A derivation by Fourier transformation of the right-hand-side for $s=1-\epsilon$ to check that it equals $k^2 \hat{u}(k)$ when $\epsilon\downarrow 0$. The coefficient $c_s$, given in <A HREF="https://en.wikipedia.org/wiki/Fractional_Laplacian">Wikipedia</A>, has the expansion 
$$c_{1-\epsilon}=2\epsilon+{\cal O}(\epsilon^2).$$

       
First rewrite the right-hand-side as
$$F_\epsilon(x)=c_{1-\epsilon}\int_{-\infty}^\infty  \frac{u(x)-u(y)}{|x-y|^{3-2\epsilon}}\,dy=\tfrac{1}{2}c_{1-\epsilon}\int_{-\infty}^\infty  \frac{2u(x)-u(x+y)-u(x-y)}{|y|^{3-2\epsilon}}\,dy,$$
and Fourier transform to obtain
$$\hat{F}_\epsilon(k)=c_{1-\epsilon}\int_{-\infty}^\infty e^{ikx}F_\epsilon(x)\,dx=c_{1-\epsilon}\,\hat{u}(k)\int_{-\infty}^\infty \frac{1-\cos ky}{|y|^{3-2\epsilon}}\,dy=$$
$$\qquad=c_{1-\epsilon}\,k^2\hat{u}(k)\int_{-\infty}^\infty \frac{1-\cos z}{|z|^{3-2\epsilon}}\,dz=c_{1-\epsilon}\,k^2\hat{u}(k)  \; 2 \cos( \pi  \epsilon) \Gamma (2\epsilon-2)$$
$$\qquad\rightarrow k^2\hat{u}(k) \;\;\text{for}\;\;\epsilon\downarrow 0,$$
since
$  2 \cos( \pi  \epsilon) \Gamma (2\epsilon-2)=\frac{1}{2\epsilon}+{\cal O}(1)$.