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http://www.uio.no/studier/emner/matnat/math/MAT2410/h14/supplementarynotes/cauchy-starshaped.pdf

I am trying to prove the Cauchy Integral formula for star shaped regions which will use one of the ideas in the notes posted.

On page 2, the last paragraph says that

If, for each fixed $ \epsilon > 0$, we let the width $\delta$ of the corridor go to zero, the contributions to the integral of the two sides of the corridor cancel due to the continuity of $g$.

This seems conceptually right, but how to write it down rigorously? Where does the continuity of $g$ come in?

Can anyone give me any idea on how to write down the explicit formulas of the integral so that I can actually see how to use the continuity of $g$?

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  • $\begingroup$ The corridor can have a $0$ width, I don't see the problem. And if you assume $f(z)$ to be continuously complex differentiable, you might want to look at those comments for showing directly $\int_{|z| = R} \frac{f(z)}{z-a} dz = \int_{|z-a| = \epsilon} \frac{f(z)}{z-a} dz$ $\endgroup$
    – reuns
    Sep 27, 2016 at 3:05

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Continuity of $g$ at $z$ basically means that for any $\epsilon > 0$, we can find a $\delta$ where all the $g(\zeta)$ are within $\epsilon$ of $g(z)$ if $\zeta$ is within $\delta$ of $z$.

Thus as the width of the corridor goes to zero, the continuity of $g$ means $g(\zeta)$ on the inner boundary and the $g(\zeta)$ on the outer boundary becomes as close as we'd like and hence the difference between the two integrals vanishes.

For a discontinuous function, you have no such guarantee. It is possible to have a jump in the function where $\zeta \to z$ does not imply $g(\zeta) \to g(z)$.

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  • $\begingroup$ Yes, those functions tend to be very close by continuity, but they are on different lines $r$ and $r'$, the sum of those two integral would be $\int _r f - \int _{r'} f$, how do I estimate this difference? $\endgroup$
    – Keith
    Sep 30, 2016 at 18:08

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