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One way to interpret the singular integral for $s=1-\epsilon$ is by Fourier transformation, to check that it tends to $k^2 \hat{u}(k)$ when $\epsilon\downarrow 0$.

I will make use of the fact that the coefficient $c_s$, given in Wikipedia, has the expansion $$c_{1-\epsilon}=2\epsilon+{\cal O}(\epsilon^2).$$

First rewrite the right-hand-side in a way that I can remove the principal value (see arXiv:1104.4345, page 13) $$F_\epsilon(x)\equiv c_{1-\epsilon}\operatorname{P.V.} \int_{-\infty}^\infty \frac{u(x)-u(y)}{|x-y|^{3-2\epsilon}}\,dy=\tfrac{1}{2}c_{1-\epsilon}\int_{-\infty}^\infty \frac{2u(x)-u(x+y)-u(x-y)}{|y|^{3-2\epsilon}}\,dy,$$ and Fourier transform to obtain $$\hat{F}_\epsilon(k)\equiv\int_{-\infty}^\infty e^{ikx}F_\epsilon(x)\,dx=c_{1-\epsilon}\,\hat{u}(k)\int_{-\infty}^\infty \frac{1-\cos ky}{|y|^{3-2\epsilon}}\,dy=$$ $$\qquad=c_{1-\epsilon}\,k^{2-2\epsilon}\hat{u}(k)\int_{-\infty}^\infty \frac{1-\cos z}{|z|^{3-2\epsilon}}\,dz=c_{1-\epsilon}\,k^{2-2\epsilon}\hat{u}(k) \; 2 \cos( \pi \epsilon) \Gamma (2\epsilon-2)$$ $$\qquad\rightarrow k^2\hat{u}(k) \;\;\text{for}\;\;\epsilon\downarrow 0,$$ since $ 2 \cos( \pi \epsilon) \Gamma (2\epsilon-2)=\frac{1}{2\epsilon}+{\cal O}(1)$.

One way to interpret the singular integral for $s=1-\epsilon$ is by Fourier transformation, to check that it tends to $k^2 \hat{u}(k)$ when $\epsilon\downarrow 0$.

I will make use of the fact that the coefficient $c_s$, given in Wikipedia, has the expansion $$c_{1-\epsilon}=2\epsilon+{\cal O}(\epsilon^2).$$

First rewrite the right-hand-side in a way that I can remove the principal value (see arXiv:1104.4345, lemma 3.2) $$F_\epsilon(x)\equiv c_{1-\epsilon}\operatorname{P.V.} \int_{-\infty}^\infty \frac{u(x)-u(y)}{|x-y|^{3-2\epsilon}}\,dy$$ $$\qquad=\tfrac{1}{2}c_{1-\epsilon}\int_{-\infty}^\infty \frac{2u(x)-u(x+y)-u(x-y)}{|y|^{3-2\epsilon}}\,dy, \tag{*} \label{PV}$$ and Fourier transform to obtain $$\hat{F}_\epsilon(k)\equiv\int_{-\infty}^\infty e^{ikx}F_\epsilon(x)\,dx=c_{1-\epsilon}\,\hat{u}(k)\int_{-\infty}^\infty \frac{1-\cos ky}{|y|^{3-2\epsilon}}\,dy=$$ $$\qquad=c_{1-\epsilon}\,k^{2-2\epsilon}\hat{u}(k)\int_{-\infty}^\infty \frac{1-\cos z}{|z|^{3-2\epsilon}}\,dz=c_{1-\epsilon}\,k^{2-2\epsilon}\hat{u}(k) \; 2 \cos( \pi \epsilon) \Gamma (2\epsilon-2)$$ $$\qquad\rightarrow k^2\hat{u}(k) \;\;\text{for}\;\;\epsilon\downarrow 0,$$ since $ 2 \cos( \pi \epsilon) \Gamma (2\epsilon-2)=\frac{1}{2\epsilon}+{\cal O}(1)$.

One way to interpret the singular integral for $s=1-\epsilon$ is by Fourier transformation, to check that it tends to $k^2 \hat{u}(k)$ when $\epsilon\downarrow 0$.

I will make use of the fact that the coefficient $c_s$, given in Wikipedia, has the expansion $$c_{1-\epsilon}=2\epsilon+{\cal O}(\epsilon^2).$$

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

One way to interpret the singular integral for $s=1-\epsilon$ is by Fourier transformation, to check that it tends to $k^2 \hat{u}(k)$ when $\epsilon\downarrow 0$.

I will make use of the fact that the coefficient $c_s$, given in Wikipedia, has the expansion $$c_{1-\epsilon}=2\epsilon+{\cal O}(\epsilon^2).$$

First rewrite the right-hand-side in a way that I can remove the principal value (see arXiv:1104.4345, page 13) $$F_\epsilon(x)\equiv c_{1-\epsilon}\operatorname{P.V.} \int_{-\infty}^\infty \frac{u(x)-u(y)}{|x-y|^{3-2\epsilon}}\,dy=\tfrac{1}{2}c_{1-\epsilon}\int_{-\infty}^\infty \frac{2u(x)-u(x+y)-u(x-y)}{|y|^{3-2\epsilon}}\,dy,$$ and Fourier transform to obtain $$\hat{F}_\epsilon(k)\equiv\int_{-\infty}^\infty e^{ikx}F_\epsilon(x)\,dx=c_{1-\epsilon}\,\hat{u}(k)\int_{-\infty}^\infty \frac{1-\cos ky}{|y|^{3-2\epsilon}}\,dy=$$ $$\qquad=c_{1-\epsilon}\,k^{2-2\epsilon}\hat{u}(k)\int_{-\infty}^\infty \frac{1-\cos z}{|z|^{3-2\epsilon}}\,dz=c_{1-\epsilon}\,k^{2-2\epsilon}\hat{u}(k) \; 2 \cos( \pi \epsilon) \Gamma (2\epsilon-2)$$ $$\qquad\rightarrow k^2\hat{u}(k) \;\;\text{for}\;\;\epsilon\downarrow 0,$$ since $ 2 \cos( \pi \epsilon) \Gamma (2\epsilon-2)=\frac{1}{2\epsilon}+{\cal O}(1)$.

One way to interpret the singular integral for $s=1-\epsilon$ is by Fourier transformation, to check that it tends to $k^2 \hat{u}(k)$ when $\epsilon\downarrow 0$.

I will make use of the fact that the coefficient $c_s$, given in Wikipedia, has the expansion $$c_{1-\epsilon}=2\epsilon+{\cal O}(\epsilon^2).$$

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

One way to interpret the singular integral for $s=1-\epsilon$ is by Fourier transformation, to check that it tends to $k^2 \hat{u}(k)$ when $\epsilon\downarrow 0$.

I will make use of the fact that the coefficient $c_s$, given in Wikipedia, has the expansion $$c_{1-\epsilon}=2\epsilon+{\cal O}(\epsilon^2).$$

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