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

The locally integrable homogeneous function $x\mapsto|x|^{-\alpha}$, $0<\alpha<n$, is a Riesz kernel. For $\widehat f(\xi)=\int e^{-ix\cdot\xi}f(x)\,dx$,
$$
\widehat{|x|^{-\alpha}}(\xi)
=2^{n-\alpha}\pi^{n/2}
\frac{\Gamma((n-\alpha)/2)}{\Gamma(\alpha/2)}
|\xi|^{\alpha-n}.
$$

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