Cauchy's integral theorem says that the integral of any holomorphic function over a closed path in a simply connected domain is always $0$.

Is there a version of Cauch's integral theorem for several complex variables?
I mean, if $f:\Omega\subset\mathbb{C}^n\rightarrow\mathbb{C}$ is holomorphic and $\Omega$ simply connected, then for any closed path $\gamma\subset\Omega$, it holds $$\int_\gamma f=0.$$ Maybe it's a trivial question, but I've never found it anywhere, so I'm starting to think it's wrong.


1 Answer 1


In one variable, you can define the integral $\int_\gamma f(z)\,dz$ by $$\int_a^bf(\gamma(t))\cdot \gamma'(t)\,dt$$ You can approximate it by a Riemann sum $$\sum_{k=0}^{n-1} f(z_k)(z_{k+1}-z_k)$$ If you have a primitive function $F(z)$, you can compute the integral as $F(b)-F(a)$, where $a$ and $b$ are the end points of the curve $\gamma$.

In several variables, all this fail because of dimensional considerations. Then $\gamma'(t)$ and $z_{k+1}-z_k$ are vectors instead of complex numbers, and there is no notion of a primitive function without specifying which variable you are differentiating with respect to. Hence the expression $\int_\gamma f$, integrating a function along a curve, is not even well defined. This is why we integrate 1-forms instead of functions. In contrast to the situation in one variable, there is no universal 1-form to use here.


In ${\bf C}^2$, use the coordinates $(z,w)$ where $z=x+iy$ and $w=u+iv$. Consider the closed curve $\gamma$ given by $x^2+u^2=1$. Note that $\gamma$ is contained in ${\bf R}\times {\bf R}\subset {\bf C}\times {\bf C}$, and that we have no complex structure on ${\bf R}\times {\bf R}$. We can compute $\int_\gamma f(z,w)\,dz$ and $\int_\gamma f(z,w)\,dw$, but they reduce to ordinary real line integrals, and there is no reason why Cauchy's integral theorem should hold here. For instance, if $f(z,w)=w$ and we use the parametrization $z=\cos(t)$, $w=\sin(t)$, we get $$\int_\gamma f(z,w)\,dz=\int_\gamma u\,dx = -\int_0^{2\pi}\sin^2(t)\,dt=-\pi$$

  • $\begingroup$ Then we can't define the integral over a curve of a complex function of several variables? $\endgroup$ Feb 2, 2022 at 13:23
  • $\begingroup$ There is nothing special about complex numbers here; the issue is the same in ${\bf R}^n$. You can define 1-dimensional integrals in various ways (for instance, with respect to arc length), but you need to make some choice. I've added an example in ${\bf C}^2$ that shows that integration with respect to one of the coordinates will not satisfy Cauchy's integral theorem. $\endgroup$ Feb 2, 2022 at 14:32
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    $\begingroup$ Thank you very much. This question came from the problem of finding a $z_m$-primitive of a function of several variables. I tried to integrate over a path in the m-slice of $\mathbb{C}^n$, but I would require the slice to be simply connected to well define the integral (i.e. the integral doesn't depend on the choice of the path). If I would allow paths in the whole $\mathbb{C}^n$, then a Cauchy's integral theorem in several variables would be necessary in order to well define the primitive. $\endgroup$ Feb 2, 2022 at 14:49

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