Let $f(x)$ is a density function, $f(x) = 0$ if $x$ in $(-\infty, 0]$. Prove that $u(x, y) = \frac{f(x + y)}{x + y}$ is a density function if $x > 0$, $y > 0$, $u(x, y) = 0$ otherwise.
I know that $\int_{0}^{+\infty}\left(\int_0^{+\infty}\frac{f(x + y)}{x + y}\,dx\right)\,dy$ should be equal to 1, but I don't know how to find it.
Try the change of variables $u=x+y, v = x$. Then $$\int^\infty_0\int^\infty_0\frac{f(x+y)}{x+y}\,dx\,dy=\int^\infty_0\int^u_0\frac{f(u)}{u} dv\,du=1$$