(15)1λ2Γ2κ∫αv∫αvlnvxκ/λ−1lnvyκ/λ−1gygxx−y2dxxdyy≤g∞21λ2Γ2κ∫αv∫αvlnvxκ/λ−1lnvyκ/λ−1x−y2dxxdyy≤g∞22lnv/ακ/λ−1Γκ+1HI1,v−κ,λv2−2HI1,v−κ,λv2.Consequently, part (i) follows from (14) and (15).To prove (ii), we can write(16)1λ2Γ2κ∫αv∫αvlnvxκ/λ−1lnvyκ/λ−1gygxx−y2dxxdyy≤supx,y∈α,vx−y2HI1,v−κ,λgv2=x−y2HI1,v−κ,λgv2.Now using (14) and (16), the part (ii) of the result follows. This completes the proof.Theorem 10.
Let the continuous r.v. beZ with p.d.f. g:α,β⟶ℝ+. Then
(i)
the inequality(17)HI1,v−κ,λgvσZ,ω,λ2+HI1,v−ω,λgvσZ,κ,λ2−EZ−EZ,κ,λv×EZ−EZ,κ,ωv≤g∞2lnv/ακ/λΓκ+1HI1,v−ω,λv2+lnv/αω/λΓκ+1HI1,v−κ,λv2−HI1,v−κ,λvHI1,v−ω,λv,holds for α<v≤β,ω≥0, and f∈L∞α,β and for all κ≥0 and
(ii)
the inequality(18)HI1,v−κ,λgvσZ,ω,λ2+HI1,v−ω,λgvσZ,κ,λ2−EZ−EZ,κ,λv×EZ−EZ,κ,ωv≤v−α2HI1,v−κ,λgvHI1,v−ω,λgv,holds for α<v≤β and ω≥0.Proof.
To prove (i), we multiply (12) by lnv/yω/λ−1/xλΓωpy for both sides and get
(19)1λ2ΓκΓω∫αv∫αvlnvxκ/λ−1lnvyκ/λ−1pypxHx,ydxxdyy=HI1,v−κ,λpvHI1,v−ω,λpvH1vH2v+HI1,v−ω,λpvHI1,v−κ,λpvH1vH2v−HI1,v−ω,λpvH2vHI1,v−λ,λpvH1v−HI1,v−κ,λpvH2vHI1,v−ω,λpvH1v.Puttingpv=gv, H1v=H2v=v−EZ, and v∈α,β in (19), we see
(20)1λ2ΓκΓω∫αv∫αvlnvxκ/λ−1lnvyω/λ−1gygxx−y2dxxdyy=HI1,v−κ,λgvHI1,v−ω,λgvv−EX2+HI1,v−ω,λgv×HI1,v−κ,λgvv−EX2−2HI1,v−κ,λgvv−EXHI1,v−ω,λgvv−EX.But we can also write(21)1λ2ΓκΓω∫αv∫αvlnvxκ/λ−1lnvyω/λ−1gygxx−y2dxxdyy≤g∞21λ2Γ2κ∫αv∫αvlnvxκ/λ−1lnvyκ/λ−1x−y2dxxdyy≤g∞2lnv/ακ/λ−1Γκ+1HI1,v−κ,λv2+lnv/αω/λΓω+1HI1,v−ω,λv2−2HI1,v−κ,λvHI1,v−ω,λv.Consequently, part (i) of the result follows from (20) and (21).To prove part (ii), we use the truth thatsupv,y∈α,vx−y2=x−α2 and get
(22)1λ2ΓκΓω∫αv∫αvlnvxκ/λ−1lnvyω/λ−1gygxx−y2dxxdyy≤x−α2HI1,v−κ,λgvHI1,v−ω,λgv.Consequently, part (ii) of the result follows by employing (20) and (21).


## 3. Applications and Examples

Application 11.
Consider the positive functionsg and h on α,β such that for every u>α, 0<HI1,v−κ,λgm,HI1,v−κ,λhm<∞ with
(23)0<K≤guhu≤L<∞,u∈α,β,where m≥1; then, for every κ>0, we have
(24)HI1,v−κ,λgmu2/m+HI1,v−κ,λhmu2/m≥K+1L+1L−2HI1,v−κ,λgmv1/mHI1,v−κ,λhmv1/m.Solution 12.
From (23), we see
(25)1L≤hugu⇒1L+1m≤hugu+1m⇒L+1mgmu≤Lmh+gmu.In a similar way, we see(26)K≤guhu⇒K+1mgmu≤h+gmu.Consequently, multiplying these equations bylnv/uκ/λ−1/uλΓκ for u∈α,v and then integrating the resulting identity over α,v with respect to u yield
(27)HI1,v−κ,λgmu1/m≤LL+1g+hmu1/m,(28)HI1,v−κ,λhmu1m≤1K+1g+hmu1m.Now on multiplying (27) and (28), we see
(29)K+1L+1LHI1,v−κ,λgmu1/mHI1,v−κ,λhmu1/m≤HI1,v−κ,λg+hmu2/m.Consequently, the result follows by using Minkowski’s integral inequality on the right-hand side of (29).Example 1.
Consider the functiongu=lnu/αη/λ−1, we see
(30)HIα,u−κ,λlnuαη/λ−1=1λΓκ∫αvlnuwκ/λ−1lnwαη/λ−1dww.
Choosingτ=lnu/w/lnu/α, for τ∈α,β with η,κ>0, we see with the help of (2) that
(31)HIα,u−κ,λlnuακ/λ−1=lnuακ+η/λ−11λΓκ∫011−τη/λ−11−τκ/λ−1dτ=lnvwκ/λ−1βκ/λ,η/λλΓκ.



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