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.
Best Answer
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 $\{|z|>R\}$.