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

Authors: Yu-Ming Chu; Miao-Kun Wang; Ye-Fang Qiu
Journal: Abstract and Applied Analysis
 (2011)

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## 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.


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## Body

## 1. Introduction

Forr∈[0,1], Lengedre's complete elliptic integrals of the first and second kind [1] are defined by
(1.1)K=K(r)=∫0π/2(1-r2sin⁡2θ)-1/2dθ,K′=K′(r)=K(r′),K(0)=π2,K(1)=∞,E=E(r)=∫0π/2(1-r2sin⁡2θ)1/2dθ,E′=E′(r)=E(r′),E(0)=π2,E(1)=1,
respectively. Here and in what follows, we set r′=1-r2. These integrals are special cases of Guassian hypergeometric function (1.2)F2(a,b;c;x)=F1(a,b;c;x)=∑n=0∞(a,n)(b,n)(c,n)xnn!(-1<x<1),
where (a,n)=∏k=0n-1(a+k). Indeed, we have (1.3)K(r)=π2F(12,12;1;r2),E(r)=π2F(-12,12;1;r2).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 (1.4)π2(arth⁡⁡rr)1/2<K(r)<π2(arth⁡⁡rr),
for r∈(0,1).In [16], Alzer and Qiu proved that the double inequality (1.5)π2(arth⁡⁡rr)α<K(r)<π2(arth⁡⁡rr)β,
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(1.6)π2(arth⁡⁡rr)3/4+α*r<K(r)<π2(arth⁡⁡rr)3/4+β*r,
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.For0<r<1, the following derivative formulas were presented in [4, Appendix  E, pages 474-475]: (2.1)dKdr=E-r′2Krr′2,dEdr=E-Kr,d(E-r′2K)dr=rK,d(K-E)dr=rEr′2.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
(2.2)f(x)-f(a)g(x)-g(a),f(x)-f(b)g(x)-g(b).
If f′(x)/g′(x) is strictly monotone, then the monotonicity in the conclusion is also strict.The following Lemma2.2 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), leth1(r)=r-r′2arthr and h2(r)=r3. Then f1(r)=h1(r)/h2(r), h1(0)=h2(0)=0 and
(2.3)h1′(r)h2′(r)=23arth⁡⁡rr.
It is well known that the functionr↦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), clearlyf2(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
(2.4)h3′(r)h4′(r)=r-r′2arth⁡⁡r2r2r′2arth⁡⁡r=12r-r′2arth⁡⁡rr3rr′2arth⁡⁡r.
It follows from Lemma2.1, 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 Lemma2.2(4), we clearly see that f3(1-)=+∞. Let h5(r)=ℰarthr-r′2𝒦arth(r)/4-3r𝒦(r)/4,
h6(r)=r5,
h7(r)=(ℰ-r′2𝒦)/(4r′2)-r𝒦arth(r)/2+3arth(r)(ℰ-r′2𝒦)/(4r), and
h8(r)=r4, then
f3(r)=h5(r)/h6(r),
h5(0)=h6(0)=h7(0)=h8(0)=0,(2.5)h5′(r)h6′(r)=15h7(r)h8(r),h7′(r)h8′(r)=14r′4r-r′2arth⁡rr3[34E(r)-r′2K(r)r2-14E(r)].
From Lemma2.2(2) and part (1), we clearly see that h7′(r)/h8′(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), leth9(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
(2.6)h9(22)=1.013⋯>0,(2.7)f4(22)=0.084⋯>0,(2.8)f4′(r)=(K-E)+rK(r)4(1+r)+rKarth⁡r4h9(r)>rKarth⁡r4h9(22)>0
for r∈(2/2,1).
Therefore, part (4) follows from (2.7) and (2.8).
For part (5), simple computations lead to(2.9)lim⁡r→0+f5(r)=0,(2.10)f5′(r)=2rlog⁡(arth⁡rr)+(34+r2)r-r′2arth⁡rrr′2arth⁡r-E-r′2Krr′2K.
Making use of parts (1)–(4), one has(2.11)r′2Karth⁡rr4f5′(r)=2r′2Karth⁡rrf2(r)+Kf1(r)-f3(r)>Kf1(r)-f3(r)>π3-f3(14)=1.040⋯>0
for r∈(0,1/4).
Therefore, part (5) follows from (2.9) and (2.11).Lemma 2.4.
Let(2.12)gc(r)=(34+cr)log⁡[arth⁡(r)r]-log⁡(2Kπ)(c∈R),
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 thatgc(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
(2.13)lim⁡r→0+g1/4(r)=0,(2.14)g1/4′(r)=14log⁡(arth⁡rr)+(34+14r)r-r′2arth⁡rrr′2arth⁡r-E-r′2Krr′2K.
We divide the proof into two cases.
Case 1 (r∈(0,2/2]).
Then, making use of Lemma2.3(1)–(3) and (2.14), we have
(2.15)r′2Karth⁡rr3g1/4′(r)=r′2Karth⁡r4rf2(r)+14K(r)f1(r)-rf3(r)>14K(r)f1(r)-rf3(r)>π12-22f3(22)=0.250⋯>0.
Case 2 (r∈(2/2,1)).
Then, making use of Lemma2.3(4) and (2.14), we get
(2.16)g1/4′(r)log⁡[arth⁡(r)/r]=14+f4(r)rr′2Karth⁡rlog⁡[arth⁡(r)/r]>0.
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 parameters1/4 and 0 are the best possible parameters in Lemma 2.4(1) and (2), respectively.
Ifc∈(0,1/4), then gc(c)=f5(c)>0 follows from Lemma 2.3(5). Moreover, let
(2.17)F(r)=gc(r)log⁡[arth⁡(r)/r]=34+cr-log⁡(2K/π)log⁡[arth⁡(r)/r],
then, using l'Hôpital's rule and Lemma 2.2(4), we get
(2.18)lim⁡r→1+F(r)=c-14<0.
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 Lemma2.4, 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(2.19)π2(arth⁡rr)3/4+α*r<K(r)<π2(arth⁡rr)3/4+β*r
holds for all r∈(0,1) with the best possible constants α*=0 and β*=1/4.



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*Source: 697547-2011-10-19.xml*