Why does $\int_{-\infty}^\infty R(x) dx$ converge iff the rational function $R(x)$ has degree of denom. at least 2 degrees higher than numerator?

Question

I am readinf Ahlfors and came across the fact that:

$\int_{-\infty}^\infty R(x) dx$, where $R(x)$ is a rational function, converges if and only if in the rational function $R(x)$ the degree of the denominator is at least two units higher than the degree of the numerator.

I am unsure of how to prove this fact rigorously. I do get that the condition on the degrees means roughly that $R(x)$ is approximately $\frac cx$ (whose integral diverges) with $c$ a constant, but could anyone post a rigorus proof of this?

Answer 1

You need an extra assumption on zeros of the denominator. For example $\frac 1 {x^{2}}$ is not integrable. Let $p$ and $q$ are polynomials and degree of $q$ is at least $2$ more than the degree of $p$ and let us assume that the denominator has no zeros on $\mathbb R$. Then there is a constant $C$ such that $|\frac p q |\leq \frac C {x^{2}}$ for $|x|$ sufficiently large and this makes $R=\frac p q$ integrable.