# Inequalities of the Type Hermite–Hadamard–Jensen–Mercer for Strong Convexity

Authors: Muhammad Adil Khan; Saeed Anwar; Sadia Khalid; Zaid Mohammed Mohammed Mahdi Sayed
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

By using the Jensen–Mercer inequality for strongly convex functions, we present Hermite–Hadamard–Mercer inequality for strongly convex functions. Furthermore, we also present some new Hermite‐Hadamard‐Mercer-type inequalities for differentiable functions whose derivatives in absolute value are convex.


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## Body

## 1. Introduction

Mathematical inequalities play a vital role in many fields of science. The field of mathematical inequalities and applications has enrolled an exponential improvement in the last two decades with a significant impact in other fields of modern mathematics including engineering [1], mathematical statistics [2], approximation theory [3, 4], information theory [5], and other disciplines [6]. In our work, we use Jensen–Mercer inequality for convex and strongly convex functions and also the well-known Hölder’s inequality which plays a significant role in functional analysis, mathematical analysis, complex analysis, numerical analysis, statistics and probability, and partial differential equations.First, we define convex and strongly convex functions ([7–10]) as follows:Definition 1.
A functionϑ:I⊂ℝ⟶ℝ is called convex if(1)ϑτm+1−τM≤τϑm+1−τϑM,holds for all m, M∈I, and all τ∈0,1.Definition 2.
A functionϑ:I⊂ℝ⟶ℝ is called strongly convex with modulus c>0 if(2)ϑτm+1−τM≤τϑm+1−τϑM−cτ1−τm−M2,holds for all m,M∈I, and τ∈0,1.Hermite–Hadamard inequality for convex ([11]) and strongly convex functions ([12]) can be stated as follows:Theorem 1.
Letϑ:m,M⟶ℝ be a convex function. Then,(3)ϑm+M2≤1M−m∫mMϑτdτ≤ϑm+ϑM2.Theorem 2.
Letϑ:m,M⟶ℝ be a strongly convex function with modulus c. Then,(4)ϑm+M2+c12M−m2≤1M−m∫mMϑτdτ≤ϑm+ϑM2−c6M−m2.For more results related to Hermite–Hadamard inequality, see [11, 13–21].Jensen–Mercer inequality ([22]) for convex functions is as follows.Theorem 3.
Letϑ:m,M⟶ℝ be a convex function. Then,(5)ϑm+M−∑i=1npixi≤ϑm+ϑM−∑i=1npiϑxi,for each xi∈m,M and pi∈0,1i=1,…,n with ∑i=1npi=1.First, Hermite–Hadamard–Mercer inequalities for fractional integrals was presented in [23]. For more recent studies, see [24–27].For strongly convex functions, the Jensen–Mercer inequality is proved in [28] as follows.Theorem 4.
Letϑ:m,M⟶ℝ be a strongly convex function with modulus c. Then,(6)ϑx1+xn−∑i=1npixi≤ϑx1+ϑxn−∑i=1npiϑxi−c2∑i=1npiλi1−λix1−xn2+∑i=1npixi−∑i=1npixi2,where ∑i=1npi=1, λi∈0,1, x1=min1≤i≤nxi, xn=max1≤i≤nxi, and xi∈m,M.Substituten=2 in (6); we obtain Jensen–Mercer inequality for strongly convex functions as follows:(7)ϑm+M−a+b2≤ϑm+ϑM−ϑa+ϑb2−cλ1−λ12+λ2−λ22M−m2−c4a−b2,where p1=p2=1/2, λ1,λ2∈0,1, and a,b∈m,M.For more details and recent results related to Theorems3 and 4, see [28–31].The following definitions for fractional integrals are important as these definitions will be used throughout this study.Definition 3.
Letϑ∈Lm,M. The Riemann–Liouville fractional integrals Jm+α and JM−α of order α>0 with m≥0 ([32, 33]) are defined by(8)Jm+αϑx=1Γα∫mxx−τα−1ϑτdτ,x>m,JM−αϑx=1Γα∫xMτ−xα−1ϑτdτ,x<M,respectively, where Γ is the gamma function.


## 2. Main Results

In this section, we obtain the Hermite–Hadamard–Mercer inequality for strongly convex functions by using the Jensen–Mercer inequality for strongly convex functions. We derive some new inequalities related to the right and left sides of the Hermite–Hadamard–Mercer type inequalities for differentiable functions whose derivatives in the absolute value are convex.Theorem 5.
Letϑ:m,M⟶ℝ be a strongly convex function with modulus c. Then, for all x,z∈m,M with x<z, m≥0 and α>0; we have(9)ϑm+M−x+z2≤ϑm+ϑM−Γα+12z−xαJx+αϑz+Jz−αϑx−cm−M2λ1−λ12+λ2−λ22−cα2−α+24α+1α+2z−x2≤ϑm+ϑM−ϑx+z2−cm−M2λ1−λ12+λ2−λ22−cα2−α+22α+1α+2z−x2.Proof.
By changing the variablesa=τx+1−τz and b=1−τx+τz for x,z∈m,M and τ∈0,1 in (7), we obtain(10)ϑm+M−x+z2≤ϑm+ϑM−ϑxτ+1−τz+ϑ1−τx+τz2−cm−M2λ1−λ12+λ2−λ22−c42τ−12z−x2.
Multiplying both sides of the above inequality byτα−1 and integrating over 0,1, we get(11)1αϑm+M−x+z2≤1αϑm+ϑM−12∫01τα−1ϑτx+1−τzdτ−12∫01τα−1ϑ1−τx+τzdτ−cαm−M2λ1−λ12+λ2−λ22−c4z−x2∫01τα−12τ−12dτ.
By substitutingu=τx+1−τz and v=1−τx+τz, we have(12)1αϑm+M−x+z2≤1αϑm+ϑM−12z−xα∫xzz−uα−1ϑudu+∫xzv−xα−1ϑvdv−cαM−m2λ1−λ12+λ2−λ22−c4z−x2α2−α+2αα+1α+2,equivalent to(13)ϑm+M−x+z2≤ϑm+ϑM−Γα+12z−xαJx+αϑz+Jz−αϑx−cM−m2λ1−λ12+λ2−λ22−cα2−α+24α+1α+2z−x2,which is the left side of (9).
To obtain the right side of (9), use the definition of strongly convex function ϑ, and we have(14)ϑx+z2=ϑτx+1−τz+τz+1−τx2≤ϑτx+1−τz2+ϑ1−τx+τz2−c42τ−12z−x2.
Multiplying both sides of the above inequality byτα−1 and integrating over 0,1, we have(15)1αϑx+z2≤12∫01τα−1ϑτx+1−τzdτ+12∫01τα−1ϑ1−τx+τzdτ−c4z−x2∫01τα−12τ−12dτ.
By substitutingu=τx+1−τz and v=1−τx+τz, we obtain(16)1αϑx+z2≤12z−xα∫xzz−uα−1ϑudu+12z−xα∫xzv−xα−1ϑvdv−c4z−x2α2−α+2αα+1α+2⇒ϑx+z2≤Γα+12z−xαJx+αϑz+Jz−αϑx−cα2−α+24α+1α+2z−x2.
From the above expression, we can write(17)ϑm+ϑM−Γα+12z−xαJx+αϑz+Jz−αϑx−cα2−α+24α+1α+2z−x2−cM−m2λ1−λ12+λ2−λ22≤ϑm+ϑM−ϑx+z2−cα2−α+22α+1α+2z−x2−cm−M2λ1−λ12+λ2−λ22.
Combining (13) and (17), we obtain (9).Theorem 6.
Let all the assumptions of Theorem5 hold. Then,(18)ϑm+M−x+z2≤Γα+12z−xαJm+M−x−αϑm+M−z+Jm+M−z+αϑm+M−x−cα2−α+24α+1α+2z−x2≤ϑm+M−x+ϑm+M−z2−c4z−x2≤ϑm+ϑM−ϑx+ϑz2−c4z−x2.Proof.
Sinceϑ is strongly convex, we can write(19)ϑm+M−a+b2=ϑm+M−a+m+M−b2≤12ϑm+M−a+ϑm+M−b−c4b−a2.
By changing the variablesm+M−a=τm+M−x+1−τm+M−z and m+M−b=1−τm+M−x+τm+M−z, such that τ∈0,1, we obtain(20)ϑm+M−x+z2≤12ϑτm+M−x+1−τm+M−z+12ϑ1−τm+M−x+τm+M−z−c42τ−12z−x2.
Multiplying both sides byτα−1, integrating over 0,1 and in the obtained expression by substituting u=τm+M−x+1−τm+M−z and v=1−τm+M−x+τm+M−z, we get the left side of (18).
Using the strongly convex functionϑ and the Jensen–Mercer inequality, we obtain(21)ϑτm+M−x+1−τm+M−z+ϑ1−τm+M−x+τm+M−z≤ϑm+M−x+ϑm+M−z−2cτ−τ2z−x2≤2ϑm+ϑM−ϑx+ϑz−2cτ−τ2z−x2.
Multiplying both sides of the above inequality byτα−1, integrating over 0,1 and in the obtained expression by substituting u=τm+M−x+1−τm+M−z and v=1−τm+M−x+τm+M−z, we get the right side of (18).Corollary 1.
If we takeα=1 in Theorem 5, then we have(22)ϑM+m−x+z2≤ϑm+ϑM−∫01ϑτx+1−τzdτ−cM−m2λ1−λ12+λ2−λ22−c12z−x2≤ϑm+ϑM−ϑx+z2−cM−m2λ1−λ12+λ2−λ22−c6z−x2.Corollary 2.
If we takeα=1 in Theorem 6, then we have(23)ϑm+M−x+z2≤1z−x∫xzϑm+M−τdτ−c12z−x2≤ϑm+ϑM−ϑx+ϑz2−c4z−x2.Remark 1.
If we takex=m and z=M in (23), we obtain the Hermite–Hadamard inequality for strongly convex function given in [12].Theorem 7.
Let all the assumptions of Theorem5 hold. Then,(24)ϑm+M−x+z2≤2α−1Γα+1z−xαJm+M−x+z/2−αϑm+M−z+Jm+M−x+z/2+αϑm+M−x−c2α+1α+2z−x2≤ϑm+ϑM−ϑx+ϑz2−cλ1−λ12+λ2−λ22m−M2−c4z−x2.Proof.
Sinceϑ is a strongly convex function, we can write(25)ϑm+M−a+b2=ϑm+M−a+m+M−b2≤ϑm+M−a2+ϑm+M−b2−c4a−b2.
Takinga=τ/2x+2−τ/2zandb=2−τ/2x+τ/2zsuch thatx,z∈m,Mandτ∈0,1, we obtain(26)ϑm+M−x+y2≤ϑm+M−τ/2x+2−τ/2z2+ϑm+M−2−τ/2x+τ/2z2−c41−τ2z−x2.
Multiplying both sides of the above inequality byτα−1, integrating over 0,1 and in the obtained expression by substituting u=m+M−τ/2x+2−τ/2z and v=m+M−2−τ/2x+τ/2z, we get the first inequality of (24).
For obtaining the second inequality of (24), use the Jensen–Mercer inequality for strongly convex function, and we get(27)ϑm+M−τ2x+2−τ2z≤ϑm+ϑM−τ2ϑx+2−τ2ϑz−cλ1−λ12+λ2−λ22m−M2−cτ21−τ2z−x2,ϑm+M−2−τ2x+τ2z≤ϑm+ϑM−2−τ2ϑx+τ2ϑz−cλ1−λ12+λ2−λ22m−M2−cτ21−τ2z−x2.
Adding these inequalities, multiplying both sides byτα−1, integrating over 0,1 and in the obtained expression by substituting u=m+M−τ/2x+2−τ/2/2z and v=m+M−2−τ/2x+τ/2z, we get the second inequality of (24).Theorem 8.
Let all the assumptions of Theorem5 hold. Then,(28)ϑm+M−x+z2≤ϑm+ϑM−2α−1Γα+1z−xαJx+z/2−αϑx+Jx+z/2+αϑz−cλ1−λ12+λ2−λ22m−M2−c2α+1α+2z−x2≤ϑm+ϑM−ϑx+z2−cλ1−λ12+λ2−λ22m−M2−cα+1α+2z−x2.Proof.
Using the Jensen–Mercer inequality for strongly convex function, we have(29)ϑm+M−a+b2≤ϑm+ϑM−ϑa+ϑb2−cλ1−λ12+λ2−λ22m−M2−c4a−b2.
By writinga=τ/2x+2−τ/2z and b=2−τ/2x+τ/2z, such that x,z∈m,M and τ∈0,1, we obtain(30)ϑm+M−x+z2≤ϑm+ϑM−ϑτ/2x+2−τ/2z+ϑ2−τ/2x+τ/2z2−cλ1−λ12+λ2−λ22m−M2−c41−τ2z−x2.
Multiplying both sides of the above inequality byτα−1, integrating over 0,1 and in the obtained expression by substituting u=τ/2x+2−τ/2z and v=2−τ/2x+τ/2z, we get the left side of (28).
For obtaining the right side of (28), use the definition of strongly convex function, and we have(31)ϑa+b2≤12ϑa+12ϑb−c4a−b2.
By writinga=τ/2x+2−τ/2z and b=2−τ/2x+τ/2z, such that x,z∈m,M and τ∈0,1, we obtain(32)ϑx+z2≤12ϑτ2x+2−τ2z+ϑ2−τ2x+τ2z−c4z−x21−τ2.
Multiplying both sides of the above inequality byτα−1, integrating over 0,1 and substituting u=τ/2x+2−τ/2z and v=2−τ/2x+τ/2z, we get(33)ϑx+z2≤2α−1Γα+1z−xαJx+z/2+αϑz+Jx+z/2−αϑx−c2α+1α+2z−x2.
From the above expression, we obtain the right side of (28).Lemma 1.
Letϑ:m,M⟶ℝ be a differentiable function, such that ϑ′∈Lm,M. Then, for all x,z∈m,M, such that x<z, α>0 and τ∈0,1, we have(34)ϑm+M−x+ϑm+M−z2−Γα+12z−xαJm+M−z+αϑm+M−x+Jm+M−x−αϑm+M−z−αcα+1α+2z−x2=z−x2∫01τα−1−ταϑ′m+M−τx+1−τz−2αcz−xτα−τα+1dτ.Proof.
We can write(35)I=∫01τα−1−ταϑ′m+M−τx+1−τz−2αcz−xτα−τα+1dτ=∫01ταϑ′m+M−τx+1−τzdτ−∫011−ταϑ′m+M−τx+1−τzdτ−2αcz−x∫01τα−τα+1dτ,that is,(36)I=I1−I2−2αcz−xI3.
Integrating each term, we obtain(37)I1=∫01ταϑ′m+M−τx+1−τzdτ=ταϑm+M−τx+1−τzz−x|01−αz−x∫01τα−1ϑm+M−τx+1−τzdτ=ϑm+M−xz−x−Γα+1z−xα+1Jm+M−x−αϑm+M−z,I2=∫011−ταϑ′m+M−τx+1−τzdτ=1−ταϑm+M−τx+1−τzz−x|01+αz−x∫011−τα−1ϑm+M−τx+1−τzdτ=−ϑm+M−xz−x−Γα+1z−xα+1Jm+M−z+αϑm+M−x,I3=∫01τα−τα+1dτ=1α+1α+2.
Substituting the values ofI1, I2, and I3 in (36) and simplifying, we obtain(38)I=ϑm+M−x+ϑm+M−zz−x−Γα+1z−xα+1Jm+M−z+αϑm+M−x+Jm+M−x−αϑm+M−z−2αcα+1α+2z−x.
Now, multiplying both sides byz−x/2, we get (34).Corollary 3.
If we takeα=1 in Lemma 1, then we obtain(39)ϑm+M−x+ϑm+M−z2−1z−x∫m+M−zm+M−xϑudu−c6z−x2=z−x2∫012τ−1ϑ′m+M−τx+1−τz−2cz−xτ−τ2dτ.Lemma 2.
Let all the assumptions of Lemma1 hold. Then,(40)2α−1Γα+1z−xαJm+M−x+z/2+αϑm+M−x+Jm+M−x+z/2−αϑm+M−z−ϑm+M−x+z2−c2α+1α+2z−x2=z−x4∫01ταϑ′m+M−2−τ2x+τ2z−ϑ′m+M−τ2x+2−τ2z−2cz−x1−τdτ.Proof.
We can write(41)z−x4∫01ταϑ′m+M−2−τ2x+τ2z−ϑ′m+M−τ2x+2−τ2z−2cz−x1−τdτ=z−x4I1−I2−2cz−xI3.
Simplifying each term separately, we obtain(42)I1=∫01ταϑ′m+M−2−τ2x+τ2zdτ.
Using integration by parts and substitution, we get(43)I1=2x−zϑm+M−x+z2+2α+1Γα+1z−xα+1Jm+M−x+z/2+αϑm+M−x.
We have(44)I2=∫01ταϑ′m+M−τ2x+2−τ2zdτ.
Using integration by parts and substitution, we get(45)I2=2z−xϑm+M−x+z2+2α+1Γα+1z−xα+1Jm+M−x+z/2−αϑm+M−z.
Similarly, we have(46)I3=1α+1α+2.
Using (43)–(46) in (41), we obtain (40).Theorem 9.
Letϑ:m,M⟶ℝ be a differentiable function, such that ϑ′ is convex on m,M. Then, for all x,z∈m,M with x<z and α>0, we have(47)|ϑm+M−x+ϑm+M−z2−Γα+12z−xαJm+M−z+αϑm+M−x+Jm+M−x−αϑm+M−z−αcα+1α+2z−x2|≤z−xα+11−12αϑ′M+ϑ′m−ϑ′x+ϑ′z2+2ααcz−x2α−1α+2.Proof.
Using Lemma1, we can write(48)|ϑm+M−x+ϑm+M−z2−Γα+12z−xαJm+M−z+αϑm+M−x+Jm+M−x−αϑm+M−z−αcα+1α+2z−x2|≤z−x2∫01τα−1−ταϑ′m+M−τx+1−τz+−2αcz−xτα−τα+1dτ=z−x2∫01τα−1−ταϑ′m+M−τx+1−τz+2αcz−xτα−τα+1dτ≤z−x2∫01τα−1−ταϑ′m+ϑ′M−tϑ′x+1−τϑ′zdτ+2αcz−x∫01τα−τα+1dτ=z−x2∫01/21−τα−ταϑ′m+ϑ′M−τϑ′x+1−τϑ′zdτ+∫1/21τα−1−ταϑ′m+ϑ′M−τϑ′x+1−τϑ′zdτ+2αcz−x∫01τα−τα+1dτ=z−x2I1+I2+2αcz−xI3.
IntegratingI1,I2, and I3, we have(49)I1=∫01/21−τα−ταϑ′m+ϑ′M−τϑ′x+1−τϑ′zdτ=ϑ′m+ϑ′M∫01/21−τα−ταdτ−ϑ′x∫01/2τ1−ταdτ−∫01/2τα+1dτ+ϑ′z∫01/21−τα+1dτ−∫01/21−τταdτ=ϑ′m+ϑ′M1α+1−1/2αα+1−ϑ′x1α+1α+2−1/2α+1α+1+ϑ′z1α+2−1/2α+1α+1,I2=∫1/21τα−1−ταϑ′m+ϑ′M−τϑ′x+1−τϑ′zdτ=ϑ′m+ϑ′M∫1/21τα−1−ταdτ−ϑ′x∫1/21τα+1dτ−∫1/21τ1−ταdτ+ϑ′z∫1/211−τταdτ−∫1/211−τα+1dτ=ϑ′m+ϑ′M1α+1−1/2αα+1−ϑ′x1α+2−1/2α+1α+1+|ϑ′z|1α+1α+2−1/2α+1α+1,I3=1α+1α+2.
Substituting the values ofI1,I2, and I3 in (48), we have(50)|ϑm+M−x+ϑm+M−z2−Γα+12z−xαJm+M−z+αϑm+M−x+Jm+M−x−αϑm+M−z−αcα+1α+2z−x2|≤z−x22ϑ′m+ϑ′M1α+1−12αα+1−ϑ′x1α+1α+2−12α+1α+1−ϑ′z1α+2−12α+1α+1−ϑ′z1α+1α+2−12α+1α+1+2αcz−xα+1α+2,and we obtain (47).Theorem 10.
Letϑ:m,M⟶ℝ be a differentiable function, such that ϑ′q is convex on m,M, where q>0 and 1/p+1/q=1. Then, for α>0 and for all x,z∈m,M, such that x<z and, we have(51)ϑm+M−x+ϑm+M−z2−Γα+12z−xαJm+M−z+αϑm+M−x+Jm+M−x−αϑm+M−z−αcα+1α+2z−x2≤z−x2Lα,p1/pϑ′mq+ϑ′Mq−ϑ′xq+ϑ′zq21/q+2αcz−xα+1α+2,where Lα,p=∫01τα−1−ταpdτ.Proof.
Using Lemma1 and applying Hölder’s inequality, we have(52)|ϑm+M−x+ϑm+M−z2−Γα+12z−xαJm+M−z+αϑm+M−x+Jm+M−x−αϑm+M−z−αcα+1α+2z−x2|≤z−x2∫01τα−1−ταpdτ1/p∫01ϑ′m+M−τx+1−τzqdτ1/q+2αcz−x∫01τα−τα+1dτ.
Using the convexity ofϑ′q and the Jensen–Mercer inequality, we get(53)|ϑm+M−x+ϑm+M−z2−Γα+12z−xαJm+M−z+αϑm+M−x+Jm+M−x−αϑm+M−z−αcα+1α+2z−x2|≤z−x2Lα,p1/p∫01ϑ′mq+ϑ′Mq−τϑ′xq+1−τϑ′zqdτ1/q+2αcz−xα+1α+2,and (51) is immediate.Theorem 11.
Ifϑ:m,M⟶ℝ is a differentiable function and ϑ′ is convex on m,M, then for all x,z∈m,M with x<z and α>0, we have(54)2α−1Γα+1z−xαJm+M−x+z/2+αϑm+M−x+Jm+M−x+z/2−αϑm+M−z−ϑm+M−x+z2−c2α+1α+2z−x2≤z−x2α+1ϑ′m+ϑ′M−ϑ′x+ϑ′z2+cα+2z−x.Proof.
From Lemma2, we can write(55)2α−1Γα+1z−xαJm+M−x+z/2+αϑm+M−x+Jm+M−x+z/2−αϑm+M−z−ϑm+M−x+z2−c2α+1α+2z−x2≤z−x4∫01ταϑ′m+M−2−τ2x+τ2zdτ+∫01ταϑ′m+M−τ2x+2−τ2zdτ+2cz−x∫01τα1−τdτ.
Using the convexity ofϑ′, we have(56)2α−1Γα+1z−xαJm+M−x+z/2+αϑm+M−x+Jm+M−x+z/2−αϑm+M−z−ϑm+M−x+z2−c2α+1α+2z−x2≤z−x4ϑ′m+ϑ′M∫01ταdτ−∫01τα2−τ2ϑ′x+τ2ϑ′zdτ+ϑ′m+ϑ′M∫01ταdτ−∫01τατ2ϑ′x+2−τ2ϑ′zdτ+2cz−x∫01ταdτ−∫01τα+1dτ=z−x42ϑ′m+ϑ′Mα+1−ϑ′xα+32α+1α+2−ϑ′z2α+2−ϑ′x2α+2−ϑ′zα+32α+1α+2+2cz−xα+1α+2,and we get (54).Theorem 12.
Let all the assumptions of Theorem10 hold. Then,(57)2α−1Γα+1z−xαJm+M−x+z/2+αϑm+M−x+Jm+M−x+z/2−αϑm+M−z−ϑm+M−x+z2−c2α+1α+2z−x2≤z−x41αp+11/pϑ′mq+ϑ′Mq−3ϑ′xq+ϑ′zq41/q+ϑ′mq+ϑ′Mq−ϑ′xq+3ϑ′zq41/q+2cz−xq+11/q.Proof.
By using Lemma2 and Hölder’s inequality, we can write(58)|2α−1Γα+1z−xαJm+M−x+z/2+αϑm+M−x+Jm+M−x+z/2−αϑm+M−z−ϑm+M−x+z2−c2α+1α+2z−x2|≤z−x4∫01ταpdτ1/p∫01ϑ′m+M−2−τ2x+τ2zqdτ1/q+∫01ϑ′m+M−τ2x+2−τ2zqdτ1/q+2cz−x∫011−τqdτ1/q.
Using the convexity ofϑ′q and simplifying, we obtain(59)2α−1Γα+1z−xαJm+M−x+z/2+αϑm+M−x+Jm+M−x+z/2−αϑm+M−z−ϑm+M−x+z2−c2α+1α+2z−x2≤z−x4∫01ταpdτ1/pϑ′mq+ϑ′Mq∫01dτ−∫012−τ2ϑ′xq+τ2ϑ′zqdτ1/q+ϑ′mq+ϑ′Mq∫01dτ−∫01τ2ϑ′xq+2−τ2ϑ′zqdτ1/q+2cz−x∫011−τqdτ1/q,and (57) is immediate.



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*Source: 5386488-2021-11-01.xml*