Assume that $$\hat f(x)= (2\pi)^{-n/2} \int_{\mathbb{R}^n} f(y) e^{-i\left<x,y\right>} dy$$ is the Fourier transform of a function $f$. What is $\hat f$ if $f(x)=|x|^{2-n}$?
-
$f(x)=|x|^{2-n}$ is not even integrable. – saz Nov 05 '14 at 19:16
2 Answers
Polar coordinates. $\int_{\mathbb R^n} |x|^{k} dx= \int_{S^{n-1}} \int_{0}^{\infty} r^{n-1}\cdot r^{k} \enspace dr d\theta$.

- 5,648
If we are not seriously interested in dismissing at numerically-divergent integrals... :) ... then we can evaluate this as a Fourier transform of a tempered distribution.
The general pattern is that the FT of $1/|x|^s$ on $\mathbb R^n$ is a constant multiple (depending on $s$) of $1/|x|^{n-s}$. When that "constant" has poles or zeros in $s$, then the interpretation is a little more complicated, but is standard. (We can have rotationally invariant residues of $1/|x|^s$, which are derivatives $\Delta^\ell \delta$ of Dirac delta...)
So, here, the FT of $|x|^{2-n}=1/|x|^{n-2}$ is a constant multiple of $1/|x|^2$. Indeed, for $n\ge 2$, the latter is locally integrable (and the constant is neither $0$ nor blows up...)
The constant can be determined by integrating against $e^{-\pi|x|^2}$, conveniently its own FT. :)

- 52,465
-
+1. Addenda: In dimension $1$ it is done here https://math.stackexchange.com/questions/3765751/what-is-the-fourier-transform-of-x/3765807. In dimension $2$, it is the Dirac delta. In higher dimensions, this is done for example in a lot of textbooks, such as the book "Functional Analysis" from Lieb and Loss. – LL 3.14 Aug 08 '22 at 10:09