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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?

Septacle
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  • Can you show how you know the answer is just the value of that integral, without $1/2$? – velut luna Oct 01 '17 at 00:39
  • OK. The function is $f(x,y)=y*\lvert x \rvert exp(-2(x^4+y^4))$. Wolfram alpha says that the value is $\pi/32$, and the value of integral above is the same. – Septacle Oct 01 '17 at 00:44
  • @MathLover can you explain more? I can't see your point. Integral with your range results in $3\pi/64$ – Septacle Oct 01 '17 at 02:28

1 Answers1

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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 $\left(\sqrt{\frac{u+v}{2}},\sqrt{\frac{v-u}{2}}\right)$.

Coolwater
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  • Thank you for your reply. But what the -u < v inequality means? If I substitute x and y I just got x^2>0. – Septacle Oct 02 '17 at 06:30
  • @Septacle For the case $x>0$ we have $x>0\Rightarrow x^2>0\Rightarrow (u+v)/2>0 \Rightarrow u+v>0\Rightarrow -u<v$ – Coolwater Oct 02 '17 at 07:52
  • O Thank you. I got it. But the problem (for me, not ur answer) is that my function is not integrable for that range. (in elementary func.). f (u,v)=e^-(u^2+v^2) – Septacle Oct 02 '17 at 17:15