Computing $\int_{-1}^{1} \frac{d}{dx}\left( \arctan\left(\frac{1}{x}\right)\right)dx$

Question

Compute the integral $$\int_{-1}^{1} \frac{d}{dx}\left( \arctan\left(\frac{1}{x}\right)\right)dx$$

I saw the above question in an Integration bee, and the answer given for the above problem is $-\frac{\pi}{2}$, which I don't understand. I did the following: $$\int_{-1}^{1} \frac{d}{dx}\left( \arctan\left(\frac{1}{x}\right)\right)dx = \lim_{\epsilon \to 0}\int_{-1}^{\epsilon}+\lim_{\delta \to 0}\int_{\delta}^{1}\frac{d}{dx}\left( \arctan\left(\frac{1}{x}\right)\right)dx$$
which is $\frac{\pi}{2}+\arctan\left(\frac{1}{\epsilon}\right)-\arctan\left(\frac{1}{\delta}\right)$, this implies that the integral does not exist. What am I doing wrong ?

Answer 1

Continuing from where where you left:

$\int_{-1}^{1} \frac{d}{dx}\left( \arctan\left(\frac{1}{x}\right)\right)dx = \lim_{\epsilon \to 0^{-}}\int_{-1}^{\epsilon}\frac{d}{dx}\left( \arctan\left(\frac{1}{x}\right)\right)dx+\lim_{\delta \to 0^+}\int_{\delta}^{1}\frac{d}{dx}\left( \arctan\left(\frac{1}{x}\right)\right)dx$

$=\frac{\pi}{2}+\lim_{\epsilon \to 0^{-}}\arctan\left(\frac{1}{\epsilon}\right)-\lim_{\delta \to 0^+}\arctan\left(\frac{1}{\delta}\right) =\frac{\pi}{2}-\frac{\pi}{2}-\frac{\pi}{2}=-\frac{\pi}{2}.$