I am playing with multivariate contour integrals, and I'm curious if there exists some sort of "residue theorem" for such integrals. Here is an example.
Consider the function $f(z,w) = \frac{1}{(1 - a_1 \frac{z}{w})(1-a_2^{-1}zw)}$ with $0 < |a_{1,2}| < 1$, and the integral $$ I = \oint_{\substack{|z|=1 \\ |w| = 1}} \frac{dz}{2\pi i}\frac{dw}{2\pi i}f(z,w) \ . $$
Naively I would expect there to be three computations that give the same result
- First compute $w$-integral (keeping $z$ arbitrary on the unit circle) by picking up two residues at $w = a_1 z$, $w = a_2z^{-1}$, then take result (which sums magically to $- a_2/z$) and compute the $z$-integral: the final answer is $I = -a_2$.
- First compute $z$-integral (keeping $w$ arbitrary on the unit circle) by picking up pole $z = a_2 w^{-1}$, then take result and compute the $w$-integral: the final answer is (happily) $I = -a_2$.
- A "multivariate residue theorem" to do the computation in one go, by first solving the "simultaneous pole equation" $1 - a_1 \frac{z}{w} = 0, 1 - a_2^{-1}zw = 0$, then identifying the solutions within the (simultaneous) inner-region $|z| < 1, |w| < 1$, and then some "magical steps" to get the $I = -a_2$.
Apparently, I could not figure out what the "magical steps" are. Even worse, the two solutions to $1 - a_1 \frac{z}{w} = 0, 1 - a_2^{-1}zw = 0$ are given by $z = \pm (a_2/a_1)^{1/2}$, $w = \pm (a_1a_2)^{1/2}$, which are in general not even in the desired region $|z| < 1, |w| < 1$.
So to summarize my questions are:
- Is there a "residue theorem" for the multivariate case?
- In the direct relation between the simultaneous pole $z = \pm (a_2/a_1)^{1/2}$, $w = \pm (a_1a_2)^{1/2}$ and the final result?
- In the first computation, the final $z$-pole is $z = 0$, which is completely invisible in $f(z,w)$ (note that $f(z = 0, w) = 1$, actually quite regular). How to understand this?
