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Is it possible to integrate by parts the fractional laplacian $(-\Delta)^su+ u=f(u)$ in $\mathbb{R}^N$, or is it true that $\int_{\mathbb{R}^N}u= \int_{\mathbb{R}^N} f(u) $ with $s\in (0, 1)$, $u\in H^s(\mathbb{R}^N)$.

The identity holds for the Laplacian operator.

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Yes, this follows from the identity $$(-\Delta^{s}u)(x)=c_{n,s}\int_{\mathbb{R}^n}\frac{u(x)-u(y)}{||x-y||^{n+2s}}\,dy,$$ $$c_{n,s}=\frac{s\,4^{s}\Gamma(s+n/2)}{\pi^{n/2}\Gamma(1-s)},$$ where the principal value of the integral should be taken. (See The Pohozaev identity for the fractional Laplacian.) Integration over $x\in\mathbb{R}^n$ gives zero because of the antisymmetry of the integrand upon interchange $x\leftrightarrow y$, so $$\int_{\mathbb{R}^n} [(-\Delta^{s}u)(x)+u(x)]\,dx=\int_{\mathbb{R}^n} u(x)\,dx=\int_{\mathbb{R}^n} f(u(x))\,dx,$$ as requested.

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