TITLE: Improper integral
QUESTION [7 upvotes]: I would like to know the value of the following improper integral:
$$\int \limits_{-\infty}^{\infty}\frac{2x}{1+x^2}dx$$
as the function $f(x)=\displaystyle\frac{2x}{1+x^2}$ satisfies $f(-x)=-f(x)$ can I immediately  conclude that the integral is equal to zero?
Thank you in advance for any suggestion.

REPLY [14 votes]: The Cauchy principal value of your integral is defined to be
$$ \lim_{a \to \infty} \int_{-a}^a {2x \over 1+x^2} \: dx $$
and this is zero by symmetry. So there is a sense in which the answer is $0$. However, if we let the limits go to infinity at different rates then we can make this integral "equal" whatever we want; for example, for any positive real constant $k$,
$$ \lim_{a \to \infty} \int_{-a}^{ka} {2x \over 1+x^2} \: dx = \lim_{a \to \infty} \log \left( {1+k^2 a^2 \over 1+a^2} \right) = 2 \log k. $$
Basically, since there is infinitely much area between this curve and the $x$-axis both in the first and third quadrants, we have to be careful about how we arrange things in order to make them "cancel out"; this is why we just throw up our hands and say that the integral doesn't converge.