TITLE: Multivariate Residue Theorem?
QUESTION [21 upvotes]: Is there an extension of the residue theorem to multivariate complex functions? Say you have a function of $n$ complex variables $s_{n}$ and you wish to integrate it over some region in $\mathbb{C}^{n}$. Can you exploit the singularities of the function as you would in the single variable case to evaluate the integral?
For example 
\begin{equation}G=\int_{\Omega} d^{n}s \frac{1}{\sum_{i} s_{i}^{2}}\end{equation}
This is singular for the $n$ roots of the equation $\sum_{i}s_{i}^{2}=0$ but I can't think of a way to extend the residue theorem to such a case. I guess you could go to 'polar' coordinates and it may simplify this one, but what about more complicated functions where this isn't possible?
I'm sorry if the question is ill-defined.

REPLY [2 votes]: You can get a higher-dimensional version of the residue theorem by applying the one-dimensional version repeatedly. For example, let $f(x_1,x_2)$ be analytic on $\bar\Omega_1 \times \bar \Omega_2$, and let $x_1 \in \Omega_1$, $x_2 \in \Omega_2$. Then, 
$$
-\frac{1}{4\pi^2}
\int_{\partial\Omega_1} \int_{\partial\Omega_2} \frac{f(t_1,t_2)}{(t_1 - x_1) \, (t_2 - x_2)} \, dt_1 \, dt_2
= 
\frac{1}{2\pi i}
\int_{\partial\Omega_1} \frac{f(t_1,x_2)}{t_1 - x_1} \, dt_1
=
f(x_1,x_2)
.
$$
The situation can get messy, however, if your pole does not have tensor-product form. For example, if we replace $(t_1 - x_1) \, (t_2 - x_2)$ with $t_1 - t_2 + x$, we get
$$
-\frac{1}{4\pi^2}
\int_{\partial\Omega_1} \int_{\partial\Omega_2} \frac{f(t_1,t_2)}{t_1 - t_2 + x} \, dt_1 \, dt_2
= 
-
\frac{1}{2\pi i}
\int_{\partial\Omega_1} f(t_1,t_1 + x) \, dt_1
,
$$
and now you have to think about residues of $g(x_1) := f(x_1,x_1+x)$ to determine the last integral.