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 $(*)$?


1 Answer 1


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.

  • 2
    $\begingroup$ what a nice and tricky answer @user90189(I would never thought about that)! Thanks so much!! a really apreciate it!!! $\endgroup$
    – user609149
    Oct 27, 2018 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy