calculus - Evaluating $int_0^inftyint_0^infty e^{-(x+y)}cdot sin(frac{picdot y}{x+y}) , dy , dx$

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 $y$

Answer

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.