&=&{{\mathrm O}}(\sqrt{v}) \qquad\mbox{as $v\to\infty$,}\label{eq:5:21}\end{aligned}$$ uniformly for $|\zeta|<\sqrt{v}$. Substituting [(\[eq:5:17\])]{} and [(\[eq:erfc\])]{} into [(\[eq:5:2\])]{} gives $$I_2(\beta,\zeta)=1-\frac{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{v\sqrt{\pi}}
\exp\left(\frac{-\zeta^2}{4{{\mathrm i}}\beta{\bar{L}}}\right)-
\frac{\zeta}{v}{\mathop{\rm erf}}\left(\frac{\zeta}{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}\right)
+\frac{R_2(\zeta)}{v^2},
\label{eq:5:22}$$ where $R_2$ is a combination of the errors in [(\[eq:5:17\])]{} and [(\[eq:5:21\])]{}. So $$|R_2(\zeta)|={{\mathrm O}}(1+\zeta^2)\qquad\mbox{as $v\to\infty$.}
\label{eq:5:222}$$

We thus have that $$\begin{aligned}
  \frac{1}{\sqrt{v}}\Bigg|\frac{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{\sqrt{\pi}}
&\exp\left(\frac{-\zeta^2}{4{{\mathrm i}}\beta{\bar{L}}}\right)+
{\zeta}{\mathop{\rm erf}}\left(\frac{\zeta}{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}\right)
-\frac{R_2(\zeta)}{v} \Bigg| \nonumber  \\
&\leq\frac{2\sqrt{\beta{\bar{L}}}}{\sqrt{\pi v}}+\frac{|\zeta|}{\sqrt{v}}
+\frac{|R_2|}{v^{3/2}} \nonumber \\
&<2\qquad\mbox{for all $\zeta\in[-\sqrt{v},\sqrt{v}]$ and $v$ sufficiently 
large.}
\label{eq:5:6}\end{aligned}$$

We are now in a position to understand the asymptotics of $(I_2(\beta,\zeta))^{v-1}$. We first consider $$\ln\left(1-\frac{a}{v}\right)^{v-1}=(v-1)\left(-\frac{a}{v}
+\frac{R_3(a)}{v^2}\right), \label{eq:5:7}$$ where $$|R_3(a)|\leq\frac{|a|^2}{1-|a|/v}$$ provided that $|a|<v$. We take $$a=\frac{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{\sqrt{\pi}}
\exp\left(\frac{-\zeta^2}{4{{\mathrm i}}\beta{\bar{L}}}\right)+
{\zeta}{\mathop{\rm erf}}\left(\frac{\zeta}{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}\right)
+\frac{R_2(\zeta)}{v}$$ which satisfies $|a|/\sqrt{v}<2$ for $v$ sufficiently large by [(\[eq:5:6\])]{} and $a={{\mathrm O}}(|\zeta|)$ as $|\zeta|\to\infty$ for $|\zeta|<\sqrt{v}$ by [(\[eq:5:222\])]{}. We see that $$\frac{|R_3(a)|}{v}={{\mathrm O}}(|a|^2/v)\qquad\mbox{as $v\to\infty$,}$$ so exponentiation of [(\[eq:5:7\])]{} gives $$\left(1-\frac{a}{v}\right)^{v-1}={{\mathrm e}}^{-a}\left(1+{{\mathrm O}}(v^{-1/2})\right)
\left(1+{{\mathrm O}}((1+\zeta^2)/v)\right),
\label{eq:5:28}$$ where the error estimates are uniform for $|\zeta|<\sqrt{v}$ as $v\to\infty$ and $${{\mathrm e}}^{-a}=\exp\left(-\frac{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{\sqrt{\pi}}
\exp\left(\frac{-\zeta^2}{4{{\mathrm i}}\beta{\bar{L}}}\right)-
{\zeta}{\mathop{\rm erf}}\left(\frac{\zeta}{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}\right) \right)
\left[1+{{\mathrm O}}((1+\zeta^2)/v)\right].$$ We can also write $$I_1(\beta,\zeta)=\frac{v}{\sqrt{\pi}}\frac{1}{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}
\exp\left(\frac{-\zeta^2}{4{{\mathrm i}}\beta{\bar{L}}}\right)\left(1+{{\mathrm O}}(v^{-2})\right).
\label{eq:5:29}$$ Thus, taking [(\[eq:5:29\])]{} together with [(\[eq:5:22\])]{} and [(\[eq:5:28\])]{} gives the required estimate.

$\square$

\[lem:5:7\] Let $|\zeta|>\sqrt{v}$, then $$|I_1(\beta,\zeta)(I_2(\beta,\zeta))^{v-1}|\leq\frac{1}{\pi}\left(
\frac{\beta{\bar{L}}}{v^2}+\frac{\zeta^2}{\beta{\bar{L}}}\right)^{-1}+{{\mathrm O}}(v\zeta^{-3})
\qquad\mbox{as $|\zeta|\to\infty$.}$$

[*Proof.*]{}

We make use of the asymptotic expression $${\mathop{\rm 