Does $\int_{-\infty}^{+\infty}\frac{2xdx}{x^2+1}$ converge?

Question

Does $$I=\int\limits_{-\infty}^{+\infty}\frac{2xdx}{x^2+1}$$ converge ? And should I use definition when calculating improper integrals ?

• $I=\ln|x^2+1|_{-\infty}^{+\infty}=\infty-\infty$ ??

• Let $x=-t$, we obtain $\displaystyle I=\int\limits_{+\infty}^{-\infty}\frac{2tdt}{t^2+1}=-I\implies I=0$

Could anybody explain the different of these solution, please ?

Answer 1

No, the integral diverges because if it did converge it would have also converged on subsets of the line, say -- $[0,\infty)$, but

$$\int_0^{\infty}\frac{2x}{x^2+1}\,dx=\lim_{M\to\infty}\int_0^{M}\frac{2x}{x^2+1}\,dx=\lim_{M\to\infty}\log(M^2+1)=\infty$$

Answer 2

The integral is undefined as written ($\infty - \infty$ is an undefined form). However, it is possible to define its Cauchy principal value (https://en.wikipedia.org/wiki/Cauchy_principal_value) as:

$$\lim_{a \to \infty} \int_{-a}^{a} \frac{2x}{x^2+1}dx$$

and the value of this is zero. Note that this is not an unqualified evaluation of the improper definite integral (which, as mentioned, is undefined), but merely a way to assign a value to it in a particular context.

Your manipulation by changing the variable (and limits) is invalid because that can only be done if the definite integral is defined (in general), which it isn't in this case. It would, however, be valid for proving that the Cauchy principal value is zero.