Vtyjkyuk

Mathematica gives $-k sqrtfracpi2 textsgn(k)$ 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 $-k sqrtfracpi2 textsgn(k)$ 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?

As you noticed, the function $frac1x^2$ cannot be identified as a tempered distribution in the usual way as it is not locally integrable.

Here the symbol "$frac1x^2$" 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
beginalign*frac1x^2:mathcalS(mathbbR)&to mathbbR\
varphi&mapsto lim_varepsilon to 0left[int_fracvarphi(x)x^2dx-2fracvarphi(0)varepsilonright]
endalign*
The function is well defined: since $varphi$ is smooth by the mean value theorem we may write
$$varphi(x)=varphi(0)+xvarphi'(xi(x)),qquad xi(x)in [0,x] $$
so that
beginalign*int_fracvarphi(x)x^2dx-2fracvarphi(0)varepsilon&=varphi(0)left(int_frac1x^2dx-frac2varepsilonright)+int_fracvarphi'(xi(x))xdx=\
&=int_fracvarphi'(xi(x))xdx
endalign*
and applying mean value again, with $varphi'(xi(x))=varphi'(0)+xi(x)varphi''(eta(x))$,
$$int_fracvarphi'(xi(x))xdx=varphi'(0)int_frac1xdx+int_varphi''(eta(x))fracxi(x)xdx=int_varphi''(eta(x))fracxi(x)xdx $$
Since $xi(x)in [0,x]$ for all $x$, we have $fracxi(x)xin [0,1]$ and so the last integral converges as $varepsilon to 0$, so that
$$frac1x^2(varphi)=int_-infty^+inftyvarphi''(eta(x))fracxi(x)xdx $$

Now to prove that the map $frac1x^2$ is a tempered distribution, suppose $varphi_nto 0$ in $mathcalS(mathbbR)$. Then in particular $varphi_n''(eta(x))to 0$ in $L^1(mathbbR)$, and so
$$left|frac1x^2(varphi_n)right|leq int_-infty^+infty|varphi_n''(eta(x))|dxto 0$$

and therefore its Fourier transform is well-defined as the Fourier transform of a tempered distribution.

I will here use the non-unitary, angular frequency Fourier transform defined as
$$hatu(nu) = mathcalF u(x) = int_-infty^infty u(x) , e^-i nu x , dx.$$

By first definition of the distribution $frac1x^2$ and then rule 106
$$
mathcalFleft frac1x^2 right
= mathcalFleft left( -frac1x right)' right
= -inu , mathcalFleft frac1x right.
$$

Here $frac1x$ is defined as the unique odd distribution satisfying $x frac1x = 1.$ Therefore, by rule 107, we have
$$
2pi , delta(nu)
= mathcalFleft 1 right
= mathcalFleft x frac1x right
= ifracddnu mathcalF left frac1x right
$$
so
$$
mathcalF left frac1x right = -i , 2pi H(nu) + C
$$
for some constant $C.$ Here $H$ is the Heaviside step function.

Since $frac1x$ is odd so must be $mathcalFleftfrac1xright.$ Therefore $C = i , pi$ and
$$
mathcalF left frac1x right
= -i , 2pi H(nu) + i pi
= -i pi operatornamesign(nu),
$$
where $operatornamesign$ is the signature function.

Thus,
$$
mathcalFleft frac1x^2 right
= -inu , mathcalFleft frac1x right
= -inu , left( -ipi operatornamesign(nu) right)
= -pi , nu , operatornamesign(nu)
= -pi , |nu|.
$$