In this page Cauchy Integral Theorem, the three Cauchy integral theorems are presented. The first is about simply connected domains and the second is a generalisation of the 1st that includes removable singularities. The third is the Cauchy Goursat theorem regarding triangles or rectangles. At one point (just before the 3rd theorem) it writes:

"Goursat's argument makes use of rectangular contour (many authors use triangles though), but the extension to an arbitrary simply-connected domain is relatively straight-forward"

How is that extension straight-forward? What is the proof of that? In other words how can I prove the 1st theorem for the 3rd (using the triangle version of Cauchy Goursat prefferably) and then how do I prove that the 1st theorem implies the 2nd? ($3\Rightarrow 1\Rightarrow 2$). It would be preferrable if someone linked a pdf file containg the proofs I am asking for. Also note that I am looking for elementary self contained answers that don't invole a lot of topological notions but basic topology. Thank you for your answers.


$(3) \Rightarrow (1)$: Proof Cauchy-Goursat-Theorem (Theorem 15.2)

$(1) \Rightarrow (2)$: It's a consequence of the residue theorem - probably you'll find a proof (using Cauchy's Theorem) in every book about analytical functions (or here: Proof 1/Proof 2 (short one))

  • $\begingroup$ Thanks for the $(3)\Rightarrow (1)$ proof, it is exactly what I asked. Now the $(1)\Rightarrow (2)$ theorem is necessary (to my knowledge) to prove Cauchy's Integral Fomula and thus the residue theorem... $\endgroup$ – Nameless Nov 10 '12 at 17:03
  • $\begingroup$ Ah, okay, you want to do it the other way round. In this case, $(1) \Rightarrow (2)$ is a simple consequence of Riemann's theorem (since the function is bounded in a neighborhood of the singularities, the singularities are removable.) $\endgroup$ – saz Nov 10 '12 at 17:26
  • $\begingroup$ Wait a minute. In the proof of $(3)\Rightarrow (1)$ the 3rd step (the one I am asking for) is pretty much omitted... $\endgroup$ – Nameless Nov 10 '12 at 17:49
  • $\begingroup$ That's true, but I doubt that there is a more detailed proof (especially since there are a lot of other (in my oppinion easier) ways to proof this theorem). But anyway: You can cover im $\gamma$ by arbritrary small squares and obtain a polygonal contour $\Gamma$ arbritary close to $\gamma$ (I tried to draw a picture ). Since $f$ is continuous it works fine. $\endgroup$ – saz Nov 10 '12 at 18:20

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