# Talk:Cauchy-Goursat Theorem

I don't think that's the problem with the proof: once you have the existence of an antiderivative; you can apply the fundamental theorem of calculus as is done here.

The problem is that to show that $F$ is well defined we want that:

$\displaystyle\int_{\gamma_1} f = \int_{\gamma_2} f$

for any two paths going between two points in the domain; that is, we want that the integral around $\gamma_1$ follows by $\gamma_2$ traversed backwards is zero; and this is precisely Cauchy's theorem for $\gamma_1 \cdot\gamma_2^{-1}$. So the statement:

"Let $F$ be an antiderivative of $f$"

needs proving; and is most naturally proved as a minor corollary of Cauchy's theorem. Unless the proof of existence of antiderivatives contains a proof of Cauchy's theorem to show well-definedness, the argument on this page becomes circular. --Linus44 (talk) 19:20, 2 May 2013 (UTC)

Moreover, the example given isn't an example of Cauchy's theorem: there is no simply connected subset of $\C$ that contains $\gamma : t \mapsto \exp(it)$ in which $1/z^2$ is holomorphic.
The fact that the integral vanishes here is an example of the residue theorem. --Linus44 (talk) 22:13, 2 May 2013 (UTC)
Linus, perhaps you can move the example to the appropriate place? Does the whole proof need an explain, questionable template? --Dfeuer (talk) 14:55, 5 May 2013 (UTC)

I just made a large edit to the page. Hopefully it clears up some of the issues with the previous version without introducing too many new problems. --Benlittle (talk) 15:14, 15 October 2018 (EDT)

Not usually a fan of replacing a proof with something completely different, but, hey ho, if the original one was unsalvageable, I suppose we haven't lost anything. No doubt when we get round to scouring a source work we'll be in a position to add to this. Needs a bit of tidying and bringing up to house style, minor stuff. In due course. --prime mover (talk) 15:28, 15 October 2018 (EDT)

## Statement of theorem

I see two problems with how the Cauchy-Goursat Theorem is phrased right now.

First, it defines $f: D \to \C$, but that does not guarantee that $f$ is defined on $\partial D$.

Second, it defines the contour of the integral as the contour bounding the simply connected set $D$. However, it is possible to define a simply connected set in the plane which boundary $\partial D$ cannot be parameterized as a Jordan curve, that is : a simply closed contour.

My solution:

$(1): \quad$ Create a page for Definition:Contour/Simple Closed Contour, and define (and prove existence of) the interior of said contour, as the simply-connected interior of the corresponding Jordan curve.
$(2): \quad$ State that the theorem talks about simple closed contours $C$ in $D$ with interior $\Int C \subseteq D$.

Any thoughts? Anghel (talk) 22:04, 26 August 2022 (UTC)

Both.
Writing it for the general contour inside a general region is essential, of course, as that's the basic result.
Then again, demonstrating that it also holds for the boundary of a region, despite the fact that the boundary itself is not actually part of the region is equally important but so much more subtle.
Good job on this. --prime mover (talk) 05:24, 12 September 2022 (UTC)