Fourier transform of product of radial power with fractional laplacian

I know that in $$\mathbb{R^n}$$, the Fourier transform of a fractional laplacian $$\widehat{(-\Delta)^s u}(\xi) = |\xi|^{2s} \hat{u}(\xi)$$. In particular, when $$s=1/2$$, $$\widehat{(-\Delta)^{\frac{1}{2}} u}(\xi) = |\xi|\hat{u}(\xi)$$. I am studying an operator $$H_s{v} =- r^{\frac{n-1}{2}}((-\Delta)^{\frac{1}{2}}u)$$, where $$u$$ is radially symmetric and $$u(r)=r^{\frac{1-n}{2}}v(r)$$. I want to solve $$H_s{v} =- v$$. Because of nice Fourier invertibility properties for $$s=1/2$$, I wanted to use Fourier analysis to solve this. I was wondering if there was a simple way to express the Fourier transform of the operator, $$\hat{H_s(v)}=\widehat{r^{\frac{n-1}{2}}(-\Delta)^{\frac{1}{2}}(r^{\frac{1-n}{2}}v)}$$ in terms of $$\hat{v}(\xi)$$.

Let $$a = {n-1}/2$$. Actually, when $$a$$ is not a negative even integer, the Fourier transform of $$\frac{1}{r^a} = \frac{1}{|x|^a}$$ is $$\mathcal{F}(\frac{1}{|x|^a}) = \frac{c_{n,a}}{|x|^{n-a}}$$ (to be interpreted as the associated Hadamard finite part distribution if $$a<0$$), so $$\mathcal{F}(r^{a}(-\Delta)^{1/2} r^{-a} v) = c_{n,a}c_{n,-a}\,\frac{1}{|\xi|^{n+a}} * \left(|\xi| \left(\frac{1}{|x|^{n-a}}* \hat{v}\right)(\xi)\right)$$ You can also rewrite that with fractional laplacians as $$\mathcal{F}(r^{a}(-\Delta)^{1/2} r^{-a} v) = (-\Delta)^{a/2}\left(|\xi| \left((-\Delta)^{-a/2} \hat{v}\right)(\xi)\right)$$ or in a more integral form $$\mathcal{F}(r^{a}(-\Delta)^{1/2} r^{-a} v)(\xi) = c_{n,a}\,(-\Delta_{\xi})^{a/2}\int_{\mathbb{R}^n} \frac{|\xi|\, \hat{v}(y)}{|\xi-y|^{n-a}}\,\mathrm{d}y$$