Why is this integral diverging? $\int\limits^{\infty}_{-\infty} \,\frac{x}{x^2+1} dx$

Solution 1:

This integral is not absolutely convergent and thus you cannot "easily prove that this integral is diverging"; such expression is usually regarded as ill-defined.

This is similar to $\int_{-\infty}^{\infty}xdx$, the "meaning" of which depends on how you interpret the symbol $\int_{-\infty}^\infty$.

If you take it as "a Cauchy principal value" $$ \lim_{N\to\infty}\int_{-N}^N xdx $$ then it is zero because the function is odd.

However, if you interpret it as an iterated limit: $$ \lim_{a\to-\infty}\left(\lim_{b\to\infty}\int_{a}^b xdx\right) $$ you already have a problem with the inner limit.

On the other hand, if a function is absolutely convergent, then all the following "interpretations" of $\int_{-\infty}^\infty$ are the same:

$\int_{-\infty}^\infty:=\lim_{a\to\infty}\int_{-a}^a$;
$\int_{-\infty}^\infty:=\lim_{a\to-\infty}\lim_{b\to\infty}\int_a^b$;
$\int_{-\infty}^\infty:=\lim_{b\to\infty}\lim_{a\to-\infty}\int_a^b$;
$\int_{-\infty}^\infty:=\int_0^\infty+\int_{-\infty}^0$

Solution 2:

You have to look more carefully at the definition of $\int_{- \infty}^{\infty}.$ Your intuition is treating it as though it means $\lim_{x \to \infty} \int_{-x}^x$, but that's not correct. It's actually $\lim_{x \to - \infty}(\lim_{y \to \infty} \int_x^y)$ (assuming the integrand is in fact defined everywhere), and for this function the inner limit does not exist for any $y$.

Solution 3:

Let $ f $ be locally integrable at $ (-\infty,+\infty)$

By définition, we say that the integral $$\int_{-\infty}^{+\infty}f(x)dx$$ is convergent if only if there exists $ c\in \Bbb R $ such that, Both integrals $$\int_{-\infty}^cf(x)dx\text{ and } \int_c^{+\infty}f(x)dx$$ Are convergent.

If ine of them is divergent, the integral will be divergent.

In your case, the integral $$\int_0^{+\infty}\frac{xdx}{1+x^2}$$ is divergent.