prove that it exist $c\in \mathbb C$ such as $\mathcal F(g)=c\delta + {1\over2i\pi}V_p({\mathcal F(f)\over x}) $
with $\mathcal F$ fourier transform , $f\in \mathbb D(\mathbb R)$ , $\delta$ dirac distribution , $g\in \mathbb D'(R)$ and :
$$ g(x)=\int_{-\infty}^xf(t)dt $$
$$ \left\langle V_p\left({u \over x}\right),\phi\right\rangle = \lim_{\epsilon\to0^+} \int_{|x|>\epsilon}{u(x)\phi(x)\over x} dx $$
i couldn't start anything here as i've never encountered this type of problems before , anything will be appreciated !
I may have some mistake in my work, but this is what apparently seems to appear, so you have to prove for any test function $\phi $
$$ \int_\mathbb{R}\hat{g}\phi dx = c\phi(0) + \lim_{\epsilon \rightarrow 0} \frac{1}{2\pi i}\int_{|x|>\epsilon} \frac{\hat{f}(x)\phi(x)}{x}dx $$ Now $ g'(x) = f(x) $, so we have $\hat{f}(x) = 2\pi ix\hat{g}(x) $, substituting this we have $$ \int_\mathbb{R}\hat{g}\phi dx = c\phi(0) + \lim_{\epsilon \rightarrow 0} \int_{|x|>\epsilon}\hat{g}(x)\phi(x)dx $$ This equivalently is $$ \lim_{\epsilon \rightarrow 0} \int_{|x|\leq\epsilon}\hat{g}(x)\phi(x)dx = c\phi(0) $$