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Suppose I have a rational function of $k$ complex variables:

$$ R(x_1,...,x_k) = P(x_1,...,x_k)/Q(x_1,...,x_k) $$

where $P$ and $Q$ are polynomials. Now I'd like to compute the integral of this function over the product of disks, $D^k$, inside $\mathbb{C}^k$, ie:

$$ I = \frac{1}{(2 \pi i)^k} \int_{D^k} dx_1...dx_k R(x_i) $$ I'd like to use a multi-variable version of Cauchy's theorem to express this as a sum over residues.

To take a trivial example, if $R(x,y,z) = 1/(xyz)$, then there is a single contribution from $x=y=z=0$, and $I=1$. More generally, I suggest the following possible strategy:

  1. Find irreducible polynomial factors of $Q(x_i)$, ie:

$$ Q(x_i) = f_1(x_i) ... f_n(x_i) $$ Each of these contributes a component on which $Q(x_i)$ vanishes, leading to a codimension one subspace on which $R(x_i)$ has a pole.

  1. Generically I expect contributions from the intersection of $k$ components, which generically intersect at a discrete points. In a neighborhood of a point, $p$, where, say, $f_1$, ..., $f_k$ vanish, I believe we can make a local linear change of variables, $x_i \rightarrow t_i$, such that we may write:

$$ R = \hat{R}(t)/(t_1^{n_1} ... {t_k}^{n_k}) $$

where $\hat{R}(t_i)$ is holomorphic near $t_i=0$.

  1. Then the residue is given by taking suitable derivatives of $\hat{R}(t)$, and summing this over all such points, $p$.

My question is, is this procedure correct? I worry what happens if $R(x_i)$ is not sufficiently generic, eg, more than $k$ components intersect at a point. And finally, is there a more elegant way to extract this integral directly from $R(x_i)$, without having to go through this algorithm?

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  • $\begingroup$ Your integral does not make sense, $D^k$ has real dimension $2k$. Do you mean $(\mathbb{S}^1)^k$? $\endgroup$
    – abx
    Commented Apr 6, 2019 at 7:59
  • $\begingroup$ Yes, sorry, I meant that the integral is over the boundary of the disks, and so by the multvariable generalization of Cauchy's theorem should receive some contribution from poles inside the product of the disks. $\endgroup$
    – user6013
    Commented Apr 6, 2019 at 16:10

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