differ by $\epsilon$ which can be made arbitrarily small, so we see that $\lim_{K\to\infty} K^{-1}\int_0^K
f_1(k{{\mathbf L}}){{\mathrm d}}k$ exists and is equal to $$\frac{1}{\pi^v}\int_{{\mathbb T}^v}
f_1({{\mathbf x}}){{\mathrm d}}{{\mathbf x}}.$$ The extension to $f_2$ and hence $f$ is obvious.

$\square$

[*Proof of theorem \[thm:2\].*]{}

We begin in the same way as in the proof of theorem \[thm:1\]. In this case $k$ is chosen uniformly from the interval $[0,K]$ with $K>0$, so that the characteristic function with respect to this uniform distribution is $${\mathbb E}_K(\exp({{\mathrm i}}\zeta Z(k,{{\mathbf L}})))=\frac{1}{K}\int_0^K \exp\left({{\mathrm i}}\zeta \textstyle \sum_{j=1}^v \tan kL_j\right) {{\mathrm d}}k.$$ By lemma \[lem:2\] we can write this integral as an average over the torus as $K\to\infty$: $$\begin{aligned}
\lim_{K\to\infty}{\mathbb E}_K(\exp({{\mathrm i}}\zeta Z(k,{{\mathbf L}})))&=&
\frac{1}{\pi^v}
\int_{{\mathbb T}^v}\exp\left( \textstyle{{\mathrm i}}\zeta\sum_{j=1}^v \tan x_j\right)
{{\mathrm d}}{{\mathbf x}}\\
&=&\left( \frac{1}{\pi}\int_0^{\pi} \exp({{\mathrm i}}\zeta\tan x){{\mathrm d}}x \right)^v.\end{aligned}$$ Following the substitution $z=\tan x$ in this final integral, we have $$\lim_{K\to\infty}{\mathbb E}_K(\exp({{\mathrm i}}\zeta Z(k,{{\mathbf L}}))={{\mathrm e}}^{-v|\zeta|}$$ and the theorem follows from the same arguments used in the end of the proof of theorem \[thm:1\].

$\square$

An equidistribution theorem {#sec:3}
===========================

Barra and Gaspard [@bar:lsd] observed that the condition for $k$ to be an eigenvalue of a graph can be written in the form $$G(k{{\mathbf L}})=0,$$ where 