Show $ \int_{C_N} \frac{\pi \sec(\pi z)}{z^3} dz \to 0$ for given contour. Let $C_N$ be a square with corners at $N(1+i)$, $N(1-i)$, $N(-1+i)$, and $N(-1-i)$. How do I show that
$$ \int_{C_N} \frac{\pi \sec(\pi z)}{z^3} dz $$
goes to zero? I'm tying to bound this to prove a larger result for Cauchy's Residue Theorem although I forget what the appropriate bound for $\sec(\pi z)$ is. I want to say something like for any point $z$ on the contour,
$$\left| \frac{\pi \sec(\pi z)}{z^3} \right| \leq \left| \frac{\pi \sec(\pi z)}{N^3} \right| \to 0$$
As $N \to \infty$. But is this appropriate with $\sec(\pi z)$ in the numerator? $\sec(\pi z)$ is defined everywhere on the contour $C_N$.

Remember that $N$ is an integer. Let $z=N+i\,t$, $-N\le t\le N$. Then
$$
|\sec(\pi\,z)|=\frac{1}{|\cos(N\,\pi+i\,\pi\,t)|}=\frac{1}{|\cos(i\,\pi\,t)|}=\frac{2}{e^{\pi t}+e^{-\pi t}}\le1.
$$
This takes care of one side of the contour. The others are dealt similarly.
