erfc}}(z)=\frac{1}{\sqrt{\pi}}{{\mathrm e}}^{-z^2}\left(\frac{1}{z}+{{\mathrm O}}(z^{-3})\right)$$ as $z\to\infty$, valid for $|\arg z|<3\pi/4$. We see that $$\begin{aligned}
{{\mathrm e}}^{\zeta/v}{\mathop{\rm erfc}}\left(\frac{\zeta}{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}+\frac{
\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{v}\right)=\frac{1}{\sqrt{\pi}}\exp\left(\frac{-\zeta^2}
{4{{\mathrm i}}\beta{\bar{L}}}+\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}\right) \\
\times\left[\left(
\frac{\zeta}{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}+\frac{
\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{v}\right)^{-1}+{{\mathrm O}}(\zeta^{-3})\right].\end{aligned}$$ Similarly, $$\begin{aligned}
{{\mathrm e}}^{-\zeta/v}{\mathop{\rm erfc}}\left(\frac{-\zeta}{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}+\frac{
\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{v}\right)=\frac{1}{\sqrt{\pi}}\exp\left(\frac{-\zeta^2}
{4{{\mathrm i}}\beta{\bar{L}}}+\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}\right) \\
\times\left[\left(
\frac{-\zeta}{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}+\frac{
\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{v}\right)^{-1}+{{\mathrm O}}(\zeta^{-3})\right].\end{aligned}$$ Adding these gives the estimate $$|I_2(\beta,\zeta)|\leq\frac{\sqrt{\beta{\bar{L}}}}{v\sqrt{\pi}}\left(
\frac{\beta{\bar{L}}}{v^2}+\frac{\zeta^2}{\beta{\bar{L}}}\right)^{-1}
+{{\mathrm O}}(\zeta^{-3})\qquad\mbox{as $|\zeta|\to\infty$.}
\label{eq:5:3}$$ Taking [(\[eq:5:3\])]{} together with the estimates $|I_2(\beta,\zeta)|\leq 1$ and $$|I_1(\beta,\zeta)|\leq\frac{v}{\sqrt{\pi\beta{\bar{L}}}}$$ gives us the estimate we require.

$\square$

\[prop:5:9\] With the notation above, $$\lim_{v\to\infty}\frac{1}{2v}
\int_{-\infty}^{\infty} I_1(\beta,\zeta) (I_2(\beta,\zeta))^{v-1}
{{\mathrm d}}\zeta=\frac{1}{2\sqrt{\pi}}\int_{-\infty}^{\infty}
\frac{1}
{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}\exp\left(\frac{-\zeta^2}{4{{\mathrm i}}\beta{\bar{L}}}\right)
M(\beta,\zeta){{\mathrm d}}\zeta.$$

[*Proof.*]{}

We split the region of integration as follows $$\begin{aligned}
\frac{1}{2v}\int_{-\infty}^{\infty} I_1(\beta,\zeta)(I_2(\beta,\zeta))^{v-1}
{{\mathrm d}}\zeta =  \frac{1}{2v}\int_{-\sqrt{v}}^{\sqrt{v}} I_1(\beta,\zeta) 
(I_2(\beta,\zeta))^{v-1} {{\mathrm d}}\zeta \nonumber \\
+\frac{1}{2v}\int_{|\zeta|>\sqrt{v}}I_1(\beta,\zeta)(I_2(\beta,\zeta))^{v-1}
{{\mathrm d}}\zeta. \label{eq:5:16}\end{aligned}$$ The function $M$ is bounded and satisfies $$|M(\beta,\zeta)|\leq\exp\left(\frac{2\sqrt{\beta{\bar{L}}}}{\sqrt{\pi}}-\frac
{|\zeta|}{2}\right)$$ for $\zeta$ sufficiently large. Integrating [(\[eq:5:30\])]{} gives $$\begin{aligned}
  \frac{1}{2v}\int_{-\sqrt{v}}^{\sqrt{v}} I_1(\beta,\zeta) 
(I_2(\beta,\zeta))^{v-1}& {{\mathrm d}}\zeta =
\frac{1}{2\sqrt{\pi}}\int_{-\sqrt{v}}^{\sqrt{v}}\frac{1}
{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}\exp\left(\frac{-\zeta^2}{4{{\mathrm i}}\beta{\bar{L}}}\right)
M(\beta,\zeta){{\mathrm d}}\zeta \nonumber \\
&+{{\mathrm O}}\left[\int_{-\sqrt{v}}^{\sqrt{v}}\exp\left(-\frac{|\zeta|}{2}\right)
\left(\frac{1}{\sqrt{v}}+\frac{\zeta^2}{v}\right){{\mathrm d}}\zeta\right]
\label{eq:5:35}\end{aligned}$$ and the integral in the remainder term converges as we let $v\to\infty$.

To deal with the integral $$\frac{1}{v}\int_{|\zeta|>\sqrt{v}}I_1(\beta,\zeta)(I_2(\beta,\zeta))^{v-1}
{{\mathrm d}}\zeta$$ we make use of the result of lemma \[lem:5:7\], giving $$\begin{aligned}
\Bigg|\frac{1}{v}\int_{|\zeta|>\sqrt{v}}I_1(\beta,\zeta)
(I_2(\beta,\zeta&))^{v-1}{{\mathrm d}}\zeta\Bigg| \nonumber \\
&\leq \frac{1}{\pi}\int_{|\zeta|>\sqrt{v}} \left(
\frac{\beta{\bar{L}}}{v^2}+\frac{\zeta^2}{\beta{\bar{L}}}\right)^{-1}{{\mathrm d}}\zeta
+{{\mathrm O}}(v^{-1}) \label{eq:5:41}\\ 
&\to0 \qquad\mbox{as 