Show that $ \int_{-∞}^{∞} \frac{x}{x^2 + 1}dx $ converges

Question

I'm having a hard time figuring how the following integral converges: $$ \int_{-∞}^{∞} \frac{x}{x^2 + 1}dx $$

When I try to calculate the same integral as $\int_{-∞}^{0} \frac{x}{x^2 + 1}dx + \int_{0}^{∞} \frac{x}{x^2 + 1}dx$, I get that it diverges. But I know that this is an odd function, and symbolab shows that the areas are cancelled out, i.e. the integral is 0. How can I show that it converges?

Answer 1

You can't. It doesn't converge. By definition, we say that the integral $\int_{-\infty}^{+\infty}\frac x{x^2+1}\,\mathrm dx$ converges if, given $a\in\mathbb R$, both integrals$$\int_{-\infty}^a\frac x{x^2+1}\,\mathrm dx\text{ and }\int_a^{+\infty}\frac x{x^2+1}\,\mathrm dx$$converge (if this happens for some $a$, then it happens for every $a$). But, as you noted, this doesn't happen when $a=0$.

Answer 2

$\displaystyle\int_{-\infty}^{\infty}\dfrac{x}{x^{2}+1}dx$ converges in the principal value sense: $(\text{p.v.})\displaystyle\int_{-\infty}^{\infty}\dfrac{x}{x^{2}+1}dx=\lim_{n\rightarrow\infty}\int_{-n}^{n}\dfrac{x}{x^{2}+1}dx=0$.

Answer 3

It is one of those $\infty -\infty$ cases where it does not converge.