Why is it that $\int_{-\infty}^{\infty} \frac{x}{1+x^2} dx $ diverges, while $\int_{-\infty}^{\infty} \frac{x}{1+x^8} dx $ converges.

Question

Why does $$ \int_{-\infty}^{\infty} \frac{x}{1+x^2} dx $$ diverge, while $$ \int_{-\infty}^{\infty} \frac{x}{1+x^8} dx $$ converges?

Answer 1

Note that, for $x\geq 1$, we have that $$\frac{1}{2x} \leq \frac{x}{1+x^2}$$ It follows that the integral $$\int_0^\infty \frac{x}{1+x^2} dx$$ diverges, and so does the original integral. As for the second integral, note that $$0 \leq \frac{x}{1+x^8} \leq \frac{1}{x^7}$$ and so $$\int_{-\infty}^\infty \frac{x}{1+x^8} dx = \int_{-\infty}^{-1} \frac{x}{1+x^8}dx + \int_{-1}^1 \frac{x}{1+x^8}dx + \int_{1}^\infty \frac{x}{1+x^8}dx$$ As we can see, the first and last terms converge by the comparison test, and the middle term is a proper integral of a continuous function and thus converges. So the whole integral converges.

Answer 2

The short answer is that $$\frac{x}{1+x^2} \sim \frac1x $$ when $x\to\infty$.

I assume (perhaps incorrectly) that you know that $$\int_1^\infty \frac1x\,\mathrm d x$$ diverges.

Answer 3

A primitive for $\frac{x}{1+x^2}$ is given by $\frac{1}{2}\log(1+x^2)$, that is an unbounded function as $x\to\pm\infty$.
On the other hand $\frac{x}{1+x^8}$ is bounded by $1$ in absolute value over the interval $[-1,1]$, and by $\frac{1}{x^7}$ on the intervals $(-\infty,-1)$ and $(1,+\infty)$. It follows that $\frac{x}{1+x^2}\not\in L^1(\mathbb{R})$ but $\frac{x}{1+x^8}\in L^1(\mathbb{R})$, and since it is an odd function its integral over $\mathbb{R}$ simply equals zero.