INTRODUCTION TO MY PROBLEM: let's suppose that we have: $\delta[g(x)]$, where $g(x)$ is a smooth function such that $g(x)=0$ at some points $(x_k)_{k=1,...,N}$ and $|g'(x_k)|\neq0 \; \forall k=1,...,N$. Well, using distributions formalism (i know how to show that) we get a formula to express: $$\delta[g(x)]=\sum_{k=1}^N \frac{\delta(x-x_k)}{|g'(x_k)|}$$
THE PROBLEM ITSELF: let's now suppose we have: $PV[g(x)]$, where g(x) is such that can be written like: $g(x)=\frac{\text{something}}{\text{something els}e}$ where $(\text{something else} = 0)$ at some points $(x_k)_{k=1,...,N}$. Is there a fancy way to express the principal value in relation to those points like happened with $\delta[g(x)]$?
MY ATTEMPT: I was trying to do the job with the test function: $PV[g(x)]\approx PV\left[\dfrac{1}{\prod_{k=1}^{N}(x-x_k)} \right]$ $$\left\langle PV[g(x)] \; \middle| \; f(x)\right\rangle=PV\left(\int_{-\infty}^{+\infty}\frac{f(x)}{\prod_{k=1}^{N}(x-x_k)}\right)=\lim_{\epsilon\;\rightarrow \;0^+}\left[\int_{-\infty}^{x_1-\epsilon}\frac{f(x)}{\prod_{k=1}^{N}(x-x_k)} \, dx \;+ \int_{x_1 + \epsilon}^{x_2-\epsilon}\frac{f(x)}{\prod_{k=1}^{N}(x-x_k)} dx+...+ \int_{x_N +\epsilon}^{+\infty}\frac{f(x)}{\prod_{k=1}^{N}(x-x_k)} \, dx\right]$$ At this point i think i should proceed doing these two steps:
For each integral i would take off many of the terms from denominator. For example: $$ \int_{-\infty}^{x_1-\epsilon}\frac{f(x)}{\prod_{k=1}^{N}(x-x_k)} \, dx \; \rightarrow \; \int_{-\infty}^{x_1-\epsilon}\frac{f(x)}{(x-x_1)} \, dx \\[6ex] \int_{x_1 + \epsilon}^{x_2-\epsilon}\frac{f(x)}{\prod_{k=1}^{N}(x-x_k)} \, dx \; \rightarrow \; \int_{x_1 + \epsilon}^{x_2-\epsilon}\frac{f(x)}{(x-x_1)} \, dx$$
Then i would regrup all terms pair-by-pair. So that each pair gives me the principal value of one of the $(x_k)$, and eventually: $$PV[g(x)]= \sum_{k=1}^{N}PV\left(\frac{1}{x-x_k}\right)$$
FINAL TOUGHTS: anyway, i think i'm in the realm of sci-fi since i cannot really imagine why i would change the dominators like that.. Moreover, i found an example where: $$\require{cancel} \cancelto{\text{**EDIT: commited a typo.**}}{PV\left(\frac{1}{x^2+1}\right) \; \rightarrow \; \frac{1}{2}\left[PV\left(\frac{1}{x-1}\right) - PV\left(\frac{1}{x+1}\right)\right]} \\[6ex] PV\left(\frac{1}{x^2-1}\right) \; \rightarrow \; \frac{1}{2}\left[PV\left(\frac{1}{x-1}\right) - PV\left(\frac{1}{x+1}\right)\right]$$ What happened here? It seems i was almost there with my intuitive path, but that minus sign really stunned me...