I'm using Donald Sarason's book as a reference.
(One version of) Cauchy's theorem states that for a holomorphic function defined on an open, convex set, $\int_\gamma f(z)\mathrm{d}z$ = 0 for any piecewise-$C^1$ closed curve $\gamma$ in this set.
On the other hand, Cauchy's formula for a circle gives one a method to compute the value of a holomorphic $f$ for any point $z$ in the interior of the circle:
$$
f(z) = \frac{1}{2 \pi i} \int_C \frac{f(\xi)}{\xi – z} \mathrm{d}\xi\ .
$$
My question is the following: why doesn't the integral on the right hand side of the above question vanish by Cauchy's theorem? To be painfully explicit, if we consider a circle centered around the origin, why is it not true that
$$
f(0) = \frac{1}{2 \pi i} \int_C \frac{f(\xi)}{\xi} \mathrm{d}\xi\ = 0
$$
for all holomorphic functions $f$?
I know I'm missing something elementary; I just can't see what it is.
Best Answer
Because the denominator of the integrand gives rise to a singularity, so the integrand is not holomorphic on any open convex set containing the circle you are integrating over.