For $u\in\mathcal E'(\mathbb R^n)$, its <Fourier transform> is the <smooth function>
$$
\widehat u(\lambda)=\langle u(x),e^{-i\lambda\cdot x}\rangle.
$$
It has at most <polynomial growth>, and its extension to complex frequency is controlled more precisely by the <Paley–Wiener–Schwartz theorem>.

Fourier transform of a compactly supported distribution

Because $0<\alpha<n$, $|x|^{-\alpha}$ is <locally integrable function>[locally integrable] at the origin and has only <polynomial growth> at infinity, so it defines a <tempered distribution>. It is a <homogeneous distribution> of degree $-\alpha$ and is <radial function>[radial]. Its Fourier transform is therefore radial and homogeneous of degree $\alpha-n$, so it must have the form $c_\alpha|\lambda|^{\alpha-n}$. In particular, $\boxed{\beta=n-\alpha}$.

To determine the constant, use the stated <Gamma integral> representation, <Fubini's theorem>, and the <Fourier transform of a Gaussian>:
$$
\begin{aligned}
\widehat u_\alpha(\lambda)
&=\frac1{\Gamma(\alpha/2)}\int_0^\infty\tau^{\alpha/2-1}
\left(\int_{\mathbb R^n}e^{-\tau|x|^2-i\lambda\cdot x}\,dx\right)d\tau\\
&=\frac{\pi^{n/2}}{\Gamma(\alpha/2)}
\int_0^\infty\tau^{(\alpha-n)/2-1}e^{-|\lambda|^2/(4\tau)}\,d\tau.
\end{aligned}
$$
The <change of variables formula> $s=|\lambda|^2/(4\tau)$ then gives
$$
\widehat u_\alpha(\lambda)
=2^{n-\alpha}\pi^{n/2}
\frac{\Gamma((n-\alpha)/2)}{\Gamma(\alpha/2)}
|\lambda|^{\alpha-n}.
$$
This is precisely the <Riesz kernel>[Fourier transform of the Riesz kernel].

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 327 / 1 / c / Solution

A function has polynomial growth when $|f(x)|\leq C(1+|x|)^N$ for some constants $C,N$. Such growth is slow enough for a <locally integrable function> to define a <tempered distribution>.

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 Incoming links: Polynomial growth