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

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.

• Note on distributions: the initial motivation of tempered distributions was to extend the Fourier transform to distributions. The motivation of distributions was to build a unified framework for both the generalized functions used in physics and mathematics, but also as a generalization of methods used to give a sense to partial differential equations. Commented Mar 8, 2022 at 4:09
• Your question as it is is quite vague as there are a lot of ways to approximate a function by a Schwartz function ... just doing a convolution by a Gaussian usually gives a Schwartz function for example. Notice that the Fourier transform of a power function follows without a lot of computations for a reason of scaling math.stackexchange.com/questions/3742640/… Commented Mar 8, 2022 at 4:10
• @LL3.14 Thanks for your comments! I understand that my question is a bit vague, but I am interested to see at least one example (which should be as simple as possible / requiring as less of a technical knowledge as possible) where Fourier transform of mentioned distribution is computed by using approximate functions as described in the question. Thanks for the example of scaling Commented Mar 8, 2022 at 5:39
• Then just do a convolution with a Gaussian converging to a Dirac delta and you have your example ... Commented Mar 8, 2022 at 6:41
• @LL3.14 Thanks for comments. It seems that you imply that such calculation is easy and obvious, but, unfortunately, I am very new to this subject and confused. Can you please help me how to show that your proposed convolution is indeed a Schwartz function and how to take Fourier transform of it? I believe we can't use result on convolution of functions as I would have to know Fourier transform of my function in the first place. Would appreciate your time Commented Mar 12, 2022 at 0:08

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.
• Thanks for your answer, appreciate your time! It is not obvious to me why your proposed $f_{\varepsilon}$ is actually in Schwartz space. Can you please help me with that? Also, you say that "if you know the Fourier transform of the Gaussian, then you know the Fourier transform of any temperate distribution". It is very difficult to believe in such a strong result. Do you have some reference on that, or maybe you can expand on that in your answer? Commented Mar 12, 2022 at 0:11