Fourier transform of $1/|x|^{\alpha}$. 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 $1/|x|^\alpha$?
 A: Denoting by $\widehat{\cdot}$ the Fourier transform in $\mathbb R^d$ and letting
$$
f_\lambda(x)\equiv f(\lambda x),
$$ 
then, by definition of the Fourier transform,
\begin{align}
\widehat {f_\lambda}(k)&=\int_{\mathbb R^n}e^{-ik\cdot x}f(\lambda x)dx\\
&=\int_{\mathbb R^n}e^{-ik\cdot x/\lambda}f(x)\lambda^{-d}dx\\
&=\lambda^{-d}\widehat f(k/\lambda).
\end{align}
Let $f(x)=F(|x|)$ be a radial function;
if $M\in SO(d)$, then $f_M(x)\equiv f(Mx)=f(x)$ and hence, since $\widehat{ f_M}(k)=\widehat f(Mk)$, we have $\widehat {f_M}= \widehat f$, which in turn implies that also $\widehat f$ is radial. Now, suppose $f(x)=|x|^{-\alpha}$, for $0<\alpha<d$, then 
$$
f_\lambda(x) = \lambda^{-\alpha}f(x)
$$
and, by the above remark,
$$
\widehat f (k/\lambda)= \lambda^{d}\widehat {f_\lambda}(k) = \lambda^{d-\alpha}\hat f(k).
$$
The only radial $\widehat f$ homogeneous of degree $-d+\alpha$ is 
$$
\widehat f (k) = |k|^{\alpha-d},
$$
up to constants.
Note that $f$ can be split as the sum of two pieces: let $B_a$ be the ball of radius $a$ centred at the origin, then
\begin{align}
f&\equiv u+v\\
&=
\chi_{B_a}(x)\frac{1}{|x|^{\alpha}}+\left(1-\chi_{B_a}(x)\right)\frac{1}{|x|^{\alpha}};
\end{align}
$u$ lies in $L^1(\mathbb R^d)$, for $0<\alpha<d$, and $v$ lies in $L^2(\mathbb R^d)$, for $\alpha>d/2$. This ensures $\widehat u \in L^\infty(\mathbb R^d)$ and $\widehat v\in L^2(\mathbb R^d)$ and that $\widehat f$ needs indeed to be a function in $L^1_{\text{loc}}(\mathbb R^d)$ from an abstract point of view.
So $f,\widehat f\in L^1_{\text{loc}}(\mathbb R^d)$.
More generally, $f$ lies in $\mathscr S'(\mathbb R^d)$, which is stable under Fourier transform.
A: 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.
