Integration by substitution in two dimensions. I want to integrate over the unit square $||(x,y)||_\infty<1$, using the substitution:
$u=\frac{1}{1-xy}$ and $v=\frac{1-x}{1-xy}$.
What is the correct expression for $dxdy$ in terms of $u,v,du,dv$?
 A: The term your looking for is $$dxdy=|\frac{\partial(x,y)}{\partial(u,v)}|dudv$$
The partial derivative-looking term is called a Jacobian and is defined as
$$\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}   \\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
\end{vmatrix}=\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}$$
In your case, $$x=1-\frac{v}{u},~~~y=\frac{u-1}{u-v}=1+\frac{v-1}{u-v}$$
And therefore
$$\frac{\partial(x,y)}{\partial(u,v)}=\frac{v}{u^2}\cdot\frac{u-1}{(u-v)^2}-\frac{-1}{u}\cdot\frac{1-v}{(u-v)^2}=\frac{1}{u^2(u-v)}$$
This can actually be generalized to as many dimensions as you please. In higher dimensions, the Jacobian generalizes to
$$\frac{\partial(y_1,y_2,\ldots y_n)}{\partial (x_1,x_2,\ldots x_n)}=\begin{vmatrix}
\frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\partial x_2} & \ldots \frac{\partial y_1}{\partial x_n}   \\
\frac{\partial y_2}{\partial x_1} & \frac{\partial y_2}{\partial x_2} & \ldots \frac{\partial y_2}{\partial x_n}   \\
\vdots & \vdots & \ddots \vdots \\
\frac{\partial y_n}{\partial x_1} & \frac{\partial y_n}{\partial x_2} & \ldots \frac{\partial y_n}{\partial x_n}
\end{vmatrix}$$
