In an exercise, I found after some time the following integral:
$$\int_{-\infty}^\infty \mathrm{d}x\,\mathrm{d}y\,f(x,y)\int_{-1}^1 \mathrm{d}z\,\delta\left(z-\frac{2-2x-2y+xy}{xy}\right)$$
In the end I should get an integral like
$$\int_{0\leq x,y\leq 1, \, x+y\geq1}\mathrm{d}x\,\mathrm{d}y\,f(x,y)$$
How can I see that they are equivalent? Or aren't they equivalent and I did a mistake in my previous calculation?
The original integral is equivalent to $\int_S\mathrm{d}x\,\mathrm{d}y\,f(x,y)$, where $S$ is the set of values of $(x,\,y)$ for which $\frac{2-2x-2y+xy}{xy}\in(-1,\,1)$. That's because the integral over $z$ is $1$ for $(x,\,y)\in S$, but $0$ otherwise. (It's actually undefined if $\frac{2-2x-2y+xy}{xy}=\pm1$, but that locus has zero measure, so doesn't affect the integral's value.) Note that$$\begin{align}\frac{2-2x-2y+xy}{xy}\in[-1,\,1]&\iff\frac{x+y-1}{xy}\in[0,\,1]\\&\iff(xy>0\land1\le x+y\le1+xy)\\&\lor(xy<0\land1-xy\ge x+y\ge1)\\&\iff(xy>0\land x+y\ge1\land(1-x)(1-y)\ge0)\\&\lor(xy<0\land x+y\le1\land(1-x)(1-y)\le0)\\&\iff(0\le x,\,y\le 1\land x+y\ge1)\\&\lor(x<0\land y\ge1-x)\\&\lor(y<0\land x\ge1-y).\end{align}$$