[(\[eq:bar:gas\])]{} can formally be written in the form $$\int_{\Sigma} 
f({{\boldsymbol \xi}}){{\mathrm d}}\nu({{\boldsymbol \xi}})
  =\frac{1}{2\pi^{v+1}
\bar{d}}\int_{-\infty}^{\infty}
\int_0^{\pi}\!\cdots\!\int_0^{\pi} f({{\mathbf x}})[{{\mathbf L}}\cdot\nabla 
G({{\mathbf x}})]{{\mathrm e}}^{{{\mathrm i}}\zeta G}{{\mathrm d}}x_1\cdots{{\mathrm d}}x_v{{\mathrm d}}\zeta,
\label{eq:states}$$ where now $f$ is a function ${\mathbb T}^v\to{\mathbb R}$ of an appropriate class. This follows from writing $\int_{\Sigma} 
f({{\boldsymbol \xi}}){{\mathrm d}}\nu({{\boldsymbol \xi}})$ in the equivalent form, $$\frac{1}{\pi^v\bar{d}}
\int_{\Sigma}\int f(t,{{\boldsymbol \xi}})[{{\mathbf L}}\cdot\nabla G] \delta(G(t,{{\boldsymbol \xi}}))
J({{\boldsymbol \xi}}){{\mathrm d}}t{{\mathrm d}}{{\boldsymbol \xi}}.
\label{eq:del:seq}$$ We then write the $\delta$-function as the limit of the sequence $$\delta_{m}(x):=\frac{1}{2\pi}\int_{-m}^{m}{{\mathrm e}}^{{{\mathrm i}}\zeta x}{{\mathrm d}}\zeta=
\frac{\sin mx}{\pi x}
\label{eq:delta}$$ as $m\to\infty$, and changing back to the usual Cartesian coordinates on ${\mathbb T}^v$. We need to show that $\delta_m$ is an appropriate $\delta$-sequence for the function $f$ that we consider. We will check this point directly in the calculations where the identity [(\[eq:states\])]{} is used. We shall also justify taking the $\zeta$-integral in [(\[eq:delta\])]{} outside the integral over ${\mathbb T}^v$.

For a star graph with $v$ bonds and the parameterisation $$\begin{aligned}
  s_i=\xi_i\qquad\mbox{$i=1,\ldots,v-1$}\\
s_v=-\tan^{-1}(\tan \xi_1+\cdots+\tan\xi_{v-1})\end{aligned}$$ $J({{\boldsymbol \xi}})$ takes the following form $$J({{\boldsymbol \xi}})=\frac{L_1\sec^2\xi_1+\cdots+L_{v-1}\sec^2{\xi_{v-1}}}
{1+(\tan \xi_1+\cdots+\tan \xi_{v-1})^2} +L_v
\label{eqn:j}$$

[*Proof.*]{}

Differentiating gives $$\frac{\partial s_i}{\partial \xi_j}=\delta_{ij}$$ for $i<v$ and $$\frac{\partial s_v}{\partial \xi_j}=\frac{-\sec^2{\xi_j}}{1+(\tan\xi_1+
\cdots+\tan\xi_{v-1})^2}.$$ For ease of notation we write $D:={1+(\tan\xi_1+
\cdots+\tan\xi_{v-1})^2}$. Thus we have $$J({{\boldsymbol 