For $s\in (0,1)$, is there are an explicit expression for
$$(-\Delta)^s \left(\frac{1}{\left(1+|x|^2\right)^s}\right)?$$


Edit: My goal is to show that for the function $u(x)=\frac{1}{\left(1+|x|^2\right)^s}$ defined on the ball $B(0,R)$ where $R>1$ we have $(-\Delta)^s u(x)\geq c(n,s) u^2(x)$ where $c(n,s)>0$ is constant dependent on $n$ and $s.$ Any references/hints on how to do this will be much appreciated.



Explicit expression for the fractional Laplacian of $1/(1+|x|^2)^s$
Unless I am making a typo, the result is:
$$ 2^{2s} \frac{\Gamma(\tfrac n2+s) \Gamma(2s)}{\Gamma(\tfrac n2)^2} {\_2F\_1}(\tfrac n2+s, 2s, \tfrac n2, -|x|^2), $$
where $n$ is the dimension and ${\_2F\_1}$ is the Gauss's hypergeometric function. See Table 1 on page 168 in my survey [1], or Corollary 2 in the original paper [2].


References:


[1] Mateusz Kwaśnicki, *Fractional Laplace Operator and its Properties*, in: A. Kochubei, Y. Luchko, *Handbook of Fractional Calculus with Applications. Volume 1: Basic Theory*, De Gruyter Reference, De Gruyter, Berlin, 2019, <https://doi.org/10.1515/9783110571622-007>


[2] Bartłomiej Dyda, Alexey Kuznetsov, Mateusz Kwaśnicki, *Fractional Laplace operator and Meijer G-function*, Constructive Approx. 45(3) (2017): 427–448, <https://doi.org/10.1007/s00365-016-9336-4>




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*Edit:* The hypergeometric function is continuous on $(-\infty, 1)$, and — as described, for example, in [this DLMF entry](https://dlmf.nist.gov/15.12#i) — we have
$$ {\_2F\_1}(a, b, c, -|x|^2) = \frac{p}{|x|^{2a}} \, {\_2F\_1}(a', b', c', |x|^{-2}) + \frac{q}{|x|^{2b}} \, {\_2F\_1}(a'', b'', c'', |x|^{-2}) $$
for some (explicit) constants $p, q, a', b', c', a'', b'', c''$. In particular,
$$ {\_2F\_1}(a, b, c, -|x|^2) \sim \frac{p}{|x|^{2a}} + \frac{q}{|x|^{2b}} $$
as $|x| \to \infty$. Therefore, as long as $n > 4s$, we have
$$ (-\Delta)^s [(1 + |x|^2)^{-2s}] \sim C\_{n,s} |x|^{-4s} $$
as $|x| \to \infty$, for some (explicit) constant $C\_{n,s}$. Obviously, $(-\Delta)^s [(1 + |x|^2)^{-2s}]$ is continuous, so we immediatiely get
$$ |(-\Delta)^s [(1 + |x|^2)^{-2s}]| \leqslant C\_{n,s}' (1 + |x|^2)^{-2s} . $$
In order to get a similar lower bound, one needs to check that $C\_{n,s} > 0$ (which seems to be the case when $n > 4s$, as long as I did not make any mistake), and that $(-\Delta)^s [(1 + |x|^2)^{-2s}]$ is never zero. I am no expert here, but [another DLMF entry](https://dlmf.nist.gov/15.13) seems to suggest that the total number of zeroes of the hypergeometric function is $$\lfloor s \rfloor + \tfrac12 + \tfrac12 \operatorname{sign}(\Gamma(-s) \Gamma(2s) \Gamma(\tfrac n2+s) \Gamma(\tfrac n2-2s)) = 0 ,$$ as long as, again, $n > 4s$.


(Double check for typos.)


