Let $\varphi$ be a test function such that $Supp(\varphi)\subseteq [-M, M]$, $a\in \mathbb R$. We define the distribution principal value of $\frac{1}{x-a}$ :
$\left\langle\operatorname{P.\!v.}\left(\frac{1}{x - a}\right),\varphi\right\rangle: =\lim_{\varepsilon\to 0^+}\int_{|x - a|\geq \varepsilon}\frac{\varphi(x)}{x-a} \, \mathrm{d}x$
How do I show that :
$$\lim_{a\to 0} \frac{1}{2a}\left(\operatorname{P.\!v.}\left(\frac{1}{x - a}\right) - \operatorname{P.\!v.}\left(\frac{1}{x + a}\right)\right) = \operatorname{F.\!p.}\left(\frac{1}{x^2}\right)$$
I proved that : $$\left\langle\operatorname{P.\!v.}\left(\frac{1}{x - a}\right) - \operatorname{P.\!v.}\left(\frac{1}{x + a}\right), \varphi \right\rangle = \lim_{\varepsilon\to 0^+}\int_{ \varepsilon}^M\frac{2\varphi(x)-\varphi(2a - x) - \varphi(x-2a)}{x-a} \, \mathrm{d}x $$
Is my reasoning correct?Thanks for any help.
If you know that $\left( \operatorname{pv}\frac{1}{x} \right)' = -\operatorname{fp}\frac{1}{x^2}$: $$ \left< \frac{1}{2a} \left( \operatorname{pv}\frac{1}{x-a} - \operatorname{pv}\frac{1}{x+a} \right), \varphi(x) \right> = \frac{1}{2a} \left( \left< \operatorname{pv}\frac{1}{x-a}, \varphi(x) \right> - \left< \operatorname{pv}\frac{1}{x+a}, \varphi(x) \right> \right) \\ = \frac{1}{2a} \left( \left< \operatorname{pv}\frac{1}{x}, \varphi(x+a) \right> - \left< \operatorname{pv}\frac{1}{x}, \varphi(x-a) \right> \right) = \left< \operatorname{pv}\frac{1}{x}, \frac{\varphi(x+a) - \varphi(x-a)}{2a} \right> \\ \to \left< \operatorname{pv}\frac{1}{x}, \varphi'(x) \right> = - \left< \left( \operatorname{pv}\frac{1}{x} \right)', \varphi(x) \right> = \left< \operatorname{fp}\frac{1}{x^2}, \varphi(x) \right> $$