Explicit expression for $(-\Delta)^s (|x|^2)$ where $x\in \mathbb{R}^n$ ($n>2s$) and $s\in (0,1)$

Question

Do we know an explicit expression for $(-\Delta)^s (|x|^2)$ where $x\in \mathbb{R}^n$ ($n>2s$) and $s\in (0,1)?$

Answer 1

A classical result in distribution theory states that if $u$ is a homogeneous distribution on $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ homogeneous distribution of degree $d\in \bR\setminus \{-n,-n-1,\dotsc\}$, then it admits a unique extension to a distribution on $\bR^n$.

For $\lambda \in \bC$ with positive real part and $s\in (0,1)$ the distribution $(-\Delta)^su_\lambda$, $u_\lambda(x):=|x|^\lambda$, is homogeneous of degree $\lambda-2s$ and rotationally symmetric so it has the form $C(s,\lambda)u_{\lambda-2s}$. Moreover the function $\lambda \mapsto C(s,\lambda)$ is holomorphic.

To find the constant $C(s,\lambda)$ use the Fourier transform $\newcommand{\eF}{\mathscr{F}}$ and we have $$ \eF\big[ (-\Delta)^su_\lambda \big](\xi)= |\xi|^{2s}\eF\big[ u_\lambda\big](\xi). $$ We deduce that $$C(s,\lambda)\eF\big[ u_{\lambda-2s}\big](\xi)= |\xi|^{2s}\eF\big[ u_\lambda\big](\xi). $$

The Fourier transform of $u_\lambda$ is known and has the form $$ \eF[u_\lambda]= K(\lambda, n)|\xi|^{\lambda-n} $$ where $K(\lambda,n)$ is an explicit constant involving the Gamma function (see e.g. volume 1 in Gelfand and Shilov's book on generalized functions).

We deduce that $$C(s,\lambda)K(\lambda-2s,n)= K(\lambda, n)\implies C(s,\lambda)= \frac{K(\lambda, n)}{K(\lambda-2s,n)}. $$