theorem for characteristic functions to deduce that the limiting density $P(y)$ exists and is given by $$\begin{aligned}
  P(y)&=&\frac{1}{4\pi^{3/2}}\int_{-\infty}^{\infty}\!\int_{-\infty}^{\infty}
{{\mathrm e}}^{-\xi^2/4}M(\beta,\xi\sqrt{{{\mathrm i}}\beta{\bar{L}}}){{\mathrm e}}^{{{\mathrm i}}\beta y}{{\mathrm d}}\xi{{\mathrm d}}\beta \nonumber \\
&=&\frac{1}{2\pi^{3/2}}{{\mathfrak{Re}}}\int_{0}^{\infty}\!\int_{-\infty}^{\infty}
{{\mathrm e}}^{-\xi^2/4}M(\beta,\xi\sqrt{{{\mathrm i}}\beta{\bar{L}}}){{\mathrm e}}^{{{\mathrm i}}\beta y}{{\mathrm d}}\xi{{\mathrm d}}\beta.\label{eq:4:66}\end{aligned}$$ The integrand is dominated by $$\exp\left(-\xi^2/4-\sqrt{\pi{\bar{L}}\beta}\right),$$ so Fubini’s theorem allows us to switch the order of integration. We quote the standard integral $$\int_0^{\infty}{{\mathrm e}}^{ax+b\sqrt{x}}{{\mathrm d}}{x}=-\frac{1}{a}-\frac{b}{2a}\sqrt
{\frac{\pi}{-a}}\exp\left(\frac{-b^2}{4a}\right){\mathop{\rm erfc}}\left(\frac{-b}
{2\sqrt{-a}}\right)$$ valid for ${{\mathfrak{Re}}}\: a<0$ and use this to perform the $\beta$ integral in [(\[eq:4:66\])]{}. This leads to the result $$P(y)=\frac{\sqrt{{\bar{L}}}}{4\pi y^{3/2}}{{\mathfrak{Re}}}\int_{-\infty}^{\infty}
\exp\left(-\frac{\xi^2}{4}-\frac{{\bar{L}}m(\xi)^2}{4y}\right)m(\xi)
{\mathop{\rm erfc}}\left(\frac{\sqrt{{\bar{L}}}m(\xi)}{2{{\mathrm i}}y}\right){{\mathrm d}}\xi,$$ which reduces to the form given in the statement of the theorem upon noticing that ${{\mathfrak{Re}}}\{ {\mathop{\rm erfc}}({{\mathrm i}}\theta)\}=1$ for all $\theta\in{\mathbb R}$.

$\square$

Value distribution of the eigenfunctions in the limit $v\to\infty$ {#sec:5}
==================================================================

To prove theorem \[thm:6\] we use a standard approximation argument. We introduce the smoothed $\delta$-function $$\delta_{\epsilon}(x):=\frac{1}{2\pi}\int_{-\infty}^{\infty}
{{\mathrm e}}^{-\epsilon|\beta|+{{\mathrm i}}\beta x}{{\mathrm d}}\beta=
\frac{\epsilon}{\pi(\epsilon^2+x^2)}.$$

\[prop:5:100\] Let $\tilde{Q}_v(\eta)$ be related to $Q_v(\eta)$ by $$\tilde{Q}_v(\eta):=\frac{1}{\eta^2}Q_v\left(\frac{1}{\eta}\right).$$ For any fixed $\epsilon$, $$\lim_{v\to\infty}\int_0^{\infty}\delta_{\epsilon}(\eta-\eta')
\tilde{Q}_v(\eta'){{\mathrm d}}\eta'=\int_0^{\infty} \delta_{\epsilon}(\eta-\eta')\tilde{Q}(\eta'){{\mathrm d}}\eta',
\label{eq:5:100}$$ uniformly for $\eta$ in compact intervals, where $$\tilde{Q}(\eta)=\frac{1}{2\pi^{3/2}\eta}{{\mathfrak{Im}}}\int_{-\infty}^{\infty}
\exp\left(-\frac{\xi^2}{4}
-\frac{{\bar{L}}m(\xi)^2}{8\eta}\right){\mathop{\rm erfc}}\left(\frac{\sqrt{{\bar{L}}}m(\xi)}
{2\sqrt{2\eta}{{\mathrm i}}}\right){{\mathrm d}}\xi.$$

[*Proof.*]{}

Let $$\tilde{Q}_{\epsilon,v}(\eta):=\int_{0}^{\infty}\delta_{\epsilon}(\eta-\eta')
\tilde{Q}_v(\eta'){{\mathrm d}}\eta'
=\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}
\delta_{\epsilon}\left(\eta-\frac{1}{v^2A_i(n,{{\mathbf L}};v)}\right).$$ We use the identity $$\delta_{\epsilon}\left(\eta-\frac{A}{B}\right)=B\delta_{\epsilon|B|}(B\eta
-A)$$ with $$A=\frac{{\bar{L}}}{v^2}\sum_{j=1}^v \sec^2{kL_j}$$ and $$B=2\sec^2kL_i.$$ 