This problem is from Basic Complex Analysis, Part 2A, by Barry Simon.
This problem will provide an alternate proof of the strong from of the Nyquist-Shannon sampling theorem (Theorem 6.6.16 of Part 1), following closely the 1920 proof of Ogura. We'll suppose $f(z)$ is an entire function obeying
$|f(z)| \leq C \exp((\pi - \epsilon)| \Im z|)$
Fix $z_0 \in \mathbb{C} \backslash \mathbb{Z}$. Let
$g(z) = \frac{f(z)}{(z - z_0)\sin(\pi z)}$
and $R_k$ be the rectangle which is the boundary of $\{x + iy\ |\ |x| \le k + \frac{1}{2},\ |y| \le k + \frac{1}{2} \}$ oriented clockwise.
Prove that
$\lim_{k \to \infty} \int_{R_k} |g(z)| d|z| = 0$
My attempt:
Choose $k \in \mathbb{Z}$ large enough such that $z_0$ is in the area bounded by $R_k$. Then let $z = x + iy$. Thus, \begin{align*} \int_{R_k} |g(z)| |dz| &\le \sup_{R_k} \left | \frac{f(z)}{(z - z_0)(\sin(\pi z)} \right | \ell(R_k)\\ & \le \sup_{R_k} \left | \frac{C \exp((\pi - \epsilon)| \Im z|)}{(\sin(\pi z))(k - |z_0|)} \right | 8(k + 1/2)\\ & = 8C \sup_{R_k} \left| e^{-\epsilon |y|} \frac{e^{\pi|y|}}{\sin(\pi z)} \right | \frac{(k + 1/2)}{k - |z_0|} \end{align*}
Then for $y > \frac{1}{2}$ we have, \begin{align*} \left | \frac{e^{\pi|y|}}{\sin(\pi z)} \right | = \left |\frac{2ie^{\pi y}}{e^{\pi i z} - e^{-\pi i z}} \right | \le \frac{2e^{\pi y}}{||e^{\pi i z}| - |e^{-\pi i z}||} = \frac{2e^{\pi y}}{e^{\pi y} - e^{-\pi y}} = \frac{2}{1 - e^{-2 \pi y}} \le \frac{2}{1 - e^{-\pi}} \end{align*} For $y < \frac{1}{2}$ we have, \begin{align*} \left | \frac{e^{\pi|y|}}{\sin(\pi z)} \right | = \left |\frac{2ie^{-\pi y}}{e^{\pi i z} - e^{-\pi i z}} \right | \le \frac{2e^{-\pi y}}{||e^{\pi i z}| - |e^{-\pi i z}||} = \frac{2e^{-\pi y}}{e^{-\pi y} - e^{\pi y}} = \frac{2}{1 - e^{2 \pi y}} \le \frac{2}{1 - e^{-\pi}} \end{align*} For $-\frac{1}{2} \le y \le \frac{1}{2}$ and $z = k + \frac{1}{2} + iy$ we have,
\begin{align*} \left | \frac{e^{\pi|y|}}{\sin(\pi z)} \right | \le \frac{e^{\pi/2}}{|\sin( \pi(k + \frac{1}{2} + iy))|} = \frac{e^{\pi/2}}{|\sin( \frac{\pi}{2} + \pi iy))|} = \frac{e^{\pi/2}}{|\cosh(\pi y ))|} \le \frac{e^{\pi/2}}{|\cosh(\pi/2 ))|} \end{align*} Let $M = \max \left \{ \frac{2}{1 - e^{-\pi}}, \frac{e^{\pi/2}}{|\cosh(\pi/2 ))|} \right \}$. Then $M$ does not depend on $k$, so \begin{align*} \int_{R_k} |g(z)| |dz| \le 8Ce^{-\epsilon |y|} M \frac{(k + 1/2)}{k - |z_0|} \end{align*} Which implies, for the top and bottom \begin{align*} \lim_{k \to \infty} \int_{R_k} |g(z)| |dz| \le \lim_{k \to \infty} 8Ce^{-\epsilon |k+1|} M \frac{(k + 1/2)}{k - |z_0|} = 0 \end{align*} But what about the sides?