# Double integral (Cauchy principal value integral)

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.

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