\frac{{{\mathrm i}}\zeta}{v}\sum_j\tan{x_j}\right)
{{\mathrm d}}{{\mathbf x}}{{\mathrm d}}\zeta. \label{jon:23}\end{aligned}$$ We have made the re-scaling $\zeta\mapsto\zeta/v$. This is a natural normalisation since $Z(k,{{\mathbf L}})$ is a sum of $v$ terms.

We exploit the symmetry in the integral in [(\[jon:23\])]{} to write $$E_v(\beta)=\frac{1}{2v}\int_{-\infty}^{\infty} I_1(\beta,\zeta) 
(I_2(\beta,\zeta))^{v-1}{{\mathrm d}}\zeta,$$ where we have replaced $\bar{d}$ by ${\bar{L}}v/\pi$ and defined the integrals $$I_1(\beta,\zeta):=\frac{1}{\pi}\int_0^{\pi}\sec^2{x}\exp\left(-
\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}
\sec^2{x}+\frac{{{\mathrm i}}\zeta}{v}\tan x\right){{\mathrm d}}x$$ and $$I_2(\beta,\zeta):=\frac{1}{\pi}\int_0^{\pi}\exp\left(-\frac{{{\mathrm i}}\beta{\bar{L}}}
{v^2}\sec^2{x}+\frac{{{\mathrm i}}\zeta}{v}\tan x\right){{\mathrm d}}x.$$ We note that $I_2$ is uniformly convergent in $\zeta$ but that $I_1$ is not.

The integrals $I_1$ and $I_2$
-----------------------------

The substitution $z=\tan x$ gives $$\begin{aligned}
  I_1(\beta,\zeta)&=&\frac{1}{\pi}\int_{-\infty}^{\infty} \exp\left(
-\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}(1+z^2)+\frac{{{\mathrm i}}\zeta z}{v}\right){{\mathrm d}}z \nonumber
\\
&=&\frac{v}{\sqrt{\pi}}\frac{1}{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}
\exp\left(-\frac{\zeta^2}{4{{\mathrm i}}\beta{\bar{L}}} -\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}\right)\end{aligned}$$ quoting a standard integral.

That $\delta_m$ defined in [(\[eq:delta\])]{} is a $\delta$-sequence for the function $\exp({{\mathrm i}}\alpha z^2)$ follows from the equality $$\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} {{\mathrm e}}^{{{\mathrm i}}\alpha z^2+{{\mathrm i}}\zeta(z-w)}{{\mathrm d}}z{{\mathrm d}}\zeta=\exp({{\mathrm i}}\alpha w^2),$$ which can be checked by direct evaluation of the integrals. This and uniform convergence of $I_2$ in $\zeta$ justifies the operations that lead to identity [(\[eq:states\])]{}.

We can treat $I_2$ in a similar manner to that in which we treated $I_1$. $$I_2(\beta,\zeta)=\frac{1}{\pi}
\int_{-\infty}^{\infty}\exp\left(-\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}(1+z^2)+\frac{{{\mathrm i}}\zeta z}{v}\right)\frac{{{\mathrm d}}z}{1+z^2}.$$ We first write $$\frac{1}{1+z^2}=\frac{1}{2{{\mathrm i}}}\left(\frac{1}{z-{{\mathrm i}}}-\frac{1}{z+{{\mathrm i}}}\right)$$ so that $I_2$ can be decomposed 