To multiply two independent RVs, Wikipedia gives the following approach:
$$f_Z(z)=\int_{-\infty}^\infty f_X(x)f_Y(z/x)\frac{1}{|x|}dx $$
https://en.wikipedia.org/wiki/Product_distribution
Maybe I'm just being thick-headed, but doesn't that integral always have a singularity at $0$ due to the factor $1/|x|$, not to mention $f_Y(z/x)$? Or is it possible this is something like renormalization theory at work, so that
$$\lim_{\delta ->0} \int_{-\infty}^{-\delta}g(x)dx + \int_{-\delta}^\infty g(x)dx$$ where $g(x)$ is the whole function inside the integral, is well-defined?
Consider an example where $X,Y$ are both Exp(1) variables. Then
$$f_Z(z)=\int_0^\infty \frac{1}{x} e^{-x} e^{-z/x} dx,z>0.$$
(Note that $P(Z=0)=P(X=0 \vee Y=0)=0$.)
You see that the integrand actually has a finite limit at $x=0$ when $z>0$; in effect $f_Y(z/x)$ is serving to remove the singularity by decaying faster than $1/x$ is growing as $x \to 0^+$.
In unusual situations (specifically, when $f_Y(y)=o(1/|y|)$ as $|y| \to \infty$ does not hold), this reasoning can break down, and then the formula is still valid but the singularity is merely integrable rather than removable. One such example would happen with
$$f_Y(x)=\sum_{n=1}^\infty T(2^{n+1}(x-2n))$$
where $T(x)=\max \{ 1-|x|,0 \}$.