3
$\begingroup$

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.

$\endgroup$
4
  • 2
    $\begingroup$ 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? $\endgroup$
    – LL 3.14
    Commented Oct 16, 2022 at 20:44
  • 1
    $\begingroup$ Did you mean $x\in \Bbb{R}$? If so then up to simple stuffs it is an incomplete gamma function. $\endgroup$
    – reuns
    Commented Oct 16, 2022 at 21:27
  • $\begingroup$ Yes $x \in \mathbb{R}.$ $\endgroup$
    – Sia-TeX
    Commented Oct 17, 2022 at 19:04
  • $\begingroup$ 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. $\endgroup$
    – Sia-TeX
    Commented Oct 17, 2022 at 19:07

1 Answer 1

1
$\begingroup$

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


\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.
$\endgroup$
3
  • $\begingroup$ Thank you very much. What is 1F_2 notation you have in the answer? $\endgroup$
    – Sia-TeX
    Commented Oct 17, 2022 at 19:09
  • $\begingroup$ $\displaystyle _{1}\!\operatorname{F}_{2}$ is a hypergeometric function $\endgroup$ Commented Oct 18, 2022 at 4:30
  • $\begingroup$ 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? $\endgroup$
    – Sia-TeX
    Commented Oct 20, 2022 at 7:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .