Precise Asymptotics for Bifurcation Curve of Nonlinear Ordinary Differential Equation

Abstract

We study the following nonlinear eigenvalue problem $-u^{{}^{″}}\left(t\right)=\lambda f\left(u\left(t\right)\right),u\left(t\right)>0,t\in I:=(-1,1),u(\pm 1)=0,$ where $f\left(u\right)=log(1+u)$ and $\lambda >0$ is a parameter. Then $\lambda$ is a continuous function of $\alpha >0$, where $\alpha$ is the maximum norm $\alpha =\parallel u_{\lambda }{\parallel }_{\infty }$ of the solution $u_{\lambda }$ associated with $\lambda$. We establish the precise asymptotic formula for $\lambda =\lambda \left(\alpha \right)$ as $\alpha \to \infty$ up to the third term of $\lambda \left(\alpha \right)$.

1. Introduction

We consider the following nonlinear eigenvalue problem

$$\begin{array}{ccc}\hfill -u^{{}^{″}}\left(t\right)& =& \lambda f\left(u\right(t\left)\right),t\in I:=(-1,1),\hfill \end{array}$$

(1)

$$\begin{array}{ccc}\hfill u\left(t\right)& >& 0,t\in I,\hfill \end{array}$$

(2)

$$\begin{array}{ccc}\hfill u(-1)& =& u\left(1\right)=0,\hfill \end{array}$$

(3)

where $\lambda >0$ is a parameter. In this paper, we mainly consider the case $f\left(u\right)=log(1+u)$. We know from [1] that, if $f\left(u\right)$ is continuous in $u\ge 0$ and positive for $u>0$, then for a given $\alpha >0$, there exists a unique classical solution pair $(\lambda ,u_{\alpha })$ of (1)–(3) satisfying $\alpha =\parallel u_{\alpha }{\parallel }_{\infty }$. Since $\lambda$ is a continuous for $\alpha >0$, we write as $\lambda =\lambda \left(\alpha \right)$ for $\alpha >0$.

Nonlinear eigenvalue and bifurcation problems have been one of the main topics in the study of nonlinear equations, and many authors have investigated the global behavior of bifurcation diagrams intensively. We refer to [2,3,4,5,6,7] and the references therein.

The purpose of this paper is to establish the asymptotic expansion formula for $\lambda \left(\alpha \right)$ when $\alpha \gg 1$ up to the third term by time-map method. Our study is motivated by the following inverse bifurcation problem.

Problem A. Consider (1)–(3). Assume that $f\left(u\right)$ is a given function with bifurcation curve $\lambda \left(\alpha \right)$, and another function $f_1\left(u\right)$ is unknown. Let ${\lambda }_1\left(\alpha \right)$ be the bifurcation curve associated with $f_1\left(u\right)$. Suppose that ${\lambda }_1\left(\alpha \right)=\lambda \left(\alpha \right)(1+{\delta }_0\left(\alpha \right))$ as $\alpha \to \infty$, where ${\delta }_0\left(\alpha \right)\to 0$ as $\alpha \to \infty$. Then can we conclude that $f_1\left(u\right)=f\left(u\right)(1+{\delta }_1\left(u\right))$ as $u\to \infty$? Here ${\delta }_1\left(u\right)$ is a function of u satisfying ${\delta }_1\left(u\right)\to 0$ as $u\to \infty$.

Since there are a few studies of inverse bifurcation problems, even the standard formulation of the inverse problems has not been established yet. Indeed, Problem A is a modified formulation which was proposed in [8]. It is natural to expect that if $\lambda \left(\alpha \right)$ and ${\lambda }_1\left(\alpha \right)$ are closer asymptotically, then $f\left(u\right)$ and $f_1\left(u\right)$ are also closer. Therefore, the first approach to Problem A is to study the “direct problems” precisely. Namely, we investigate the precise asymptotic formula for $\lambda \left(\alpha \right)$ as $\alpha \to \infty$ for $f\left(u\right)=log(1+u)$ here. Then, if we obtain the asymptotic behavior of ${\lambda }_1\left(\alpha \right)$ corresponding to $f_1\left(u\right)=f\left(u\right)+g\left(u\right)$, which was perturbed by some $g\left(u\right)$, then we will obtain the good evidence to propose the improved formulation of Problem A.

We introduce some known results about global behavior of bifurcation curve. One of the most famous results was shown for the one-dimensional Gelfand problem, namely, the Equations (1)–(3) with $f\left(u\right)=e^u$. It was shown in [9] that it has the exact solution

$$\begin{array}{c}\hfill u_{\alpha }\left(t\right)=\alpha +log\left({\mathrm{sech}}^2\left(\frac{\sqrt{2\lambda \left(\alpha \right)}}{2}t{e}^{\alpha /2}\right)\right),\end{array}$$

(4)

where $\mathrm{sech}x=1/coshx$. Then by time-map method, we explicitly obtain that for $\alpha >0$ (cf. [10]),

$$\begin{array}{ccc}\hfill \lambda \left(\alpha \right)& =& \frac{1}{2e^{\alpha }}{\left[log\left(2e^{\alpha }+2\sqrt{e^{\alpha }(e^{\alpha }-1)}-1\right)\right]}^2.\hfill \end{array}$$

(5)

Unfortunately, however, such explicit expression of bifurcation curves as (5) cannot be expected in general. For example, we introduce the following result.

Theorem 1

([11]). Let $f\left(u\right)=u+sinu$ and consider (1)–(3). Then as $\alpha \to \infty$,

$$\begin{array}{c}\hfill \lambda \left(\alpha \right)=\frac{{\pi }^2}{4}-\frac{{\pi }^{3/2}}{\sqrt{2}}{\alpha }^{-3/2}sin\left(\alpha -\frac{\pi }{4}\right)+o\left({\alpha }^{-3/2}\right).\end{array}$$

(6)

The case $f\left(u\right)=u+sinu$ was introduced in [12] first, which was inspired by [13]. We see from Theorem 1 that the oscillatory property of the nonlinear term $f\left(u\right)$ gives explicit effect to its bifurcation curve, as expected. Moreover, we understand the convergent behavior of $\lambda \left(\alpha \right)$ to the line $\lambda ={\pi }^2/4$, which is the “bifurcation curve” of (1)–(3) with $f\left(u\right)=u$.

As far as the author knows, however, the exact solution $u_{\alpha }\left(t\right)$ of the equation in Theorem 1 is not known, although it is quite simple, and it also seems impossible to obtain the explicit formula for $\lambda \left(\alpha \right)$ as (5). From this point of view, one of the standard approaches for the better understanding of the global structure of $\lambda \left(\alpha \right)$ is to establish precise asymptotic expansion formula for $\lambda \left(\alpha \right)$ as $\alpha \to \infty$. In some cases, the asymptotic expansion formulas for $\lambda \left(\alpha \right)$ up to the second term like (6) have been obtained. We refer to [11,14,15,16] and the references therein. However, to obtain the third term of $\lambda \left(\alpha \right)$, we need a very long and complicated calculation in general. We overcome this difficulty and establish the following results.

Theorem 2.

Let $f\left(u\right)=log(1+u)$ and consider (1)–(3). Then as $\alpha \to \infty$,

$$\begin{array}{cc}\sqrt{\lambda \left(\alpha \right)}=\sqrt{\frac{2\alpha }{log(1+\alpha )}}\{1-\hfill & \frac{1}{2}\left(4log2-3\right)\frac{1}{log(1+\alpha )}\hfill \\ & +\frac{3}{8}\left(5-8log2+C_1\right)\frac{1}{{(log(1+\alpha ))}^2}\}+R_3,\hfill \end{array}$$

(7)

where $R_3$ is the remainder term satisfying

$$\begin{array}{c}\hfill b_3{(log(1+\alpha ))}^{-3}\le |R_3|\le b_3^{-1}({(log(1+\alpha ))}^{-3},\end{array}$$

(8)

and $0<b_3<1$ is a constant independent of $\alpha \gg 1$.

By Taylor expansion, it is easy to see that if $0<u\ll 1$, then $log(1+u)=u+o\left(u\right)$. Therefore, (1) is approximated by $-u_{\alpha }^{{}^{″}}\left(t\right)=\lambda \left(\alpha \right)u_{\alpha }\left(t\right)$ when $\alpha =\parallel u_{\alpha }{\parallel }_{\infty }\ll 1$. So $\lambda \left(\alpha \right)$ starts from $(\alpha ,\lambda )=(0,{\pi }^2/4)$.

Remark 1.

From a view point of asymptotic expansion formula for $\lambda \left(\alpha \right)$, it is natural to expect that the following asymptotic formula for $\sqrt{\lambda \left(\alpha \right)}$ holds.

$$\begin{array}{ccc}\hfill \sqrt{\lambda \left(\alpha \right)}& =& \sqrt{\frac{2\alpha }{log(1+\alpha )}}\left\{1+\sum _{n=1}^{\infty }{b}_n{(log(1+\alpha ))}^{-n}\right\}.\hfill \end{array}$$

(9)

Here, $\left\{b_n\right\}$ ($n=1,2,\cdots )$ are expected to be constants, which are determined by induction. However, if the readers look at Section 3 below, then they understand immediately that it seems quite difficult to prove (9), since the calculation are quite long to obtain even the third term of (7).

The proof of Theorem 2 depends on the time-map method and Taylor expansion formula.

2. Second Term of $\mathit{\lambda }\left(\mathit{\alpha }\right)$ in Theorem 2

In this section, let $\alpha \gg 1$. In what follows, we denote by C the various positive constants independent of $\alpha$. It is known that if $(u_{\alpha },\lambda \left(\alpha \right))\in C^2\left(\overline{I}\right)\times {\mathbb{R}}_{+}$ satisfies (1)–(3), then

$$\begin{array}{c}{u}_{\alpha }\left(t\right)=u_{\alpha }(-t),0\le t\le 1,\hfill \end{array}$$

(10)

$$\begin{array}{c}{u}_{\alpha }\left(0\right)=\underset{-1\le t\le 1}{max}{u}_{\alpha }\left(t\right)=\alpha ,\hfill \end{array}$$

(11)

$$\begin{array}{c}{u}_{\alpha }^{\prime }\left(t\right)>0,-1<t<0.\hfill \end{array}$$

(12)

By (1), we have

$$\begin{array}{c}\hfill \left\{u_{\alpha }^{″}\left(t\right)+\lambda \left(log(1+u_{\alpha }\left(t\right)\right)\right\}{u}_{\alpha }^{\prime }\left(t\right)=0.\end{array}$$

By this, (11) and putting $t=0$, we obtain

$$\begin{array}{c}\frac{1}{2}{u}_{\alpha }^{\prime }{\left(t\right)}^2+\lambda \left\{u_{\alpha }\left(t\right)log(1+u_{\alpha }\left(t\right))-u_{\alpha }\left(t\right)+log(1+u_{\alpha }\left(t\right))\right\}=\mathrm{const}.\hfill \\ =\lambda \left\{\alpha log(1+\alpha )-\alpha +log(1+\alpha )\right\}.\hfill \end{array}$$

This along with (12) implies that for $-1\le t\le 0$,

$$\begin{array}{c}\hfill u_{\alpha }^{\prime }\left(t\right)=\sqrt{2\lambda }\sqrt{\alpha log(1+\alpha )-u_{\alpha }\left(t\right)log(1+u_{\alpha }\left(t\right))+\xi \left(u_{\alpha }\left(t\right)\right)},\end{array}$$

(13)

where

$$\begin{array}{c}\hfill \xi \left(u\right):=log(1+\alpha )-log(1+u)-(\alpha -u).\end{array}$$

(14)

By this and putting $u_{\alpha }\left(t\right)=\alpha s^2$, we obtain

$$\begin{array}{ccc}\sqrt{2\lambda }\hfill & =& {\int }_{-1}^0\frac{u_{\alpha }^{\prime }\left(t\right)}{\sqrt{\alpha log(1+\alpha )-u_{\alpha }\left(t\right)log(1+u_{\alpha }\left(t\right))+\xi \left(u_{\alpha }\left(t\right)\right)}}dt\hfill \\ & =& {\int }_0^1\frac{2\alpha s}{\sqrt{\alpha (1-s^2)log(1+\alpha )+\alpha s^2{A}_{\alpha }\left(s\right)+\xi \left(\alpha s^2\right)}}ds\hfill \\ & :=& \frac{2\sqrt{\alpha }}{\sqrt{log(1+\alpha )}}{\int }_0^1\frac{s}{\sqrt{1-s^2}}\frac{1}{\sqrt{1+g_{\alpha }\left(s\right)}}ds,\hfill \end{array}$$

(15)

where

$$\begin{array}{ccc}\hfill A_{\alpha }\left(s\right)& :=& log(1+\alpha )-log(1+\alpha s^2),\hfill \end{array}$$

(16)

$$\begin{array}{ccc}\hfill g_{\alpha }\left(s\right)& :=& \frac{1}{log(1+\alpha )}\frac{s^2}{(1-s^2)}{A}_{\alpha }\left(s\right)+\frac{\xi \left(\alpha s^2\right)}{\alpha (1-s^2)log(1+\alpha )}.\hfill \end{array}$$

(17)

For $0\le s\le 1$, we have

$$\begin{array}{ccc}\left|\frac{1}{log(1+\alpha )}\frac{s^2}{(1-s^2)}{A}_{\alpha }\left(s\right)\right|\hfill & \le & \frac{s^2}{1-s^2}\frac{1}{log(1+\alpha )}{\int }_{\alpha s^2}^{\alpha }\frac{1}{1+x}dx\hfill \\ & \le & \frac{1}{log(1+\alpha )}\frac{\alpha s^2}{1+\alpha s^2}\hfill \\ & \le & \frac{1}{log(1+\alpha )}\ll 1,\hfill \end{array}$$

(18)

$$\begin{array}{ccc}\hfill \left|\frac{\xi \left(\alpha s^2\right)}{\alpha (1-s^2)log(1+\alpha )}\right|& \le & \frac{2}{log(1+\alpha )}\ll 1.\hfill \end{array}$$

(19)

By this, (15) and Taylor expansion, we obtain

$$\begin{array}{ccc}\sqrt{2\lambda }\hfill & =& \frac{2\sqrt{\alpha }}{\sqrt{log(1+\alpha )}}{\int }_0^1\frac{s}{\sqrt{1-s^2}}\left\{1+\sum _{n=1}^{\infty }{(-1)}^n\frac{(2n-1)!!}{n!2^n}{g}_{\alpha }{\left(s\right)}^n\right\}ds\hfill \\ & =& \frac{2\sqrt{\alpha }}{\sqrt{log(1+\alpha )}}{\int }_0^1\frac{s}{\sqrt{1-s^2}}\left\{1-\frac{1}{2}{g}_{\alpha }\left(s\right)+\sum _{n=2}^{\infty }{(-1)}^n\frac{(2n-1)!!}{n!2^n}{g}_{\alpha }{\left(s\right)}^n\right\}ds,\hfill \end{array}$$

(20)

where $(2n-1)!!=(2n-1)(2n-3)\cdots 3·1,(-1)!!=1$. We see from (20) that the second term of $\lambda \left(\alpha \right)$ in Theorem 2 follows from Lemma 1 below.

Lemma 1.

As $\alpha \to \infty$,

$$\begin{array}{ccc}\hfill L:={\int }_0^1\frac{s}{\sqrt{1-s^2}}{g}_{\alpha }\left(s\right)ds& =& \frac{1}{log(1+\alpha )}\left(4log2-3\right)+O\left(\frac{1}{\alpha log(1+\alpha )}\right).\hfill \end{array}$$

(21)

The proof of Lemma 1 is a conclusion of Lemmas 2 and 3 below. By (17), we have

$$\begin{array}{c}\hfill L=L_1+L_2,\end{array}$$

(22)

where

$$\begin{array}{ccc}\hfill L_1& :=& \frac{1}{log(1+\alpha )}{\int }_0^1\frac{s^3}{{(1-s^2)}^{3/2}}{A}_{\alpha }\left(s\right)ds,\hfill \end{array}$$

(23)

$$\begin{array}{ccc}\hfill L_2& :=& \frac{1}{\alpha log(1+\alpha )}{\int }_0^1\frac{s}{{(1-s^2)}^{3/2}}\xi \left(\alpha s^2\right)ds.\hfill \end{array}$$

(24)

Lemma 2.

As $\alpha \to \infty$,

$$\begin{array}{ccc}\hfill L_1& =& \left(4log2-2\right)\frac{1}{log(1+\alpha )}+O\left(\frac{1}{\alpha log(1+\alpha )}\right).\hfill \end{array}$$

(25)

Proof. 

We put $s=sin\theta$ in (23). Then by integration by parts,

$$\begin{array}{ccc}{L}_1\hfill & =& \frac{1}{log(1+\alpha )}{\int }_0^{\pi /2}\frac{1}{{cos}^2\theta }\left\{{sin}^3\theta A_{\alpha }(sin\theta )\right\}d\theta \hfill \\ & =& \frac{1}{log(1+\alpha )}{\left[tan\theta \left\{{sin}^3\theta A_{\alpha }(sin\theta )\right\}\right]}_0^{\pi /2}\hfill \\ & -& \frac{1}{log(1+\alpha )}{\int }_0^{\pi /2}tan\theta {\left\{{sin}^3\theta A_{\alpha }(sin\theta )\right\}}^{\prime }d\theta \hfill \\ & :=& Q_1+Q_2+Q_3,\hfill \end{array}$$

(26)

where

$$\begin{array}{ccc}\hfill Q_1& :=& \frac{1}{log(1+\alpha )}{\left[tan\theta \left\{{sin}^3\theta A_{\alpha }(sin\theta )\right\}\right]}_0^{\pi /2},\hfill \end{array}$$

(27)

$$\begin{array}{ccc}\hfill Q_2& :=& -\frac{3}{log(1+\alpha )}{\int }_0^{\pi /2}{sin}^3\theta A_{\alpha }(sin\theta )d\theta ,\hfill \end{array}$$

(28)

$$\begin{array}{ccc}\hfill Q_3& :=& \frac{2}{log(1+\alpha )}{\int }_0^{\pi /2}{sin}^3\theta \frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }d\theta .\hfill \end{array}$$

(29)

Then by l’Hôpital’s rule, we obtain

$$\begin{array}{c}\hfill \underset{\theta \to \pi /2}{lim}\frac{A_{\alpha }(sin\theta )}{cos\theta }=\underset{\theta \to \pi /2}{lim}\frac{2\alpha cos\theta }{1+\alpha {sin}^2\theta }=0.\end{array}$$

(30)

By this, we see that $Q_1=0$. We next calculate $Q_2$. For $0\le s\le 1$, by Taylor expansion, we obtain

$$\begin{array}{ccc}{A}_{\alpha }\left(s\right)\hfill & =& log(1+\alpha )-log\alpha -log\left(\frac{1}{\alpha }+s^2\right)+2logs-2logs\hfill \\ & =& \frac{1}{\alpha }+O\left({\alpha }^{-2}\right)-log\left(\frac{1}{\alpha }+s^2\right)+2logs-2logs.\hfill \end{array}$$

(31)

Then by direct calculation, we obtain the following (32) and (33).

$$\begin{array}{ccc}\hfill 0& \le & sin\theta A_{\alpha }(sin\theta )=-2sin\theta log(sin\theta )+O\left(\frac{1}{\sqrt{\alpha }}\right),\hfill \end{array}$$

(32)

$$\begin{array}{ccc}\hfill 0& \le & {sin}^2\theta A_{\alpha }(sin\theta )=-2{sin}^2\theta log(sin\theta )+O\left(\frac{1}{\alpha }\right).\hfill \end{array}$$

(33)

The proofs of (32) and (33) are elementaty but long and complicated. Therefore, the precise proofs of (32) and (33) will be given in Appendix A. For $0\le \theta \le \pi /2$, we have

$$\begin{array}{c}\hfill {sin}^2\theta -\frac{1}{\alpha }={sin}^2\theta \left(1-\frac{1}{\alpha {sin}^2\theta }\right)\le {sin}^2\theta \left(\frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }\right)\le {sin}^2\theta .\end{array}$$

(34)

By (34), we obtain

$$\begin{array}{ccc}{Q}_2\hfill & =& -\frac{3}{log(1+\alpha )}{\int }_0^{\pi /2}{sin}^3\theta A_{\alpha }(sin\theta )d\theta \hfill \\ & =& -\frac{3}{log(1+\alpha )}{\int }_0^{\pi /2}sin\theta \left\{-2{sin}^2\theta log(sin\theta )+O\left(\frac{1}{\alpha }\right)\right\}d\theta \hfill \\ & =& \frac{6}{log(1+\alpha )}{\int }_0^{\pi /2}{sin}^3\theta log(sin\theta )d\theta +O\left(\frac{1}{\alpha log(1+\alpha )}\right)\hfill \\ & =& \frac{1}{log(1+\alpha )}\left(4log2-\frac{10}{3}\right)+O\left(\frac{1}{\alpha log(1+\alpha )}\right).\hfill \end{array}$$

(35)

By (33), we have

$$\begin{array}{ccc}{Q}_3\hfill & =& \frac{2}{log(1+\alpha )}{\int }_0^{\pi /2}{sin}^3\theta \frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }d\theta \hfill \\ & =& \frac{2}{log(1+\alpha )}{\int }_0^{\pi /2}{sin}^3\theta d\theta +O\left(\frac{1}{\alpha log(1+\alpha )}\right)\hfill \\ & =& \frac{4}{3log(1+\alpha )}+O\left(\frac{1}{\alpha log(1+\alpha )}\right).\hfill \end{array}$$

(36)

By this, (30) and (35), we obtain (25). Thus the proof is complete. □

Lemma 3.

As $\alpha \to \infty$,

$$\begin{array}{ccc}\hfill L_2& =& -\frac{1}{log(1+\alpha )}+O\left(\frac{1}{\alpha log(1+\alpha )}\right).\hfill \end{array}$$

(37)

Proof. 

We have

$$\begin{array}{ccc}{L}_2\hfill & =& \frac{1}{\alpha log(1+\alpha )}{\int }_0^1\frac{s\{A_{\alpha }\left(s\right)-\alpha (1-s^2)\}}{{(1-s^2)}^{3/2}}ds\hfill \\ & :=& L_{21}+L_{22}.\hfill \end{array}$$

(38)

Then

$$\begin{array}{ccc}\hfill L_{22}& =& -\frac{1}{log(1+\alpha )}{\int }_0^1\frac{s}{{(1-s^2)}^{1/2}}ds=-\frac{1}{log(1+\alpha )}.\hfill \end{array}$$

(39)

By putting $s=sin\theta$, (32), l’Hôpital’s rule and integration by parts, we have

$$\begin{array}{ccc}{L}_{21}\hfill & =& \frac{1}{\alpha log(1+\alpha )}{\int }_0^{\pi /2}\frac{1}{{cos}^2\theta }\{sin\theta A_{\alpha }(sin\theta )\}d\theta \hfill \\ & =& \frac{1}{\alpha log(1+\alpha )}{\left[tan\theta sin\theta A_{\alpha }(sin\theta )\right]}_0^{\pi /2}\hfill \\ & & -\frac{1}{\alpha log(1+\alpha )}{\int }_0^{\pi /2}sin\theta A_{\alpha }(sin\theta )d\theta +\frac{2}{\alpha log(1+\alpha )}{\int }_0^{\pi /2}\frac{\alpha {sin}^3\theta }{1+\alpha {sin}^2\theta }d\theta \hfill \\ & =& O\left(\frac{1}{\alpha log(1+\alpha )}\right).\hfill \end{array}$$

(40)

By this, (38) and (39), we obtain (37). Thus the proof is complete.□

By Lemmas 2 and 3, we obtain Lemma 1.

3. The Third Term of $\mathbf{\lambda }\left(\mathbf{\alpha }\right)$ in Theorem 2

By (20), to obtain the third term of $\lambda \left(\alpha \right)$, we calculate

$$\begin{array}{c}\hfill M:={\int }_0^1\frac{s}{\sqrt{1-s^2}}{g}_{\alpha }{\left(s\right)}^{2}ds.\end{array}$$

(41)

We have

$$\begin{array}{ccc}{g}_{\alpha }{\left(s\right)}^2\hfill & =& \frac{1}{{(log(1+\alpha ))}^2}\frac{s^4}{{(1-s^2)}^2}{A}_{\alpha }{\left(s\right)}^2+\frac{1}{{\alpha }^2{(log(1+\alpha ))}^2}\frac{\xi {\left(\alpha s^2\right)}^2}{{(1-s^2)}^2}\hfill \\ & & +2\frac{1}{\alpha {(log(1+\alpha ))}^2}\frac{s^2}{{(1-s^2)}^2}{A}_{\alpha }\left(s\right)\xi \left(\alpha s^2\right).\hfill \end{array}$$

(42)

We put

$$\begin{array}{ccc}\hfill I_1& :=& \frac{1}{{(log(1+\alpha ))}^2}{J}_1=\frac{1}{{(log(1+\alpha ))}^2}{\int }_0^1\frac{s^5}{{(1-s^2)}^{5/2}}{A}_{\alpha }{\left(s\right)}^{2}ds,\hfill \end{array}$$

(43)

$$\begin{array}{ccc}\hfill I_2& :=& \frac{1}{{\alpha }^2{(log(1+\alpha ))}^2}{J}_2=\frac{1}{{\alpha }^2{(log(1+\alpha ))}^2}{\int }_0^1\frac{s\xi {\left(\alpha s^2\right)}^2}{{(1-s^2)}^{5/2}}ds,\hfill \end{array}$$

(44)

$$\begin{array}{ccc}\hfill I_3& :=& \frac{2}{\alpha {(log(1+\alpha ))}^2}{J}_3=\frac{2}{\alpha {(log(1+\alpha ))}^2}{\int }_0^1\frac{s^3}{{(1-s^2)}^{5/2}}{A}_{\alpha }\left(s\right)\xi \left(\alpha s^2\right)ds.\hfill \end{array}$$

(45)

Then $M=I_1+I_2+I_3$.

Lemma 4.

As $\alpha \to \infty$,

$$\begin{array}{c}\hfill I_1=C_1\frac{1}{{(log(1+\alpha ))}^2}+O\left(\frac{1}{\alpha {(log(1+\alpha ))}^2}\right),\end{array}$$

(46)

where

$$\begin{array}{ccc}\hfill C_1& :=& -\frac{4}{3}{C}_{11}+\frac{1}{3}(C_{12}+C_{13}+C_{14}+C_{15}+C_{16}),\hfill \end{array}$$

(47)

$$\begin{array}{ccc}\hfill C_{11}& :=& 20{\int }_0^{\pi /2}{sin}^7\theta {(log(sin\theta ))}^{2}d\theta -8{\int }_0^{\pi /2}{sin}^7\theta log(sin\theta )d\theta ,\hfill \end{array}$$

(48)

$$\begin{array}{ccc}\hfill C_{12}& :=& 100{\int }_0^{\pi /2}{sin}^5\theta (2{cos}^2\theta +1){(log(sin\theta ))}^{2}d\theta ,\hfill \end{array}$$

(49)

$$\begin{array}{ccc}\hfill C_{13}& :=& 40{\int }_0^{\pi /2}{sin}^7\theta (2{cos}^2\theta +1)log(sin\theta )d\theta ,\hfill \end{array}$$

(50)

$$\begin{array}{ccc}\hfill C_{14}& :=& 8{\int }_0^{\pi /2}{sin}^5\theta (2{cos}^2\theta +1)d\theta ,\hfill \end{array}$$

(51)

$$\begin{array}{ccc}\hfill C_{15}& :=& 56{\int }_0^{\pi /2}{sin}^7\theta (2{cos}^2\theta +1)log(sin\theta )d\theta ,\hfill \end{array}$$

(52)

$$\begin{array}{ccc}\hfill C_{16}& :=& -16{\int }_0^{\pi /2}{sin}^5\theta (2{cos}^2\theta +1)log(sin\theta )d\theta .\hfill \end{array}$$

(53)

Proof. 

We recall that

$$\begin{array}{c}\hfill \frac{1}{{cos}^4\theta }=\frac{d}{d\theta }\left(\frac{sin\theta }{3{cos}^3\theta }(2{cos}^2\theta +1)\right).\end{array}$$

(54)

We put $s=sin\theta$. Then we have

$$\begin{array}{ccc}{J}_1\hfill & =& {\int }_0^{\pi /2}\frac{1}{{cos}^4\theta }{sin}^5\theta A_{\alpha }{(sin\theta )}^{2}d\theta \hfill \\ & =& {\int }_0^{\pi /2}\frac{d}{d\theta }\left(\frac{sin\theta }{3{cos}^3\theta }(2{cos}^2\theta +1\right){sin}^5\theta A_{\alpha }{(sin\theta )}^{2}d\theta \hfill \\ & =& {\left[\frac{sin\theta }{3{cos}^3\theta }(2{cos}^2\theta +1){sin}^5\theta A_{\alpha }{(sin\theta )}^2\right]}_0^{\pi /2}\hfill \\ & & -{\int }_0^{\pi /2}\frac{sin\theta }{3{cos}^2\theta }(2{cos}^2\theta +1)\left\{5{sin}^4\theta A_{\alpha }{(sin\theta )}^2-4\alpha A_{\alpha }(sin\theta ){sin}^6\theta \frac{1}{1+\alpha {sin}^2\theta }\right\}d\theta .\hfill \end{array}$$

(55)

By using l’Hôpital’s rule, we easily see that the first term of (55) is equal to 0. Then

$$\begin{array}{cc}{J}_1\hfill & =-\frac{1}{3}{\int }_0^{\pi /2}{(tan\theta )}^{\prime }(2{cos}^2\theta +1)\hfill \\ & \times \left\{5{sin}^5\theta A_{\alpha }{(sin\theta )}^2-4\alpha A_{\alpha }(sin\theta ){sin}^7\theta \frac{1}{1+\alpha {sin}^2\theta }\right\}d\theta \hfill \\ & =-\frac{1}{3}{\left[tan\theta (2{cos}^2\theta +1)\left\{5{sin}^5\theta A_{\alpha }{(sin\theta )}^2-4\alpha A_{\alpha }(sin\theta ){sin}^7\theta \frac{1}{1+\alpha {sin}^2\theta }\right\}\right]}_0^{\pi /2}\hfill \\ & -\frac{4}{3}{\int }_0^{\pi /2}{sin}^2\theta \left\{5{sin}^5\theta A_{\alpha }{(sin\theta )}^2-4\alpha A_{\alpha }(sin\theta ){sin}^7\theta \frac{1}{1+\alpha {sin}^2\theta }\right\}d\theta \hfill \\ & +\frac{1}{3}{\int }_0^{\pi /2}sin\theta (2{cos}^2\theta +1)\left\{25{sin}^4\theta A_{\alpha }{\left(s\right)}^2-20\alpha A_{\alpha }(sin\theta ){sin}^6\theta \frac{1}{1+\alpha {sin}^2\theta }\right.\hfill \\ & +8{\alpha }^2{sin}^8\theta \frac{1}{{(1+\alpha {sin}^2\theta )}^2}-28\alpha {sin}^6\theta A_{\alpha }(sin\theta )\frac{1}{1+\alpha {sin}^2\theta }\hfill \\ & \left.+8{\alpha }^2{sin}^8\theta A_{\alpha }(sin\theta )\frac{1}{{(1+\alpha {sin}^2\theta )}^2}\right\}d\theta \hfill \\ & :=-\frac{1}{3}{J}_{10}-\frac{4}{3}{J}_{11}+\frac{1}{3}(J_{12}+J_{13}+J_{14}+J_{15}+J_{16}).\hfill \end{array}$$

(56)

By l’Hôpital’s rule, we see that $J_{10}=0$. By (33), (34) and (56), we have

$$\begin{array}{ccc}{J}_{11}\hfill & =& 5{\int }_0^{\pi /2}{sin}^7\theta A_{\alpha }{(sin\theta )}^{2}d\theta -4{\int }_0^{\pi /2}{sin}^5\theta \left({sin}^2\theta A_{\alpha }(sin\theta )\right)\frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }d\theta \hfill \\ & =& 20{\int }_0^{\pi /2}{sin}^3\theta ({sin}^2\theta {(log(sin\theta ))}^{2}d\theta -8{\int }_0^{\pi /2}{sin}^7\theta log(sin\theta )d\theta +O\left({\alpha }^{-1}\right)\hfill \\ & :=& C_{11}+O\left({\alpha }^{-1}\right).\hfill \end{array}$$

(57)

By (33), (34) and (56), we obtain

$$\begin{array}{ccc}{J}_{12}\hfill & =& 25{\int }_0^{\pi /2}{sin}^5\theta (2{cos}^2\theta +1)A_{\alpha }{(sin\theta )}^{2}d\theta \hfill \\ & =& 100{\int }_0^{\pi /2}{sin}^5\theta (2{cos}^2\theta +1){(log(sin\theta ))}^{2}d\theta +O\left({\alpha }^{-1}\right)\hfill \\ & :=& C_{12}+O\left({\alpha }^{-1}\right),\hfill \end{array}$$

(58)

$$\begin{array}{ccc}{J}_{13}\hfill & =& -20{\int }_0^{\pi /2}sin\theta (2{cos}^2\theta +1)\left({sin}^2\theta A_{\alpha }(sin\theta )\right)\left({sin}^2\theta \left(\frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }\right)\right)d\theta \hfill \\ & =& 40{\int }_0^{\pi /2}{sin}^7\theta (2{cos}^2\theta +1)log(sin\theta )d\theta +O\left({\alpha }^{-1}\right):=C_{13}+O\left({\alpha }^{-1}\right),\hfill \end{array}$$

(59)

$$\begin{array}{ccc}{J}_{14}\hfill & =& 8{\int }_0^{\pi /2}{sin}^5\theta (2{cos}^2\theta +1){\left(\frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta )}\right)}^{2}d\theta \hfill \\ & =& 8{\int }_0^{\pi /2}{sin}^5\theta (2{cos}^2\theta +1)d\theta +O\left({\alpha }^{-1}\right):=C_{14}+O\left({\alpha }^{-1}\right),\hfill \end{array}$$

(60)

$$\begin{array}{ccc}{J}_{15}\hfill & =& -28{\int }_0^{\pi /2}{sin}^3\theta (2{cos}^2\theta +1)\left({sin}^2\theta A_{\alpha }(sin\theta )\right)\left(\frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }\right)d\theta \hfill \\ & =& 56{\int }_0^{\pi /2}{sin}^5\theta (2{cos}^2\theta +1)log(sin\theta )d\theta +O\left({\alpha }^{-1}\right)\hfill \\ & :=& C_{15}+O\left({\alpha }^{-1}\right).\hfill \end{array}$$

(61)

By (33) and (34), we have

$$\begin{array}{ccc}{J}_{16}\hfill & =& 8{\int }_0^{\pi /2}(2{cos}^2\theta +1){sin}^4\theta {\left(\frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }\right)}^2(sin\theta A_{\alpha }(sin\theta ))d\theta \hfill \\ & =& 8{\int }_0^{\pi /2}(2{cos}^2\theta +1)({sin}^4\theta +O\left({\alpha }^{-1}\right))(sin\theta A_{\alpha }(sin\theta ))d\theta \hfill \\ & =& -16{\int }_0^{\pi /2}{sin}^5\theta (2{cos}^2\theta +1)log(sin\theta )d\theta +O\left({\alpha }^{-1}\right)\hfill \\ & :=& C_{16}+O\left({\alpha }^{-1}\right).\hfill \end{array}$$

(62)

By (55)–(62), we obtain (46). Thus the proof is complete. □

Lemma 5.

As $\alpha \to \infty$,

$$\begin{array}{c}\hfill I_2=\frac{1}{{(log(1+\alpha ))}^2}+O\left(\frac{1}{\alpha {(log(1+\alpha ))}^2}\right).\end{array}$$

(63)

Proof. 

By (14) and (44), we have

$$\begin{array}{ccc}{J}_2\hfill & =& {\int }_0^1\frac{s}{{(1-s^2)}^{5/2}}{(A_{\alpha }\left(s\right)-\alpha (1-s^2))}^{2}ds\hfill \\ & =& J_{21}+J_{22}+J_{23}\hfill \\ & :=& {\int }_0^1\frac{s}{{(1-s^2)}^{5/2}}{A}_{\alpha }{\left(s\right)}^{2}ds-2\alpha {\int }_0^1\frac{s}{{(1-s^2)}^{3/2}}{A}_{\alpha }\left(s\right)ds+{\alpha }^2{\int }_0^1\frac{s}{\sqrt{1-s^2}}ds.\hfill \end{array}$$

(64)

It is clear that $J_{23}={\alpha }^2$. By (54), putting $s=sin\theta$, integration by parts and l’Hôpital’s rule, we obtain

$$\begin{array}{cc}{J}_{21}\hfill & ={\int }_0^{\pi /2}\frac{sin\theta }{{cos}^4\theta }{A}_{\alpha }{(sin\theta )}^{2}d\theta \hfill \\ & ={\left[\frac{sin\theta }{3{cos}^3\theta }(2{cos}^2\theta +1)sin\theta A_{\alpha }{(sin\theta )}^2\right]}_0^{\pi /2}\hfill \\ & -{\int }_0^{\pi /2}\frac{sin\theta }{3{cos}^2\theta }(2{cos}^2\theta +1)\left\{A_{\alpha }{(sin\theta )}^2-4A_{\alpha }(sin\theta )\frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }\right\}d\theta \hfill \\ & ={\left[-\frac{1}{3}tan\theta sin\theta (2{cos}^2\theta +1)\left\{A_{\alpha }{(sin\theta )}^2-4A_{\alpha }(sin\theta )\frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }\right\}\right]}_0^{\pi /2}\hfill \\ & +\frac{1}{3}{\int }_0^{\pi /2}sin\theta (2{cos}^2\theta -4{sin}^2\theta +1)\left\{A_{\alpha }{(sin\theta )}^2-4A_{\alpha }(sin\theta )\frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }\right\}d\theta \hfill \\ & +\frac{1}{3}{\int }_0^{\pi /2}sin\theta (2{cos}^2\theta +1)\left\{-4\frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }{A}_{\alpha }(sin\theta )\right.\hfill \\ & \left.+8{\left(\frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta )}\right)}^2-8A_{\alpha }(sin\theta )\frac{\alpha {sin}^2\theta }{{(1+\alpha {sin}^2\theta )}^2}\right\}d\theta \hfill \\ & =O\left(1\right).\hfill \end{array}$$

(65)

By integration by parts and l’Hôpital’s rule, we obtain

$$\begin{array}{ccc}{J}_{22}\hfill & =& -2\alpha {\int }_0^{\pi /2}\frac{1}{{cos}^2\theta }sin\theta A_{\alpha }(sin\theta )d\theta \hfill \\ & =& {\left[-2\alpha tan\theta sin\theta A_{\alpha }(sin\theta )\right]}_0^{\pi /2}\hfill \\ & & +2\alpha {\int }_0^{\pi /2}\left\{sin\theta A_{\alpha }(sin\theta )-2sin\theta \frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }\right\}d\theta \hfill \\ & =& O\left(\alpha \right).\hfill \end{array}$$

(66)

By this and (65), we obtain (63). Thus the proof is complete. □

Lemma 6.

As $\alpha \to \infty$,

$$\begin{array}{ccc}\hfill I_3& =& \frac{4}{{(log(1+\alpha ))}^2}\left(-2log2+1\right)+O\left(\frac{1}{\alpha {(log(1+\alpha ))}^2}\right).\hfill \end{array}$$

(67)

Proof. 

By putting $s=sin\theta$, we have

$$\begin{array}{ccc}{J}_3\hfill & =& {\int }_0^1\frac{s^3}{{(1-s^2)}^{5/2}}{A}_{\alpha }\left(s\right)\xi \left(\alpha s^2\right)ds\hfill \\ & =& {\int }_0^{\pi /2}\frac{1}{{cos}^4\theta }{sin}^3\theta A_{\alpha }{(sin\theta )}^{2}d\theta -\alpha {\int }_0^{\pi /2}\frac{1}{{cos}^2\theta }{sin}^3\theta A_{\alpha }(sin\theta )d\theta \hfill \\ & :=& J_{31}-J_{32}.\hfill \end{array}$$

(68)

By (54) and integration by parts, we obtain

$$\begin{array}{ccc}{J}_{31}\hfill & =& {\left[\left(\frac{sin\theta }{3{cos}^3\theta }(2{cos}^2\theta +1)\right){sin}^3\theta A_{\alpha }{(sin\theta )}^2\right]}_0^{\pi /2}\hfill \\ & & -{\int }_0^{\pi /2}\frac{1}{{cos}^2\theta }(2{cos}^2\theta +1){sin}^3\theta A_{\alpha }{(sin\theta )}^{2}d\theta \hfill \\ & & +\frac{4}{3}\alpha {\int }_0^{\pi /2}\frac{1}{{cos}^2\theta }(2{cos}^2\theta +1){sin}^5\theta \frac{1}{1+\alpha {sin}^2\theta }{A}_{\alpha }(sin\theta )d\theta \hfill \\ & =& J_{310}-J_{311}+J_{312}.\hfill \end{array}$$

(69)

By l’Hôpital’s rule, we see that $J_{310}=0$. By integration by parts, (32)–(34), we obtain

$$\begin{array}{ccc}{J}_{311}\hfill & =& {\left[tan\theta (2{cos}^2\theta +1){sin}^3\theta A_{\alpha }{(sin\theta )}^2\right]}_0^{\pi /2}\hfill \\ & & -{\int }_0^{\pi /2}tan\theta {\left\{{sin}^3\theta (2{cos}^2\theta +1)A_{\alpha }{(sin\theta )}^2\right\}}^{\prime }d\theta \hfill \\ & =& {\int }_0^{\pi /2}(-6sin\theta {cos}^2\theta +4{sin}^3\theta -3sin\theta ){(sin\theta A_{\alpha }(sin\theta ))}^{2}d\theta \hfill \\ & & +4{\int }_0^{\pi /2}(2{cos}^2\theta +1)(sin\theta A_{\alpha }(sin\theta )){sin}^2\theta \frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }d\theta \hfill \\ & =& 4{\int }_0^{\pi /2}(-6sin\theta {cos}^2\theta +4{sin}^3\theta -3sin\theta ){(sin\theta log(sin\theta ))}^{2}d\theta +O\left({\alpha }^{-1/2}\right)\hfill \\ & & -8{\int }_0^{\pi /2}{sin}^2\theta (2{cos}^2\theta +1)(sin\theta log(sin\theta ))d\theta +O\left({\alpha }^{-1}\right)\hfill \\ & =& O\left(1\right).\hfill \end{array}$$

(70)

By (32)–(34) and integration by parts and l’Hôpital’s rule, we have

$$\begin{array}{ccc}{J}_{312}\hfill & =& \frac{4}{3}\alpha {\left[tan\theta {sin}^5\theta (2{cos}^2\theta +1)A_{\alpha }(sin\theta )\frac{1}{1+\alpha {sin}^2\theta }\right]}_0^{\pi /2}\hfill \\ & & -\frac{4}{3}\alpha {\int }_0^{\pi /2}tan\theta {\left\{{sin}^5\theta (2{cos}^2\theta +1)A_{\alpha }(sin\theta )\frac{1}{1+\alpha {sin}^2\theta }\right\}}^{\prime }d\theta \hfill \\ & =& -\frac{4}{3}{\int }_0^{\pi /2}(10{sin}^2\theta {cos}^2\theta -4{sin}^4\theta +5{sin}^2\theta )(sin\theta A_{\alpha }(sin\theta ))\frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }d\theta \hfill \\ & & +\frac{8}{3}{\int }_0^{\pi /2}{sin}^3\theta (2{cos}^2\theta +1){\left(\frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }\right)}^{2}d\theta \hfill \\ & & +\frac{8}{3}{\int }_0^{\pi /2}{sin}^2\theta (2{cos}^2\theta +1)(sin\theta A_{\alpha }(sin\theta )){\left(\frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }\right)}^{2}d\theta \hfill \\ & =& O\left(1\right).\hfill \end{array}$$

By integration by parts, we obtain

$$\begin{array}{ccc}{J}_{32}\hfill & =& \alpha {\left[tan\theta {sin}^3\theta A_{\alpha }(sin\theta )\right]}_0^{\pi /2}-\alpha {\int }_0^{\pi /2}tan\theta {\left\{{sin}^3\theta A_{\alpha }(sin\theta )\right\}}^{\prime }d\theta \hfill \\ & :=& J_{320}+J_{321}.\hfill \end{array}$$

(71)

By l’Hôpital’s rule, we obtain that $J_{320}=0$. By (33) and (34), we obtain

$$\begin{array}{ccc}{J}_{321}\hfill & =& -3\alpha {\int }_0^{\pi /2}sin\theta \left({sin}^2\theta A_{\alpha }(sin\theta )\right)d\theta +2\alpha {\int }_0^{\pi /2}{sin}^3\theta \frac{\alpha {sin}^2\theta }{1+\alpha {sin}^2\theta }d\theta \hfill \\ & =& 6\alpha {\int }_0^{\pi /2}{sin}^3\theta log(sin\theta )d\theta +2\alpha {\int }_0^{\pi /2}{sin}^3\theta d\theta +O\left(1\right).\hfill \end{array}$$

(72)

By this and (69)–(71), we obtain

$$\begin{array}{ccc}{I}_3\hfill & =& \frac{-4}{{(log(1+\alpha ))}^2}\left(3{\int }_0^{\pi /2}{sin}^3\theta log(sin\theta )d\theta +{\int }_0^{\pi /2}{sin}^3\theta d\theta \right)+O\left(\frac{1}{\alpha {(log(1+\alpha ))}^2}\right)\hfill \\ & =& \frac{4}{{(log(1+\alpha ))}^2}\left(-2log2+1\right)+O\left(\frac{1}{\alpha {(log(1+\alpha ))}^2}\right).\hfill \end{array}$$

(73)

This implies (20). Thus the proof is complete. □

By Lemmas 4–6, we obtain

$$\begin{array}{ccc}\hfill M& =& (C_1+5-8log2)\frac{1}{{(log(1+\alpha ))}^2}+O\left(\frac{1}{\alpha {(log(1+\alpha ))}^2}\right).\hfill \end{array}$$

(74)

This along with (20) and Lemma 1, we obtain (7).

4. Remainder Estimate

To complete the proof of Theorem 2, in this section, we prove (8). Let $n\ge 3$. By (17), we have

$$\begin{array}{c}\hfill g_{\alpha }{\left(s\right)}^n:=\frac{1}{{(log(1+\alpha ))}^n{(1-s^2)}^n}{\eta }_{\alpha }{\left(s\right)}^n,\end{array}$$

(75)

where

$$\begin{array}{c}\hfill {\eta }_{\alpha }\left(s\right):=s^2{A}_{\alpha }\left(s\right)+\frac{1}{\alpha }\xi \left(\alpha s^2\right).\end{array}$$

(76)

By direct calculation, we see that ${\eta }_{\alpha }^{\prime }\left(s\right)=2s{A}_{\alpha }\left(s\right)\ge 0$ and find that ${\eta }_{\alpha }\left(s\right)$ is increasing in $0\le s\le 1$. We have ${\eta }_{\alpha }\left(0\right)=\frac{1}{\alpha }log(1+\alpha )-1=-1+o\left(1\right)$, ${\eta }_{\alpha }\left(1\right)=0$. Let an arbitrary $0<ϵ\ll 1$ be fixed. Then there exists a constant $\delta >0$ independent of $\alpha \gg 1$ such that for $s\in [0,ϵ]$ and $\alpha \gg 1$,

$$\begin{array}{ccc}-1\hfill & \le & {\eta }_{\alpha }\left(s\right)\le {\eta }_{\alpha }\left(ϵ\right)\hfill \\ & =& ({ϵ}^2-1)+\frac{\alpha {ϵ}^2+1}{\alpha }log\frac{1+\alpha }{1+\alpha {ϵ}^2}\hfill \\ & =& ({ϵ}^2-1)-({ϵ}^2+o\left(1\right))log({ϵ}^2+o\left(1\right))\hfill \\ & <& -\delta <0.\hfill \end{array}$$

(77)

By this, if n is odd, then we have

$$\begin{array}{ccc}{\int }_0^1\frac{s}{\sqrt{1-s^2}}{g}_{\alpha }{\left(s\right)}^{n}ds\hfill & =& \frac{1}{{(log(1+\alpha ))}^n}{\int }_0^1\frac{s}{\sqrt{1-s^2}}\frac{1}{{(1-s^2)}^n}{\eta }_{\alpha }{\left(s\right)}^{n}ds\hfill \\ & \le & \frac{1}{{(log(1+\alpha ))}^n}{\int }_0^{ϵ}\frac{s}{\sqrt{1-s^2}}\frac{1}{{(1-s^2)}^n}{(-\delta )}^{n}ds\hfill \\ & \le & -{\delta }^n\frac{1}{{(log(1+\alpha ))}^n}.\hfill \end{array}$$

(78)

If n is even, then by the same argument as that in (78), we have

$$\begin{array}{c}\hfill {\int }_0^1\frac{s}{\sqrt{1-s^2}}{g}_{\alpha }{\left(s\right)}^{n}ds\ge {\delta }^n\frac{1}{{(log(1+\alpha ))}^n}.\end{array}$$

(79)

By (16) and (19), for $n=3,4,\cdots$, we have

$$\begin{array}{c}\hfill |g_{\alpha }{\left(s\right)|}^n\le \frac{3^n}{{(log(1+\alpha ))}^n}.\end{array}$$

(80)

By this, (78) and (79), we obtain

$$\begin{array}{c}\hfill C\frac{1}{{(log(1+\alpha ))}^n}\le \left|\sum _{n=3}^{\infty }{\int }_0^1\frac{s}{\sqrt{1-s^2}}{(-1)}^n\frac{(2n-1)!!}{n!2^n}{g}_{\alpha }{\left(s\right)}^{n}ds\right|\le C^{-1}\frac{1}{{(log(1+\alpha ))}^n}.\end{array}$$

(81)

This implies (8). Thus the proof is complete.