So my question went like this:
If $X_1\sim\text{Unif(0,a)}$ and $X_2\sim\text{Unif(0,a)}$ independently, then find the distribution of $Z=X_1X_2$.
I tried the following:
Let $Z_1=\frac{X_1}{a}\sim\text{Unif(0,1)}$ and $Z_2=\frac{X_2}{a}\sim\text{Unif(0,1)}$. Thus $Z_1$ and $Z_2$ are independent.
Also, we know that, $-2\log_eZ_1\sim\chi_2^2$ and $-2\log_eZ_2\sim\chi_2^2$.
This implies,
$-2\log_e{Z_1Z_2}\sim\chi_4^2\implies-2\log_e{\frac{X_1X_2}{a^2}}\sim\chi_4^2\implies-2\log_e{\frac{Z}{a^2}}\sim\chi_4^2$.
Let $Z^*=-2\log_e{\frac{Z}{a^2}}\sim\chi_4^2$.
Thus, applying the Jacobian Transformation, I found the pdf of $Z$ as: $$f_Z(z)=\begin{cases}\frac{\left(-\log_e{z/a^2}\right)e^{-\log_e{z/a^2}}}{2z} & 0\leq z\leq a^2\\ 0 & \text{otherwise}\end{cases}$$ Is what I did, right?
Actually, before proceeding like this, I tried using the direct method using the Jacobian transformation but I was getting something like this:
$$\begin{align}f_Z(z)&=\int_0^a{f_{X_1}\left(\frac{z}{x_2}\right)f_{X_2}\left(x_2\right)\frac{1}{x_2}}.dx_2\\ &=\frac{1}{a^2}\int_0^a\frac{1}{x_2}.dx_2\end{align}$$ But this integral is not defined.
So does this mean that the direct formula of the Jacobian Transformation is invalid in this case?
Just to simplify the notation I set $Z=XY$.
To use the jacobian Method set
$$\begin{cases} z=xy \\ v=x \end{cases}$$
The jacobian is $|J|=\frac{1}{v}$ thus
$$f_{UZ}(u,z)=\frac{1}{a^2v}\mathbb{1}_{(0;a^2)}(z)\cdot\mathbb{1}_{(\frac{z}{a};a)}(v)$$
Thus
$$f_Z(z)=\frac{1}{a^2}\int_{\frac{z}{a}}^a \frac{1}{v}dv=\frac{2\log a-\log z}{a^2}$$