Talk:Fundamental Theorem of Contour Integration

I'm sorry, I may be half-remembering something that I only half-learned, but does this result apply only when the function in question is holomorphic, i.e. satisfies the Cauchy-Riemann Equations? Just being continuous may not be enough (btw the link for "continuous function" goes to a disambiguation page).

The whole point here is that when a function is holomorphic (or whatever, I may have got that wrong, may be "analytical" is the word I need, I haven't checked), the value of the integral is independent of the path taken by the contour. If it is *not* analytical, then this is not the case.

Sorry, but it seems to me there's a lot of work that still needs to be done in order to get to where we want to go. I may of course be completely wrong, and the terminology of complex analysis has so completely changed in the last 8 years that I can't recognise it any more. If this is the case, then apologies. --prime mover (talk) 08:38, 3 May 2013 (UTC)


 * The function is holomorphic: since $F'(z) = f(z)$ for all $z \in D$ it follows that $F$ is analytic in $D$ and therefore so is $F' = f$. The essential statement here is that if $f$ has an antiderivative, the integral depends only on the endpoints. --Linus44 (talk) 09:29, 3 May 2013 (UTC)


 * See here (even has the same proof!) --Linus44 (talk) 09:31, 3 May 2013 (UTC)


 * Can we sort out the links then? They don't actually go to anywhere relevant. --prime mover (talk) 10:02, 3 May 2013 (UTC)
 * Or do they? They may already have been fixed. Wotever. Sorry. --prime mover (talk) 10:03, 3 May 2013 (UTC)


 * Almost, chain rule was still the wrong page. --Linus44 (talk) 10:06, 3 May 2013 (UTC)