Integration of $\tan(\frac1x)$

Question

$$\int_{-1}^{1}\tan\left(\frac1x\right) dx$$ How do I proceed? Please help.

Answer 1

Hint. The function $\tan(1/x)$ is not integrable in a neighbourhood of $\frac{2}{(2k+1)\pi}\in [-1,1]$ for all $k\in\mathbb{Z}$ because as $x\to \frac{2}{(2k+1)\pi}$ $$\tan(1/x)\sim \frac{C_k}{x-\frac{2}{(2k+1)\pi}}.$$

Answer 2

Note that for $x=\frac{2}{\pi}+h$ with $h\to 0$

$$\tan\frac1x=\tan\frac1{\frac{2}{\pi}+h}=\tan\left[\left(\frac{2}{\pi}+h\right)^{-1}\right]=\\\tan\left[\frac{\pi}2\left(1+\frac{\pi h}{2}\right)^{-1}\right]=\tan\left[\frac{\pi}2\left(1-\frac{\pi h}{2}+o(h)\right)\right]=\tan{\left(\frac{\pi}2-\frac{\pi^2 h}{4}+o(h)\right)}=\frac1{\tan{\left(\frac{\pi^2 h}{4}+o(h)\right)}}\sim\frac1{\frac{\pi^2 h}{4}}$$

and thus the integral does not converge.