How can I compute the spectral fractional Laplacian of $(ax)_+^\alpha$ on $\Omega = (0,1)$? Here the operator is defined as $$(\Delta)^s u = c_{N,s} \int_0^\infty (e^{t\Delta_N}u(x)  u(x)) t^{1  s} dt ,$$ where $e^{t\Delta_N}u$ is the solution of the heat equation with homogeneus Neumann boundary conditions on $\Omega$.

$\begingroup$ I would be surprised if a nice closedform expression existed. A natural approach would be to expand $(ax)_+^\alpha$ in the Fourier cosine series $\sum a_n \cos(n \pi x)$ and use the fact that $(\Delta_N)^s [\cos(n \pi x)] = (n \pi)^{2s} \cos(n \pi x)$. I do not know if there is a simple expression for $a_n$, though. $\endgroup$– Mateusz KwaśnickiFeb 16 at 1:23

$\begingroup$ Right hand integral looks like an operator with respect to $t$: $\int^\infty_0(u(x,t)u(x,0))t^{1s}dt$ with parameter $x$. $\endgroup$– user171871Feb 16 at 8:53
The operator $(∆)^s$ is the Fourier multiplier $\vert\xi\vert^{2s}$ and thus deserves to be denoted by $\vert D\vert^{2s}$. Now the distribution $u_\alpha(x)=x_+^\alpha$ is homogeneous with degree $\alpha$ and thus has a Fourier transform homogeneous with degree $1\alpha$. We have in fact $$ \hat{u}_\alpha(\xi)=c_\alpha (\xii0)^{1\alpha}. $$ You are interested in $v_\alpha(x)=u_\alpha(xt)$ so that $$ \vert\xi\vert^{2s}\hat{v}_\alpha(\xi)=e^{it \xi}c_\alpha (\xii0)^{1\alpha}\vert\xi\vert^{2s}, $$ and $ (\vert D\vert^{2s} v_\alpha)(x)=\int e^{i(xt) \xi}c'_\alpha (\xii0)^{1\alpha}\vert\xi\vert^{2s}d\xi =(\vert D\vert^{2s} u_\alpha)(xt). $ Translations commute with derivations, including fractional versions.

1$\begingroup$ I think Jay is asking about a different operator: the fractional power of the Neumann Laplacian on $(0,1)$. $\endgroup$ Feb 16 at 1:21
Let us denote the corresponding eigensystem $\{\lambda_n,\phi_n\}$. If $u(x)=\sum\limits_nu_n\phi_n(x)$, then we have $u(x,t)=\sum\limits_n\exp(\lambda_n t)u_n\phi_n(x)$. Since $\int\limits^\infty_0(\exp(\lambda_n t)1)t^{1s}=\lambda^s_n\frac{\Gamma(1s)}{s}$ we obtain $$ \text{R.H.I.}=C_{s,N}\frac{\Gamma(1s)}{s}\sum\limits_n\lambda^s_nu_n\phi_n(x)$$