On Alzer and Qiu's Conjecture for Complete Elliptic Integral and Inverse Hyperbolic Tangent Function

Abstract

We prove that the double inequality (𝜋/2)(arth𝑟/𝑟)^{3/4+𝛼∗𝑟}<𝒦(𝑟)<(𝜋/2)(arth𝑟/𝑟)^{3/4+𝛽∗𝑟} holds for all 𝑟∈(0,1) with the best possible constants 𝛼^∗=0 and 𝛽^∗=1/4, which answer to an open problem proposed by Alzer and Qiu. Here, 𝒦(𝑟) is the complete elliptic integrals of the first kind, and arth is the inverse hyperbolic tangent function.

1. Introduction

For 𝑟∈[0,1], Lengedre's complete elliptic integrals of the first and second kind [1] are defined by 𝒦=𝒦𝒦(𝑟)=^0𝜋/2(1−𝑟²sin²𝜃)^−1/2𝑑𝜃,_𝒦(0)==𝒦𝜋^(𝑟)=𝒦𝑟^,ℰ=ℰ(𝑟)=2,_0𝜋/2𝒦(1)=∞,ℰ(1−𝑟²sin²𝜃)^1/2𝑑𝜃,_ℰ(0)==ℰ𝜋^(𝑟)=ℰ𝑟^,2,ℰ(1)=1,(1.1) respectively. Here and in what follows, we set 𝑟^=√1−𝑟². These integrals are special cases of Guassian hypergeometric function 𝐹₂(𝑎,𝑏;𝑐;𝑥)=𝐹₁(𝑎,𝑏;𝑐;𝑥)=^∞_𝑛=0(𝑎,𝑛)(𝑏,𝑛)(𝑐,𝑛)𝑥𝑛^𝑛!(−1<𝑥<1),(1.2) where (𝑎,𝑛)=∏^{𝑛−1𝑘=0}(𝑎+𝑘). Indeed, we have 𝒦(𝑟)=𝜋2𝐹12,12;1;𝑟²,ℰ(𝑟)=𝜋2𝐹−12,21;1;𝑟².(1.3)

It is well known that the complete elliptic integrals have many important applications in physics, engineering, geometric function theory, quasiconformal analysis, theory of mean values, number theory, and other related fields [2–13].

Recently, the complete elliptic integrals have been the subject of intensive research. In particular, many remarkable properties and inequalities can be found in the literature [3, 10–18].

In 1992, Anderson et al. [15] discovered that 𝒦 can be approximated by the inverse hyperbolic tangent function, arth, and proved that 𝜋2art𝑟h𝑟^1/2<𝒦(𝑟)<𝜋2art𝑟h𝑟,(1.4) for 𝑟∈(0,1).

In [16], Alzer and Qiu proved that the double inequality 𝜋2art𝑟h𝑟^𝛼<𝒦(𝑟)<𝜋2art𝑟h𝑟^𝛽,(1.5) holds for all 𝑟∈(0,1) with the best possible constants 𝛼=3/4 and 𝛽=1 and proposed an open problem as follows.

Open Problem #
The double inequality 𝜋2art𝑟h𝑟^{3/4+𝛼∗𝑟}<𝒦(𝑟)<𝜋2art𝑟h𝑟^{3/4+𝛽∗𝑟},(1.6) holds for all 𝑟∈(0,1) with the best possible constants 𝛼^∗=0 and 𝛽^∗=1/4.

It is the aim of this paper to give a positive answer to the open problem #.

2. Lemmas and Theorem

In order to establish our main result, we need several formulas and lemmas, which we present in this section.

For 0<𝑟<1, the following derivative formulas were presented in [4, Appendix  E, pages 474-475]: 𝑑𝒦𝑑𝑟=ℰ−𝑟^2𝒦𝑟𝑟^2,𝑑ℰ𝑑ℰ−𝑟_2𝑑𝑟=ℰ−𝑟𝒦,𝒦𝑑𝑟=𝑟𝒦,𝑑(𝒦𝑑𝑟−ℰ)=𝑟𝑟ℰ^2.(2.1)

Lemma 2.1 (see [4, Theorem  1.25]). For −∞<𝑎<𝑏<∞, let 𝑓,𝑔∶[𝑎,𝑏]→ℝ be continuous on [𝑎,𝑏] and be differentiable on (𝑎,𝑏), let 𝑔^(𝑥)≠0 be on (𝑎,𝑏). If 𝑓^(𝑥)/𝑔^(𝑥) is increasing (decreasing) on (𝑎,𝑏), then so are 𝑓(𝑥)−𝑓(𝑎)𝑔(𝑥)−𝑔(𝑎),𝑓𝑔((𝑥𝑥))−−𝑓𝑔((𝑏𝑏)).(2.2) If 𝑓^(𝑥)/𝑔^(𝑥) is strictly monotone, then the monotonicity in the conclusion is also strict.

The following Lemma 2.2 can be found in [9, Lemma  3(1)] and [4, Theorem  3.21(1) and Exercise 3.43(30) and (46)].

Lemma 2.2. (1)  [(𝑟^)^𝑐arth𝑟]/𝑟 is strictly decreasing in (0,1) if and only if 𝑐≥2/3;
(2)  (ℰ−𝑟^2𝒦)/𝑟² is strictly increasing from (0,1) onto (𝜋/4,1);
(3)  (ℰ−𝑟^2𝒦)/(𝑟²𝒦) is strictly decreasing from (0,1) onto (0,1/2);
(4)  𝑟𝒦/arth𝑟 is strictly decreasing from (0,1) onto (1,𝜋/2).

Lemma 2.3. (1)  𝑓₁(𝑟)=[𝑟−𝑟^2arth𝑟]/𝑟³ is strictly increasing from (0,1) onto (2/3,1);
(2)  𝑓₂(𝑟)=(log[arth(𝑟)/𝑟])/𝑟² is strictly increasing from (0,1) onto (1/3,∞);
(3)  𝑓₃(𝑟)=[ℰarth𝑟−𝑟^2𝒦arth(𝑟)/4−3𝑟𝒦/4]/𝑟⁵ is strictly increasing from (0,1) onto (𝜋/480,∞);
(4)  𝑓₄(𝑟)=(3/4+𝑟/4)(𝑟−𝑟^2arth𝑟)𝒦−(ℰ−𝑟^2𝒦)arth𝑟 is positive and strictly increasing in (√2/2,1);
(5)  𝑓₅(𝑟)=(3/4+𝑟²)log[arth(𝑟)/𝑟]−log(2𝒦/𝜋) is positive and strictly increasing on (0,1/4).

Proof. For part (1), let ℎ₁(𝑟)=𝑟−𝑟^2arth𝑟 and ℎ₂(𝑟)=𝑟³. Then 𝑓₁(𝑟)=ℎ₁(𝑟)/ℎ₂(𝑟), ℎ₁(0)=ℎ₂(0)=0 and ℎ^1(𝑟)ℎ^2(𝑟)=23art𝑟h𝑟.(2.3)
It is well known that the function 𝑟↦arth(𝑟)/𝑟 is strictly increasing from (0,1) onto (1,∞). Therefore, from (2.3) and Lemma 2.1 together with l'Hôpital's rule, we know that 𝑓₁(𝑟) is strictly increasing in (0,1), 𝑓₁(0⁺)=2/3 and 𝑓₁(1⁻)=1.
For part (2), clearly 𝑓₂(1⁻)=+∞. Let ℎ₃(𝑟)=log[arth(𝑟)/𝑟] and ℎ₄(𝑟)=𝑟², then 𝑓₂(𝑟)=ℎ₃(𝑟)/ℎ₄(𝑟), ℎ₃(0)=ℎ₄(0)=0, and ℎ^3(𝑟)ℎ^4(𝑟)=𝑟−𝑟^2arth𝑟2𝑟²𝑟^2arth𝑟=12𝑟−𝑟^2arth𝑟𝑟³𝑟𝑟^2arth𝑟.(2.4)
It follows from Lemma 2.1, Lemma 2.2(1), part (1), (2.4), and l'Hôpital's rule that 𝑓₂(𝑟) is strictly increasing in (0,1) and 𝑓₂(0⁺)=1/3.
For part (3), from Lemma 2.2(4), we clearly see that 𝑓₃(1⁻)=+∞. Let ℎ3₅a(r𝑟t)h(=𝑟)(ℰℰa−rt𝑟h_2𝑟−𝑟^2𝒦arth(𝑟)/4−3𝑟𝒦(𝑟)/4,ℎ₆(𝑟)=𝑟⁵,ℎ₇(𝑟)=(ℰ−𝑟^2𝒦)/(4𝑟^2)−𝑟𝒦arth(𝑟)/2+𝒦)/(4𝑟),andℎ₈(𝑟)=𝑟⁴,then𝑓₃(𝑟)=ℎ₅(𝑟)/ℎ₆(𝑟),ℎ₅(0)=ℎ₆(0)=ℎ₇(0)=ℎ₈(0)=0,ℎ^5(𝑟)ℎ^6(𝑟)=15ℎ⁷(𝑟)ℎℎ₈(𝑟)_7,(𝑟)ℎ^8(𝑟)=4𝑟1^4𝑟−𝑟^2arth𝑟𝑟³43ℰ(𝑟)−𝑟^2𝒦(𝑟)𝑟²−14ℰ(𝑟).(2.5)
From Lemma 2.2(2) and part (1), we clearly see that ℎ^7(𝑟)/ℎ^8(𝑟) is strictly increasing in (0,1). Thus, the monotonicity of 𝑓₃(𝑟) can be obtained from (2.5) and Lemma 2.1. Moreover, making use of l'Hôpital's rule, we have 𝑓₃(0⁺)=𝜋/480.
For part (4), let ℎ₉(𝑟)=2(1+𝑟)−ℰ/(𝑟𝒦)−3(ℰ−𝑟^2𝒦)/(𝑟²𝒦). Then, Lemma 2.2(3) leads to the conclusion that ℎ₉(𝑟) is strictly increasing in (0,1). Note that ℎ₉√2𝑓2=1.013⋯>0,(2.6)₄√2𝑓2=0.084⋯>0,(2.7)_4(𝑟)=(𝒦−ℰ)+𝑟𝒦(𝑟)4(1+𝑟)+𝑟𝒦a4rth𝑟ℎ⁹(𝑟)>𝑟𝒦arth𝑟4ℎ⁹√22>0(2.8) for 𝑟∈(√2/2,1).
Therefore, part (4) follows from (2.7) and (2.8).
For part (5), simple computations lead to 𝑓lim_𝑟→0+𝑓₅(𝑟)=0,(2.9)_5(𝑟)=2𝑟logarth𝑟𝑟+34+𝑟²𝑟−𝑟^2arth𝑟𝑟𝑟^2arth𝑟−ℰ−𝑟^2𝒦𝑟𝑟^2𝒦.(2.10)
Making use of parts (1)–(4), one has 𝑟^2𝒦arth𝑟𝑟⁴𝑓^5(𝑟)=2𝑟^2𝒦arth𝑟𝑟𝑓²>𝐾𝑓₁(𝑟)+𝒦𝑓₁(𝑟)−𝑓₃(𝑟)(𝑟)−𝑓₃(𝑟)>𝜋3−𝑓³41=1.040⋯>0(2.11) for 𝑟∈(0,1/4).
Therefore, part (5) follows from (2.9) and (2.11).

Lemma 2.4. Let 𝑔_𝑐(𝑟)=34+𝑐𝑟logarth𝑟(𝑟)−log2𝜋𝒦(𝑐∈ℝ),(2.12) then the following statements are true:
(1)  𝑔_𝑐(𝑟)>0 for all 𝑟∈(0,1) if and only if 𝑐∈[1/4,∞);
(2)  𝑔_𝑐(𝑟)<0 for all 𝑟∈(0,1) if and only if 𝑐∈(−∞,0].

Proof. Firstly, we prove that 𝑔_𝑐(𝑟)>0 for 𝑐∈[1/4,∞). Since 𝑔_𝑐(𝑟) is continuous and strictly increasing with respect to 𝑐∈ℝ for fixed 𝑟∈(0,1), it suffices to prove that 𝑔_1/4(𝑟)>0 for all 𝑟∈(0,1). Note that 𝑔lim_𝑟→0+𝑔_1/4(𝑟)=0,(2.13)_1/4(𝑟)=14logart𝑟h𝑟+34+14𝑟𝑟−𝑟^2arth𝑟𝑟𝑟^2arth𝑟−ℰ−𝑟^2𝒦𝑟𝑟^2𝒦.(2.14) We divide the proof into two cases.
Case 1 (𝑟∈(0,√2/2]). Then, making use of Lemma 2.3(1)–(3) and (2.14), we have 𝑟^2𝒦arth𝑟𝑟³𝑔^1/4(𝑟)=𝑟^2𝒦arth𝑟4𝑟𝑓²(𝑟)+14𝒦(𝑟)𝑓¹>1(𝑟)−𝑟𝑓₃(𝑟)4𝒦(𝑟)𝑓¹(𝑟)−𝑟𝑓₃(𝑟)>𝜋12−√22𝑓³√2=0.250⋯>0.2(2.15)
Case 2 (𝑟∈(√2/2,1)). Then, making use of Lemma 2.3(4) and (2.14), we get 𝑔^1/4(𝑟)log[arth(𝑟)/𝑟]=41+𝑓⁴(𝑟)𝑟𝑟^2𝒦arth𝑟log[arth(𝑟)/𝑟]>0.(2.16)
Inequalities (2.15) and (2.16) imply that 𝑔_1/4(𝑟) is strictly increasing in (0,1). Therefore, 𝑔_1/4(𝑟)>0 follows from (2.13) and the monotonicity of 𝑔_1/4(𝑟).
On the other hand, inequality (1.5) leads to the conclusion that 𝑔_𝑐(𝑟)<0 for all 𝑟∈(0,1) and 𝑐∈(−∞,0].
Next, we prove that the parameters 1/4 and 0 are the best possible parameters in Lemma 2.4(1) and (2), respectively.
If 𝑐∈(0,1/4), then 𝑔_𝑐(𝑐)=𝑓₅(𝑐)>0 follows from Lemma 2.3(5). Moreover, let 𝐹(𝑟)=𝑔^𝑐(𝑟)log[arth(𝑟)/𝑟]=43+𝑐𝑟−lolgog[a(r2th𝒦(/𝑟𝜋))/𝑟],(2.17) then, using l'Hôpital's rule and Lemma 2.2(4), we get lim_𝑟→1+𝐹(𝑟)=𝑐−14<0.(2.18)
Inequality (2.18) implies that there exists 𝛿=𝛿(𝑐)>0 such that 𝐹(𝑟)<0 for all 𝑟∈(1−𝛿,1). Therefore, 𝑔_𝑐(𝑟)<0 for 𝑟∈(1−𝛿,1) follows from (2.17).

From Lemma 2.4, we clearly see that the following Theorem 2.5 holds, which give a positive answer to the open problem #.

Theorem 2.5. The double inequality 𝜋2art𝑟h𝑟^{3/4+𝛼∗𝑟}<𝒦(𝑟)<𝜋2art𝑟h𝑟^{3/4+𝛽∗𝑟}(2.19) holds for all 𝑟∈(0,1) with the best possible constants 𝛼^∗=0 and 𝛽^∗=1/4.

Acknowledgments

This paper was supported by the Natural Science Foundation of China under Grant 11071069 and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.