# What is $\dfrac{i}{4-π} \int_{|z|=4} \dfrac{dz}{z \cos z}$?

How to evaluate the integral $$\dfrac{i}{4-π} \int_{|z|=4} \dfrac{dz}{z \cos z}?$$

Here $$f(z)=\dfrac{1}{z \cos z}$$ has poles at $$0$$ and $$\frac{\pm π}{2}$$ .

Residue at $$z=0$$ is 1 and residues at remaining poles add up to give 0. So the integral using Cauchy integral formula is $$2π(4-π)$$.

I think I am wrong. How to get the integral?

It's not true that residues add up to $$0$$. We have $$\cos z = \sin(\frac{\pi}{2}-z) = \sin(\frac{\pi}{2}+z)$$ so $${\rm Res}_{z=0} \frac{1}{z\cos z} = \lim_{z\rightarrow 0}\frac{1}{\cos z} = 1$$ $${\rm Res}_{z=\frac{\pi}{2}} \frac{1}{z\cos z} = \lim_{z\rightarrow \frac{\pi}{2}}\frac{(z-\frac{\pi}{2})}{z \sin(\frac{\pi}{2}-z)} = -\frac2\pi$$ $${\rm Res}_{z=-\frac{\pi}{2}} \frac{1}{z\cos z} = \lim_{z\rightarrow -\frac{\pi}{2}}\frac{(z+\frac{\pi}{2})}{z \sin(\frac{\pi}{2}+z)} = -\frac2\pi$$ and $$\sum {\rm Res} = 1 - \frac{4}{\pi} = \frac{\pi-4}{\pi}$$ so the final result is $$\frac{i}{4-\pi}\cdot 2\pi i\frac{\pi-4}{\pi} = 2$$

You made a mistake in your residues. The residues of $$1/(z\cos z)$$ at $$z=\pm\frac\pi2$$ are both $$-\frac2\pi$$.

• ohh...yes....thank you Jun 13 '19 at 16:20

The function $$f(z)=\frac{1}{z\cos(z)}$$ has simple poles at $$z=0$$ and $$z= (2n-1)\pi/2$$ for $$n\in \mathbb{Z}$$. The poles that are inside the circle $$|z|=4$$ are at $$z=0$$ and $$z=\pm \pi/2$$.

The residues of $$f$$ at the implicated poles are

\begin{align} \text{Res}\left(\frac{1}{z\cos(z)}, z=0\right)&=1\\\\ \text{Res}\left(\frac{1}{z\cos(z)}, z=\pi/2\right)&=-\frac2\pi\\\\ \text{Res}\left(\frac{1}{z\cos(z)}, z=-\pi/2\right)&=-\frac2\pi \end{align}

Therefore we have

$$\frac{i}{4-\pi}\oint_{|z|=4}f(z)\,dz=\frac{i}{4-\pi}\times 2\pi i\times\left(1-\frac{4}{\pi}\right)=2$$