# Computing the fractional Laplacian of power function

Is it possible to compute explicitly the fractional Laplacian (in $$\mathbb R^n$$) of a power function $$|x|^p$$?

• Yes: $(-\Delta)^{\alpha/2} [|x|^p] = 2^\alpha \Gamma((p+n)/2) \Gamma((\alpha-p)/2) (\Gamma((p+n-\alpha)/2) \Gamma((-p/2))^{-1} |x|^{p-n-\alpha}$ whenever defined. May 22, 2020 at 11:29
• This follows easily from the composition rule for the Riesz potential kernel. See Theorem 3.1 in my survey [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], DOI:10.1515/9783110571622-007 for a rigorous statement and discussion. May 22, 2020 at 11:32
• @MateuszKwaśnicki Thanks! Could you clarify the $x$ dependence in your result? I expect it to be something like $-C_{N,\alpha} |x|^q$.
– Hiro
May 22, 2020 at 11:38
• Ah, I noticed a typo in my previous comment: should be $|x|^{p-\alpha}$, not $|x|^{p-n-\alpha}$, sorry. This is exactly as you say, with $q = p - \alpha$, but the constant depends on $n$, $\alpha$ and $p$. May 22, 2020 at 11:47
• @MateuszKwaśnicki Thanks again. Oh, now I understand; in my browser the formula inside the comment is not properly displayed (I see it divided in two pieces with "whenever defined" in the middle). For the sake of clarity, could you please write it in an answer?
– Hiro
May 22, 2020 at 11:50

Proposition: Let $$\alpha \in (0, \infty)$$, $$p \in (-n, \alpha)$$, and $$f(x) = |x|^p .$$ Then $$(-\Delta)^{\alpha/2} f(x)$$ is well-defined for $$x \ne 0$$, and $$(-\Delta)^{\alpha/2} f(x) = 2^\alpha \frac{\Gamma(\frac{p+n}{2}) \Gamma(\frac{\alpha-p}{2})}{\Gamma(\frac{p+n-\alpha}{2}) \Gamma(-\tfrac{p}{2})} \, |x|^{p - \alpha} .$$ Here we understand the right-hand side is zero if $$\tfrac{p+n-\alpha}{2}$$ or $$-\tfrac{p}{2}$$ is a non-positive integer.