that the scaling transformations of the integration variables, first $z\to z/\sqrt{\Omega}$ and then $\Omega\to\Omega/X$, and interchanging the order of integrations over $z$ and $\Omega$, we may cast Eq. (\[p\_M(X)int\]) in the following form $$\begin{gathered}
\label{app1}
  {\mathcal{P}^{\mathrm{pf}}}_M(X) = \frac{2\pi X^{M/2-1}}{\Gamma(M/2)} \int\limits_{0}^{\infty}\frac{dz\, z^M}{\sinh(\pi z)} \\ {\times} \int\limits_{-i\infty}^{+i\infty}\frac{d\Omega}{2\pi i}\Omega^{-M/2}e^{\Omega-Xz^2/\Omega} \,.\end{gathered}$$ To calculate here the last integral over $\Omega$, we expand $e^{-Xz^2/\Omega}$ into a series and evaluate the result termwise $$\begin{gathered}
\label{app2}
  \sum_{k=0}^{\infty}\frac{(-Xz^2)^k}{k!}\int_{-i\infty}^{+i\infty}\limits\frac{d\Omega}{2\pi i}\Omega^{-(M/2+k)}e^{\Omega} \\ = \sum_{k=0}^{\infty}\frac{[-(\sqrt{X}z)^2]^k}{k!\Gamma(\frac{M}{2}+k)}\,,\end{gathered}$$ where we have used $\int_{-i\infty}^{+i\infty}\frac{d\Omega}{2\pi i}\Omega^{-\nu}e^{\Omega}=1/\Gamma(\nu)$. Making now use of the well-known series representation for the Bessel function [@Abr65], one can immediately recognize the r.h.s. of (\[app2\]) to be equal to $(\sqrt{X}z)^{1-M/2}J_{M/2-1}(2\sqrt{X}z)$. Collecting all the factors together, we finally arrive at Eq. (\[p\_M(X)bessel\]).

Further progress is possible in the case of odd $M$. It is instructive first to start with the case of $M=1$, which turns out to play the central rôle in this calculation. We may use the known relation $J_{-1/2}(z)=\sqrt{2z/\pi}\cos(z)/z$ in this case [@Abr65], thus $(\sqrt{X}z)^{1/2}J_{-1/2}(2\sqrt{X}z)=\frac{1}{\sqrt{\pi}}\cos(2\sqrt{X}z)$, that allows us to perform the integration in Eq. (\[app1\]) analytically: $$\begin{gathered}
\label{app3}
   \int_{0}^{\infty}\limits dz \frac{z}{\sinh(z\pi)} \int_{-i\infty}^{+i\infty}\limits \frac{d\Omega}{2\pi i} \Omega^{-1/2}e^{\Omega-Xz^2/\Omega} \\= \int_{0}^{\infty}\limits dz \frac{z\cos(2\sqrt{X}z)}{\sinh(z\pi)\sqrt{\pi}} = \frac{1}{4\sqrt{\pi}}\frac{1}{\cosh^2(\sqrt{X})} \,.\end{gathered}$$ Taking now into account the (omitted) factor $2\sqrt{\pi/X}$, we obtain ${\mathcal{P}^{\mathrm{pf}}}_1(X)$, Eq. 