Mathematica gives Fourier transform of $\frac{1}{x^2}$ as $-\pi t \operatorname{sgn}(t)$ or $-\pi|t|$. At $t=0$ it is zero. But how this can be the case, given that at zero it should go to infinity because the function is even and its integral from $-\infty$ to $\infty$ is infinite?
[Math] Fourier transform of $\frac{1}{x^2}$, why it is zero at zero
fourier analysisfourier transformintegral-transforms
Best Answer
Since $1/x^2$ is not a locally integrable function, it does not define a tempered distribution and therefore does not have a Fourier transform of its own. Instead one considers a tempered distribution that agrees with $1/x^2$ when $x\ne 0$. A natural choice is to begin with $f(x)=-\log|x|$, which is a tempered distribution, and let $T = f''$ in the distributional sense, meaning that $$ T(\phi) = \int_{\mathbb{R}} -\log|x| \phi''(x)\,dx$$ for any test function $\phi$. The tempered distribution $T$ looks like $1/x^2$ outside of $0$, meaning that if $0\notin \operatorname{supp} \phi$, then (via integration by parts) $$T(\phi) = \int_{\mathbb{R}} \frac{1}{x^2}\phi(x)\,dx$$ Yet in other aspects $T$ is not like $1/x^2$, for example it is not true that $T(\phi)\ge 0$ for nonnegative $\phi$.
Think of $T$ as $1/x^2$ de-fanged; regularization takes out the singularity at $0$ without changing the function for $x\ne 0$.
The Fourier transform of $\log|x|$ can be computed as here. Taking derivative twice amounts to multiplying the Fourier transform it by $\xi^2$ (possibly with some $2\pi$ in there, depending on FT convention). This leads to the result that Mathematica returns.