Double integral using jacobian

Question

I have to evaluate following integral $\int_0^\infty\int_{-y}^yf(x,y)dxdy$
using $u=x^2-y^2, v=x^2+y^2$.
Actual form of function f is not important here, except for that it's odd function of y.
So the area in uv plane is $\int_0^\infty\int_{-\infty}^0f(u,v)/\lvert J(u,v/x,y)\rvert dudv$
But I feel like that the above integral is somewhat incomplete, cuz it includes also y<0 case. However the answer is just the value of above integral. Can someone explain what's wrong in my logic?

Answer 1

The reparametrization isn't injective, so you have to split the domain into 2 pieces such that the restriction to each piece is injective. What you need to consider seperately is $x>0$ and $x<0$, so it's actually the fact that $f$ is a even function of $x$ that allows to consider just one case, and multiply the result by $2$.

If $x>0$ then
$\quad\quad\quad\quad\quad\quad\quad u=x^2-y^2\ ,\ v=x^2+y^2$ and $0<y\ ,-y<x<y$
can be rewritten to
$\quad\quad\quad\quad\quad\quad\quad (x,y) = \left(\sqrt{\frac{u+v}{2}},\sqrt{\frac{v-u}{2}}\right)$ with $u<0$ and $-u<v$.
Hence $$\int_0^\infty\int_{-y}^yf(x,y)dxdy = 2\int_ {-\infty}^0 \int_ {-u}^\infty f\left(\sqrt{\frac{u+v}{2}},\sqrt{\frac{v-u}{2}}\right)\frac{1/4}{\sqrt{(v-u) (u+v)}}dv\ du$$ where $\frac{1/4}{\sqrt{(v-u) (u+v)}}$ is the absolute determinant of the Jacobian of $\left(\sqrt{\frac{u+v}{2}},\sqrt{\frac{v-u}{2}}\right)$.