For a function $f$ in $L^1(\mathbb{R}^n)$, it is natural to define the Fourier transform as $$\mathscr{F}(f)(\xi)=\int_{\mathbb{R}^n}f(x)e^{-ix\cdot \xi}dx.$$ And the we may extend it to rapidly decreasing functions, Schwartz functions (distributions). And it is easy to verify that $$\mathscr{F}(1)=\delta,$$ the Dirac $\delta$ function.
My question is that how to prove that $$\mathscr{F}\left(\frac{x_i^2}{|x|^2}\right)=\frac{1}{n}\delta.$$
My intuition is to use the rotational invariance. But I could not proceed.
Let $$f(\alpha,\xi) = \int_{\mathbb R^n} e^{-\frac\alpha2 |x|^2} e^{-i x\cdot\xi} \frac{x_1^2}{|x|^2} \, dx .$$ Consider $\frac d{d\alpha} f(\alpha,\xi)$. It splits as the product of lots of one-dimensional integrals, each of which can be explicitly computed. Now get $$ \mathcal F\left(\frac{x_1^2}{|x|^2}\right) = f(0,\xi) = - \int_0^\infty \frac d{d\alpha} f(\alpha,\xi) \, d\alpha .$$ You will probably get something like $$ \mathcal F\left(\frac{x_1^2}{|x|^2}\right) = \text{principle value} \left(\frac{c}{|\xi|^n} - \frac{c n \xi_1^2}{|\xi|^{n+2}}\right) + \theta,$$ where $c$ is some explicit constant, and $\theta$ is only at the origin.
To figure out what happens at the origin, sum both sides from $i=1$ to $n$, to get $$ \mathcal F (1) = n \theta = \delta ,$$ so $ \theta = \frac1n \delta$.