contour encloses a pole (figure \[fig:2\]). In this case, $$\int_{-\infty}^{\infty}\frac{\exp(-{{\mathrm i}}\beta{\bar{L}}y^2/v^2)}
{y-{{\mathrm i}}-\zeta v/2\beta{\bar{L}}}{{\mathrm d}}y=2\pi{{\mathrm i}}R +\int_{-\infty}^{\infty}
\frac{{{\mathrm e}}^{-r^2}}{r-\frac{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{v}({{\mathrm i}}+\frac{\zeta v}
{2\beta{\bar{L}}})}{{\mathrm d}}r$$ where $R$ is the residue at the pole $$\begin{aligned}
  R&=&\left.\exp\left(-\frac{{{\mathrm i}}\beta{\bar{L}}y^2}{v^2}\right)\right|_{y={{\mathrm i}}+
\zeta v /2\beta{\bar{L}}}\\
&=&\exp\left(-\frac{{{\mathrm i}}\zeta^2}{4\beta{\bar{L}}}+\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}+
\frac{\zeta}{v}\right),\end{aligned}$$ so that we also get in this case $$I_2^{-}(\beta,\zeta)=\frac{1}{2}\exp\left(\frac{\zeta}{v}\right)
{\mathop{\rm erfc}}\left(\frac{\zeta}{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}+\frac{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}
{v}\right).$$

Treating $I_2^{+}$ in a similar way, yields an expression for $I_2$, $$I_2(\beta,\zeta)=\frac{1}{2}{{\mathrm e}}^{\zeta/v}{\mathop{\rm erfc}}\left(
\frac{\zeta}{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}+\frac{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{v}\right)
+ \frac{1}{2}{{\mathrm e}}^{-\zeta/v}{\mathop{\rm erfc}}\left(
\frac{-\zeta}{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}+\frac{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{v}\right).
\label{eq:5:2}$$

Some estimates
--------------

Let $-\sqrt{v}\leq\zeta\leq\sqrt{v}$, then $$\begin{aligned}
  I_1(\beta,\zeta)(I_2(\beta,\zeta))^{v-1}=\frac{v}{\sqrt{\pi}}\frac{1}
{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}\exp\left(\frac{-\zeta^2}{4{{\mathrm i}}\beta{\bar{L}}}\right)
M(\beta,\zeta)\nonumber \\
\times\left(1+{{\mathrm O}}(v^{-1/2})+{{\mathrm O}}(\zeta^2/v)\right)
\label{eq:5:30}\end{aligned}$$ as $v\to\infty$, where $$M(\beta,\zeta):=\exp\left(-\frac{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{\sqrt{\pi}}
\exp\left(\frac{-\zeta^2}{4{{\mathrm i}}\beta{\bar{L}}}\right)-
{\zeta}{\mathop{\rm erf}}\left(\frac{\zeta}{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}\right) \right).$$

[*Proof.*]{}

A key step will be to make a uniform expansion of $I_2(\beta,\zeta)$ as $v\to\infty$. By Taylor’s theorem, $$\exp\left(\pm\frac{\zeta}{v}\right)=1\pm\frac{\zeta}{v}+{{\mathrm O}}(
\zeta^2{v^{-2}}) \qquad\mbox{as $v\to\infty$.}\label{eq:5:17}$$ A second application of Taylor’s theorem yields $${\mathop{\rm erfc}}\left(\frac{\pm\zeta}{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}+
\frac{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{v}\right)={\mathop{\rm erfc}}\left(\frac{\pm\zeta}{2
\sqrt{{{\mathrm i}}\beta{\bar{L}}}}\right)-\frac{2}{\sqrt{\pi}}\frac{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}
{v}\exp\left(-\frac{\zeta^2}{4{{\mathrm i}}\beta{\bar{L}}}\right)+\frac{R_{1}(\zeta)}
{v^2},
\label{eq:erfc}$$ where the remainder term is $$R_1(\zeta)=-{{{\mathrm i}}\beta{\bar{L}}}\int_0^1 \left.\frac{{{\mathrm d}}^2}
{{{\mathrm d}}z^2}{\mathop{\rm erfc}}(z)\right|_{z=\frac{\pm\zeta}{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}+\sigma
\frac{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{v}}(1-\sigma){{\mathrm d}}\sigma.$$ The second derivative of ${\mathop{\rm erfc}}$ is $$\frac{{{\mathrm d}}^2}{{{\mathrm d}}z^2}{\mathop{\rm erfc}}(z)=\frac{4}{\sqrt{\pi}}z{{\mathrm e}}^{-z^2}.$$ This allows us to estimate $$\begin{aligned}
  |R_1|&\leq&\frac{4\beta{\bar{L}}}{\sqrt{\pi}}\int_0^1 \left(\frac{|\zeta|}
{2\sqrt{\beta{\bar{L}}}}+\sigma\frac{|\sqrt{\beta{\bar{L}}}|}{v}\right)
{{\mathrm e}}^{\mp\sigma\zeta/v}(1-\sigma){{\mathrm d}}\sigma \nonumber\\
&\leq&\frac{4\beta{\bar{L}}}{\sqrt{\pi}}{{\mathrm e}}^{|\zeta|/v}\left(\frac{|\zeta|}
{2\sqrt{\beta{\bar{L}}}}+{{\mathrm O}}(v^{-1})\right) \nonumber \\
