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?

  • $\begingroup$ According to Wiki, $1/x^2$ is defined in this context as the distributional derivative $$ -\frac{d^2}{dx^2}\log|x|. $$ It also refers to two books that should contain the derivation of this transform, namely Erdélyi (1954), and Kammler (2000). $\endgroup$ – MSDG Aug 31 '18 at 7:36
  • $\begingroup$ @Sobi Thank you. Then the notation $1/x^2$ should not be interpreted as a usual function $\mathbb{R} \setminus 0 \to \mathbb{R} : x \mapsto 1/x^2$. The distribution you gave is different from the usual fuction $1/x^2$ in the sense of equality between distributions. Am I right? $\endgroup$ – Balbadak Aug 31 '18 at 7:54
  • $\begingroup$ They should coincide in the distributional sense, I think. $\endgroup$ – MSDG Aug 31 '18 at 8:02
  • $\begingroup$ @Sobi Do you mean $\int_{-\infty}^\infty \frac{1}{x^2} \phi(x) dx = \left< -\frac{d^2}{dx^2} \log|x|, \phi \right>$ with test function $\phi$? $\endgroup$ – Balbadak Aug 31 '18 at 8:05
  • $\begingroup$ I don't feel confident enough to answer, but I would guess that $$ -\left\langle \frac{d^2}{dx^2}\log|x|, \phi \right\rangle = \lim_{\epsilon \to 0^+} \int_{|x|>\epsilon} \frac{1}{x^2} \phi(x) \, dx.$$ $\endgroup$ – MSDG Aug 31 '18 at 8:10

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.

  • $\begingroup$ How to extend this reasoning for any natural number $n$ to find the Fourier Transform of $\frac{1}{x^n}$? $\endgroup$ – Dr Potato Jul 17 '20 at 16:04

I will here use the non-unitary, angular frequency Fourier transform defined as $$\hat{u}(\nu) = \mathcal{F}\{ u(x) \} = \int_{-\infty}^{\infty} u(x) \, e^{-i \nu x} \, dx.$$

By first definition of the distribution $\frac{1}{x^2}$ and then rule 106 $$ \mathcal{F}\left\{ \frac{1}{x^2} \right\} = \mathcal{F}\left\{ \left( -\frac{1}{x} \right)' \right\} = -i\nu \, \mathcal{F}\left\{ \frac{1}{x} \right\}. $$

Here $\frac{1}{x}$ is defined as the unique odd distribution satisfying $x \frac{1}{x} = 1.$ Therefore, by rule 107, we have $$ 2\pi \, \delta(\nu) = \mathcal{F}\left\{ 1 \right\} = \mathcal{F}\left\{ x \frac{1}{x} \right\} = i\frac{d}{d\nu} \mathcal{F} \left\{ \frac{1}{x} \right\} $$ so $$ \mathcal{F} \left\{ \frac{1}{x} \right\} = -i \, 2\pi H(\nu) + C $$ for some constant $C.$ Here $H$ is the Heaviside step function.

Since $\frac{1}{x}$ is odd so must be $\mathcal{F}\left\{\frac{1}{x}\right\}.$ Therefore $C = i \, \pi$ and $$ \mathcal{F} \left\{ \frac{1}{x} \right\} = -i \, 2\pi H(\nu) + i \pi = -i \pi \operatorname{sign}(\nu), $$ where $\operatorname{sign}$ is the signature function.

Thus, $$ \mathcal{F}\left\{ \frac{1}{x^2} \right\} = -i\nu \, \mathcal{F}\left\{ \frac{1}{x} \right\} = -i\nu \, \left( -i\pi \operatorname{sign}(\nu) \right) = -\pi \, \nu \, \operatorname{sign}(\nu) = -\pi \, |\nu|. $$

  • $\begingroup$ How to extend this reasoning for any natural number $n$ to find the Fourier Transform of $\frac{1}{x^n}$? $\endgroup$ – Dr Potato Jul 17 '20 at 15:52
  • 1
    $\begingroup$ You missed to add a constant multiple of the Dirac Delta at zero in the final result. $\endgroup$ – Dr Potato Jul 17 '20 at 16:03
  • $\begingroup$ @DrPotato. I don't think that I have missed any term. What makes you think that I have missed a term? $\endgroup$ – md2perpe Jul 17 '20 at 16:19
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    $\begingroup$ @DrPotato. Since $\frac{1}{x^n} = \frac{(-1)^{n-1}}{(n-1)!} \frac{d^{n-1}}{dx^{n-1}} \left( \frac{1}{x} \right)$ (did I get this right?) one has $$\mathcal{F}\left\{ \frac{1}{x^n} \right\} = \frac{(-1)^{n-1}}{(n-1)!} (i\nu)^{n-1} \mathcal{F}\left\{ \frac{1}{x} \right\}.$$ $\endgroup$ – md2perpe Jul 17 '20 at 16:24
  • $\begingroup$ The correct result is $2\pi\delta(0)-\pi|\nu|$ $\endgroup$ – Anixx May 8 at 18:25

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