Just like in the title:
How to compute the CDF of $X\cdot Y$ if $X,Y$ are independent random variables, uniformly distributed over $(-1,1)$?
I tried using the next formula: the density of $X\cdot Y$ is the integral of $f(u/v)\cdot g(v)$ where $f$ is the density of $X$ and $g$ is the density of $Y$ but it has no sense.
Let $Z=X\cdot Y$ and $f_Z,f_Y,f_X$ be the pdfs of $X,Y,Z$.
$f_Z$ is quite trivially an even function, supported on $[-1,1]$.
So, let we compute, for any $p\in(0,1)$:
$$\mathbb{P}[0\leq Z\leq p]=2\int_{0}^{1}\int_{0}^{\min\left(1,\frac{p}{x}\right)}\frac{1}{2}\cdot\frac{1}{2}\,dy\,dx=\frac{1}{2}\int_{0}^{1}\min\left(1,\frac{p}{x}\right)\,dx\tag{1}$$ that gives: $$\mathbb{P}[0\leq Z\leq p]=\frac{1}{2}\left(\int_{0}^{p}1\,dx+\int_{p}^{1}\frac{p}{x}\,dx\right)=\frac{p-p\log p}{2}\tag{2}$$ hence: $$ \int_{0}^{p}f_Z(t)\,dt = \frac{p-p\log p}{2}\tag{3} $$ gives, through differentiation, $f_Z(p)=-\frac{\log p}{2}$. At last, $$ f_Z(z) = -\frac{\log|z|}{2}\cdot\mathbb{1}_{(-1,1)}(z),\\ F_Z(z)=\left[\frac{1}{2}-\frac{(1-|z|)+|z|\log |z|}{2}\text{Sign}(z)\right]\cdot\mathbb{1}_{(-1,1)}(z)\tag{4}$$ are the pdf and cdf of $Z$.
Hint : Try jacobian transformation with $u=xy$ and $v=y$ then integrate on v between(-1,1).
