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 $(*)$?
 A: 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.
