Definition of $\frac{1}{|\xi|^2}$ as Distribution in $\mathbb{R}^2$

Question

I'm trying to understand the Fundamental solution of the Laplacian in two dimensions using Fourier Transform with Definition $$\mathcal{F}(f)(\xi) = \frac{1}{(2\pi)^{n/2}} \int_{x \in \mathbb{R}^n} f(x) e^{-i \xi \cdot x} dx .$$ So in theory all I have to do is calculate the (inverse) Fourier transform of $f(\xi) = \frac{1}{|\xi|^2}$, but how is $T_f(\varphi)$ defined, if $f$ is not locally integrable? Like if $\varphi(\xi) \geq 0 $ and $\varphi(\xi) = 1$ if $\xi \leq 1$ then the Integral over the unit ball does not converge.

Answer 1

$\bullet$ You can define it as the derivative in the sense of distributions of a locally integrable function, i.e. you can define the distribution $T$ by $$ T := \mathrm{pf}\left(\frac{1}{|x|^2}\right) := \mathrm{div}\left(\frac{\ln(|x|)}{|x|^2} \,x\right). $$ Then, $T(x) = \tfrac{1}{|x|^2}$ if $x≠ 0$ and as already detailed in my answer in The Fourier transform of $1/p^3$, it acts on test functions as $$ \begin{align*} \langle T,\varphi\rangle &= \int_{|x|\leq 1} \frac{\varphi(x)-\varphi(0)}{|x|^2}\,\mathrm{d}x + \int_{|x|> 1} \frac{\varphi(x)}{|x|^2} \,\mathrm{d}x, \end{align*} $$ and its Fourier transform in your convention is then $$ \mathcal{F}\left(\frac{1}{|x|^2}\right) = \ln(2)-\gamma - \ln(|x|). $$

$\bullet$ In your case, it is important to notice that this is indeed an inverse of $|x|^2$ since $$ \begin{align*} \langle |x|^2\,T,\varphi\rangle &= \langle T,|x|^2\,\varphi\rangle \\ &= \int_{|x|\leq 1} \frac{|x|^2\,\varphi(x)-|0|^2\varphi(0)}{|x|^2}\,\mathrm{d}x + \int_{|x|> 1} \frac{|x|^2\,\varphi(x)}{|x|^2} \,\mathrm{d}x \\ &= \int_{\mathbb R^2} \varphi(x)\,\mathrm d x = \langle 1,\varphi\rangle \end{align*} $$ so $|x|^2\,T = 1$ in the sense of distributions. Notice however that it is not the unique solution, as one can add $a\,\delta_0 + b\cdot\nabla\delta_0$. This is not a problem when solving the Laplace equation since solutions of the Laplace equation are also defined up to $a + b·x$.