I am trying to calculate the distribution of the sum of two independent log-uniform distributions but something doesn't add up.
Suppose $a \sim \mathrm{uni}(0,1)$ and $b \sim \mathrm{uni}(0,1)$. Thus, $u=\log(a)$ has an exponential distribution of the form $e^u$, which is defined for values for which $u<0$ (the same applies to $v=\log(b)$ ).
Now, define a new r.v $z=u+v$. I have tried to compute the new distribution via the convolution formula, but I get a non-converging integral. Can anyone help?