# Functional Inequalities for Generalized Complete Elliptic Integrals with Two Parameters

Authors: Xiangkai Dou; Li Yin; Xiu-Li Lin
Journal: Journal of Function Spaces
 (2019)

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## Abstract

In this paper, we establish some functional inequalities for generalized complete elliptic integrals with two parameters, such as estimation of bounds and mean inequalities. Our main results give(p,q)-analogues to the early results for classical complete elliptic integrals.


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## Body

## 1. Introduction

For the given complex numbersa,b,c with c≠0,-1,-2,⋯, theGaussian hypergeometric function is defined by (1)Fa,b;c;z=2F1a,b;c;z=∑n⩾0a,nb,nc,nznn!,z<1.Here (a,0)=1 for (a≠0), and (a,n) for n∈N is the shifted factorial or Appell symbol (a,n)=a(a+1)⋯(a+n-1). The following is the integral representation formula of the Gaussian hypergeometric function:(2)Fa,b;c;z=ΓcΓbc-b∫01tb-11-tc-b-11-zt-adtRe(c)>Re(a)>0,|arg⁡(1-z)|<π; see [1]. For more information on the history, background, properties, and applications, please refer to [1, 2] and related references. Here, we give the following definitions of some classical functions. For real number x,y>0, the Euler gamma function Γ and its logarithmic derivative ψ, the so-called digamma function, are defined by (cf. [1, 3])(3)Γx=∫0∞tx-1e-tdt,Bx,y=ΓxΓyΓx+y,ψx=Γ′xΓx.respectively.Recently, Takeuchi [4] studied the (p,q)-trigonometric functions depending on two parameters. For p=q, these functions reduce to the so-called p-trigonometric functions introduced by Lindqvist [5]. The following (p,q)-eigenvalue problem with Dirichlét boundary condition was considered by Drábek and Manásevich [3]. Let ϕp(x)=|x|p-2x. For T,λ>0 and p,q>1(4)ϕpu′′+λϕqu=0,t∈0,T,u0=uT=0,They found the complete solution to this problem. The solution to this problem also appears in [4, Thm 2.1]. In particular, for T=πp,q the function ut=sinp,q⁡(t) is a solution to this problem with λ=p/q(p-1), where(5)πp,q=∫011-tq-1/pdt=2qB1-1p,1q.For p=q, πp,q reduces to πp; see [6]. In order to give the definition of the function sinp,q, first of all we define its inverse function arcsinp,q and then the function itself. For x∈[0,1], set(6)Fp,qx=arcsinp,q⁡x=∫0x1-tq-1/pdt.The function Fp,q:[0,1]→[0,πp,q/2] is an increasing homeomorphism, and(7)sinp,q=Fp,q-1is defined on the interval [0,πp,q/2]. The function sinp,q can be extended to [0,πp,q] by(8)sinp,q⁡x=sinp,q⁡πp,q-x,x∈πp,q2,πp,q.By oddness, the further extension can be made to [-πp,q,πp,q]. Finally, the function sinp,q is extended to whole R by 2πp,q-periodicity; see [7]. Similarly, the other generalized inverse trigonometric and hyperbolic functions that appeared in the current paper are defined as follows:(9)arctanp⁡x=∫0x1+tp-1dtarcsin⁡hpx=∫0x1+tp-1/pdtarctan⁡hpx=∫0x1-tp-1dt.For the expression of the above generalized inverse trigonometric and hyperbolic functions in terms of hypergeometric functions, we have the following formula [8, 9]:(10)arcsinp⁡x=xF1p,1p;1+1p;xparctan⁡hpx=xF1,1p;1+1p;xparcsinp,q⁡x=xF1p,1q;1+1q;xq.Based on the above definitions, we simply introduce generalized elliptic integrals defined by Takeuchi [10] with two parameters. For all p,q∈(1,∞) and r∈(0,1), the complete (p,q)-elliptic integrals of the first and second kinds [11] are defined by(11)Kp,qr≔∫0πp,q/21-rqsinp,qq⁡t1/p-1dt,Kp,q′=Kp,q′r=Kp,qr′and(12)Ep,qr≔∫0πp,q/21-rqsinp,qq⁡t1/pdt,Ep,q′=Ep,q′r=Ep,qr′,respectively. Here, p,q>1,r∈(0,1) and r′=(1-rq)1/q. The complete (p,q)-elliptic integrals can be expressed in terms of hypergeometric functions as follows:(13)Kp,qr=πp,q2F1q,1-1p;1-1p+1q;rp,and(14)Ep,qr=πp,q2F1q,-1p;1-1p+1q;rp.If p=q=2, we can obtain the classically complete elliptic integrals as follows: for r∈(0,1),(15)Kr=∫0π/21-r2sin2⁡t-1/2dt=π2F12,12;1;r2,Er=∫0π/21-r2sin2⁡t1/2dt=π2F-12,12;1;r2.If p=q, we can obtain the generalized complete elliptic integrals with single parameters(16)Kpr=∫0πp/21-rpsinp⁡t-1/pdt=πp2F1p,1-1p;1;rp,Epr=∫0πp/21-rpsinp⁡t1/pdt=πp2F-1p,1p;1;rp.It is worth noting that Bhayo and Yin also gave new (p,q)-complete elliptic integrals in [12]. For more on this topic, the readers can see related references [12–25]. Some recent results regarding the case of semielliptical crack in a cylindrical rod for viscoelastic medium can be found in [26, 27].


## 2. Lower and Upper Bounds for Generalized Complete Elliptic Integrals

In 1992, Anderson et al. [28] discovered that the complete elliptic integral of the first kind can be approximated by the inverse hyperbolic tangent function:(17)arctan⁡hr=rF12,1;32;r2.For r∈(0,1), they proved(18)π2arctan⁡hrr1/2<Kr<π2arctan⁡hrr.Later, Qi and Huang [29] obtained the following inequality by using Chebyshev inequality:(19)π2arcsin⁡rr<Kr<π2arctan⁡hrr.In 2004, Alzer and Qiu [30] proved that, for r∈(0,1), we have(20)π2arctan⁡hrrα<Kr<π2arctan⁡hrrβwith the best possible constants α=3/4 and β=1. Here, we shall show some new inequalities for the generalized complete elliptic integrals.Lemma 1 (see [31, Lemma 1]).
Consider the power seriesf(x)=∑n≥0anxn and g(x)=∑n≥0bnxn, where an∈R and bn>0 for all n∈N\{0}, and suppose that both converge on (-r,r),r>0. If the sequence an/bnn≥0 is increasing (decreasing), then the function x↦f(x)/g(x) is increasing (decreasing) on (0,r).Theorem 2.
Letp,q>1,r∈(0,1) and let Kp,q,Ep,q be as (11) and (12). Then we have the following:
(1) The functionrEp,q(r)/arcsinp,q⁡(r) is strictly decreasing from (0,1) onto (21)πp,q2Γ1+1/p+1/qΓ1+1/qΓ1-1/p2,πp,q2
(2) The functionrKp,q(r)/arctan⁡hq(r) is strictly decreasing from (0,1) onto 1,πp,q/2
(3) The functionrEp,q(r)/arctan⁡hq(r) is strictly decreasing from (0,1) onto 0,πp,q/2
(4) The functionEp,q(r)/2/1+1-xq1/2 is strictly decreasing from (0,1) onto (22)Γ1-1/p+1/q2Γ1+1/qΓ1-1/p,1Proof.
The proofs of assertion (1)-(4) are similar to each other. Here we mainly prove (1) for the sake of simplicity. First of all, we consider the functionαp,q:(0,1)→(0,∞) defined by(23)αp,qr=πp,q2F-1/p,1/q;1-1/p+1/q;rqF1/p,1/q;1+1/q;rq=πp,q2∑n=0∞-1/pn1/qn/1+1/q-1/pnrqn/n!∑n=0∞1/pn1/qn/1+1/qnrqn/n!.Considering Lemma 1, we only need to discuss the monotonicity of the sequence ωnn≥0 defined by (24)ωn=-1/pn1+1/qn1/pn1+1/q-1/pn.Since(25)ωn+1ωn=-1/p+n1+1/q+n1/p+n1-1/p+1/q+n≤1⇔2-1p1p+2pq+np>0,we get that the sequence ωnn≥0 is strictly decreasing. On the other hand, making use of Gauss formula (26)Fa,b;c;1=ΓcΓc-a-bΓc-aΓc-b,we may conclude that (27)αp,q0+=πp,q2,αp,q1-=πp,q2Γ1+1/p+1/qΓ1+1/qΓ1-1/p2.The proof of assertion (1) is complete. Due to (4), we only note the known identity (28)21+1-xq1/2=F14,34;32,xq.Similar method can complete the proof. Here we omit the details.As a direct result of Theorem2, we have the following Corollary 3.Corollary 3.
Forp,q>1 and r∈(0,1), we have
(1)πp,q/2Γ1+1/p+1/q/Γ1+1/qΓ1-1/p2arcsinp,q⁡(r)/r<Ep,q(r)<πp,q/2arcsinp,q⁡(r)/r
(2)arctan⁡hq(r)/r<Kp,q(r)<πp,q/2arctan⁡hq(r)/r
(3)Ep,q(r)<πp,q/2arctan⁡hq(r)/r
(4)Γ1-1/p+1/q/Γ1+1/qΓ1-1/p1/1+1-rq<Ep,q(r)<2/1+1-rq
where the constants in the above inequalities are the best possible.Theorem 4.
Letp≥2,q>1, and r∈(0,1). The function rKp,q(r)/arcsinp,q⁡(r) is strictly increasing from (0,1) onto πp,q/2,∞. As a result, we have the following inequality:(29)Kp,qr>πp,q2arcsinp,q⁡rr.Proof.
Define(30)βp,qr=πp,q2F1-1/p,1/q;1-1/p+1/q;rpF1/p,1/q;1+1/q;rp=πp,q2∑n=0∞1-1/pn1/qn/1+1/q-1/pnrn/n!∑n=0∞1/pn1/qn/1+1/qnrn/n!.And Ωnn≥0 is defined by (31)Ωn=1-1/pn1+1/qn1/pn1+1/q-1/pn.Simple computation yields(32)Ωn+1Ωn=1-1/p+n1+1/q+n1/p+n1-1/p+1/q+n≥1⇔1-1pn+1+1q1-2p+1p2>0.This implies that the sequence Ωnn≥0 is strictly increasing. Using Lemma 1, we obtain that the function βp,q is strictly increasing on (0,1). The limiting value follows easily. This completes the proof.Remark 5.
Corollary3 and Theorem 4 give (p,q)-analogues to the right of (18), (20), and inequality (19). Considering the left of (20), we naturally pose the following open question: for p,q≥2 and r∈(0,1), the following inequality holds true:(33)Kp,qr>πp,q2arctan⁡hqrrp+q-1/pq.Theorem 6.
Letp,q>1 and r∈(0,1). Then(34)1-2log⁡1-rqqπp,q<Kp,qr<1-p-1log⁡1-rqpq+p-q.Proof.
Using the known fact [32], for a,b>0, the function 1-F(a,b;a+b;x)/log⁡(1-x) is strictly increasing from (0,1) onto ab/a+b,1/B(a,b). We easily complete the proof by taking a=1-1/p,b=1/q,x=rq.Theorem 7.
Let1<p≤4/3,q>1 and r∈(0,1). Then(35)πp,q2<Kp,qr<πp,q21-rq1/4.Proof.
We apply the following fact [32]: for a,b>0, the function (1-x)1/4F(a,b;a+b;x) is strictly decreasing on (0,1) if and only if 4ab≤a+b. By taking a=1-1/p,b=1/q,x=rq, we easily verify the condition based on 1<p≤4/3,q>1. Due to limiting value, we have(36)limr→0+⁡1-rq1/4Kp,qr=πp,q2,limr→1-⁡1-rq1/4Kp,qr=0,by using the formula F(a,b;a+b;x)~-1/Ba,blog⁡(1-x),x→1. So, the proof is complete.


## 3. Mean Inequalities for Generalized Complete Elliptic Integrals

For two distinct positive real numbersx and y, the arithmetic mean, geometric mean, logarithmic mean, and identric mean are, respectively, defined by(37)Ax,y=x+y2,Gx,y=xy,Lx,y=x-ylog⁡x-log⁡y,x≠y,Hx,y=1A1/x,1/y,Ix,y=1exxyy1/x-y,x≠y.Letf:I→(0,∞) be continuous, where I is a subinterval of (0,∞). Let M and N be the means defined above; then we call that the function f is MN-convex (concave) if(38)fMx,y≤≥Nfx,fyfor all x,y∈I. Recently, generalized convexity/concavity with respect to general mean values has been studied by Anderson et al. in [28]. In [33], Baricz studied that if the function f is differentiable, then it is (a,b)-convex (concave) on I if and only if x1-af′(x)/f(x)1-b is increasing (decreasing). In [34], Bhayo and Vuorinen established all kinds of mean inequalities for the generalized trigonometric functions. In this section, we shall show logarithmic and identic means inequalities for the generalized complete elliptic integrals by using Chebyshev inequality.Lemma 8 (see [29]).
Letf,g:[a,b]→R be integrable functions, both increasing or both decreasing. Furthermore, let p:[a,b]→R be a positive, integrable function. Then(39)∫abpxfxdx·∫abpxgxdx≤∫abpxdx·∫abpxfxgxdx.If one of the functions f or g is nonincreasing and the other is nondecreasing, then the inequality in (39) is reversed.Lemma 9 (see [35]).
Fora≤y<x≤b, then(40)φx+y2≤1x-y∫yxφudu≤φx+φy2if the function φ(x) is convex on [a,b].Theorem 10.
Letp,q>1 and r,s∈(0,1). Then(41)Kp,qLr,s≤LKp,qr,Kp,qs.Proof.
Let us supposer≤s without loss of generality. Define (42)fr=11-rqsinp,q⁡xq1-1/p.Simple computation yields(43)log⁡fr′=1-1pqrq-1sinp,q⁡xq1-rqsinp,q⁡xq,log⁡fr′′=1-1pqsinp,q⁡xqq-1rq-21-rqsinp,q⁡xq+qr2q-2sinp,q⁡xq1-rqsinp,q⁡xq2>0.This implies that the function f(r) is strictly log-convex on (0,1). By using the fact that the integral preserves monotonicity and log-convexity, we get that the function Kp,q(r) is strictly increasing and log-convex on (0,1). The substitution t=Kp,q(u) results in(44)LKp,qr,Kp,qs=Kp,qr-Kp,qslog⁡Kp,qr-log⁡Kp,qs=∫Kp,qrKp,qs1dt∫Kp,qrKp,qs1/tdt=∫rsKp,qu′du∫rsKp,qu′/Kp,qudu.Since the functions Kp,q(u) and Kp,q(u)′/Kp,q(u) are strictly increasing, we may obtain(45)∫rsKp,qudu∫rsKp,qu′Kp,qudu≤∫rs1du∫rsKp,qu′duby taking p(u)=1,f(u)=Kp,q(u),g(u)=Kp,q(u)′/Kp,q(u) in Lemma 8. This is equivalent to (46)∫rsKp,qu′du∫rsKp,qu′/Kp,qudu≥∫rsKp,qudus-r.Noting that log-convexity implies the convexity and Lemma 9, we have(47)∫rsKp,qudus-r≥Kp,q∫rsudus-r=Kp,qAr,s≥Kp,qLr,swhere we apply the known inequality L(r,s)≤A(r,s).Theorem 11.
Letp,q>1 and r,s∈(0,1). Then(48)Kp,qIr,s≤IKp,qr,Kp,qs.Proof.
This proof is similar to Theorem10. We still suppose r≤s. Direct calculation yields(49)ln⁡IKp,qr,Kp,qs=Kp,qrlog⁡Kp,qr-Kp,qslog⁡Kp,qs-1Kp,qr-Kp,qs=∫Kp,qrKp,qsln⁡tdt∫Kp,qrKp,qs1dt=∫rslog⁡Kp,quKp,qu′du∫rsKp,qu′duSince the functions log⁡Kp,q(u) and Kp,q(u)′ are strictly increasing, we have(50)∫rsKp,qu′du∫rslog⁡Kp,qudu≤∫rs1du∫rslog⁡Kp,quKp,qu′duby taking p(u)=1,f(u)=Kp,qu′,g(u)=log⁡Kp,q(u) in Chebyshev inequality of Lemma 8. Applying Lemma 9 and the known inequality I(r,s)≤A(r,s), we easily prove (51)log⁡IKp,qr,Kp,qs≥∫rslog⁡Kp,qudus-r≥log⁡Kp,q∫rsudus-r=log⁡Kp,qAr,s≥log⁡Kp,qIr,s.The proof is complete.Remark 12.
Our method may be an effective way to deal with Baricz’s conjecture: ifm1,m2 are two-variable means, i.e., for i=1,2 and for all x,y,α>0, we have(52)mix,y=miy,x,mix,x=x,miαx,αy=αmix,yand x<mi(x,y)<y whenever x<y; then find conditions on a1,a2>0 and c>0 for which the inequality(53)m1Fa1r,Fa2r≤≥Fm2a1,a2rholds true for all r∈(0,1), where Fa(r)=F(a,c-a;c;r).


## 4. Grünbaum Type Inequalities for Generalized Complete Elliptic Integrals

Lemma 13 (see [9, Lemma 3, p246]).
Let us consider the functionf:(a,∞)→R, where a≥0. If the function g, defined by g(x)=1/x(f(x)-1), is increasing on (a,∞), then, for the function h, defined by h(x)=f(x2), we have the following Grünbaum type inequality:(54)1+hz≥hx+hy,where x,y≥a and z2=x2+y2. If the function g is decreasing, then inequality (54) is reversed.Lemma 14.
As a function ofr∈(0,1), the function Kp,qr+clog⁡r is increasing if and only if c≥0. In particular, the function Kp,q(r) is increasing on r∈(0,1).Proof.
Simple calculation yields(55)ddrKp,qr+clog⁡r=Ep,qr-r′qKp,qrrr′q+cr≥0⇔inf⁡Ep,qr-r′qKp,qrr′q=0≥-c⇔c≥0.This completes the proof.Theorem 15.
Letx,y,c>0 and x2+y2=z2. Then(56)z2Kp,qz2+cx2log⁡zx2+cy2log⁡zy2≥x2Kp,qx2+y2Kp,qy2.Proof.
By definingf(x)=xKp,qx+clog⁡x+1 and applying Lemmas 13 and 14, the proof can easily be completed.Remark 16.
SinceKp,q(r) is increasing on (0,1) and Ep,q(r) is decreasing on (0,1), we also obtain the following Grünbaum type inequalities by applying Lemma 13:(57)z2Kp,qz2≥x2Kp,qx2+y2Kp,qy2,z2Ep,qz2≤x2Ep,qx2+y2Ep,qy2.


## 5. Conclusion

We mainly established some functional inequalities for generalized complete elliptic integrals with two parameters. First of all, our results give(p,q)-analogues to the early results for classical complete elliptic integrals. Moreover, we show logarithmic and identic means inequalities by using Chebyshev inequality. Furthermore, we show a Grünbaum type inequality.



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