Definite integral $\int\limits_{-\infty}^{\infty} x (\arctan x)' dx$

Question

I need to calculate definite integral $$I = \int\limits_{-\infty}^{\infty} x (\arctan x)' dx$$

I try it as usual but I have indeterminacy $\ln |x^2| _{+\infty} - \ln |x^2| _{-\infty} = \infty - \infty$. How to accurately calculate it? Thanks.

Answer 1

Notice:

$$f(x) = x(\arctan x)'=x \cdot \frac{1}{1+x^2}$$

This is an odd function (i.e. $f(x) = -f(-x)$), and thus the integration result is $0$.

EDIT per comment

The integration does not exist unless we consider Cauchy principle value:

$$\int_{-\infty}^{+\infty}f(x) \,dx =: \lim_{R \rightarrow +\infty}\int_{-R}^Rf(x)\,dx $$

And in our case, the Cauchy PV is $0$.