Let $C_n$ denote the boundary of the square formed by the lines x= $\pm$ $N$$\pi$ and y = $\pm$ $N$$\pi$ where $N$$\in$ $\mathbb{N}$, and let the orientation of $C_n$ be counterclockwise. Show that $\exists$ $M>0$ not depending on $N$ such that $\mid$ $\int_{C_n}$ $\frac{dz}{z^2cosz}$ $\mid$ $\leq$ $\frac{M}{N}$.
I know that this problem is about Estimation Lemma. But I can't find M which is not depending on $N$. How can I slove this problem?
Let $f(z)=z^{-2}\sec z$.
By Cauchy’s theorem $$I:=\int_{C_N}f(z)dz=\int_{|z|=N\pi}f(z)dz$$
$$|I|=\left\vert \int_{|z|=N\pi} fdz\right\vert=\left\vert\int^{2\pi}_0f(N\pi e^{it})iN\pi e^{it}dt\right\vert$$
By triangle inequality, this is bounded by $$\int^{2\pi}_0 |f(N\pi e^{it})|N\pi dt=\int^{2\pi}_0\frac{|\sec(N\pi e^{it})|}{N\pi}dt$$
Moreover,
$$|\sec(a+bi)|=\frac2{|e^{ia}e^{-b}+e^{-ia}e^{b}|}=O\left(\frac1{e^{|b|}}\right)$$
Thus, the integral is further bounded by $$C\cdot\frac{e^{-N\pi}}{N\pi}$$ for some positive constant $C$. The upper bound of the magnitude of integral clearly tends to zero as $N\to\infty$, thus so as $I$.
I skipped a few details. You may ask me to add them if they would help your understanding.