PDF of product of uniform variables I have encountered a problem in computing the PDF of a variable (call it $y_n$) that is the product of n uniformly distributed random variables $x$: 
$y_n=\prod_i^n x_i.$
In https://math.stackexchange.com/a/2812234 there is the solution for the case $x \in (0,1),$ but in my case the random variables are distributed in the interval (0.2,1.8), or more generally in the interval (a,b). I have not been able to translate the formula for the product in the interval (0,1) to my case; if I follow the procedure used to retrieve the formula (see link above) I have a problem since the integrand has no pole, hence no residue. 
Can anyone point out any suggestions on how to proceed, or if there is any reference to books or articles where this kind of case is treated?
 A: It’s not true that the integrand has no pole if you take the approach in the linked answer. It does, it’s just that the residue is complicated and accordingly we don’t get a nice result as in the case when the range starts at $0$.
You can map your interval $(0.2,1.8)$ to $\left(\frac19,1\right)$ by taking a factor of $1.8$ out of each variable. Then the probability density function of $\log x$ is $\mathrm e^y[-r\le y\le0]$ with $r=\log9$, and thus its Fourier transform is
$$
\int_{-r}^0\mathrm e^{-2\pi\mathrm iyt}\mathrm e^y\mathrm dy=\frac1{1-2\pi\mathrm it}\left(1-\mathrm e^{-(1-2\pi\mathrm it)r}\right)\;.
$$
Then the probability density function for the sum of $n$ values is
\begin{eqnarray}
\sigma_n(y)
&=&
\int_{-\infty}^\infty\frac{\mathrm e^{2\pi\mathrm iyt}}{(1-2\pi\mathrm it)^n}\left(1-\mathrm e^{-(1-2\pi\mathrm it)r}\right)^n\mathrm dt
\\
&=&
\frac{\mathrm e^y}{2\pi\mathrm i}\int_{1-\mathrm i\infty}^{1+\mathrm i\infty}\frac{\mathrm e^{-yz}}{z^n}\left(1-\mathrm e^{-rz}\right)^n\;.
\end{eqnarray}
The residue at $z=0$ is the coefficient of the $z^{n-1}$ term of $\mathrm e^{-yz}\left(1-\mathrm e^{-rz}\right)^n$, which unfortunately leads to quite a messy expression.
