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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?

Anixx
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    In order to take the Fourier transform of $f(x)=1/x^2$ one must interpret it as a tempered distribution. The integral $\int_{-\infty}^\infty|f(x)|,dx$ is divergent so one cannot expect the usual formula to give the distributional Fourier transform. – Angina Seng Apr 21 '17 at 09:51
  • @Lord Shark the Unknown $\int_{-\infty}^\infty 1 dx$ is also divergent, yet the Fourier transform gives Dirac delta which is infinite at zero. – Anixx May 03 '17 at 08:27
  • Indeed, these are also tempered distributions. – Angina Seng May 03 '17 at 16:50
  • See THIS ANSWER, which evaluates the Fourier Transform of $|x|^\alpha$ for all real values of $\alpha$ including $\alpha =-2$. The issue is that we need a distribution for $|x|^\alpha$, $\alpha\le -1$. For $\alpha=-2$ we can use the following $$\langle |x|^{-2}, \phi \rangle \equiv \int_{-\infty}^\infty \frac{\phi(x) -\left(\phi(0)+\phi'(0)x\right)\xi_{[-1,1]}(x)}{|x|^2},dx$$ – Mark Viola May 08 '21 at 17:31

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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.