$v\to\infty$.} \nonumber\end{aligned}$$

Hence substituting [(\[eq:5:35\])]{} and [(\[eq:5:41\])]{} into [(\[eq:5:16\])]{} and taking the limit $v\to\infty$ gives the required result.

$\square$

Properties of $M(\beta,\zeta)$
------------------------------

We wish to rotate the variable $\zeta$. However, we need to check that the function $M$ does not blow up for large $|\zeta|$.

If $\zeta=R{{\mathrm e}}^{{{\mathrm i}}\theta}$, then $${\mathop{\rm erf}}\left(\frac{\zeta}{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}\right)={\mathop{\rm erf}}\left(
\frac{R}{2\sqrt{\beta{\bar{L}}}}{{\mathrm e}}^{{{\mathrm i}}(\theta-\pi/4)}\right)=1+{{\mathrm O}}(R^{-1})
\label{eq:7:7}$$ as $R\to\infty$, provided that $0<\theta<\pi/2$ [@abr:ste]. So, $$\begin{aligned}
  \left|\exp\left(-\zeta{\mathop{\rm erf}}\left(\frac{\zeta}{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}\right)
\right)\right|&=&\exp\left(-R\;{{\mathfrak{Re}}}\left[{{\mathrm e}}^{{{\mathrm i}}\theta}{\mathop{\rm erf}}\left(\frac{\zeta}{2\sqrt{{{\mathrm i}}\beta
{\bar{L}}}}\right)\right]\right)\\
&=&{{\mathrm e}}^{-R\cos\theta}{{\mathrm O}}(1)\\
&\to&0\qquad\mbox{as $R\to\infty$ provided $0<\theta<\pi/2$}\end{aligned}$$ and convergence is exponentially fast. Similarly, $$\begin{aligned}
  \left|\exp\left(-\frac{2\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{\sqrt{\pi}}\exp
\left(-\frac{\zeta^2}{4{{\mathrm i}}\beta{\bar{L}}}\right)\right)\right|
&=&\exp\left(\frac{-2\sqrt{\beta{\bar{L}}}}{\sqrt{\pi}}\exp\left(-\frac{R^2}{4\beta
{\bar{L}}}\sin2\theta\right)\right.\\
& &\left.\times\cos\left(\frac{R^2}{4\beta{\bar{L}}}\cos2\theta+\frac{\pi}
{4}\right)\right)\\
&\to&1\qquad\mbox{as $R\to\infty$} \end{aligned}$$ provided $0<\theta<\pi/2$.

Hence, if $0<\theta<\pi/4$ then $$\lim_{R\to\pm\infty}  |RM(\beta,R{{\mathrm e}}^{{{\mathrm i}}\theta})|=0$$ and we can make the change of variables $\zeta=\xi\sqrt{{{\mathrm i}}\beta{\bar{L}}}$: $$\int_{-\infty}^{\infty}
\frac{1}
{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}\exp\left(\frac{-\zeta^2}{4{{\mathrm i}}\beta{\bar{L}}}\right)
M(\beta,\zeta){{\mathrm d}}\zeta
=
\int_{-\infty}^{\infty}{{\mathrm e}}^{-\xi^2/4}M(\beta,\xi\sqrt{{{\mathrm i}}\beta{\bar{L}}}){{\mathrm d}}\xi. \label{eq:5:10}$$ We observe that $$\begin{aligned}
  M(\beta,\xi\sqrt{{{\mathrm i}}\beta{\bar{L}}})&=&\exp\left(-\frac{2}{\sqrt{\pi}}
\sqrt{{{\mathrm i}}\beta{\bar{L}}}{{\mathrm e}}^{-\xi^2/4}-\xi\sqrt{{{\mathrm i}}\beta{\bar{L}}}{\mathop{\rm erf}}\left(\frac{\xi}{2}\right)\right)\\
&=&\exp\left(-\sqrt{{{\mathrm i}}\beta{\bar{L}}}m(\xi)\right)\end{aligned}$$ where $$m(\xi):=\frac{2}{\sqrt{\pi}}{{\mathrm e}}^{-\xi^2/4}+\xi{\mathop{\rm erf}}(\xi/2).$$

[*Proof of theorem \[thm:4\]*]{}
--------------------------------

$m$ satisfies the bound $m(\xi)\geq2/\sqrt{\pi}$, so $M( \beta,\xi\sqrt{{{\mathrm i}}\beta{\bar{L}}})$ is bounded for all $\beta$. By the Weierstrass $M$-test the integral in [(\[eq:5:10\])]{} is uniformly convergent and hence $$\lim_{v\to\infty} E_v(\beta)$$ is a continuous function of $\beta$. We appeal, once again, to the continuity 