Q:
$(X,Y)$ has the following joint p.d.f $$f_{X,Y} (x,y)= \begin{cases} \frac{2}{(1+x+y)^3}, \ &x,y > 0 \\ 0, \ &\text{elsewhere.} \end{cases}$$ Show that $$ f_{X-Y} (u)= \frac{1}{2(1+|u|)^2}, \ -\infty < v < \infty. $$
Attempted solution:
Using change-of-variables, $ \begin{cases} u = x-y \\ v= y \end{cases}$ gives the Jacobian $J = 1$. Hence $$f_{U,V}(u,v) = f_{X,Y}(u+v,v) = \begin{cases} \frac{2}{(1+u+2v)^3}, \ &v > 0 \ \land v > -u \\ 0, \ &\text{elsewhere.} \end{cases}$$
The p.d.f for $f_{X-Y} = f_U $ is then
\begin{equation} \begin{split} &\int_{0}^{\infty} \frac{2}{(1+u+2v)^3} dv, \ u >0 \\ &\int_{-u}^{\infty} \frac{2}{(1+u+2v)^3} dv, \ u < 0\end{split} \end{equation}
I'm having trouble solving the integrals, which yields a feeling that I might've taken the wrong route to the solution. Any hints?
