Is there a good integration by parts formula to compute $$\int_{0}^\infty f \ H (f') dx,$$ where $H$ denotes the Hilbert transform and $f$ is a smooth function?
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$\begingroup$ are you sure you want the integral fron 0 to $\infty$, not from $-\infty$ to $\infty$ ? $\endgroup$– Carlo BeenakkerCommented Jan 20, 2021 at 18:40
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$\begingroup$ @CarloBeenakker Yes, I'd like to compute the integral from 0. $\endgroup$– JunCommented Jan 20, 2021 at 19:14
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1$\begingroup$ We have $H f' = (-\Delta)^{1/2} f$, so if $f$ is smooth and supported in $(0,\infty)$, the integral should be equal to $\int_0^\infty f (-\Delta)^{1/2} f = \int_{-\infty}^\infty f (-\Delta)^{1/2} f = \int_{-\infty}^\infty ((-\Delta)^{1/4} f)^2$. I doubt this is what you are looking for, though. $\endgroup$– Mateusz KwaśnickiCommented Jan 20, 2021 at 20:02
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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 $\infty$, $$\frac{1}{\pi}\text{P.V.}\,\int_{0}^\infty \int_{0}^\infty \frac{f(x)f'(y)}{x-y}\,dxdy$$ $$\qquad=\frac{1}{\pi}\int_{0}^\infty \int_{0}^\infty f(x)f'(y)\frac{d}{dx}\log|x-y|\,dxdy$$ $$\qquad=-\frac{1}{\pi}f(0)\int_{0}^\infty f'(y)\log y\,dy-\frac{1}{\pi}\int_{0}^\infty \int_{0}^\infty f'(x)f'(y)\log|x-y|\,dxdy.$$
answered Jan 20, 2021 at 19:10
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$\begingroup$ Thank you very much! Yes, I actually wanted the integral from 0 to $\infty$ (in which case we also define the Hilbert transform with the integral $pv \int_0^\infty......$. Is there anything we can do in this case? $\endgroup$– JunCommented Jan 20, 2021 at 19:13
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$\begingroup$ Thank you very much! That looks great. Can this be generalized if we replace $f'$ with $f^{(k)}$ ($k$-th derivative of $f$)? $\endgroup$– JunCommented Jan 20, 2021 at 19:42
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2$\begingroup$ I don't think it's clear that the step from the first to the second line is justified. The principal value has disappeared. (It is true that $PV(1/x)=(\log |x|)'$, but this is in distributional sense, not as functions.) $\endgroup$ Commented Jan 20, 2021 at 19:52
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$\begingroup$ @ChristianRemling --- since we are integrating, the distributional equality $\text{PV}(1/x)=d/dx \log |x|$ should hold for smooth $f(x)$, you don't agree? $\endgroup$ Commented Jan 20, 2021 at 20:39
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$\begingroup$ @CarloBeenakker Also, could you add some detail on the last identity (from second to third line) of the half-line case? How do you get the expression in line 3 from integrating by parts in line 2? $\endgroup$– JunCommented Jan 20, 2021 at 21:03