Computing Fourier transform of $|x|^{-\alpha}$ in $\mathbb{R}^n$ by approximating it with Schwartz functions

Question

Motivated by questions/answers in here (1) and here (2), I am interested in understanding whether there is a reasonable method of computing Fourier Transforms of $|x|^{-\alpha}$ by approximating this function (and the corresponding tempered distribution) using functions form the Schwartz space.

For example, in answer here (1), it is shown using the usual definition of Fourier Transform of tempered distribution that for $0 < \alpha < n$,

$$ F(|x|^{-\alpha}) = \frac{\Gamma(\frac{n-\alpha}{2})}{\Gamma(\frac{n}{2})} 2^{n-\alpha}\pi^{n/2} \frac{1}{|k|^{n-\alpha}}. $$

However, in answer here (2), in order to compute Fourier transform of $f(x) = \log(|x|)$, instead of the distributional Fourier transform definition, one took a family of functions $f_{\varepsilon} = e^{-\varepsilon|x|}\log(|x|)$ such that $f_{\varepsilon} \to f$ as $\varepsilon \to 0$ (even though, functions $f_{\varepsilon}$ are not Schwartz as they are not smooth). Then, Fourier transform is computed for all test functions $\phi$ as $\lim \limits_{\varepsilon \to 0} \langle F(f_{\varepsilon}), \phi \rangle$.

So, my question is whether it is possible to find $f_{\varepsilon}$ in Schwartz space such that $f_{\varepsilon} \to |x|^{-\alpha}$ as $\varepsilon \to 0$, and it is easy (practically) to compute Fourier transforms of $f_{\varepsilon}$ and easy (practically) to compute $\lim \limits_{\varepsilon \to 0} \langle F(f_{\varepsilon}), \phi \rangle$? If yes, what is the intuition for the construction of $f_{\varepsilon}$? If not, why?

You might wonder why not just use known results obtained by using Fourier transform definition of tempered distributions? I am interested in this as such way of "approximating" distribution is often encountered in physics, and I think this was the initial motivation for the theory of tempered distributions. Also, I am interested for $f_{\varepsilon}$ being in Schwartz space and not just $L^1$ (which would make Fourier transform well defined) because I feel like definitions of tempered distributions are always motivated assuming approximations by Schwartz space.

Answer 1

Nice question.

In fact if you look carefully at the proof you referenced as here (1), you will see that the approach is entirely based on the representation known in physics as the Schwinger trick $$ \frac{1}{|x|^\alpha}= \frac{1}{\Gamma(\frac{\alpha}{2})}\int_0^\infty\frac{dt}{t}t^{\frac{\alpha}{2}}e^{-t|x|^2}\ . $$ If you let $$ f_{\varepsilon}(x)= \frac{1}{\Gamma(\frac{\alpha}{2})}\int_{\varepsilon}^{\frac{1}{\varepsilon}}\frac{dt}{t}t^{\frac{\alpha}{2}}e^{-t|x|^2} $$ then $f_{\varepsilon}$ gives you an expample of approximation of an element of $\mathscr{S}'(\mathbb{R}^n)$ by elements of the sequentially dense subspace $\mathscr{S}(\mathbb{R}^n)$.

In physics terminology the $1/\varepsilon$ upper bound in the integral is an ultraviolet cutoff whereas the $\varepsilon$ lower bound is an infrared cutoff.

A key point also here is that if you know the Fourier transform of the Gaussian, then you know the Fourier transform of any function or temperate distribution. This is because you can approximate them with linear combinations of translates and rescalings of Gaussians. The same principle works when doing analysis over $p$-adic fields with the indicator of $\mathbb{Z}_p$ replacing the Gaussian.