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$\newcommand{\PV}{\operatorname{P.V.}}$I have a doubt about the Cauchy Principal Value of real rational functions.

$f: \mathbb{R}\rightarrow \mathbb{R}$ is a rational function with $\deg(\text{denominator})>\deg(\text{numerator})$.

$\{x_1,x_2,\ldots,x_n \} \subset \mathbb{R} $ is the set of all $f$ poles


$\PV$ exists and it is

$$\PV \int_{-\infty}^{+\infty} f(x) \ dx=\pi i \left( \sum_{k=1}^n \operatorname{Res}(f,x_k) \right) $$


$$\pi i \left( \sum_{k=1}^n \operatorname{Res}(f,x_k) \right) \in \mathbb{I}$$

So: $$\PV \int_{-\infty}^{+\infty} f(x) \ dx=0$$ because: $$\operatorname{Re} \left( \pi i \left( \sum_{k=1}^n \operatorname{Res}(f,x_k) \right) \right)=0$$


Is it true?


$\PV \int_{-\infty}^{+\infty} f(x) \ dx$ doesn't exist if $\deg(\text{numerator}) \ge \deg(\text{denominator})$, does it?


In general, which are the conditions of existence of P V?



Is it correct?


Thanks

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    If you don't require all poles to be simple, the principal value of the integral need not exist, consider $f(x) = \dfrac{x+1}{x^2(x-1)}$. – 2017-02-11
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    @DanielFischer , I have done these calculations: $\frac{x+1}{x^2 (x-1)}=\frac{1}{x^2}+\frac{-2}{x}+\frac{3}{x-1}$ $\lim_{R\rightarrow +\infty} \int_{-R}^{R} \frac{1}{x^2}=\lim_{R \rightarrow \infty} -\frac{2}{R}=0$ $\lim_{R\rightarrow +\infty} \int_{-R}^{R} \frac{-2}{x}=0$ (odd function) $\lim_{R\rightarrow +\infty} \int_{-R}^{R} \frac{3}{x-1}=\lim_{R \rightarrow \infty} 3 \ln \rvert \frac{R-1}{-R-1} \rvert=0$ Is there any mistake? – 2017-02-15

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Daniel Fischer already notes in the comments that some condition must be imposed on $f$ to assure that the principal value exists. For example, $\int_{-\infty}^\infty dx/x^2 = \infty$, principal value or no: the integrand is positive, so there's no cancellation when we remove small symmetrical neighborhoods of the pole at $x=0$.

Assume that the poles of $f$ are simple. Then the argument that OP Francesco Serie gave is essentially correct, though one must include the pole at infinity if the difference between the denominator's and numerator's degree is only $1$ (because the differential $f(x)\,dx$ then has a simple pole at infinity). Indeed it is known that for any rational function $f \in {\bf C}(x)$ the sum of the residues of $f(x)\,dx$ at all its poles on the Riemann sphere vanishes.

A simpler argument is to expand $f$ in partial fractions. Under our hypothesis, $f$ is an $\bf R$-linear combination of the functions $1/(x-x_k)$; the principal-value integral is a linear map to $\bf R$, and it is elementary that the principal value of each $\int_{-\infty}^\infty dx/(x-x_k)$ is zero, so the same is true of $\int_{-\infty}^\infty f(x) \, dx$.

This also suggests the generalization to functions that might have multiple roots: the principal-value integral exists iff $f$ is a linear combination of functions $1/(x-x_k)^{e_k}$ with each $e_k$ odd, and then the P.V. integral again equals zero.