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It might be a good idea to distinguish between Cauchy´s integral Theorem and Cauchy Integrals.
Missing information
[edit]The last few lines that speak about the book by Marsden-Tromba are unclear; I have the latest edition (5th) and this is not cited on the given page. More specific reference should be given. It would be better to mention the applications of the topics along with its theory to make them more interesting. Its always a welcome feeling if you know why you are learning a particular thing.
The term "loop" suddenly makes an appearance, within the discussion, with neither a definition in the text nor a hyperlink or a reference linking to a definition. — Preceding unsigned comment added by 83.217.170.175 (talk) 17:53, 17 June 2015 (UTC)
"general Cauchy theorem"
[edit]I guess that Tetvesdugo was actually thinking of a slight variation on what he/she wrote: if the two paths have the same start points and the same end points, but are not necessarily closed, then the integrals along them are the same. Of course it is still equivalent to the closed-contour version, but the equivalence is a tiny bit less trivial. McKay 07:30, 1 February 2007 (UTC)
- OK, I see. But keeping with the standard (and a little less general form) is I think better. As to the more general version, I think it can be proved by making a closed path aout of those two paths with the same starting and ending points (by reversing one of the paths). Oleg Alexandrov (talk) 15:41, 1 February 2007 (UTC)
there is a somewhat different Cauchy integral theorem: if f is holomorphic on an open connected star-shaped region D, then f has a complex anti-derivative. (so the simple-connectedness of the domain is replaced by star-shapedness and continuity on its closure is not needed.) slightly more generally, if D has star center c and f is holomorphic on the punctured set D - c and continuous on D, the same is true.
instead of borrowing Green's from real variable calculus, it can be shown using Goursat's integral lemma. (Cauchy-Riemann gives the continuity of the partial derivatives? if not invoking Green's as the article does might be a problem.) someone correct me if i'm wrong but i think this reflects the historical developement of the integral theorem better than the version in article. Mct mht 20:41, 3 August 2007 (UTC)
The proof is wrong
[edit]The proof presented here is wrong. It relies on Green's theorem, but the latter requires the partial derivatives to be continuous - and this continuity has not been established. Of course, it will follow once we have shown that the derivative of a complex function is differentiable, but the proof of this fact typically relies on Cauchy's integral theorem to begin with. This circular reasoning is not unique to Wikipedia, but I think we should be a little better than that and give a valid proof. -- Meni Rosenfeld (talk) 13:04, 29 October 2007 (UTC)
- Quite right. I am going to delete it. Also there is a fashion emerging to move proofs to sub pages.Billlion 09:39, 31 October 2007 (UTC)
The proof performed by Cauchy requires that the partial derivatives be continuous. Goursat proved it without this requirement and the theorem is more often called the Cauchy-Goursat theorem. —Preceding unsigned comment added by 69.134.94.144 (talk) 07:06, 20 November 2008 (UTC)
Move it?
[edit]Shouldn't we move it to an article called Cauchy-Goursat theorem ? --LQST (talk) 17:25, 23 March 2009 (UTC)
Style!
[edit]The introduction is very poorly worded. The definition of "winding number" seems to start in the middle of a sentence. (Collin237) —Preceding unsigned comment added by 32.178.64.127 (talk) 10:16, 28 November 2010 (UTC)
+1. 128.112.110.117 (talk) 20:40, 26 October 2011 (UTC)
IMO the whole section is extraordinarily dense. My recollection is that the theorem is very simply stated in terms of poles, which are but vaguely referenced in the side-bar. I also dispute that the path of integration needs to be rectifiable, since that implies a finite path:
"A rectifiable curve is a curve with finite length." https://en.wikipedia.org/wiki/Curve#Lengths_of_curves (Redirected from Rectifiable path)
Many uses of the path loop around the entire real axis, which is not especially finite!
can anyone with a suitable authority make this stuff clearer? A1jrj (talk) 18:32, 23 November 2018 (UTC)
The contour has to be rectifiable, for the integral cannot be defined on general curves (think of the Peano curve). Using as path loop the entire real axis is usally not done directly, but is obtained as the result of a limiting process, similarly to the improper riemann integral.--ClerVen (talk) 13:13, 7 April 2020 (UTC)
easy proof with continuous derivatives and convex domain
[edit]I think this proof for convex domains is very intuitive. Assume is continuously complex differentiable on a convex domain containing , is a (piecewise) closed rectifiable curve , and let (so that ).
Then
So with we have and .
It is obvious in (1) how the fact that is fundamental in the proof.
78.196.93.135 (talk) 12:59, 25 September 2016 (UTC)
Homotopy WITHIN the domain not emphasized enough.
[edit]I added (within U) somewhere, but maybe it will help if we reword to emphasize that the homotopy in plane always exists and the subtlety is if such a homotopy can be done completely within U. I also would not say U is a "space". 130.234.241.216 (talk) 21:36, 20 December 2022 (UTC)
Homology
[edit]Is no one going to mention homology? Indeed the homology version of Cauchy's theorem is stronger. There are loops that are null-homologous in a some open set but not null-homotopic. (The concept behind why it is possible to hang a picture frame from two nails so that removal of either one will allow the frame to drop!) 2001:14BB:101:8B11:549D:FA50:72E0:61C3 (talk) 19:25, 6 February 2023 (UTC)