Is it true that $$\int\_{\mathbb{R}^N}(-\Delta)^su(x) dx = 0,$$
where $(-\Delta)^s$ is the [fractional Laplacian](https://www.ma.utexas.edu/mediawiki/index.php/Fractional_Laplacian)?



Integral of fractional Laplacian is zero
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](http://www.pitt.edu/~hajlasz/Notatki/Harmonic%20Analysis4.pdf) (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).


