Multidimensional complex integral of a holomorphic function with no poles

Question

I am looking for a generalization of the Cauchy integral theorem. I know that there are generalizations of the Cauchy integral formula (eg the Bochner-Martinelli formula), but I do not know if this simplifies as I would hope. In single-variable complex analysis, we have the Cauchy integral theorem: $$ \oint_{\gamma}f(z) = 0 $$ if $f(z)$ is a holomorphic function with no poles within the region that $\gamma$ encloses.

If $f(z_1\cdots,z_n)$ is a holomorphic function in $n$ complex variables with no poles within a domain $D$, is it true that $$ \oint_{\partial D}f(z_1,\cdots,z_n) = 0 $$

Answer 1

The "local" version of Cauchy's integral theorem would be something like: Assume that $f$ is holomorphic on (a neighbourhood) of the polydisc $\Omega = \mathbb{D}(z_1,r_1) \times \mathbb{D}(z_2,r_2) \times \cdots \times \mathbb{D}(z_n,r_n)$. Then $$ \int_{|\zeta-z_1|=r_1} \int_{|\zeta-z_2|=r_2}\cdots\int_{|\zeta_n-z_n|=r_n} f(\zeta_1, \zeta_2, \ldots, \zeta_n)\,d\zeta_1\cdots d\zeta_n. $$ But for this to hold, it's even enough to assume that $f$ is holomorphic in one of the variables (integrate first with respect to this variable to get zero...) so it's not very useful.

More useful is the straight-forward generalization of Cauchy's integral formula to polydiscs. If $f$ is holomorphic on (a neighbourhood) of the polydisc $\Omega = \mathbb{D}(z_1,r_1) \times \mathbb{D}(z_2,r_2) \times \cdots \times \mathbb{D}(z_n,r_n)$, and $a \in \Omega$, then

$$ f(a) = \frac{1}{(2\pi i)^n} \int_{|\zeta-z_1|=r_1} \int_{|\zeta-z_2|=r_2}\cdots\int_{|\zeta_n-z_n|=r_n} \frac{f(\zeta_1, \zeta_2, \ldots, \zeta_n)}{(\zeta_1-a_1)(\zeta_2-a_2)\cdots(\zeta_n-a_n)}\,dz_1\cdots dz_n. $$

The really fascinating thing is that we are integrating just on a tiny part of the boundary of $\Omega$.

Answer 2

Following up on @mrf's answer: Another way to go is to generalize the Residue Theorem. If $f$ is meromorphic in a neighborhood of $a\in\Bbb C$, you can write $$\frac1{2\pi i}\int_{|z|=r} \frac{df}{f} = \text{Res}(f,a).$$ Generalizing to $n$ dimensions, you can consider $n+1$ holomorphic functions $f_1,\dots,f_n,g$ in a neighborhood of $a\in \Bbb C^n$ with $\{a\} = f_1^{-1}(0)\cap\dots\cap f_n^{-1}(0)$. Then, for small $r_1,\dots,r_n>0$, let $\gamma = f_1^{-1}(r_1)\cap\dots\cap f_n^{-1}(r_n)$; note that $\gamma$ is a smooth $n$-dimensional cycle. Let $\omega = \frac{g\,df_1\wedge\dots\wedge df_n}{f_1\cdots f_n}$. Then we have the Residue Theorem $$\frac1{(2\pi i)^n}\int_\gamma \omega = \text{Res}(\omega,a)\,,$$ provided we orient $\gamma$ correctly. Corresponding to the meromorphic $n$-form $\omega$, the Dolbeault isomorphism now gives a smooth $(n,n-1)$-form $\psi$ defined on a punctured neighborhood of $a$, and $$\text{Res}(\omega,a) = \int_{S^{2n-1}(\epsilon)} \psi$$ for small $\epsilon>0)$.

A very readable source, if you want to follow up on this, is the beginning of Chapter 5 of Griffiths and Harris.