I'm making some mistake in my reasoning, which leads to $Pv\frac{1}{x} = 0$ (in distributional sense):
$$<Pv\frac{1}{x},\phi> = lim_{\epsilon \rightarrow 0}(\int_{-a}^{-\epsilon}\frac{\phi}{x}dx+\int_{\epsilon}^{a}\frac{\phi}{x}dx)$$ Where I made use of the fact that $\phi$ is a test function with compact support in $[-a,a]$. If I define: $$f(\epsilon) = (\int_{-a}^{-\epsilon}\frac{\phi}{x}dx+\int_{\epsilon}^{a}\frac{\phi}{x}dx)$$ Then: $$<Pv\frac{1}{x},\phi> = lim_{\epsilon \rightarrow 0}f(\epsilon)$$ I now reason that for all $\epsilon$, I can write: $$f(\epsilon) \leq \vert\vert\phi\vert\vert_{[-a,a]}(\int_{-a}^{-\epsilon}\frac{1}{x}dx+\int_{\epsilon}^{a}\frac{1}{x}dx) \triangleq \vert\vert\phi\vert\vert_{[-a,a]}g(\epsilon) $$ Where $\vert\vert\phi\vert\vert_{[-a,a]}$ is the supremum norm of $\phi$. Also: $$f(\epsilon) \geq -\vert\vert\phi\vert\vert_{[-a,a]}(\int_{-a}^{-\epsilon}\frac{1}{x}dx+\int_{\epsilon}^{a}\frac{1}{x}dx) \triangleq \vert\vert\phi\vert\vert_{[-a,a]}g(\epsilon) $$ If I use now that $$\lim_{\epsilon \rightarrow 0}g(\epsilon) = Pv\int_{-a}^{a}\frac{1}{x} = 0$$ then $$\lim_{\epsilon \rightarrow 0}f(\epsilon) = 0 = <Pv\frac{1}{x},\phi>$$ Which can't be right. I would be glad if someone could find the mistake for me, because I don't see it.