Evaluating $\int_0^\infty\int_0^\infty e^{-(x+y)}\cdot \sin(\frac{\pi\cdot y}{x+y}) \, dy \, dx$ 
Evaluating
  $$
\int_0^\infty \int_0^\infty
    e^{-(x+y)}\cdot
\sin\left(\frac{\pi\cdot y}{x+y}\right) \, dy \, dx
$$

Using the transformation $x+y=u$ and $y=v$ (Is there a better transformation than this?)
The absolute value of Jacobian $|J| = 1$, therefore $dx\cdot dy = du \cdot dv$
$$\int_0^\infty\int_0^\infty e^{-(x+y)}\cdot \sin\left(\frac{\pi\cdot y}{x+y}\right) \, dy \, dx  = \iint_R e^{-(u)}\cdot \sin\left(\frac{\pi\cdot v}{u} \right) \, du \, dv$$
Now this integration is easy but the problem I'm facing is difficulty figuring out the new region $E$ and setting the limits of $u$ and $v$. This is particularly because $\infty$ is involved in the limits of $x$ and $y$
 A: Let us use the change of variables $(x,y)\to(u,v)$ such that $u=x+y$ and $v=\frac{y}{x+y}$. 
Then $x=u(1-v),y=uv$ which means $x,y\ge0\implies u\ge0$ and $0\le v\le1$. 
Absolute value of jacobian of transformation is $u$. The integral then simplifies as
$$I=\int_0^1\int_0^\infty ue^{-u}\sin(\pi v)\,du\,dv=\int_0^1\sin(\pi v)\,dv\int_0^\infty ue^{-u}\,du$$
I think this transformation works better here.
A: The change of variable proposed by StubbornAtom is very elegant, but yours is not bad. With your transformation 
$$
u=x+y,\qquad v=y, $$ 
which has inverse
$$
x=u-v, \qquad y=v, $$ 
the integration set $\{x\ge 0, y\ge 0\}$ transforms into $\{u-v\ge 0, v\ge 0\}$, so the integral becomes 
$$
\int_0^\infty \left( \int_v^\infty e^{-u}\sin\left(\pi\frac v u\right)\, du \right)\,dv, $$ 
which is a little hard to compute. Arranging $u, v$ in the other way, that is,$\{u\ge 0, 0\le v\le u\}$, we get 
$$
\int_0^\infty e^{-u}\left(\int_0^u\sin\pi\frac{v}{u}\, dv\right)\, du=\int_0^\infty e^{-u}\frac{1-\cos\pi u}{\pi}\, du=\frac{\pi^2}{\pi(1+\pi^2)}.$$ 
