That is, given some function $f:\mathbb{C}^n \to \mathbb{C}$ entire/ sufficiently holomorphic, if we have two domains $D,D'$ in $\mathbb{C}^n$ with the same boundary i.e $\delta D = \delta D' $, will we have $$\int_D f = \int_{D'} f$$ I'm aware there is an analogue for cauchy's theorem on a polydisc, which I would hope we could use in proving this. I've found a proof for the $n=2$ case using Stoke's theorem, but I'm lacking knowledge in the generalised form for stokes and not sure how I could apply it here.
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D.Dog
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You mean the same boundary modulo homotopy ? If so the question should be that you are integrating a closed-form – reuns Mar 17 '19 at 01:11
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@reuns my knowledge of homotopies is relatively slim but I meant to have the same boundary curve describing the surface. E.g a flat disc in C^2 versus a half sphere with an open base. – D.Dog Mar 17 '19 at 01:26
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First of all it is not $f$ that you integrate but some $k$-forms created from $f$. For example with $\gamma : [a,b] \to U \subset \mathbb{C}^n$ a curve then $f$ holomorphic on $U$ simply connected implies for each $j$ there is $F_j$ such that $f(z) = \partial_{z_j} F_j(z)$ (think to $f$ being a power series in several variables) so $\int_\gamma f(z) dz_j = F_j(\gamma(b))-F_j(\gamma(a))$. Equivalently $f(z)dz_j$ is closed and $\gamma \cup \gamma_2^-$ is homotopic to identity so $\int_{\gamma \cup \gamma_2^-} f(z)dz_j=0$ – reuns Mar 18 '19 at 00:40
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Could you direct me to some further reading on this? This seems very interesting and isn't something I know much about. – D.Dog Mar 18 '19 at 15:47
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You can see this for something closely related. – mrf Mar 26 '19 at 12:11