is defined by [(\[eqn:j\])]{}. Since $\sec^2x>0$ for all $x\in{\mathbb R}$, it follows that $P_v(y)=0$ for $y<0$.

$\square$

[*Proof of theorem \[thm:5\].*]{}

In this case, we take as the function $f$ in theorem \[thm:bar:gas\], $$f({{\mathbf x}})={{1\!\!1}}_{[0,R]}\left(\frac{2v^2\sec^2x_i}{\sum_j L_j\sec^2{x_j}}\right).$$ We take as the parameterisation of $\Sigma$ $$\begin{aligned}
  s_j&=&\xi_j,\quad\mbox{$1\leq j<i$}\\
  s_i&=&-\tan^{-1}(\tan\xi_1+\cdots+\tan\xi_{v-1}) \\
  s_j&=&\xi_{j-1},\quad\mbox{$i<j\leq v$}.\end{aligned}$$ This does not change the form of $J({{\boldsymbol \xi}})$ from that in [(\[eqn:j\])]{}, but introduces extra symmetry in $f$. Then define, as before, $$\int_0^R P_v(\eta){{\mathrm d}}\eta=\frac{1}{\pi^v\bar{d}}
\int_0^{\pi}\!\cdots\!\int_0^{\pi}f(s_1({{\boldsymbol \xi}}),\ldots,s_v({{\boldsymbol \xi}}))
J({{\boldsymbol \xi}}){{\mathrm d}}\xi_1\cdots{{\mathrm d}}\xi_{v-1}.$$ Since $f({{\mathbf s}}({{\boldsymbol \xi}}))$ is symmetric in $\xi_1,\ldots,\xi_{v-1}$ we see that $P_v(\eta)$ is independent of the choice of bond $i$.

$\square$

Value distribution of $Z'(k_n)$ in the limit $v\to\infty$ {#sec:4}
=========================================================

We take as the function $f$ in [(\[eq:states\])]{} the characteristic function for the value distribution of the derivative of $Z(k,{{\mathbf L}})$, $$Z'(k,{{\mathbf L}})=\sum_{j=1}^v L_j\sec^2 kL_j.
\label{eq:5:1}$$ Since we are stipulating that $v\Delta L\to0$ as $v\to\infty$, we replace $L_j$ by ${\bar{L}}$ where it does not multiply $k$ in [(\[eq:5:1\])]{} and take as our $f$ $$f({{\mathbf x}})=\exp\left(-\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}\sum_{j} \sec^2x_j\right).$$ Let the quantity in which we are interested be denoted $E_v(\beta)$. Then $$\begin{aligned}
  E_v(\beta)&:=&\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}
\exp\left(-\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}\sum_j \sec^2 k_nL_j\right) \nonumber\\
 &=&\frac{{\bar{L}}}{2\pi\bar{d}v}\int_{-\infty}^{\infty}\frac{1}{\pi^v}
\int_0^{\pi}\!\cdots\!\int_0^{\pi}
\left(\sum_j \sec^2{x_j}\right) \nonumber \\
& &\times\prod_{j=1}^v\exp\left(-\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}\sec^2 x_j+
