Confused about the Jacobian matrix

Find $$\int_0^{\infty} \int_0^{\infty} e^{-2xy} \, \mathrm d x \mathrm dy$$ using $$u = x^2 - y^2$$ and $$v=2xy$$.

I have tried using the Jacobian matrix to obtain the Jacobian of the transformation. However, confusion arises since I do not know what should be kept constant. Do I directly differentiate $$u$$ with respect to $$x$$ while keeping $$y$$ constant, or do I substitute $$y$$ from $$v=2xy$$, and then differentiate $$u$$ with respect to $$x$$ while keeping $$v$$ constant?

• When finding $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$, just differentiate $u$ as a function of $(x,y)$, ignore the $v$. Similarly, when differentiating $v$, don't worry about $u$. So for example, $\frac{\partial u}{\partial x} = 2x$. – Minus One-Twelfth Mar 10 at 11:55
• However, the Jacobian matrix says that we should derive x with respect to u, x wrt v, y wrt u, and y wrt v. Therefore, it is not deriving u wrt to x and y, but it's the other way around. – user205891 Mar 10 at 11:59
• There is a trick that you can use if you want: the Jacobian determinant of the inverse map is the inverse (reciprocal) of the Jacobian determinant of the forward map. Therefore, you can just do it the way I described, get the Jacobian determinant of that, and then take the reciprocal. (This will get you something in terms of $x$ and $y$, which you will need to express in terms of $u$ and $v$ for the integral.) – Minus One-Twelfth Mar 10 at 12:01

$$J=J\frac{(x,y)}{(u,v)}=\begin{vmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{vmatrix}^{-1}=\begin{vmatrix}2x&-2y\\2y&2x\end{vmatrix}^{-1}=\frac1{4(x^2+y^2)}$$