Abstract and Applied Analysis

Abstract and Applied Analysis / 2011 / Article

Research Article | Open Access

Volume 2011 |Article ID 697547 | https://doi.org/10.1155/2011/697547

Yu-Ming Chu, Miao-Kun Wang, Ye-Fang Qiu, "On Alzer and Qiu's Conjecture for Complete Elliptic Integral and Inverse Hyperbolic Tangent Function", Abstract and Applied Analysis, vol. 2011, Article ID 697547, 7 pages, 2011. https://doi.org/10.1155/2011/697547

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

Academic Editor: Dirk Aeyels
Received23 May 2011
Accepted16 Aug 2011
Published19 Oct 2011

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=𝒦(𝑟)=0𝜋/2(1−𝑟2sin2𝜃)−1/2𝒦𝑑𝜃,=ğ’¦î…žî€·ğ‘Ÿ(𝑟)=ğ’¦î…žî€¸,𝜋𝒦(0)=2,𝒦(1)=∞,ℰ=ℰ(𝑟)=0𝜋/2(1−𝑟2sin2𝜃)1/2ℰ𝑑𝜃,=â„°î…žî€·ğ‘Ÿ(𝑟)=ℰ,𝜋ℰ(0)=2,ℰ(1)=1,(1.1) respectively. Here and in what follows, we set ğ‘Ÿî…ž=√1−𝑟2. These integrals are special cases of Guassian hypergeometric function 𝐹2(ğ‘Ž,𝑏;𝑐;𝑥)=𝐹1(ğ‘Ž,𝑏;𝑐;𝑥)=âˆžî“ğ‘›=0(ğ‘Ž,𝑛)(𝑏,𝑛)𝑥(𝑐,𝑛)𝑛(𝑛!−1<𝑥<1),(1.2) where ∏(ğ‘Ž,𝑛)=𝑛−1𝑘=0(ğ‘Ž+𝑘). Indeed, we have 𝜋𝒦(𝑟)=2𝐹12,12;1;𝑟2𝜋,ℰ(𝑟)=2𝐹−12,12;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arth𝑟𝑟1/2𝜋<𝒦(𝑟)<2arth𝑟𝑟,(1.4) for 𝑟∈(0,1).

In [16], Alzer and Qiu proved that the double inequality 𝜋2arth𝑟𝑟𝛼𝜋<𝒦(𝑟)<2arth𝑟𝑟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arth𝑟𝑟3/4+𝛼∗𝑟𝜋<𝒦(𝑟)<2arth𝑟𝑟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]: 𝑑𝒦=ğ‘‘ğ‘Ÿâ„°âˆ’ğ‘Ÿî…ž2ğ’¦ğ‘Ÿğ‘Ÿî…ž2,𝑑ℰ=𝑑𝑟ℰ−𝒦𝑟,ğ‘‘î€·â„°âˆ’ğ‘Ÿî…ž2𝒦𝑑𝑟=𝑟𝒦,𝑑(𝒦−ℰ)=ğ‘‘ğ‘Ÿğ‘Ÿâ„°ğ‘Ÿî…ž2.(2.1)

Lemma 2.1 (see [4, Theorem  1.25]). For −∞<ğ‘Ž<𝑏<∞, let 𝑓,𝑔∶[ğ‘Ž,𝑏]→ℝ be continuous on [ğ‘Ž,𝑏] and be differentiable on (ğ‘Ž,𝑏), let ğ‘”î…ž(𝑥)≠0 be on (ğ‘Ž,𝑏). If ğ‘“î…ž(𝑥)/ğ‘”î…ž(𝑥) is increasing (decreasing) on (ğ‘Ž,𝑏), then so are 𝑓(𝑥)−𝑓(ğ‘Ž),𝑔(𝑥)−𝑔(ğ‘Ž)𝑓(𝑥)−𝑓(𝑏)𝑔(𝑥)−𝑔(𝑏).(2.2) If ğ‘“î…ž(𝑥)/ğ‘”î…ž(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)  [(ğ‘Ÿî…ž)𝑐arth𝑟]/𝑟 is strictly decreasing in (0,1) if and only if 𝑐≥2/3;
(2)  (â„°âˆ’ğ‘Ÿî…ž2𝒦)/𝑟2 is strictly increasing from (0,1) onto (𝜋/4,1);
(3)  (â„°âˆ’ğ‘Ÿî…ž2𝒦)/(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)=[ğ‘Ÿâˆ’ğ‘Ÿî…ž2arth𝑟]/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)=[ℰarthğ‘Ÿâˆ’ğ‘Ÿî…ž2𝒦arth(𝑟)/4−3𝑟𝒦/4]/𝑟5 is strictly increasing from (0,1) onto (𝜋/480,∞);
(4)  𝑓4(𝑟)=(3/4+𝑟/4)(ğ‘Ÿâˆ’ğ‘Ÿî…ž2arth𝑟)𝒦−(â„°âˆ’ğ‘Ÿî…ž2𝒦)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)=ğ‘Ÿâˆ’ğ‘Ÿî…ž2arth𝑟 and ℎ2(𝑟)=𝑟3. Then 𝑓1(𝑟)=ℎ1(𝑟)/ℎ2(𝑟), ℎ1(0)=ℎ2(0)=0 and ℎ1(𝑟)ℎ2(=2𝑟)3arth𝑟𝑟.(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)ğ‘Ÿâˆ’ğ‘Ÿî…ž2arth𝑟2𝑟2ğ‘Ÿî…ž2=1arth𝑟2ğ‘Ÿâˆ’ğ‘Ÿî…ž2arth𝑟𝑟3ğ‘Ÿğ‘Ÿî…ž2.arth𝑟(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 ℎ5(𝑟)=ℰarthğ‘Ÿâˆ’ğ‘Ÿî…ž2𝒦arth(𝑟)/4−3𝑟𝒦(𝑟)/4,ℎ6(𝑟)=𝑟5,ℎ7(𝑟)=(â„°âˆ’ğ‘Ÿî…ž2𝒦)/(4ğ‘Ÿî…ž2)−𝑟𝒦arth(𝑟)/2+3arth(𝑟)(â„°âˆ’ğ‘Ÿî…ž2𝒦)/(4𝑟),andℎ8(𝑟)=𝑟4,then𝑓3(𝑟)=ℎ5(𝑟)/ℎ6(𝑟),ℎ5(0)=ℎ6(0)=ℎ7(0)=ℎ8(0)=0,ℎ5(𝑟)ℎ6=1(𝑟)5ℎ7(𝑟)ℎ8,ℎ(𝑟)7(𝑟)ℎ8=1(𝑟)4ğ‘Ÿî…ž4ğ‘Ÿâˆ’ğ‘Ÿî…ž2arth𝑟𝑟334ℰ(𝑟)âˆ’ğ‘Ÿî…ž2𝒦(𝑟)𝑟2−14.ℰ(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(â„°âˆ’ğ‘Ÿî…ž2𝒦)/(𝑟2𝒦). Then, Lemma 2.2(3) leads to the conclusion that ℎ9(𝑟) is strictly increasing in (0,1). Note that ℎ9√22𝑓=1.013⋯>0,(2.6)4√22𝑓=0.084⋯>0,(2.7)4(𝑟)=(𝒦−ℰ)+𝑟𝒦(𝑟)+4(1+𝑟)𝑟𝒦arth𝑟4ℎ9(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 lim𝑟→0+𝑓5𝑓(𝑟)=0,(2.9)5(𝑟)=2𝑟logarth𝑟𝑟+34+𝑟2î‚ğ‘Ÿâˆ’ğ‘Ÿî…ž2arthğ‘Ÿğ‘Ÿğ‘Ÿî…ž2−arthğ‘Ÿâ„°âˆ’ğ‘Ÿî…ž2ğ’¦ğ‘Ÿğ‘Ÿî…ž2𝒦.(2.10)
Making use of parts (1)–(4), one has ğ‘Ÿî…ž2𝒦arth𝑟𝑟4ğ‘“î…ž5(𝑟)=2ğ‘Ÿî…ž2𝒦arth𝑟𝑟𝑓2(𝑟)+𝒦𝑓1(𝑟)−𝑓3(𝑟)>𝐾𝑓1(𝑟)−𝑓3𝜋(𝑟)>3−𝑓314=1.040⋯>0(2.11) for 𝑟∈(0,1/4).
Therefore, part (5) follows from (2.9) and (2.11).

Lemma 2.4. Let 𝑔𝑐3(𝑟)=4+𝑐𝑟logarth(𝑟)𝑟−log2𝒦𝜋(𝑐∈ℝ),(2.12) then the following statements are true:
(1)  𝑔𝑐(𝑟)>0 for all 𝑟∈(0,1) if and only if 𝑐∈[1/4,∞);
(2)  𝑔𝑐(𝑟)<0 for all 𝑟∈(0,1) if and only if 𝑐∈(−∞,0].

Proof. Firstly, we prove that 𝑔𝑐(𝑟)>0 for 𝑐∈[1/4,∞). Since 𝑔𝑐(𝑟) is continuous and strictly increasing with respect to 𝑐∈ℝ for fixed 𝑟∈(0,1), it suffices to prove that 𝑔1/4(𝑟)>0 for all 𝑟∈(0,1). Note that lim𝑟→0+𝑔1/4𝑔(𝑟)=0,(2.13)1/41(𝑟)=4logarth𝑟𝑟+34+14ğ‘Ÿî‚ğ‘Ÿâˆ’ğ‘Ÿî…ž2arthğ‘Ÿğ‘Ÿğ‘Ÿî…ž2−arthğ‘Ÿâ„°âˆ’ğ‘Ÿî…ž2ğ’¦ğ‘Ÿğ‘Ÿî…ž2𝒦.(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 ğ‘Ÿî…ž2𝒦arth𝑟𝑟3ğ‘”î…ž1/4𝑟(𝑟)=2𝒦arth𝑟𝑓4𝑟21(𝑟)+4𝒦(𝑟)𝑓1(𝑟)−𝑟𝑓3>1(𝑟)4𝒦(𝑟)𝑓1(𝑟)−𝑟𝑓3𝜋(𝑟)>−√1222𝑓3√22=0.250⋯>0.(2.15)
Case 2 (√𝑟∈(2/2,1)). Then, making use of Lemma 2.3(4) and (2.14), we get ğ‘”î…ž1/4(𝑟)[]=1logarth(𝑟)/𝑟4+𝑓4(𝑟)ğ‘Ÿğ‘Ÿî…ž2[]𝒦arth𝑟logarth(𝑟)/𝑟>0.(2.16)
Inequalities (2.15) and (2.16) imply that 𝑔1/4(𝑟) is strictly increasing in (0,1). Therefore, 𝑔1/4(𝑟)>0 follows from (2.13) and the monotonicity of 𝑔1/4(𝑟).
On the other hand, inequality (1.5) leads to the conclusion that 𝑔𝑐(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 𝑔𝑐(𝑐)=𝑓5(𝑐)>0 follows from Lemma 2.3(5). Moreover, let 𝑔𝐹(𝑟)=𝑐(𝑟)[]=3logarth(𝑟)/𝑟4+𝑐𝑟−log(2𝒦/𝜋)[],logarth(𝑟)/𝑟(2.17) then, using l'Hôpital's rule and Lemma 2.2(4), we get lim𝑟→1+1𝐹(𝑟)=𝑐−4<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, 𝑔𝑐(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arth𝑟𝑟3/4+𝛼∗𝑟𝜋<𝒦(𝑟)<2arth𝑟𝑟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.

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