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Consider the path $\gamma: [0,2\pi]\rightarrow \mathbb{C}$ given by $\gamma(t) = e^{it}$.

Let $f(z) = \frac{1}{z}, z\neq 0$.

I worked out the integral $$\int_\gamma f = \int_{0}^{2\pi} f(\gamma(t))\gamma'(t) dt = 2\pi i$$

However, $f$ is continuous on $f(\gamma([0,2\pi]))$, so I was wondering why this integral is nonzero, since I thought that the integral over a closed path that takes $[a,b] \rightarrow G$, where $f$ is continous on $G$ and $f$ has a primitive on $G$ should be $0$. What is going wrong here?

InsideOut
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fosho
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  • There's nothing wrong. The key idea here is that there is hole inside the loop at the origin, and the integral is detecting it. I'm pretty sure someone else will give a more detailed and insightful answer. You could try calculating the same integral along a loop that doesn't contain the origin and see what happens. – user347489 Sep 06 '16 at 09:04
  • But what is wrong in what I said at the bottom then? Surely something there is wrong? Maybe that f doesnt have a primitive on G? – fosho Sep 06 '16 at 09:06
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    The function $f(z)=1/z$ does not have an antiderivative on $\mathbb{C}\setminus{0}$. – egreg Sep 06 '16 at 09:09

3 Answers3

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Point 1: You have not said what $G$ is; however, $f$ does not have a primitive on the unit circle. $\log(z)$ cannot be continuously well-defined on the unit circle. This is why any definition of $\log(z)$ has branch cuts.

Point 2: Since $\frac1z$ has a singularity at $z=0$, we cannot continuously contract the unit circle to a point in $\mathbb{C}\setminus\{0\}$ without crossing $0$.

robjohn
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  • The existence of a singularity is not conclusive; the same would apply to $g(z)=1/z^2$. – egreg Sep 06 '16 at 10:09
  • No. Just because some singularity has residue $0$ does not mean we can blindly contract a path over any singularity. We need to justify whether we can contract a path over a given singularity. – robjohn Sep 06 '16 at 10:18
  • I meant it is not conclusive with respect to $f$ not having an antiderivative. – egreg Sep 06 '16 at 10:27
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    The fact that $\frac1z$ has a singularity was not used to show that $\log(z)$ cannot be continuously well-defined on the unit circle. Only to show that we cannot continuously contract the unit circle to a point, and so the integral along the unit circle need not be $0$. Perhaps this was not clear. I have tried to separate the two points further to avoid confusion. – robjohn Sep 06 '16 at 10:36
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The following is a fundamental result in complex analysis (see the books of Ahlfors or Rudin).

Theorem. Let $f$ be a holomorphic function on a region $\Omega$; then $f$ has an antiderivative on $\Omega$ if and only if $$ \int_\gamma f(z)\,dz=0 $$ for every closed path in $\Omega$.

(A region is an open and connected subset of the complex plane).

Since the integral you're considering is not zero, it follows that $1/z$ has no antiderivative on a region $\Omega$ that contains a circle centered at $0$, because the integral over this circle is $2\pi i\ne 0$.

It is also well known and basic in complex analysis that the integral of a holomorphic function on a contractible path is zero, so the result follows that a holomorphic function defined on a simply connected region has an antiderivative. So, for instance, $1/z$ does have an antiderivative on a region obtained from the complex plane by removing a ray with origin at $0$ (the main branch cut of the complex logarithm is defined as the antiderivative of $1/z$ over the region obtained by removing the nonpositive real numbers).

On the other hand, simple connectedness is only a sufficient condition; for example, the function $1/z^2$ has an antiderivative on $\mathbb{C}\setminus\{0\}$, which is a non simply connected region.

egreg
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The integral should be zero if the path is contractible. Since $G=\Bbb C\setminus \{0\}$, every path around the origin cannot be deform to a point, so every time you turn around the origin you pay $2\pi i$.

InsideOut
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    The integral on the unit circle of $1/z^2$ is zero, but the path is not contractible. Your “if and only if” should be “if”. – egreg Sep 06 '16 at 10:10