The joint probability density function of random variables $ X$ and $ Y$ is given by $$ f_{X,Y}(x, y) = 2(1-y) \quad \text{if} \quad 0 \leq x \leq 1, 0 \leq y \leq 1 $$ $$ f_{X,Y}(x, y) = 0 \quad \text{otherwise} $$ Determine the probability distribution function of $Z = XY$.
I know that this is a duplicate but there is never a full explanation of how to evaluate the double integral required to solve this question, or if there is an easier way. My working so far: $$ \int_0^1\int_0^{\frac{z}{x}}2(1-y)dydx = \left(\frac{z^2}{x}+2z\ln(x)\right)|_0^1 $$ This is where I fail to continue, because of terms like $\ln(0)$. What do I do? I expect an answer of $F_Z(z)=z^2−2z\ln(z)$ from this answer. I know that $F_Z = 0$ at $z=0$ since that is a right way to define it, but regardless I am stuck with getting an equation for $F_Z(z)$.