Mathematica gives $-k \sqrt{\frac{\pi}{2}} \text{sgn}(k)$ as the Fourier transform of $1/x^2$ (i.e., the result of the command FourierTransform[1/x^2, x, k]). And the Fourier transform of $-k \sqrt{\frac{\pi}{2}} \text{sgn}(k)$ is $1/x^2$ (FourierTransform[-k Sqrt[Pi/2] Sign[k], k, x]). I found this by chance while playing with Mathematica, but I cannot understand the result. The definition of Fourier transform $\int_{-\infty}^\infty f(x) e^{-ikx} dx$ does not converge for the functions, and even Mathematica reports that the integrals does not converge. I read the documentation of FourierTransform but could not find any relevant information. In what sense are they Fourier transform of each other?
I found an almost same question (What does the Fourier transform of $1/x^2$ mean?), but the answers looks not so complete. The answers mention "tempered distribution" and so I read some basic materials about it, but $1/x^2$ is not a tempered distribution as some comments on the answers already pointed. Is the notion of the Fourier transform of tempered distributions really relevant to my question?
Best Answer
As you noticed, the function $\frac{1}{x^2}$ cannot be identified as a tempered distribution in the usual way as it is not locally integrable.
Here the symbol "$\frac{1}{x^2}$" has another meaning in the context of tempered distributions, which is formalized by the concept of Hadamard finite-part integral.
That is we define a linear map \begin{align*}\frac{1}{x^2}:\mathcal{S}(\mathbb{R})&\to \mathbb{R}\\ \varphi&\mapsto \lim_{\varepsilon \to 0}\left[\int_{|x|\geq \varepsilon}\frac{\varphi(x)}{x^2}dx-2\frac{\varphi(0)}{\varepsilon}\right] \end{align*} The function is well defined: since $\varphi$ is smooth by the mean value theorem we may write $$\varphi(x)=\varphi(0)+x\varphi'(\xi(x)),\qquad \xi(x)\in [0,x] $$ so that \begin{align*}\int_{|x|\geq \varepsilon}\frac{\varphi(x)}{x^2}dx-2\frac{\varphi(0)}{\varepsilon}&=\varphi(0)\left(\int_{|x|\geq \varepsilon}\frac{1}{x^2}dx-\frac{2}{\varepsilon}\right)+\int_{|x|\geq \varepsilon}\frac{\varphi'(\xi(x))}{x}dx=\\ &=\int_{|x|\geq \varepsilon}\frac{\varphi'(\xi(x))}{x}dx \end{align*} and applying mean value again, with $\varphi'(\xi(x))=\varphi'(0)+\xi(x)\varphi''(\eta(x))$, $$\int_{|x|\geq \varepsilon}\frac{\varphi'(\xi(x))}{x}dx=\varphi'(0)\int_{|x|\geq \varepsilon}\frac{1}{x}dx+\int_{|x|\geq \varepsilon}\varphi''(\eta(x))\frac{\xi(x)}{x}dx=\int_{|x|\geq \varepsilon}\varphi''(\eta(x))\frac{\xi(x)}{x}dx $$ Since $\xi(x)\in [0,x]$ for all $x$, we have $\frac{\xi(x)}{x}\in [0,1]$ and so the last integral converges as $\varepsilon \to 0$, so that $$\frac{1}{x^2}(\varphi)=\int_{-\infty}^{+\infty}\varphi''(\eta(x))\frac{\xi(x)}{x}dx $$
Now to prove that the map $\frac{1}{x^2}$ is a tempered distribution, suppose $\varphi_n\to 0$ in $\mathcal{S}(\mathbb{R})$. Then in particular $\varphi_n''(\eta(x))\to 0$ in $L^1(\mathbb{R})$, and so $$\left|\frac{1}{x^2}(\varphi_n)\right|\leq \int_{-\infty}^{+\infty}|\varphi_n''(\eta(x))|dx\to 0$$
and therefore its Fourier transform is well-defined as the Fourier transform of a tempered distribution.