Integral of fractional Laplacian is zero Is it true that $$\int_{\mathbb{R}^N}(-\Delta)^su(x) dx = 0,$$
where $(-\Delta)^s$ is the fractional Laplacian?
 A: Noam D. Elkies and Piotr Hajlasz give a positive answer to the original question, with two different proofs. While their arguments are completely fine, they are somewhat restrictive: the integral of $(-\Delta)^{\alpha/2} f$ is zero if $f$ is integrable (and regular enough). The general answer is no!

Suppose that $\alpha < n$, and let $g$ be any continuous, compactly supported (or merely integrable) function. Let
$$ f(x) = c_{n,\alpha} \int_{\mathbb{R}^n} \frac{g(y)}{|x - y|^{n - \alpha}} \, dy $$
be the Riesz potential of $g$. Then $(-\Delta)^{\alpha/2} f(x) = g(x)$ for every $x \in \mathbb{R}^n$ (for the proof, see, for example, Samko's book Hypersingular Integrals and Their Applications), and so the integral of $(-\Delta)^{\alpha/2} f$ need not be zero.

A nice explicit example is $$f(x) = (1 + |x|^2)^{\alpha/2 - n/2},$$ which corresponds to $$g(x) = c_{n,\alpha} (1 + |x|^2)^{-\alpha/2 - n/2}.$$ This follows, for example, from Corollary 1 here.
A: Yes, under certain assumptions. 
For $0<\alpha<2$ and $\varphi\in\mathscr{S}_n$ we have
\begin{eqnarray*}
(-\Delta)^{\alpha/2}\varphi 
& = &
\frac{1}{2\gamma(-\alpha)}
\int_{\mathbb R^n}
\frac{\varphi(x+y)+\varphi(x-y)-2\varphi(x)}{|y|^{n+\alpha}}\, dy,
\end{eqnarray*}
where
$$
\gamma(-\alpha)=
\frac{\pi^{\frac{n}{2}}2^{-\alpha}\Gamma\left(-\frac{\alpha}{2}\right)}{\Gamma\left(\frac{n+\alpha}{2}\right)}\,.
$$
One can show that
$$
(x,y)\mapsto\frac{\varphi(x+y)+\varphi(x-y)-2\varphi(x)}{|y|^{n+\alpha}}\in L^1(\mathbb R^n\times\mathbb R^n).
$$
Then integrating $(-\Delta)^{\alpha/2}\varphi$ over $\mathbb R^n$ and applying Fubini theorem yields zero. The same applies to $s=\frac{\alpha}{2}+k$, where $k$ is a positive integer since
\begin{eqnarray*}
(-\Delta)^{\frac{\alpha}{2}+k}\varphi 
& = &
\frac{(-1)^k}{2\gamma(-\alpha)}
\int_{\mathbb R^n}
\frac{\Delta^k\varphi(x+y)+\Delta^k\varphi(x-y)-2\Delta^k\varphi(x)}{|y|^{n+\alpha}}\, dy.
\end{eqnarray*}
You can find proofs for example

here (Theorem 7.13 and Corollary 7.16). The argument does not apply to negative powers of the Laplacian. The negative powers of compactly supported smooth functions are rarely integrable (I think).
