We want to derive a distribution from two Gaussian distributions using the definition of a compound distribution $$ \qquad \qquad \qquad \quad f(z) = \int g(z|\theta) h(\theta) \,\text{d}\theta \qquad \qquad \qquad (*) $$ More specific, we would like to use $(*)$ when $g(z|\sigma)$ is the pdf of a 1D complex normal distribution (ie $Z \sim \mathcal{CN}(0, |\sigma|^2)$), where $\Sigma \sim N(0, \tilde{\sigma}^2)$ with pdf $h(\sigma)$. The complex normal distribution will be defined $$ g(z|\sigma) = \frac{1}{\pi \sigma^2}e^\frac{|z|^2}{\sigma^2}, \qquad z\in\mathbb{C}, $$ where $\sigma$ is normally distributed $$ h(\sigma) = \frac{1}{\sqrt{2\pi}\tilde{\sigma}} \exp\left(-\frac{1}{2\tilde{\sigma}^2} \sigma^2\right), \qquad \sigma\in\mathbb{R}. $$
From the definition in $(*)$ we then get an integral $$ \begin{align} \int_{-\infty}^\infty \frac{1}{\pi \sigma^2}e^\frac{|z|^2}{\sigma^2} \frac{1}{\sqrt{2\pi}\tilde{\sigma}} \exp\left(-\frac{1}{2\tilde{\sigma}^2} \sigma^2\right) \, \text{d} \sigma = 2 \int_{0}^\infty \frac{1}{\pi \sigma^2}e^\frac{|z|^2}{\sigma^2} \frac{1}{\sqrt{2\pi}\tilde{\sigma}} \exp\left(-\frac{1}{2\tilde{\sigma}^2} \sigma^2\right) \, \text{d} \sigma, \end{align} $$ where the last step comes from that the function is even.
Hence, we end up with an integral similar to the one below. $$ \int_0^\infty \exp\left(-\frac{1}{\sigma^2} - \sigma^2\right)\frac{1}{\sigma^2}\, \text{d} \sigma. $$ And if we make the substitution $x = 1/\sigma$ => $\text{d}x = -1/\sigma^2 \text{d} \sigma$, we get $$ -\int_{-\infty}^0 \exp\left(-x^2-\frac{1}{x^2}\right)\, \text{d} x $$ According to integral calculator this should have a solution (but not an analytical solution) consisting of erfc factors. I suppose you can rewrite this as a normal distribution but I cant understand how exactly.
