2. Case I: $x=\cosh \beta >1$ with $\beta >0$

The saddle points are $t_\pm =e^{\pm \beta }/2$ with $0< t_-< t_+$. Moreover, $f''(t_-)=2(1-e^{2\beta })<0$ and $f''(t_+)=2(1-e^{-2\beta })>0$. We shall deform the contour of integration $\Gamma$ to the path of steepest descent passing through the saddle point $t_-$. To describe this steepest descent contour, we shall introduce the analytic function

(2.1)\begin{equation} w(z)=[f(z)-f(t_-)]^{1/2}, \end{equation}

where the branch of the square root function is chosen so that

\begin{align*} w(z)& ={-}i(z-t_-)\sqrt{-f''(t_-)/2}+\mathcal{O}((z-t_-)^{2})\\ & ={-}i(z-t_-)\sqrt{e^{2\beta}-1}+\mathcal{O}((z-t_-)^{2}), \end{align*}

for $z$ in a small complex neighbourhood of $t_-$. Let $z=re^{i\theta }$ with $r>0$ and $\theta \in (-\pi,\,\pi )$. We investigate the equation ${\rm Im} [f(z)-f(t_-)] = {\rm Im}\ f(z)=0$; namely,

(2.2)\begin{equation} r^{2}\sin(2\theta)-2 r\cosh \beta \sin\theta+\frac{\theta}{2}=0. \end{equation}

Clearly, $\theta =0$ satisfies the above equation. By symmetry, we can restrict $\theta \in (0,\,\pi )$ and find two solutions

(2.3)\begin{align} r_-(\theta)& =\frac{\theta/(2\sin\theta)}{ \cosh \beta+\sqrt{\cosh^{2} \beta-\theta/\tan\theta}},\quad\theta\in(0,\pi), \nonumber\\ r_+(\theta)& =\frac{\cosh \beta+\sqrt{\cosh^{2} \beta-\theta/\tan\theta}}{2\cos\theta},\quad\theta\in(0,\pi/2). \end{align}

It is readily seen that $r_\pm (\theta )\to t_\pm =e^{\pm \beta }/2$ as $\theta \to 0$. For consistency, we define $r_\pm (0)=t_\pm$. Consequently, we can regard $r_-(\theta )$ as a symmetric function on $(-\pi,\,\pi )$ and $r_+(\theta )$ a symmetric function on $(-\pi /2,\,\pi /2)$. Define two contours

\[ \Gamma_- = \{r_-(\theta)e^{i\theta}:\theta\in(-\pi,\pi)\},\quad \Gamma_+ = \{r_+(\theta)e^{i\theta}:\theta\in(-\pi/2,\pi/2)\}. \]

It follows that the solution of ${\rm Im}\ f(z)=0$ is the union $\Gamma _-\cup \Gamma _+\cup \mathbb {R}^{+}$, where $\mathbb {R}^{+}$ is the positive real line. The contours $\Gamma _\pm$ are depicted in figure 2.

FIGURE 2.

The contours $\Gamma _\pm$. The arrows indicate the increasing direction of the phase function $f$.

Since $f'(z)$ has no zeros other than $t_\pm$, it follows from the asymptotic behaviour $f(z)\sim z^{2}$ as $z\to \infty$ that $\Gamma _-$ is a steepest descent contour passing through $t_-$, while $\Gamma _+$ is a steepest ascent contour passing through $t_+$. Moreover, the solutions of $f(z)-f(t_-)=0$ are real and positive. Consider $f(t)=t^{2}-2t\cosh \beta +(\ln t)/2$ for $t>0$. Since $f(t)\to -\infty$ as $t\to 0^{+}$ and $f(t)\to \infty$ as $t\to \infty$, we obtain from $f'(t_\pm )=0$ that the equation $f(t)-f(t_-)=0$ has exactly two positive roots: one is $t_-$; and the other, denoted by $t_c$, is larger than $t_+$. Hence, we can choose $[t_c,\,\infty )$ as a branch cut and extend the definition of $w(z)$ on $\mathbb {C}\setminus (-\infty,\,0]\cup [t_c,\,\infty )$ by analytic continuation.

Now, we deform the contour of integration from $\Gamma$ to the steepest descent contour $\Gamma _-$ with counter-clockwise orientation and then make a change of variable $u=w(t)$:

(2.4)\begin{align} H_n(\sqrt N\cosh \beta)& =\frac{n!e^{{-}Nf(t_-)}}{2\pi iN^{n/2}}\int_{\Gamma_-} e^{{-}N[f(t)-f(t_-)]}t^{{-}1/2}dt \notag\\ & =\frac{n!e^{{-}Nf(t_-)}}{2\pi iN^{n/2}}\int_\mathbb{R} e^{{-}Nu^{2}}\frac{[w^{{-}1}(u)]^{{-}1/2}}{ w'(w^{{-}1}(u))}du. \end{align}

Recall that $\Gamma _+$ is the steepest ascent contour passing through $t_+$. We also denote by $\Gamma _0$ a clockwise oriented contour encircling $(-\infty,\,0]$; namely,

\[ \Gamma_0=\{\lambda+i\delta:\lambda<0\}\cup\{\delta e^{i\theta}:-\pi/2\le\theta\le\pi/2\}\cup\{\lambda-i\delta:\lambda<0\}, \]

where $\delta >0$ is a small positive number such that $\delta < t_c$. In fact, $\Gamma _0$ is identical to $\Gamma$ but has the opposite orientation. Since $|w(z)|/|z|\to 1$ as $z\to \infty$, we have from the Cauchy integral formula

\[\frac{[w^{{-}1}(u)]^{{-}1/2}}{ w'(w^{{-}1}(u))}=\frac{1}{2\pi i}\int_{\Gamma_0\cup\Gamma_+}\frac{z^{{-}1/2}}{ w(z)-u}dz \]

for any $u\in \mathbb {R}$. Substituting this expression into (2.4) gives a double integral representation

\[ H_n(\sqrt N\cosh \beta)=\frac{n!e^{{-}Nf(t_-)}}{(2\pi i)^{2}N^{n/2}}\int_\mathbb{R} e^{{-}Nu^{2}}\int_{\Gamma_0\cup\Gamma_+}\frac{z^{{-}1/2}}{ w(z)-u}dzdu. \]

For any $p\ge 0$, it follows from

\[ \frac{1}{w(z)-u}=\sum_{k=1}^{2p}\frac{u^{k-1}}{ w(z)^{k}}+\frac{u^{2p}}{ w(z)^{2p}[w(z)-u]} \]

and

(2.5)\begin{equation} \int_\mathbb{R} e^{{-}Nu^{2}}u^{k-1}du= \begin{cases} 0, & ~k=2j,\\ \Gamma(j+1/2)/N^{j+1/2}, & ~k=2j+1 \end{cases} \end{equation}

that

(2.6)\begin{equation} \begin{aligned} H_n(\sqrt N\cosh \beta) & = \frac{n!e^{{-}Nf(t_-)}}{(2\pi i)^{2}N^{n/2}} \Bigg\{\sum_{k=1}^{2p}\int_\mathbb{R} e^{{-}Nu^{2}}u^{k-1}du\int_{\Gamma_0\cup\Gamma_+}\frac{z^{{-}1/2}}{ w(z)^{k}}dz \\ & \quad+\int_\mathbb{R} e^{{-}Nu^{2}}u^{2p}\int_{\Gamma_0\cup\Gamma_+}\frac{z^{{-}1/2}}{ w(z)^{2p}[w(z)-u]}dzdu\Bigg\} \\ & =\frac{n!e^{{-}Nf(t_-)}}{2\sqrt{\pi\sinh\beta}N^{(n+1)/2}}\left\{\sum_{j=0}^{p-1}\frac{A_j(\coth \beta)}{N^{j}} + \varepsilon_p(N,\beta)\right\}, \end{aligned} \end{equation}

where

(2.7)\begin{align} A_j(\coth \beta) & ={-}\frac{\sqrt{\sinh\beta} \, \Gamma(j+1/2)}{2\pi^{3/2}}\int_{\Gamma_0\cup\Gamma_+}\frac{z^{{-}1/2}}{ w(z)^{2j+1}}dz, \end{align}

(2.8)\begin{align} \varepsilon_p(N,\beta)& ={-}\frac{\sqrt{N\sinh\beta}}{2\pi^{3/2}}\int_\mathbb{R} e^{{-}Nu^{2}}u^{2p}\int_{\Gamma_0\cup\Gamma_{+}}\frac{z^{{-}1/2}}{ w(z)^{2p}[w(z)-u]}dzdu. \end{align}

We show in appendix A.1 that the coefficients $A_j(\coth \beta )$ are polynomials in $\coth \beta$ of degree $3j$ with rational coefficients, and can be calculated using a recursive formula.

To find an upper bound for the remainder $\varepsilon _p(N,\,\beta )$, we need to estimate the integral

(2.9)\begin{equation} c_p(u)=\int_{\Gamma_0\cup\Gamma_+}\frac{z^{{-}1/2}}{ w(z)^{2p}[w(z)-u]}dz,\quad u\in\mathbb{R}. \end{equation}

If there exists $C_p>0$ such that $|c_p(u)|\le C_p$ for all $u\in \mathbb {R}$, then we obtain from (2.5) with $k=2p+1$ and (2.8) that

(2.10)\begin{equation} |\varepsilon_p(N,\beta)|\le \frac{\sqrt{\sinh\beta}}{2\pi^{3/2}} \frac{C_p\Gamma(p+1/2)}{N^{p}}. \end{equation}

First, we estimate the integral

\[ I_0=\int_{\Gamma_0}\frac{z^{{-}1/2}}{ w(z)^{2p}[w(z)-u]}dz. \]

Recall that the contour $\Gamma _0$ encircles the negative real line having a distance $\delta >0$ from it. By analyticity of the integrand, the value $I_0$ does not change if we let $\delta \to 0^{+}$. Hence,

(2.11)\begin{align} |I_0|\le \int_0^{+\infty} \frac{s^{{-}1/2}}{ |w(se^{i\pi})|^{2p}|w(se^{i\pi})-u|}ds +\int_0^{+\infty} \frac{s^{{-}1/2}}{ |w(se^{{-}i\pi})|^{2p}|w(se^{{-}i\pi})-u|}ds, \end{align}

where $w(se^{\pm i\pi })$ is defined as the limit of $w(se^{i\theta })$ as $\theta \to \pm \pi$. Note from (1.2) and (2.1) that

\[ w(se^{{\pm} i\pi})^{2}=s^{2}+2s\cosh \beta+(\ln s)/2-f(t_-)\pm i\pi/2. \]

Thus, we obtain

(2.12)\begin{equation} |w(se^{{\pm} i\pi})|=\{[s^{2}+2s\cosh \beta +(\ln s)/2-f(t_-)]^{2}+(\pi/2)^{2}\}^{1/4}=:R(s), \end{equation}

where the last equality defines $R(s)$, the modulus of $w(se^{\pm i\pi })$. Let $\theta _\pm (s)$ be the phase of $w(se^{\pm i\pi })$. We then have

\[ R(s)^{2}\cos[2\theta_\pm(s)]={\rm Re}[w(se^{{\pm} i\pi})^{2}]=s^{2}+2s\cosh \beta +(\ln s)/2-f(t_-), \]

which implies that

\begin{align*} |w(se^{{\pm} i\pi})-u|& \ge|{\rm Im}[w(se^{{\pm} i\pi})]|=R(s)|\sin[\theta_\pm(s)]| =R(s)\sqrt{\frac{1-\cos[2\theta_\pm(s)]}{2}}\nonumber\\ & =\sqrt{\frac{R(s)^{2}-[s^{2}+2s\cosh \beta +(\ln s)/2-f(t_-)]}{2}}\nonumber\\ & =\sqrt{\frac{(\pi/2)^{2}/2}{R(s)^{2}+[s^{2}+2s\cosh \beta +(\ln s)/2-f(t_-)]}}\nonumber\\ & \ge\sqrt{\frac{(\pi/2)^{2}}{4R(s)^{2}}}=\frac{\pi}{4R(s)}. \end{align*}

Assume that $p\geq 1$. Substituting this inequality into (2.11) yields

\[ |I_0|\le \frac{8}{\pi}\int_0^{+\infty}\frac{s^{{-}1/2}}{ R(s)^{2p-1}}ds. \]

To estimate the integral on the right-hand side of the above inequality, we note from (2.12) that $R(s)\ge \sqrt {\pi /2}$ for $s\in [0,\,1]$ and $R(s)\ge \sqrt {2s\cosh \beta }$ for $s\ge 1>t_-$. Thus,

(2.13)\begin{align} |I_0|& \le \frac{8}{\pi}\int_0^{1}\frac{s^{{-}1/2}}{(\pi/2)^{p-1/2}}ds+\frac{8}{\pi}\int_1^{+\infty}\frac{s^{{-}1/2}}{(2s\cosh \beta)^{p-1/2}}ds \notag\\ & =\frac{2^{p+7/2}}{\pi^{p+1/2}}+\frac{1}{\pi(p-1)2^{p-7/2}(\cosh \beta)^{p-1/2}}, \end{align}

where we have assumed $p\ge 2$ to ensure convergence of the second integral.

It remains to estimate the integral

\[ I_c=\int_{\Gamma_+}\frac{z^{{-}1/2}}{ w(z)^{2p}[w(z)-u]}dz. \]

For $z\in \Gamma _+$, we have ${\rm Im}\ f(z)=0$ and ${\rm Re} [f(z)-f(t_-)]< f(t_+)-f(t_-)<0$. It then follows from (2.1) that $w(z)$ is purely imaginary and $|w(z)-u|\ge |w(z)|$ for $z\in \Gamma _+$ and $u\in \mathbb {R}$. Now, we parameterize $\Gamma _+$ as $z=z_+(\theta )=r_+(\theta )e^{i\theta }$ with $\theta \in (-\pi /2,\,\pi /2)$, where $r_+(\theta )$ is given in (2.3). Since $dz/d\theta =[r_+'(\theta )+ir_+(\theta )]e^{i\theta }$ and $|w(z_+(\theta ))|^{2}=|f(z_+(\theta ))-f(t_-)|=f(t_-)-f(z_+(\theta ))$, we deduce

\begin{align*} |I_c|& \le\int_{-\pi/2}^{\pi/2}\frac{r_+(\theta)^{{-}1/2}\sqrt{[r_+'(\theta)]^{2}+[r_+(\theta)]^{2}}}{|w(z_+(\theta))|^{2p+1}}d\theta\\ & =\int_0^{\pi/2}\frac{2r_+(\theta)^{{-}1/2}\sqrt{[r_+'(\theta)]^{2}+[r_+(\theta)]^{2}}}{[f(t_-)-f(z_+(\theta))]^{p+1/2}}d\theta. \end{align*}

As both $\theta /\tan \theta$ and $\cos \theta$ are decreasing functions for $\theta \in (0,\,\pi /2)$, it follows from (2.3) that $r_+(\theta )$ is an increasing function for $\theta \in (0,\,\pi /2)$; namely, $r_+'(\theta )\ge 0$. To estimate $r_+'(\theta )$, we apply an implicit differentiation on (2.2) and obtain

\[ r_+'(\theta)=\frac{-2r_+(\theta)^{2}\cos(2\theta)+2r_+(\theta) \cosh\beta \cos\theta-1/2}{2r_+(\theta)\sin(2\theta)-2\cosh\beta\sin\theta}. \]

A simple calculation together with (2.3) gives

\begin{align*} 2r_+(\theta)\sin(2\theta)-2\cosh\beta\sin\theta& = 2\sin\theta[2r_+(\theta)\cos\theta-\cosh\beta]\\ & =2\sin\theta\sqrt{\cosh^{2}\beta-\theta/\tan\theta} \end{align*}

and

(2.14)\begin{align}& 2 r_+(\theta)\cosh\beta\cos\theta-2r_+(\theta)^{2}\cos^{2}\theta \notag\\ & =\cosh\beta(\cosh\beta+\sqrt{\cosh^{2}\beta-\theta/\tan\theta})-\frac{(\cosh\beta+\sqrt{\cosh^{2}\beta-\theta/\tan\theta})^{2}}{2}\notag\\ & =\frac{\theta}{2\tan\theta}. \end{align}

Consequently,

\begin{align*} 0& \le r_+'(\theta)=\frac{2r_+(\theta)^{2}\sin^{2}\theta+\theta/(2\tan\theta)-1/2}{2\sin\theta\sqrt{\cosh^{2}\beta-\theta/\tan\theta}}\\ & \le \frac{r_+(\theta)^{2}\sin\theta}{\sqrt{\cosh^{2}\beta-\theta/\tan\theta}}\le\frac{r_+(\theta)\tan\theta}{\tanh \beta}, \end{align*}

and

(2.15)\begin{equation} |I_c|\le\frac{2}{\tanh \beta}\int_0^{\pi/2}\frac{r_+(\theta)^{1/2}/\cos\theta}{ [f(t_-)-f(z_+(\theta))]^{p+1/2}}d\theta. \end{equation}

Let $\theta _0=\arccos (1/4)$. If $\theta \in (0,\,\theta _0)$, then $\cos \theta >1/4$, $f(t_-)-f(z_+(\theta ))>f(t_-)-f(t_+)=\sinh (2\beta )/2-\beta$ and $r_+(\theta )<4\cosh \beta$. Hence,

(2.16)\begin{equation} \int_0^{\theta_0}\frac{r_+(\theta)^{1/2}/\cos\theta}{ [f(t_-)-f(z_+(\theta))]^{p+1/2}}d\theta\le\frac{2\pi(4\cosh \beta)^{1/2}}{[\sinh(2\beta)/2-\beta]^{p+1/2}}. \end{equation}

If $\theta \in (\theta _0,\,\pi /2)$, then $r_+(\theta )>2e^{\beta }=4t_+>2\cosh \beta >2$. This together with (2.14) implies that

\begin{align*} -f(z_+(\theta))& ={-}r_+(\theta)^{2}\cos(2\theta)+2r_+(\theta)\cosh \beta \cos\theta-\frac{1}{2}\ln r_+(\theta) \\ & =r_+(\theta)^{2}+\frac{\theta}{2\tan\theta}-\frac{1}{2}\ln r_+(\theta)\nonumber\\ & \ge\frac{4}{5}r_+(\theta)^{2}+\frac{1}{5}r_+(\theta)^{2}-\frac{1}{2}\ln r_+(\theta)\ge\frac{4}{5}r_+(\theta)^{2}, \end{align*}

and

\[{-}f(t_-)=\frac{e^{{-}2\beta}}{4}+{\beta}{2}+\frac{1+\ln 2}{2}\le\frac{e^{2\beta}}{4}+\frac{19}{20}\le t_+^{2}+\frac{19}{80}r_+(\theta)^{2}\le\frac{3}{10}r_+(\theta)^{2}. \]

Coupling the above two inequalities gives

\[ f(t_-)-f(z_+(\theta))\ge\frac{1}{2}r_+(\theta)^{2}. \]

Therefore,

(2.17)\begin{align} & \int_{\theta_0}^{\pi/2}\frac{r_+(\theta)^{1/2}/\cos\theta}{ [f(t_-)-f(z_+(\theta))]^{p+1/2}}d\theta\le \int_{\theta_0}^{\pi/2}\frac{r_+(\theta)^{1/2}\cdot r_+(\theta)/t_+}{ r_+(\theta)^{2p+1}/2^{p+1/2}}d\theta\nonumber\\ & =\frac{2^{p+1/2}}{ t_+}\int_{\theta_0}^{\pi/2}r_+(\theta)^{{-}2p+1/2}d\theta \le\frac{2^{p+1/2}}{ t_+}\int_{\pi/3}^{\pi/2}(4t_+)^{{-}2p+1/2}d\theta =\frac{2^{{-}3p+1/2}\pi}{3 t_+^{2p+1/2}}, \end{align}

where we have assumed $p\ge 1$. Substituting (2.16) and (2.17) into (2.15) yields

(2.18)\begin{equation} |I_c|\le \frac{2}{\tanh\beta}\Bigg\{\frac{4\pi \sqrt{\cosh\beta}}{[\sinh(2\beta)/2-\beta]^{p+1/2}}+\frac{2^{{-}p+1}\pi}{3 e^{(2p+1/2)\beta}}\Bigg\},\quad p\ge 1. \end{equation}

A combination of (3.5), (2.13) and (2.18) gives

\begin{align*} |c_p(u)|\le|I_0|+|I_c|& \le \frac{2^{p+7/2}}{\pi^{p+1/2}}+\frac{1}{\pi(p-1)2^{p-7/2}(\cosh\beta)^{p-1/2}}\\ & \quad+\frac{2}{\tanh\beta}\Bigg\{\frac{4\pi\sqrt{\cosh\beta}}{[\sinh(2\beta)/2-\beta]^{p+1/2}}+\frac{2^{{-}p+1}\pi}{3 e^{(2p+1/2)\beta}}\Bigg\}, \end{align*}

for $p\ge 2$. Let $C_p$ be the number on the right-hand side of the above inequality. The error bound (2.10) holds for $p\ge 2$. Note from (2.6) that

\[ \varepsilon_p(N,\beta)=\frac{A_p(\coth \beta) }{ N^{p}}+\varepsilon_{p+1}(N,\beta). \]

Hence, we may improve the error bound (2.10) to

\[ |\varepsilon_p(N,\beta)|\le \frac{|A_p(\coth \beta)| }{ N^{p}}+\frac{\sqrt{\sinh\beta}}{2\pi^{3/2}} \frac{C_{p+1}\Gamma(p+3/2)}{ N^{p+1}}=\frac{\widetilde C_p }{ N^{p}}, \]

for $p\ge 1$, where

\[ \widetilde C_p=|A_p(\coth \beta)|+\frac{\sqrt{\sinh\beta}}{2\pi^{3/2}} \frac{C_{p+1}\Gamma(p+3/2)}{ N}. \]

Using the explicit value

\[{-}f(t_-) = \frac{e^{{-}2\beta}}{4}+\frac{\beta}{2}+\frac{1+\ln 2}{2} \]

in (2.6), we can summarize our results as follows.

Theorem 2.1 Let $\beta >0$ and $N=2n+1 \geq 1$. Then for any $p\ge 1,$ we have

(2.19)\begin{equation} H_n(\sqrt N\cosh \beta)=\frac{2^{n} n!e^{N(e^{{-}2\beta}+2\beta+2)/4}}{\sqrt{2\pi\sinh\beta}N^{(n+1)/2}} \left\{ \sum_{j=0}^{p-1}\frac{A_j(\coth \beta)}{N^{j}} + \varepsilon_p(N,\beta)\right\}, \end{equation}

where

(2.20)\begin{equation} |\varepsilon_p(N,\beta)|\le \frac{\widetilde C_p }{ N^{p}}, \end{equation}

with

\[ \widetilde C_p=|A_p(\coth \beta)|+\frac{\sqrt{\sinh\beta}}{2\pi^{3/2}} \frac{C_{p+1}\Gamma(p+3/2)}{ N}, \]

and

\begin{align*} C_{p+1}& =\frac{2^{p+9/2}}{\pi^{p+3/2}}+\frac{1}{\pi p2^{p-5/2}(\cosh\beta)^{p+1/2}} \\ & \quad +\frac{2}{\tanh\beta}\Bigg\{\frac{4\pi\sqrt{\cosh\beta}}{[\sinh(2\beta)/2-\beta]^{p+3/2}}+\frac{2^{{-}p}\pi}{3 e^{(2p+5/2)\beta}}\Bigg\}. \end{align*}

In particular, $\widetilde C_p\to |A_p(\coth \beta )|$ as $N\to +\infty$; namely, the error bound is close to the absolute value of the first neglected term when $N$ is large. The coefficients $A_j(\coth \beta )$ are polynomials in $\coth \beta$ of degree $3j$ with rational coefficients, and these polynomials can be calculated using a recursive formula given in appendix A.1.

3. Case II: $x=\cos \alpha \in [0,\,1)$ with $\alpha \in (0,\,\pi /2]$

The saddle points are $t_\pm =e^{\pm i\alpha }/2$, and we have

\[ f''(t_\pm)=2-2e^{{\mp} 2i\alpha}=4\sin\alpha e^{{\pm} i(\pi/2-\alpha)}. \]

Now, we define an analytic function $w(z)$ via

(3.1)\begin{equation} w(z)=[f(z)-f(t_+)]^{1/2}, \end{equation}

where the branch of the square root function is chosen so that

\[ w(z)=\sqrt{2\sin\alpha}e^{i(\pi/4-\alpha/2)}(z-t_+)+\mathcal{O}((z-t_+)^{2}) \]

for $z$ in a small complex neighbourhood of $t_+$. The steepest descent contour passing through $t_+$ is described via the equation ${\rm Im}\ [f(z)-f(t_+)]=0$. By using polar coordinates $z=re^{i\theta }$, this equation becomes

\[ r^{2}\sin(2\theta)-2r\cos\alpha\sin\theta+\frac{\theta-\theta_0}{2}=0, \]

with

\[ \theta_0=\alpha-\sin(2\alpha)/2\in(0,\alpha). \]

We are only interested in solutions in the upper half-plane; namely, $\theta \in (0,\,\pi )$. By the quadratic formula, there are exactly two solutions:

\[ r_{d+}(\theta)=\begin{cases} \displaystyle\frac{(\theta-\theta_0)/(2\sin\theta)}{\cos\alpha+\sqrt{\cos^{2}\alpha-(\theta-\theta_0)/\tan\theta}}, & ~\theta\in[\alpha,\pi),\\\\ \displaystyle\frac{\cos\alpha+\sqrt{\cos^{2}\alpha-(\theta-\theta_0)/\tan\theta}}{2\cos\theta}, & ~\theta\in(0,\alpha], \end{cases} \]

and

\[ r_{a+}(\theta)=\begin{cases} \displaystyle\frac{\cos\alpha+\sqrt{\cos^{2}\alpha-(\theta-\theta_0)/\tan\theta}}{2\cos\theta}, & ~\theta\in[\alpha,\pi/2),\\\\ \displaystyle\frac{\cos\alpha-\sqrt{\cos^{2}\alpha-(\theta-\theta_0)/\tan\theta}}{2\cos\theta}, & ~\theta\in(\theta_0,\alpha], \end{cases} \]

where $r_{d+}(\theta )$ and $r_{a+}(\theta )$ are defined as piecewise functions so that they are differentiable at $\theta =\alpha$. This is because the function $\sqrt {\cos ^{2}\alpha -(\theta -\theta _0)/\tan \theta }$ is not differentiable at $\theta =\alpha$. For any $\theta$ in a small real neighbourhood of $\alpha$, we have, by Taylor expansion, $\cos ^{2}\alpha -(\theta -\theta _0)/\tan \theta =(\theta -\alpha )^{2}+\mathcal {O}((\theta -\alpha )^{3})$ and $\sqrt {\cos ^{2}\alpha -(\theta -\theta _0)/\tan \theta }=|\theta -\alpha |+\mathcal {O}((\theta -\alpha )^{2})$. Hence,

\begin{align*} r_{d+}(\theta)& =\frac{\cos\alpha-(\theta-\alpha)}{2[\cos\alpha-(\theta-\alpha)\sin\alpha]}+\mathcal{O}((\theta-\alpha)^{2}) \\ & =\frac{1}{2}-\frac{1-\sin\alpha}{2\cos\alpha}(\theta-\alpha)+\mathcal{O}((\theta-\alpha)^{2}),\\ r_{a+}(\theta)& =\frac{\cos\alpha+(\theta-\alpha)}{2[\cos\alpha-(\theta-\alpha)\sin\alpha]}+\mathcal{O}((\theta-\alpha)^{2}) \\ & =\frac{1}{2}+\frac{1+\sin\alpha}{2\cos\alpha}(\theta-\alpha)+\mathcal{O}((\theta-\alpha)^{2}). \end{align*}

In particular, we have

\[ r_{d+}(\alpha)=r_{a+}(\alpha)=\frac{1}{2},\quad r_{d+}'(\alpha)={-}\frac{1}{2}\tan\left(\frac{\pi}{4}-\frac{\alpha}{2}\right),\ r_{a+}'(\alpha)=\frac{1}{2}\cot\left(\frac{\pi}{4}-\frac{\alpha}{2}\right). \]

Now, we define two contours

\[ \Gamma_{d+}=\{r_{d+}(\theta)e^{i\theta}:\theta\in(0,\pi)\},\quad \Gamma_{a+}=\{r_{a+}(\theta)e^{i\theta}:\theta\in(\theta_0,\pi/2)\}. \]

Recall that $-Nf(t)$ is the phase function in the integral representation of $H_n(\sqrt N \cos \alpha )$. It is readily seen from the asymptotic behaviours $f(z)\sim z^{2}$ as $z\to \infty$ and $2f(z)\sim \ln z$ as $z\to 0$ that $\Gamma _{d+}$ is a steepest descent contour and $\Gamma _{a+}$ is a steepest ascent contour passing through the saddle point $t_+=e^{i\alpha }/2$. Moreover, the only zero of $f(z)-f(t_+)$ in the upper half-plane is $t_+$. Hence, $w(z)$ defined in (3.1) is analytic for all ${\rm Im}\ z>0$ and positive $z>0$. For $z=se^{i\pi }$ with $s>0$, we define

\[ w(se^{i\pi})=\lim_{\theta\to\pi^{-}}w(se^{i\theta}). \]

We can define in a similar manner the contours $\Gamma _{d-}$ and $\Gamma _{a-}$ which, together with $\Gamma _{d+}$ and $\Gamma _{a+}$, are illustrated in figure 3.

FIGURE 3.

The contours $\Gamma _{d\pm }$ and $\Gamma _{a\pm }$. The arrows indicate the increasing direction of the phase function $f$.

Now, we return to the integral representation of $H_n(\sqrt N \cos \alpha )$ in (1.1) and deform the contour of integration to the steepest descent contours $\Gamma _{d+}\cup \Gamma _{d-}$ oriented with increasing parameter $\theta$ (i.e., counter-clockwise direction):

\begin{align*} H_n(\sqrt N \cos\alpha)& =\frac{n!}{2\pi iN^{n/2}}\int_{\Gamma_{d+}} e^{{-}Nf(t)}t^{{-}1/2}dt\\ & \quad +\frac{n!}{2\pi iN^{n/2}}\int_{\Gamma_{d-}} e^{{-}Nf(t)}t^{{-}1/2}dt=: I_+ + I_-, \end{align*}

where the last equality defines the two integrals $I_+$ and $I_-$, respectively. Since the contours $\Gamma _{d+}$ and $\Gamma _{d-}$ are symmetric with respect to the real axis, have opposite orientations, and $f(t) = \overline {f(\bar t)}$, the integrals $I_+$ and $I_-$ are complex conjugates of each other. Therefore, it suffices to study the integral $I_+$. First, we introduce the change of variable $u=w(t)$:

\[ I_+ = -\frac{n!e^{{-}Nf(t_+)}}{2\pi iN^{n/2}}\int_{\mathbb{R}} e^{{-}Nu^{2}}\frac{[w^{{-}1}(u)]^{{-}1/2}}{ w'(w^{{-}1}(u))}du. \]

Since $w(z)$ is analytic in the upper half-plane and $|w(z)|/|z|\to 1$ as $z\to \infty$, we can infer from the Cauchy integral formula that

\[ \frac{[w^{{-}1}(u)]^{{-}1/2}}{ w'(w^{{-}1}(u))}=\frac{1}{2\pi i}\int_{\Gamma_0^{+}\cup\mathbb{R}^{+}}\frac{z^{{-}1/2}}{ w(z)-u}dz, \]

where $\mathbb {R}^{+}$ is the positive real line and $\Gamma _0^{+}$ is the negative real line with argument $\pi$:

\[ \Gamma_0^{+}=\lim_{\delta\to 0^{+}}\{\lambda+i\delta:\lambda<0\}=\{se^{i\pi}:s>0\}. \]

Both $\mathbb {R}^{+}$ and $\Gamma _0^{+}$ are oriented from left to right. Now, we have the following double integral representation:

\[ I_+ = -\frac{n!e^{{-}Nf(t_+)}}{(2\pi i)^{2}N^{n/2}}\int_{\mathbb{R}} e^{{-}Nu^{2}}\int_{\Gamma_0^{+}\cup\mathbb{R}^{+}}\frac{z^{{-}1/2}}{ w(z)-u}dzdu. \]

For any $p\ge 0$, we use the identities (2.5) and

\[ \frac{1}{ w(z)-u}=\sum_{k=1}^{2p}\frac{u^{k-1}}{ w(z)^{k}}+\frac{u^{2p}}{ w(z)^{2p}[w(z)-u]} \]

to derive

(3.2)\begin{equation} I_+ = \frac{n!e^{{-}Nf(t_+)}}{2\sqrt{\pi\sin\alpha}N^{(n+1)/2}}\left\{\sum\limits_{j=0}^{p-1}\frac{A_j(i\cot\alpha)}{N^{j}}e^{\pi i/4}+\widetilde{\varepsilon}_p(N,\alpha)\right\}, \end{equation}

where

(3.3)\begin{align} A_j(i\cot\alpha)& =e^{-\pi i/4}\frac{\sqrt{\sin\alpha}\, \Gamma(j + 1/2)}{2\pi^{3/2}}\int_{\Gamma_0^{+}\cup\mathbb{R}^{+}}\frac{z^{{-}1/2}}{ w(z)^{2j+1}}dz, \end{align}

(3.4)\begin{align} \widetilde{\varepsilon}_p(N,\alpha) & =\frac{\sqrt{N\sin \alpha}}{2\pi^{3/2}}\int_\mathbb{R} e^{{-}Nu^{2}}u^{2p}\int_{\Gamma_0^{+}\cup\mathbb{R}^{+}}\frac{z^{{-}1/2}}{ w(z)^{2p}[w(z)-u]}dzdu. \end{align}

The coefficients $A_j(i\cot \alpha )$ are polynomials in $i\cot \alpha$ and these polynomials are identical to those appearing in the expansion (2.19) (see appendix A.1 for details).

To find an upper bound for the remainder $\widetilde {\varepsilon }_p(N,\,\alpha )$, we need to estimate the integral

(3.5)\begin{equation} c_p(u)=\int_{\Gamma_0^{+}\cup\mathbb{R}^{+}}\frac{z^{{-}1/2}}{ w(z)^{2p}[w(z)-u]}dz,\quad u\in\mathbb{R}. \end{equation}

By expressing $w(z)$ in polar coordinates, one can easily show that

\[ |w(z)-u|\ge|{\rm Im}\ w(z)|\ge\frac{|{\rm Im}\ w(z)^{2}|}{2|w(z)|}. \]

Note from (3.1) that

\[ |{\rm Im}\ w(z)^{2}|=|{\rm Im}\ f(z)-\theta_0/2|=\begin{cases} \theta_0/2, & ~z\in\mathbb{R}^{+},\\ (\pi-\theta_0)/2, & ~z\in\Gamma_0^{+}. \end{cases} \]

Assume that $p\geq 1$. Substituting the above two formulas into (2.9) yields

\[ |c_p(u)|\le \frac{4}{\theta_0}\int_0^{+\infty} \frac{s^{{-}1/2}}{|w(s)|^{2p-1}}ds+\frac{4}{\pi-\theta_0}\int_0^{+\infty} \frac{s^{{-}1/2}}{|w(se^{i\pi})|^{2p-1}}ds. \]

For $s\in (0,\,2)$, we have

\begin{align*} |w(s)|& \ge\sqrt{|{\rm Im}\ w(s)^{2}|}=\sqrt{\theta_0/2},\\ |w(se^{i\pi})|& \ge \sqrt{|{\rm Im}\ w(se^{i\pi})^{2}|}=\sqrt{(\pi-\theta_0)/2}, \end{align*}

and for $s>2$, we have

\begin{align*} |w(s)|^{2}& \ge|{\rm Re}\ w(s)^{2}|=s^{2}-2s\cos \alpha +\frac{\ln s}{2}+\frac{\cos^{2}\alpha +\ln 2+1/2}{2}\ge(s-1)^{2},\\ |w(se^{i\pi})|^{2}& \ge |{\rm Re}\ w(se^{i\pi})^{2}|=s^{2}+2s\cos \alpha +\frac{\ln s}{2}+\frac{\cos^{2}\alpha +\ln 2+1/2}{2}\ge s^{2}. \end{align*}

Accordingly,

\begin{align*} |c_p(u)| & \le \frac{4}{\theta_0}\left\{\int_0^{2}\frac{s^{{-}1/2}}{(\theta_0/2)^{p-1/2}}ds+\int_2^{+\infty}\frac{s^{{-}1/2}}{ (s-1)^{2p-1}}ds\right\}\\ & \quad+\frac{4}{\pi-\theta_0}\left\{\int_0^{2} \frac{s^{{-}1/2}}{[(\pi-\theta_0)/2]^{p-1/2}}ds+\int_2^{+\infty}\frac{s^{{-}1/2}}{ s^{2p-1}}ds\right\}\\ & \le \frac{4}{\theta_0}\left\{\frac{2^{3/2}}{(\theta_0/2)^{p-1/2}}+\frac{1}{2p-3/2}\right\}\\ & \quad +\frac{4}{\pi-\theta_0}\left\{\frac{2^{3/2}}{[(\pi-\theta_0)/2]^{p-1/2}}+\frac{2^{3/2-2p}}{2p-3/2}\right\}. \end{align*}

Denote the right-hand side of the above inequality by $C_p$. It then follows from (2.5), (3.4) and (2.9) that

\[ |\widetilde{\varepsilon}_p(N,\alpha)|\le \frac{\sqrt{\sin\alpha}}{2\pi^{3/2}}\frac{C_p\Gamma(p+1/2)}{ N^{p}}, \]

provided $p \geq 1$. We define

\[ \varepsilon_p(N,\alpha ) := {\rm Re}(\widetilde{\varepsilon}_p(N,\alpha)e^{{-}iN\theta_0/2}), \]

and observe that

\begin{align*} \left|\varepsilon_p(N,\alpha)\right| & \le \left|{\rm Re}\left(A_p(i\cot\alpha)e^{{-}i(\theta_0 N-\pi/2)/2}\right)\right|\frac{1}{N^{p}} + \left|{\rm Re}\widetilde{\varepsilon}_{p+1}(N,\alpha)\right| \\ & \le \left|{\rm Re}\left(A_p(i\cot\alpha)e^{{-}i(\theta_0 N-\pi/2)/2}\right)\right|\frac{1}{N^{p}} + \left|\widetilde{\varepsilon}_{p+1}(N,\alpha)\right| \end{align*}

for any $p\geq 0$. Using the explicit value

\[{-}f(t_+) =\frac{\cos(2\alpha)}{4}+\frac{1+\ln 2}{2} - i\frac{\theta_0}{2} \]

in (3.2), and the fact that $H_n(\sqrt N \cos \alpha )=I_+ + I_- = I_+ +\overline {I_+}=2{\rm Re}\ I_+$, we can summarize our results as follows.

Theorem 3.1 Let $\alpha \in (0,\,\pi /2],$ $N=2n+1 \geq 1$ and $\theta _0 =\alpha -\sin (2\alpha )/2$. Then for any $p\ge 0,$ we have

\begin{align*} H_n(\sqrt N \cos \alpha) & =\frac{2^{n+1}n!e^{N(\cos(2\alpha )+2)/4}}{\sqrt{2\pi\sin\alpha} N^{(n + 1)/2}}\\ & \quad \times \left\{\sum\limits_{j=0}^{p-1} \frac{{\rm Re}\left(A_j(i\cot\alpha)e^{{-}i(\theta_0 N-\pi/2)/2}\right)}{N^{j}}+\varepsilon_p(N,\alpha) \right\} \\ & = \frac{2^{n+1}n!e^{N(\cos(2\alpha )+2)/4}}{\sqrt{2\pi\sin\alpha} N^{(n + 1)/2}}\left\{\cos\left(\frac{\theta_0 N}{2}-\frac{\pi}{4}\right)\sum\limits_{j=0}^{p-1}\frac{{\rm Re}(A_j(i\cot\alpha))}{N^{j}} \right.\\ & \quad\left.+\sin\left(\frac{\theta_0 N}{2}-\frac{\pi}{4}\right) \sum\limits_{j=1}^{p-1} \frac{{\rm Im}(A_j (i\cot \alpha ))}{N^{j}}+\varepsilon_p(N,\alpha) \right\}, \end{align*}

where

(3.6)\begin{equation} |\varepsilon_p (N,\alpha)|\le \frac{\widetilde C_p}{ N^{p}}, \end{equation}

with

\[ \widetilde C_p=\left|{\rm Re}\left(A_p(i\cot\alpha)e^{{-}i(\theta_0 N-\pi/2)/2}\right)\right|+\frac{\sqrt{\sin\alpha}}{2\pi^{3/2}}\frac{C_{p+1} \Gamma(p+3/2)}{ N}, \]

and

\[ C_{p+1}=\frac{4}{\theta_0}\left\{\frac{2^{3/2}}{(\theta_0/2)^{p+1/2}}+\frac{1}{2p+1/2}\right\} +\frac{4}{\pi-\theta_0}\left\{\frac{2^{3/2}}{[(\pi-\theta_0)/2]^{p+1/2}}+\frac{2^{{-}1/2-2p}}{2p+1/2}\right\}. \]

In particular, $\widetilde C_p\to |{\rm Re} (A_p(i\cot \alpha )e^{-i(\theta _0 N-\pi /2)/2})|$ as $N\to +\infty ;$ namely, the error bound is close to the absolute value of the first omitted term for large $N$. The coefficients $A_j(i\cot \alpha )$ are polynomials in $i\cot \alpha$ of degree $3j$ with rational coefficients, and these polynomials can be calculated using a recursive formula given in appendix A.1.

4. Case III: $x=1$

The saddle points coincide $t_s=t_\pm =1/2$. We have $f(t_s)=-3/4-(\ln 2)/2$, $f'(t_s)=f''(t_s)=0$ and $f'''(t_s)=8$. We introduce the analytic function

(4.1)\begin{equation} w(z)=[f(z)-f(t_s)]^{1/3}, \end{equation}

with the branch of the cube root being chosen so that

(4.2)\begin{equation} w(z)=(4/3)^{1/3}(z-t_s)e^{{-}2i\pi/3}+\mathcal{O}((z-t_s)^{2}) \end{equation}

for $z$ in a small complex neighbourhood of $t_s$. We shall extend the definition of $w(z)$ by analytic continuation to $\mathbb {C}\setminus (-\infty,\,0]$. First, we investigate the equation ${\rm Im} [f(z)-f(t_s)]={\rm Im}\ f(z)=0$, which in polar coordinates $z=re^{i\theta }$ is

(4.3)\begin{equation} r^{2}\sin(2\theta)-2r\sin\theta+\theta/2=0. \end{equation}

This equation is clearly satisfied by all $z>0$. There are two further solutions:

\[ r_+(\theta)=\begin{cases} \displaystyle\frac{\theta/(2\sin\theta)}{1+\sqrt{1-\theta\cot\theta}}, & ~\theta\in(0,\pi),\\ \displaystyle\frac{1}{2}, & ~\theta=0,\\ \displaystyle\frac{1+\sqrt{1-\theta\cot\theta}}{2\cos\theta}, & ~\theta\in(-\pi/2,0), \end{cases} \]

and

\[ r_-(\theta)=\begin{cases} \displaystyle\frac{1+\sqrt{1-\theta\cot\theta}}{2\cos\theta}, & ~\theta\in(0,\pi/2),\\ \displaystyle\frac{1}{2}, & ~\theta=0,\\ \displaystyle\frac{\theta/(2\sin\theta)}{1+\sqrt{1-\theta\cot\theta}}, & ~\theta\in(-\pi,0), \end{cases} \]

where we have defined $r_\pm (\theta )$ as piecewise functions so that they are differentiable at $\theta =0$. Note that since $1-\theta \cot \theta =\theta ^{2}/3+\mathcal {O}(\theta ^{3})$ for small $|\theta |>0$, we have

\[ \sqrt{1-\theta\cot\theta}=\frac{|\theta|}{\sqrt 3}+\mathcal{O}(\theta^{2}) \]

and

\[ r_\pm(\theta)=\frac{1}{2}\mp\frac{\theta}{2\sqrt 3}+\mathcal{O}(\theta^{2}) \]

for small values of $\theta$. Especially,

(4.4)\begin{equation} r_\pm(0)=\frac{1}{2},\quad r_\pm'(0)={\mp}\frac{1}{2\sqrt 3}. \end{equation}

Now, we introduce two contours

\[ \Gamma_1 =\{r_+(\theta)e^{i\theta}:\theta\in(-\pi/2,\pi)\},\quad \Gamma_2 =\{r_-(\theta)e^{i\theta}:\theta\in(-\pi,\pi/2)\}, \]

which we illustrate in figure 4.

FIGURE 4.

The contours $\Gamma _1$ and $\Gamma _2$. The arrows indicate the increasing direction of the phase function $f$.

We have the following lemma.

Proof. For small $\theta$, we have from $f'(t_s)=f''(t_s)=0$, $f'''(t_s)=8$ and (4.4) that

\[ z_+(\theta)-t_s=[r_+'(0)+ir_+(0)]\theta+\mathcal{O}(\theta^{2})=\frac{e^{2i\pi/3}}{\sqrt 3}\theta+\mathcal{O}(\theta^{2}), \]

and

\[ f(z_+(\theta))-f(t_s)=\frac{4}{3}(z_+(\theta)-t_s)^{3}+\mathcal{O}((z_+(\theta)-t_s)^{4}) =\frac{4}{9\sqrt 3}\theta^{3}+\mathcal{O}(\theta^{4}). \]

Hence, the function $f(z_+(\theta ))-f(t_s)$ is increasing for small $\theta$. We claim that $f(z_+(\theta ))-f(t_s)$ is increasing for all $\theta \in (-\pi /2,\,\pi )$. Assume to the contrary that

\[ 0=\frac{d}{ d\theta}f(z_+(\theta))=f'(z_+(\theta))[r_+'(\theta)+ir_+(\theta)]e^{i\theta} \]

for some $\theta \in (-\pi /2,\,0)\cup (0,\,\pi )$. Since $f'(z)\neq 0$ for any $z\neq t_s$, we have $r_+'(\theta )=r_+(\theta )=0$. This contradicts to the equation (4.3). Hence, the function $f(z_+(\theta ))-f(t_s)$ is increasing for all $\theta \in (-\pi /2,\,\pi )$. Note that $r_+(\theta )\to \infty$ as $\theta$ tends to $-\pi /2$ from the right or $\pi$ from the left. It is readily seen that $f(z_+(\theta ))-f(t_s)=r_+(\theta )^{2}\cos (2\theta )-2r_+(\theta )\cos \theta +(1/2)\ln r_+(\theta )-f(t_s)$ increases from $-\infty$ to $+\infty$ as $\theta$ increases from $-\pi /2$ to $\pi$.

The second statement of the lemma follows in a similar manner.

Since the solutions of ${\rm Im}\ f(z)=0$ are the union $\Gamma _1\cup \Gamma _2\cup \mathbb {R}^{+}$, the equation $f(z)-f(t_s)=0$ has a unique solution $z=t_s$ in $\mathbb {C}\setminus (-\infty,\,0]$. This implies that $w(z)$ defined in (4.1) can be analytically continued to $\mathbb {C}\setminus (-\infty,\,0]$. Moreover, it follows from (4.2) that $w(z_+(\theta ))$ increases from $-\infty$ to $\infty$ as $\theta$ increases from $-\pi /2$ to $\pi$.

Next, we let $\Gamma _{1+}$ be the portion of $\Gamma _1$ in the upper half-plane and $\Gamma _{2-}$ be the portion of $\Gamma _2$ in the lower half-plane; namely,

\[ \Gamma_{1+}=\{r_+(\theta)e^{i\theta}:\theta\in[0,\pi)\},\quad \Gamma_{2-}=\{r_-(\theta)e^{i\theta}:\theta\in(-\pi,0]\}. \]

Both $\Gamma _{1+}$ and $\Gamma _{2-}$ are oriented with increasing $\theta$ (counter clockwise direction). We now deform the contour of integration for $H_n(\sqrt N)$ in (1.1) from $\Gamma$ to $\Gamma _{1+}\cup \Gamma _{2-}$:

\begin{align*} H_n(\sqrt N)& = \frac{n!}{2\pi iN^{n/2}}\int_{\Gamma_{1+}} e^{{-}Nf(t)}t^{{-}1/2}dt\\ & \quad + \frac{n!}{2\pi iN^{n/2}}\int_{\Gamma_{2-}} e^{{-}Nf(t)}t^{{-}1/2}dt =:I_+ + I_-, \end{align*}

where the last equality defines the two integrals $I_+$ and $I_-$, respectively. Since the contours $\Gamma _{1+}$ and $\Gamma _{2-}$ are symmetric with respect to the real axis, have opposite orientations, and $f(t) = \overline {f(\bar t)}$, the integrals $I_+$ and $I_-$ are complex conjugates of one another. Thus, it is sufficient to study the integral $I_+$. First, we introduce the change of variable $u = w(t)$:

(4.5)\begin{equation} I_+ = \frac{n!e^{{-}Nf(t_s)}}{2\pi iN^{n/2}} \int_0^{+\infty} e^{{-}Nu^{3}}\frac{[w^{{-}1}(u)]^{{-}1/2}}{ w'(w^{{-}1}(u))}du. \end{equation}

We denote by $\Gamma _0^{\pm }$ the negative real lines with arguments $\pm \pi$ and orientation from left to right. For convenience, we denote by $\Gamma _*$ the union of $\Gamma _0^{+}$ oriented to the right and $\Gamma _0^{-}$ oriented to the left. Since $w(z)$ is analytic in $\mathbb {C}\setminus (-\infty,\,0]$ and $|w(z)|/|z|^{2/3}\to 1$ as $z\to \infty$, we can assert from the Cauchy integral formula that

\[ \frac{[w^{{-}1}(u)]^{{-}1/2}}{ w'(w^{{-}1}(u))} =\frac{1}{2\pi i}\int_{\Gamma_*}\frac{z^{{-}1/2}}{ w(z)-u}dz. \]

Substituting this into (4.5) gives

\[ I_+ = \frac{n!e^{{-}Nf(t_s)}}{(2\pi i)^{2}N^{n/2}}\int_0^{+\infty} e^{{-}Nu^{3}} \int_{\Gamma_*}\frac{z^{{-}1/2}}{ w(z)-u}dzdu. \]

For any $p\ge 1$, we use the identities

\[ \frac{1}{ w (z)-u}=\sum_{j=1}^{p-1}\frac{u^{j-1}}{ w (z)^{j}}+\frac{u^{p-1}}{ w (z)^{p-1}[w (z)-u]} \]

and

(4.6)\begin{equation} \int_0^{+\infty} e^{{-}Nu^{3}}u^{j-1}du=\frac{\Gamma(j/3)}{3N^{j/3}} \end{equation}

to find

(4.7)\begin{equation} I_+ = \frac{2^{1/2}n!e^{{-}Nf(t_s)}}{6\pi N^{n/2}}\left\{ \sum\limits_{j = 1}^{p-1} \left(\frac{3}{4}\right)^{j/3} D_j e^{2\pi ij/3-\pi i/2}\frac{\Gamma(j/3)}{N^{j/3}} + \widetilde{\varepsilon}_p(N) \right\}, \end{equation}

where

(4.8)\begin{align} D_j & = \left(\frac{4}{3}\right)^{j/3}\frac{e^{{-}2\pi ij/3}}{2^{3/2}\pi i}\int_{\Gamma_*}\frac{z^{{-}1/2}}{ w(z)^{j}}dz, \end{align}

(4.9)\begin{align} \widetilde{\varepsilon}_p(N) & ={-}\frac{3}{2^{3/2}\pi}\int_\mathbb{R} e^{{-}Nu^{3}}u^{p-1}\int_{\Gamma_*}\frac{z^{{-}1/2}}{ w(z)^{p-1}[w(z)-u]}dzdu. \end{align}

It is shown in appendix A.2 that the coefficients $D_j$ are rational numbers which can be calculated via a recurrence relation.

To find an upper bound for the remainder $\widetilde {\varepsilon }_p(N)$, we need to estimate the integral

(4.10)\begin{equation} c_p(u)=\int_{\Gamma_*}\frac{z^{{-}1/2}}{ w(z)^{p-1}[w(z)-u]}dz,\quad u\in\mathbb{R}. \end{equation}

By expressing $w(z)$ in polar coordinates, one can easily show that

\[ |w(z)-u|\ge|{\rm Im}\ w(z)|\ge\frac{|{\rm Im}\ w(z)^{3}|}{3|w(z)|^{2}}. \]

We also note from (4.1) that

\[ |{\rm Im}\ w(z)^{3}|=|{\rm Im}\ f(z)|=\pi/2, \]

for $z=se^{\pm i\pi }\in \Gamma _*$. Assume that $p\ge 4$. Substituting the above two formulas into (4.10) yields

\[ |c_p(u)|\le \frac{6}{\pi}\int_0^{+\infty} \frac{s^{{-}1/2}}{|w(se^{i\pi})|^{p-3}}ds+\frac{6}{\pi}\int_0^{+\infty} \frac{s^{{-}1/2}}{|w(se^{{-}i\pi})|^{p-3}}ds. \]

For $z=se^{\pm i\pi }\in \Gamma _*$ with $s=|z|\in (0,\,1)$, we have

\[ |w(z)|\ge|{\rm Im}\ w(z)^{3}|^{1/3}=(\pi/2)^{1/3}. \]

For $z=se^{\pm i\pi }\in \Gamma _*$ with $s=|z|>1$, we have

\[ |w(z)|^{3}\ge|{\rm Re}\ w(z)^{3}|=s^{2}+2s+\frac{\ln s}{2}+\frac{\ln 2+3/2}{2}\ge s^{2}. \]

Consequently,

\begin{align*} |c_p (u)|& \le \frac{12}{\pi}\left\{\int_0^{1}{s^{{-}1/2}}{(\pi/2)^{p/3-1}}ds+\int_1^{+\infty}\frac{s^{{-}1/2}}{ s^{2p/3-2}}ds\right\}\\ & \le\frac{12}{\pi}\left\{\frac{2}{(\pi/2)^{p/3-1}}+\frac{6}{4p-15}\right\}. \end{align*}

Denote the right-hand side of the above inequality by $C_p$. It then follows from (4.6), (4.9) and (4.10) that

\[ \left|\widetilde{\varepsilon}_p(N)\right| \le \frac{1}{2^{3/2}\pi}\frac{C_p\Gamma(p/3)}{N^{p/3}}, \]

provided $p \geq 4$. We define

\[ \varepsilon_p (N) := {\rm Re}\ \widetilde{\varepsilon}_p (N), \]

and observe that

\begin{align*} \left|\varepsilon_p(N)\right| & \le \left|\left(\frac{3}{4}\right)^{p/3}D_p\sin\left(\frac{2\pi p}{3}\right)\right|\frac{\Gamma(p/3)}{N^{p/3}} + \left|{\rm Re}\widetilde{\varepsilon}_{p+1}(N)\right| \\ & \le \left|\left(\frac{3}{4}\right)^{p/3} D_p \sin\left(\frac{2\pi p}{3}\right)\right|\frac{\Gamma(p/3)}{N^{p/3}} + \left|\widetilde{\varepsilon}_{p + 1}(N)\right| \end{align*}

for any $p\ge 3$. Using the explicit value $-f(t_s) = 3/4+(\ln 2)/2$ in (4.7), and the fact that $H_n(\sqrt N)=I_+ + I_- = I_+ +\overline {I_+}=2{\rm Re}\ I_+$, we can summarize our results as follows.

Theorem 4.2 Denote $N=2n+1 \geq 1$. Then for any $p\ge 3,$ we have

\[ H_n(\sqrt N)= \frac{2^{n + 1}n!e^{3N/4}}{3\pi N^{n/2}}\left\{\sum\limits_{j=1}^{p-1} \left(\frac{3}{4}\right)^{j/3} D_j \sin \left(\frac{2\pi j}{3}\right)\frac{\Gamma(j/3)}{N^{j/3}} + \varepsilon_p (N)\right\}, \]

where

(4.11)\begin{equation} \left|\varepsilon_p(N)\right| \le \frac{\widetilde C_p}{N^{p/3}}, \end{equation}

with

\[ \widetilde C_p = \left|\left(\frac{3}{4}\right)^{p/3} D_p\sin\left(\frac{2\pi p}{3}\right)\right|\Gamma(p/3)+\frac{1}{2^{3/2}\pi}\frac{C_{p+1}\Gamma((p+1)/3)}{N^{1/3}}, \]

and

\[ C_{p+1}=\frac{12}{\pi}\left\{\frac{2}{(\pi/2)^{(p-2)/3}}+\frac{6}{4p-11}\right\}. \]

In particular, if $p$ is not divisible by $3$, $\widetilde C_p \to \left |(3/4)^{p/3}D_p\sin (2\pi p/3)\right |\Gamma (p/3)$ as $N \to +\infty ;$ namely, the error bound is close to the absolute value of the first neglected term when $N$ is large. The coefficients $D_j$ are rational numbers and can be calculated using a recurrence relation given in appendix A.2.

5. Numerical examples

In tables 1, 2 and 3, we present some numerical results that demonstrate the accuracy of our error bounds given in theorems 2.1, 3.1 and 4.2, respectively. We can infer from the tables that the bounds are rather realistic, that is, they do not seriously overestimate the actual error.

TABLE 1.

Bounds for $\left |\varepsilon _p(N,\,\beta )\right |$ with various $N=2n+1$, $\beta$ and $p$, using (2.20)

TABLE 2.

Bounds for $\left |\varepsilon _p(N,\,\alpha )\right |$ with various $N=2n+1$, $\alpha$ and $p$, using (3.6)

TABLE 3.

Bounds for $\left |\varepsilon _p(N)\right |$ with various $N=2n+1$ and $p$, using (4.11)