Integrate $\oint_{C} \sec z \mathrm{d}z$

Question

The literature I read says that

$$\oint_C \sec z \mathrm{d}z = 0$$

This is what I know; From the Cauchy integral theorem $\oint_C f(z)\mathrm{d}z = 0$ if $f(z)$ is analytic in a simple connected domain $D$ for every simple closed path $C$ in $D$.

But $f(z) = \sec z$ is not analytic at $\frac{(2n + 1) \pi}{2}$, $n \in \mathbb{Z}$

Answer 1

Actually, if you have to evaluate the following integral $~\oint_C \sec(z)~dz~$ in counter clockwise direction around the unit circle $C:|z|=1$, then you get the value of the integral is zero. The reason behind this is as follows:

Clearly, $~\sec(z)~$ is analytic at all pint except at $~\cos(z)=0~$ i.e., at $~z=\frac{(2n + 1) \pi}{2}~, n \in \mathbb{Z}.$

So poles are at $~z_0=\cdots,~-\frac{3\pi}{2},~-\frac{\pi}{2},~\frac{\pi}{2},~\frac{3\pi}{2},~ \cdots~$

But none of these poles lie inside the unit circle $C:|z|=1$.

Hence the sum of the residues at the poles is zero.

Therefore by residue theorem, $$I=2\pi i\left[\cdots+\text{Res}(f, -3\pi/2)+\text{Res}(f, -\pi/2)+\text{Res}(f, 3\pi/2)+\text{Res}(f, \pi/2)+\cdots\right]=0$$