# Fourier transform of $\frac {x} {(x^2+y^2)}$

I was trying to compute the Fourier transform of $$f(x,y)=\frac{x}{x^2+y^2}$$.

I saw in a paper I was reading that $$\hspace{4cm}\hat{f}(\xi_1,\xi_2)=Const.\frac{\xi_1}{\xi_1^2+\xi_2^2}\hspace{4cm} (*)$$ (i.e. $$f$$ works like a eigenvetor for the Fourier transform)

[my progress]

1) I know that If $$u$$ is homogeneous of degree $$r$$, then $$\hat{u}$$ is homogeneous of degree $$−r − n$$ (here, $$n$$ is the space dimension). So, as $${f}$$ is homogeneous of degree $$-1$$, $$\hat{f}$$ also must be homogeneous of degree $$-1$$.

2) I also showed (using some fourier proprieties) that if we define $$T:\mathbb{R}^2\to\mathbb{R}^2$$ by $$T(x,y)=(x,-y)$$, then $$\hat{f}(\xi)=\hat{f}(T(\xi))$$.

Can I use 1) and 2) to show $$(*)$$? there's any other way to show $$(*)$$?

You've done a good job verifying some properties of $$\hat{f}$$. In the following, I'll proceed formally, that is, manipulating the form without taking care of rigurosity, which can be given using distributions.

The trick I learnt is: $$|x|^{-2} = \int_0^\infty e^{-t|x|^2}\,dt$$, where $$x=(x_1,x_2)$$, so we want to calculate the Fourier transform of $$f=x_1\int_0^\infty e^{-t|x|^2}\,dt$$, that is,

$$\hat{f}(\xi)=\int_0^\infty\int_{\mathbb{R}^2} x_1e^{-t|x|^2} e^{-2\pi i x\cdot \xi}\,dxdt,$$

but $$e^{-2\pi i x\cdot\xi}=\hat{\delta_\xi}(x)$$, where $$\delta_\xi$$ is the Dirac delta centered at $$\xi$$; also, $$x_1e^{-t|x|^2}=-(2t)^{-1}\partial_{x_1}(e^{-t|x|^2})$$ and $$[\partial_{x_1}(e^{-t|x|^2})]^\wedge(\eta)=2\pi i\eta_1e^{-|\eta|^2/t}/t$$ (I'm sure I'm missing some constants here). Replacing above we get

\begin{align}\hat{f}(\xi)&=\frac{\pi}{i}\int_0^\infty\frac{1}{t^2}\int \mathcal{F}^{-1}(\eta_1 e^{-|\eta|^2/t}) \hat{\delta_\xi} \,dxdt \\ &= \frac{\pi}{i}\int_0^\infty\frac{1}{t^2}\int \eta_1 e^{-|\eta|^2/t} \delta_\xi(\eta) \,d\eta dt \\ &= \frac{\pi}{i}\int_0^\infty\frac{1}{t^2} \xi_1 e^{-|\xi|^2/t} \,dt \\ &= c\frac{\xi_1}{|\xi|^2}, \end{align}

where $$\mathcal{F}^{-1}$$ is the inverse Fourier transform (unfortunately, the command \check doesn't work). There are other methods to calculate this Fourier transform.

• what a nice and tricky answer @user90189(I would never thought about that)! Thanks so much!! a really apreciate it!!!
– user609149
Oct 27, 2018 at 20:14