Šeba billiards in section \[sec:7\].

The value distribution of $Z({{\mathbf L}},k)$ {#sec:2}
==============================================

\[lem:1\] Let ${\bar{L}}>0$ and $\zeta$ be real constants, then $$\lim_{k\to\infty}\frac{1}{\Delta L}{\int_{{\bar{L}}}^{{\bar{L}}+\Delta L}}\exp({{\mathrm i}}\zeta\tan kL){{\mathrm d}}L
={{\mathrm e}}^{-|\zeta|}$$ uniformly for $k\Delta L\to\infty$ as $k\to\infty$.

[*Proof.*]{}

By the periodicity of the integrand we may shift the range of integration by multiples of $\pi/k$ so that without loss of generality we may take ${\bar{L}}$ in the range $0\leq{\bar{L}}\leq \pi/k$. We write $\Delta L=\pi(n+p)/k$ where $n\in{\mathbb Z}$ and $0\leq p < 1$. Then by the periodicity of the integrand, $$\begin{aligned}
\int_{{\bar{L}}}^{{\bar{L}}+\pi n/k+\pi p/k}\exp&({{\mathrm i}}\zeta\tan kL){{\mathrm d}}L \\
&= \int_{{\bar{L}}}^{{\bar{L}}+p\pi/k}\exp({{\mathrm i}}\zeta\tan kL){{\mathrm d}}L
+n\int_0^{\pi/k}\exp({{\mathrm i}}\zeta\tan{kL}){{\mathrm d}}L 
\\
&= \int_{{\bar{L}}}^{{\bar{L}}+p\pi/k}\exp({{\mathrm i}}\zeta\tan kL){{\mathrm d}}L
+\frac{n}{k}\int_{-\infty}^{\infty} 
\frac{{{\mathrm e}}^{{{\mathrm i}}\zeta z}}{1+z^2}{{\mathrm d}}z\end{aligned}$$ where the substitution $z=\tan kL$ has been made. We note now that $n/k\Delta L\to\pi^{-1}$ as $k\to\infty$ and $$\left|\frac{1}{\Delta L}\int_{{\bar{L}}}^{{\bar{L}}+p\pi/k}\exp({{\mathrm i}}\zeta\tan{kL})
{{\mathrm d}}L\right| \leq \frac{\pi}{k\Delta L} \to 0\quad \mbox{as $k\to\infty$.}$$ A simple application of Cauchy’s residue theorem allows us to evaluate the integral $$\int_{-\infty}^{\infty}\frac{{{\mathrm e}}^{{{\mathrm i}}\zeta z}}{1+z^2}{{\mathrm d}}z=
\pi{{\mathrm e}}^{-|\zeta|}.
\label{eqn:10}$$

$\square$

[*Proof of theorem \[thm:1\].*]{}

We use here the characteristic function. With bond lengths chosen from a uniform distribution, $$\begin{aligned}
{\mathbb E}_{\rm L}(\exp({{\mathrm i}}\zeta Z(k,{{\mathbf L}})))&=&\frac{1}{(\Delta L)^v}
{\int_{{\bar{L}}}^{{\bar{L}}+\Delta L}}\mspace{-20mu}
\cdots\!{\int_{{\bar{L}}}^{{\bar{L}}+\Delta L}}\mspace{-15mu}\exp\left({{\mathrm i}}\zeta\textstyle\sum_{j=1}^v 
\tan kL_j\right){{\mathrm 