# Improper integral of a rational function whose denominator is of degree at least two greater than that of the numerator

There's a technique in complex analysis (involving residue calculus) to solve the improper integral (from $-\infty$ to $\infty$) of a rational function whose denominator is of degree at least $2$ greater than that of the numerator.

My question is, this technique merely gives the Cauchy principal value of this improper integral, but it does not tell us if the integral actually exists, correct? Or is the existence of the Cauchy principal value in such cases sufficient for the improper integral to actually exist?

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If the degree of the denominator is at least $2$ more than that of the numerator (and there are no poles on the real line), the improper integral does exist; the proof of that is a separate estimate (though very similar to the one used in showing that the integral over the "return arc" goes to $0$). If the degree of the denominator is $1$ more than that of the numerator (and there are no poles on the real line) the Cauchy principal value exists, but the improper integral does not exist.
The integrand is $c/z + O(1/z^2)$ as $z \to \infty$ for some nonzero constant $c$. – Robert Israel Apr 4 at 4:17
The Cauchy principal value of $\int_{-1}^1 \frac{dx}{x^3}$ exists, but the integral does not. This is also true of $\int_{-1}^{1}\frac{dx}{x(1+x^2)}$.