\xi}})=\left|
  \begin{array}{ccccc}
L_1 & L_2 & \cdots & L_{v-1} & L_v \\
1   &  0  & \cdots & 0 & \frac{-\sec^2\xi_1}{D} \\
0   &  1  & \cdots & 0 & \frac{-\sec^2\xi_2}{D} \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & 1 & \frac{-\sec^2\xi_{v-1}}{D}
  \end{array}\right|.$$ To complete the proof we employ the identity $$\left|  \begin{array}{ccccc}
\alpha_1 & \alpha_2 & \cdots & \alpha_{n-1} & \alpha_n \\
1   &  0  & \cdots & 0 & \beta_1 \\
0   &  1  & \cdots & 0 & \beta_2 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & 1 & \beta_{n-1}
  \end{array}\right|
=(-1)^n\left( -\alpha_n+\sum_{k=1}^{n-1}\alpha_k\beta_k\right),$$ which may be readily checked by induction.

$\square$

We note in passing that the explicit form of $J({{\boldsymbol \xi}})$ given above together with theorem \[thm:bar:gas\] provides a convenient representation for use numerical studies of eigenvalues and eigenfunctions because the zeros of $Z(k,{{\mathbf L}})$ do not need to be computed explicitly.

[*Proof of theorem \[thm:3\].*]{}

We take as the function $f$ in theorem \[thm:bar:gas\] $$f({{\mathbf x}})={{1\!\!1}}_{(-\infty,R]}\left(\frac{1}{v^2}\sum_{j=1}^vL_j\sec^2{x_j}\right)$$ Then we define $P_v(y)$ by $$\begin{aligned}
\int_{-\infty}^R P_v(y){{\mathrm d}}y=\frac{1}{\pi^v\bar{d}}\int_0^{\pi}
\!\cdots\!\int_0^{\pi}f(\xi_1,\ldots,\xi_{v-1},-\tan^{-1}(\tan \xi_1+
\cdots+\tan \xi_{v-1}))\\ \times J({{\boldsymbol \xi}}){{\mathrm d}}\xi_1\cdots{{\mathrm d}}\xi_{v-1},\end{aligned}$$ where $J({{\boldsymbol \xi}})$ 