Fourier Transform of $\frac{1}{|x|^\delta}$ I am looking for the Fourier transform of the following function:
\begin{equation}
{\rm f}\left(x\right)= \begin{cases} 
1, & \mbox{for} \hspace{1 mm}
\left\vert x\right\vert < 1, 
\\[2mm] 
\frac{1}{\rule{0pt}{4mm}\left\vert x\right\vert^{\,\delta}} & \text{ otherwise, }
\end{cases}
\end{equation}
where $\delta>0.$ Is there a neat formula for $\hat{\rm f}\ ?$.
If $\delta <1$ then $\hat{\rm f}$ would blow up at zero, in that case we can use smooth function like $\exp\left(-\left(x/X\right)^{2}\right)$, for some large $X$, and look for the Fourier transform of
$$
\exp\left(-\,\left(x \over X\right)^{2}\right)\
{\rm f}\left(x\right).
$$
I am specially interested when $\delta$ is small, even approaching zero.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
\hat{\on{f}}\pars{k} & \equiv
\color{#44f}{\int_{-\infty}^{\infty}\on{f}\pars{x}\expo{\ic kx}\dd x}
\\[5mm] & =
\int_{-\infty}^{-1}
{1 \over \verts{x}^{\delta}}\expo{\ic kx}\dd x +
\int_{-1}^{1}\expo{\ic kx}\dd x +
\int_{1}^{\infty}
{1 \over \verts{x}^{\delta}}\expo{\ic kx}\dd x
\\[5mm] & =
2\,{\sin\pars{k} \over k} +
2\int_{1}^{\infty}
{\cos\pars{kx} \over x^{\delta}}\dd x
\\[5mm] & =\quad
\bbx{\color{#44f}{\begin{array}{l}
\ds{2\,{\sin\pars{k} \over k} -
2\,{_{1}\!\on{F}_{\,2}\pars{%
\left.\begin{array}{c}
\ds{1/2 - \delta/2}
\\
\ds{1/2\quad 3/2 - \delta/2}
\end{array}\right\vert -k^{2}/4} \over 1 - \delta}}
\\[2mm]
\ds{-2\,{\Gamma\pars{1 - \delta} \over \verts{k}^{1 - \delta}}\sin\pars{{\pi \over 2}\delta}}
\end{array}}} \\ &
\end{align}
The last integration was evaluated with a CAS.
