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Let $f$ be an entire function. We know that $f$ can be written as a series centered in $0$ $$ \sum_{k=0}^{\infty} a_k z^k $$

where the coefficients $a_k=f^{(k)}(0)/k!$ are given by

$$ \frac{1}{2\pi i}\int_{C} \frac{f(\zeta)}{\zeta^{k+1}}\,\mathrm{d}\zeta $$

where $C$ is a (simple) circle of radius $R>0$ centered in $0$. By hypothesis $f$ is entire and the integrand seems entire too, then by Cauchy's theorem the integral should be $0$ for each $k>1$, and so each of the $a_k$ is $0$... but I know it is wrong. Where is my mistake?

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The integrand is not entire, look at what it is at zero. –  Gerry Myerson Feb 25 '13 at 2:33

1 Answer 1

The integrand $\frac{f(\zeta)}{\zeta^{k+1}}$ is not generally going to be entire since it has a pole at $\zeta =0$ (at least it has such a pole if $f(\zeta )$ does not have a zero of order $k+1$ or higher at $\zeta =0$). Just consider the case where $f(\zeta)=1.

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