Three dimensional Fourier Transform of a rational function

Question

Given $s\in(0,2)$, can we find an explicit representation for the (three-dimensional) Fourier transform of the function $$f_s(x):=\frac{|x|^s}{1+|x|^2},$$ or at least some pointwise bound?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \hat{\mrm{f}}_{s}\pars{\vec{k}} & \equiv \iiint_{\large\mathbb{R}^{3}}{r^{s} \over 1 + r^{2}}\, \expo{-\ic\vec{k}\cdot\vec{r}}\dd^{3}\vec{r} = \int_{0}^{\infty}{r^{s} \over 1 + r^{2}}\, 4\pi r^{2}\ \overbrace{\int_{\Omega_{\large\vec{r}}} \expo{-\ic\vec{k}\cdot\vec{r}}{\dd\Omega_{\vec{r}} \over 4\pi}}^{\ds{\sin\pars{kr} \over kr}} \\[5mm] &= {4\pi \over k}\int_{0}^{\infty}{r^{s + 1} \over 1+ r^{2}}\,\sin\pars{kr}\,\dd r = {4\pi \over k^{s + 1}}\int_{0}^{\infty}{r^{s + 1} \over r^{2} + k^{2}}\,\sin\pars{r}\,\dd r \end{align}

It's still a cumbersome task but some CAS yields

\begin{align} \hat{\mrm{f}}_{s}\pars{\hat{k}} & = {1 \over 2}\,\sec\pars{\pi s \over 2}\bracks{% \sin\pars{\pi s}\Gamma\pars{s}\,{}_{1}\mrm{F}_{2}\pars{1;{1 \over 2} - {s \over 2}, 1 - {s \over 2};{k^{2} \over 4}} - \pi\verts{k}^{s}\sinh\pars{\verts{k}}} \\[5mm] & \mbox{with}\quad-3 < s < 1 \end{align}

${}_{1}\mrm{F}_{2}$ is the Generalized Hypergeometric PFQ.