Is there a good integration by parts formula to compute
$$\int\_{0}^\infty f \ H (f') dx,$$
where $H$ denotes the [Hilbert transform](https://en.wikipedia.org/wiki/Hilbert_transform) and $f$ is a smooth function?



Integration by parts with Hilbert transform
Perhaps something like this? (with integration from $-\infty$ to $\infty$ to arrive at a nicely symmetric answer):
$$\int\_{-\infty}^\infty f \ H (f') dx=\frac{1}{\pi}\text{P.V.}\,\int\_{-\infty}^\infty \int\_{-\infty}^\infty \frac{f(x)f'(y)}{x-y}\,dxdy$$
$$\qquad=\frac{1}{\pi}\int\_{-\infty}^\infty \int\_{-\infty}^\infty f(x)f'(y)\frac{d}{dx}\log|x-y|\,dxdy$$
$$\qquad=-\frac{1}{\pi}\int\_{-\infty}^\infty \int\_{-\infty}^\infty f'(x)f'(y)\log|x-y|\,dxdy.$$
Alternatively, with both integrals from $0$ to $\infty$,
$$\frac{1}{\pi}\text{P.V.}\,\int\_{0}^\infty \int\_{0}^\infty \frac{f(x)f'(y)}{x-y}\,dxdy$$
$$\qquad=\frac{1}{\pi}\int\_{0}^\infty \int\_{0}^\infty f(x)f'(y)\frac{d}{dx}\log|x-y|\,dxdy$$
$$\qquad=-\frac{1}{\pi}f(0)\int\_{0}^\infty f'(y)\log y\,dy-\frac{1}{\pi}\int\_{0}^\infty \int\_{0}^\infty f'(x)f'(y)\log|x-y|\,dxdy.$$


