I am currently reading a book of Hilbert transforms and I have found with the following equality:

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{isx}\int_{-\infty}^{\infty^*}\frac{\phi(y)}{x-y}\text{ d}y\text{ d}x=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{isx}\phi(x)\text{ d}x\int_{-\infty}^{\infty^*}\frac{e^{isy}}{y}\text{ d}y,$$

where $\phi \in L^2\left(-\infty,\infty\right)$.

The $^*$ above the integral means that it is a Cauchy principal value integral (because of the poles at $x=y$ in the left side and at $y=0$ in the right side).

My guess was to try a change of variable, $w=x-y$ and I guess Fubini can be used in some later step but I am not sure how to use it properly. Any help/hint is welcomed. Thanks.


Right ok so here's what I'm going to do:

Do Fubini to swap $dx$ and $dy$.

Then we have

$\int_{-\infty}^{\infty} e^{iny}\phi(y)\int_{-\infty}^{\infty}\frac{e^{in(x-y)}}{x-y}dxdy$

Set $\tilde{y} = x-y$ in place of $dx$ with $d\tilde{y} = dx$. The limits stay the same.

Set $\tilde{x} = y$ i.e. $d\tilde{x} = dy$

Then we are done.

  • 1
    $\begingroup$ You may wish to justify the Fubini properly. $\endgroup$ – fGDu94 Apr 1 '19 at 22:23

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