# Contour Integral of $f(z) \; \cot(\pi z)$

In lecture, my professor stated that $$\lim_{N \to \infty} \int_{\gamma_N} f(z) \cot(\pi z) \; dz = 0$$

where $\gamma_N$ is the square contour with vertices at $\pm (N + \frac12) \pm i(N + \frac12)$ and $f$ is complex-valued function (not necessarily entire) such that $|z f(z)| < M$ for sufficiently large $|z|$.

I can't see how to prove this. If $|z^2f(z)|$ is bounded, then you can bound the cotangent on the contour and get the desired limit, but I don't know how to manage it with the weaker constraint given. Any advice would be appreciated. Thanks.

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Is $f$ merely a continuous function or even worse? – 23rd Nov 15 '12 at 17:13
The statement wasn't clear on what properties $f$ has. However, for the cases in which I want to use the statement, $f$ is meromorphic with only finitely many poles, so I'm assuming that. – ec92 Nov 15 '12 at 19:03
I think this is exactly the sort of thing that you should ask your professor instead of us. He knows what $f$ was when he said this, and we don't know that. – GEdgar Nov 16 '12 at 15:00

Assume that there exist $R>0$, $M>0$ and a discrete set $E\subset \mathbb{C}$, such that on $\{|z|>R\}\setminus E$, $f$ is holomorphic and $|zf(z)|\le M$. Then $f$ is holomorphic on $\{|z|>R\}$, and the Laurent expansion of $f$ on $\{|z|>R\}$ must have the form $f(z)=\sum_{n=1}^\infty \frac{a_n}{z^n}$. Express $f$ as $f(z)=\frac{a_1}{z}+g(z)$. On the one hand, a direct calculation shows that for every $N\ge 1$, $\int_{\gamma_N}\frac{\cot\pi z}{z} dz=0$; on the other hand, $|z^2g(z)|$ is bounded on $\{|z|>R\}$.