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.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$
\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.