Fourier Transform of $\frac{1}{|x|^\delta}$

I am looking for the Fourier transform of the following function:

$$$${\rm f}\left(x\right)= \begin{cases} 1, & \mbox{for} \hspace{1 mm} \left\vert x\right\vert < 1, \\[2mm] \frac{1}{\rule{0pt}{4mm}\left\vert x\right\vert^{\,\delta}} & \text{ otherwise, } \end{cases}$$$$

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.

• I think there will be no simple formula, except things such as $$\mathcal F(f) = \frac{1}{|x|^{d-\delta}} * J + \delta_0 - J$$ where $J = \mathcal F(\mathbf{1}_{B_1})$ is a Bessel function and $\mathcal F(\mathbf{1}_{B_1^c}) = \delta_0 - J$, but I suppose you are more looking for properties of this function more than an exact formula, right? If yes, what property would you like to know? Commented Oct 16, 2022 at 20:44
• Did you mean $x\in \Bbb{R}$? If so then up to simple stuffs it is an incomplete gamma function. Commented Oct 16, 2022 at 21:27
• Yes $x \in \mathbb{R}.$ Commented Oct 17, 2022 at 19:04
• I am interested to see what are differnces between this function and a function like exp(x/X) for some large X, for example. Do we expect the FT to be supported very close to zero, as well, etc. Commented Oct 17, 2022 at 19:07

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ $$\ds{\bbox[5px,#ffd]{}}$$
\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.
• $\displaystyle _{1}\!\operatorname{F}_{2}$ is a hypergeometric function Commented Oct 18, 2022 at 4:30
• Ok, the thing is I need to know, what is $\hat{f}$ near zero when I change $\delta$. For example, what is $\hat{f}(1/X)$ when $\delta$ is like $1/\sqrt{X}.$ So how can I use your formula for these types of calculation? Commented Oct 20, 2022 at 7:12