Consider a distribution $f(t)$. I can take the convolution of this with itself as $$(f * f) = \int^{+\infty}_{-\infty} f(\tau)f(t-\tau) \ d\tau$$ If we then do a second stage of convolution as $$((f * f) * f) = \int^{+\infty}_{-\infty} (f * f)(\tau)f(t-\tau) \ d\tau$$
To make this more compact I relabel $(f * f)$ as $F_{N=1} = (f * f)$, as the convolution iteration cycle as been done once. I would therefore relabel $((f * f) * f)$ as $F_{N=2} = ((f * f) * f)$ and so on.
Writing this as a general form $$F_{N} = \int^{+\infty}_{-\infty} F_{N-1}(\tau) f(t - \tau) \ d\tau$$
Is it possible to solve such an integral explicitly?
What I am trying to show is that as $N\rightarrow \infty$ I should recover the Gaussian distribution as per the central limit theorem.
Specifically I want to show this for when my $f(x)$ distribution is the Rayleigh distribution: $$f(x) = \frac{x}{\sigma^{2}} e^{-x^{2}/\left(2 \sigma^{2} \right)}$$
I would ideally like to derive a distribution where I have a two parameter distribution $R(\sigma, N)$ where $$RR(\sigma, N) \lim_{N\rightarrow\infty} = G(x,\sigma)$$ $G(x,\sigma)$ being the Gaussian distribution.