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An Inequality for Generalized Complete Elliptic Integral Li Yin1*, Li-Guo Huang1,Yong-Liwang2 and Xiu-Li Lin3

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An Inequality for Generalized Complete Elliptic Integral Li Yin1*, Li-Guo Huang1,Yong-Liwang2 and Xiu-Li Lin3 Yin et al. Journal of Inequalities and Applications (2017)2017:303 DOI 10.1186/s13660-017-1578-6 R E S E A R C H Open Access An inequality for generalized complete elliptic integral Li Yin1*, Li-Guo Huang1,Yong-LiWang2 and Xiu-Li Lin3 *Correspondence: [email protected] Abstract 1Department of Mathematics, Binzhou University, Binzhou City, In this paper, we show an elegant inequality involving the ratio of generalized Shandong Province 256603, China complete elliptic integrals of the first kind and generalize an interesting result of Alzer. Full list of author information is available at the end of the article MSC: 33E05 Keywords: generalized complete elliptic integrals; psi function; hypergeometric function; inequality 1 Introduction The generalized complete elliptic integral of the first kind is defined for r ∈ (0, 1) by πp 2 dθ 1 dt Kp(r)= 1 = 1 1 , 0 p p 1– p 0 p p p p 1– p (1 – r sinp θ) (1 – t ) (1 – r t ) where sinp θ is the generalized trigonometric function and 1 dt 2 1 1 πp =2 1 = B ,1– . 0 (1 – tp) p p p p The function sinp θ and the number πp play important roles in expressing the solu- tions of inhomogeneous eigenvalue problem of p-Laplacian –(|u|p–2u) = λ|u|p–2u with a boundary condition. These functions have some applications in the quasi-conformal the- ory, geometric function theory and the theory of Ramanujan modular equation. Báricz [1] established some Turán type inequalities for a Gauss hypergeometric function and for a generalized complete elliptic integral and showed a sharp bound for the generalized com- plete elliptic integral of the first kind in 2007. In 2012, Bhayo and Vuorinen [2]dealtwith generalized elliptic integrals and generalized modular functions. Several new inequalities are given for these and related functions. For more details on monotonicity, inequalities and convexity and concavity of these functions, the reader may refer to [3–5]and[6] and the references therein. In 1990, Anderson et al. [7] presented the following inequality: K(r) 1 √ > for r ∈ (0, 1). (1.1) K( r) 1+r © The Author(s) 2017. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro- vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Yin et al. Journal of Inequalities and Applications (2017)2017:303 Page 2 of 6 Inspired by this work, Alzer and Richards [6] gave the refinement of (1.1): for all r ∈ (0, 1), the following inequality K√(r) 1 > r (1.2) K( r) 1+ 4 holds true. It is natural how inequality (1.2) is generalized to Kp(r). Our main result reads as follows. Theorem 1.1 For r ∈ (0, 1) and p ∈ [1, 2], we have 1 K (r) 1 < p√ < , (1.3) 1+λpr Kp( r) 1+upr 1 1 where the constants λp = p (1 – p ) and up =0are the best possible. 2 Lemmas 1+ax Lemma 2.1 The function (x)= 1+bx (1 + bx =0)is strictly increasing (decreasing) in (0, ∞) if and only if a – b >0(a – b <0). Proof Simple computation yields d 1+ax a – b = . dx 1+bx (1 + bx)2 The proof is complete. (x) ∞ Lemma 2.2 (Lemma 2.1 in [8]) The psi function ψ(x)= (x) is strictly concave on (0, ) and satisfies the duplication formula 1 1 1 ψ(2x)= ψ(x)+ ψ x + + log 2 (2.1) 2 2 2 for x >0. Lemma 2.3 (Lemma 3 in [9]) For x >0,we have 1 1 ln x – < ψ(x)<ln x – . (2.2) x 2x Lemma 2.4 For x >0and p ∈ [1, 2], we have 1 1 1 1 1 1 ψ x + + ψ x +1– > ψ x + – + ψ x + + . (2.3) p p 2 2p 2 2p Proof Using Lemma 2.3, we only need to prove the following inequality: 1 1 (x + p )(x +1– p ) 1 1 ln 1 1 1 1 + 1 – 1 >0. (x + 2 – 2p )(x + 2 + 2p ) x +1– p x + p Yin et al. Journal of Inequalities and Applications (2017)2017:303 Page 3 of 6 ∈ 1 ≥ 1 For p [1, 2], we easily obtain p 1– p .So,wehave 1 1 1 – 1 >0. x +1– p x – p On the other hand, (x + 1 )(x +1– 1 ) p p ≥ 1 1 1 1 1 (x + 2 – 2p )(x + 2 + 2p ) 1 1 1 1 1 ⇔ 1– ≥ 1– 1+ p p 4 p p 1 2 1 ⇔ 3 –4 +1≤ 0. p p So,wecompletetheproof. Lemma 2.5 We have K (r) lim p√ = 1. (2.4) → p r 1 Kp( r) Proof Applying the asymptotic formula ([10], equality (2)) 1 F(a, b; a + b; x) ∼ – log(1 – x)(x → 1) (2.5) B(a, b) and expression [10] π 1 1 K (r)= p F ,1– ;1,rp , (2.6) p 2 p p where F(a; b; c; z)andB(x, y) denote a classical hypergeometric function and a beta func- tion, respectively, we obtain F(a, b; a + b; rp) lim = 1. (2.7) r→1 F(a, b; a + b; r) 1 1 Putting a = p and b =1– p , we complete the proof. 3 Proof of Theorem 1.1 Define 2 1 1 fp(r)= 1+ 1– r Kp(r) πp p p and √ 2 p gp(r)= Kp r . πp Yin et al. Journal of Inequalities and Applications (2017)2017:303 Page 4 of 6 By applying (2.6), we get ∞ pn fp(r)= (1 + λpr)r n=0 and ∞ 2n gp(r)= (a2n + a2n+1r)r , n=0 1 1 ( p )n(1– p )n ··· where an = (n!)2 and (r)n = r(r +1) (r + n –1). Because of 1 ≤ p ≤ 2, we have ∞ pn fp(r) n=0(1 + λpr)anr ≥ ∞ . pn gp(r) n=0(a2n + a2n+1r)r Let (1 + λpr)an θp,n(r)= . a2n + a2n+1r Simple computation results in (1 + 1 (1 – 1 )r) p p · an θp,n(r)= 1 1 ( p +2n)(1– p +2n) a2n 1+ (2n+1)2 (1 + 1 (1 – 1 )r) ≥ p p · an 1 1 ( p +2n)(1– p +2n) a2n 1+ (2n+1)2 by using Lemma 2.1. (In fact, we easily know 1 1 1 1 1 ( +2n)(1 – +2n) 1– ≤ ≤ p p p p 4 (2n +1)2 1 1 ⇔ 4n2 +4n +1≤ 4 +2n 1– +2n . p p Next, considering 1 ≤ p ≤ 2, we only need to prove 12n2 +4n –1≤ 0. It is obvious.) Setting ( 1 + x)(1 – 1 + x)2(2x +1) Q (x)= p p , p 2 1 1 (x +1)( p +2x)(1 – p +2x) we have Q (x) 1 1 p = ψ + x + ψ 1– + x +4ψ(2x +1)–2ψ(x +1) Qp(x) p p 1 1 –2ψ 2x + –2ψ 2x +1– . 2 p Yin et al. Journal of Inequalities and Applications (2017)2017:303 Page 5 of 6 Using Lemma 2.2,weeasilyget Q (x) 1 1 1 1 p = ψ x + + ψ x +1– +2ψ x + – ψ x + Qp(x) p p 2 2p 1 1 1 1 1 – ψ x + + – ψ x + – – ψ x +1– . 2 2p 2 2p 2p Applying Lemma 2.2 again, we have 1 1 1 1 ψ x + + ψ x +1– ≤ ψ x + . (3.1) 2 2p 2p 2 Hence, we have Q (x) p >0. Qp(x) It follows that the function Qp(x)isincreasinginx ∈ (0, ∞). So, we have a2n 4 ≥ 16 = Qp(n)>Qp(1) = 1 1 , a2n+1 ( p + 1)(2 – p ) 9 where we apply 1 1 1 1 +1+2– 2 3 2 9 +1 2– ≤ p p = = . p p 2 2 4 Hence, we obtain 1+ 1 (1 – 1 ) ≥ 16 p p Qp,n(r) 1 1 9 ( p +2n)(1– p +2n) 1+ (2n+1)2 1+ 1 ≥ 16 4 4n+1 2 9 1+(4n+2 ) 32n2 + 104n +35 =1+ 9(32n2 +24n +5) >1, and fp(r)>gp(r). On the other hand, since the function Kp(r) is strictly increasing on r ∈ (0, 1), we have Kp√(r) p <1. Kp( r) Hence, we rewrite formula (1.3)as √ p Kp( r) K (r) –1 u < u (r)= p < λ. p r Yin et al. Journal of Inequalities and Applications (2017)2017:303 Page 6 of 6 Simple calculation leads to 1 1 ∞ ( p )n(1– p )n rn 1 n=0 (1)n n! up(r)= 1 1 –1 r ∞ ( p )n(1– p )n rpn n=0 (1)n n! and 1 1 lim up(r)= 1– , r→0 p p lim up(r)=0. r→1 The proof is complete. 4Conclusions We show an elegant inequality involving the ratio of generalized complete elliptic integrals of the first kind and generalize an interesting result of Alzer. Acknowledgements The authors are grateful to anonymous referees for their careful corrections to and valuable comments on the original version of this paper. The authors were supported by NSFC 11401041, the Science Foundation of Binzhou University under grant number BZXYL1704, and by the Science and Technology Foundation of Shandong Province J16li52. Competing interests The authors declare that they have no competing interests. Authors’ contributions All authors contributed equally to the manuscript and read and approved the final manuscript.
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