Help on proving the following equality:
$$K(-\operatorname{sgn})=S$$ where $K$ is the operator defined by $K(f)=\mathcal{F}^{−1}f\mathcal{F}$
$\mathcal{F}$=Fourier transform, $f$=any function, sgn is the sign function and $S$ is the Cauchy singular integral operator given by
$$\left(Sf\right)\left(x\right)=\lim_{\varepsilon \to 0}{\left(\frac{1}{\pi i}\int_{{\mathbb R}\backslash ({\rm x-}\varepsilon {\rm ,x+}\varepsilon {\rm )}}{\frac{f\left(t\right)}{t-x}dt\ }\right)},\quad x\in {\mathbb R}.$$
Let $u(x)$ be, say, a Schwartz function (smooth and rapidly decaying at infinity together with derivatives). Then $$ Ku=-F^{-1}_{p\to x}\operatorname{sgn}(p)F_{x\to p}u(x) =-\frac1{2\pi}\int e^{ipx}\left(\operatorname{sgn}(p)\int e^{-ipy}u(y)dy\right)dp\\ =-\lim_{\delta\to+0} \frac1{2\pi}\int e^{ipx-\delta|p|}\left(\operatorname{sgn}(p)\int e^{-ipy}u(y)dy\right)dp\\ =-\lim_{\delta\to+0} \frac1{2\pi}\iint e^{ip(x-y)-\delta|p|}\operatorname{sgn}(p)u(y)dydp\\ =-\lim_{\delta\to+0} \frac1{2\pi}\int \left(\int e^{ip(x-y)-\delta|p|}\operatorname{sgn}(p)dp\right)u(y)dy\\ =-\lim_{\delta\to+0} \frac1{2\pi i}\int \left(\frac{1}{x-y+i\delta}+\frac{1}{x-y-i\delta}\right)u(y)dy\\ =-\frac1{\pi i} v.p.\int \frac{1}{x-y}u(y)dy, $$ because (Sokhotskii-Plemelj) $$ \lim_{\delta\to+0}\left(\frac{1}{x+i\delta}+\frac{1}{x-i\delta}\right)=2v.p.\frac1x $$