TITLE: Fourier transform of $1/|x|^{\alpha}$.
QUESTION [9 upvotes]: My problem is to prove the following identity:
$$C_{\alpha}\int_{\mathbb R^n} \frac{1}{|x|^\alpha} \phi(x) dx = C_{n-\alpha}\int_{\mathbb R^n} \frac{1}{|x|^{n-\alpha}} \widehat{\phi}(x) dx$$
where $\phi:\mathbb R^n \to \mathbb C$ is on the Schwartz space  , $0<\alpha <n$, $C_{\beta} = \Gamma(\beta/2)/\pi^{\beta/2}$ and $$\widehat{\phi}(x) = \int_{\mathbb R^n} \phi(y)e^{-2\pi i x\cdot y}dy.$$ 
In other words, how to calculate fourier transform of $1/|x|^\alpha$?

REPLY [3 votes]: Following Brightsun solution we have $\widehat f (\xi) = k/|x|^{n-\alpha}$. Let $g(x) = e^{-\pi x^2}$, we have $\widehat g = g$. Since 
$$ \int_{\mathbb R^n} \widehat{f}(x)g(x)dx = \int_{\mathbb R^n} f(x)\widehat{g}(x)dx,$$
we have
$$ \int_{\mathbb R^n} \frac{k}{|x|^{n-\alpha}}e^{-\pi x^2}dx = \int_{\mathbb R^n} \frac{1}{|x|^\alpha}e^{-\pi x^2}dx.$$
In polar coordinates we have:
$$\int_{\mathbb R^n} \frac{1}{|x|^{\alpha}}e^{-\pi x^2}dx = \sigma(\mathbb S^{n-1})\int_{0}^{\infty} \frac{e^{-\pi r^2}r^{n-1}}{r^{\alpha}} = \sigma(S^{n-1})\int_{0}^{\infty} e^{-\pi r^2}r^{n-1-\alpha} dr,$$
where $\sigma$ is the measure on the sphere. Taking $u = \pi r^2$ we obtain 
$$\sigma(S^{n-1})\int_{0}^{\infty} e^{-\pi r^2}r^{n-1-\alpha} dr = \frac{\sigma(S^{n-1})}{\pi^{\alpha/2}}\int_{0}^{\infty}u^{\alpha/2 -1} e^{-u}du = \frac{\sigma(S^{n-1})\Gamma(\alpha/2)}{\pi^{\alpha/2}}.$$
Analogously, we can deduce 
$$\int_{\mathbb R^n} \frac{k}{|x|^{n-\alpha}}e^{-\pi x^2}dx = \frac{k\sigma(S^{n-1})\Gamma((n-\alpha)/2)}{\pi^{(n-\alpha)/2}}$$
and we have $$k=\frac{\Gamma(\alpha/2)\pi^{(n-\alpha)/2}}{\Gamma((n-\alpha)/2)\pi^{\alpha/2}}.$$
Therefore, the identity 
$$C_{\alpha}\int_{\mathbb R^n} \frac{1}{|x|^\alpha} \phi(x) dx = C_{n-\alpha}\int_{\mathbb R^n} \frac{1}{|x|^{n-\alpha}} \widehat{\phi}(x) dx$$
holds.