If I ask Mathematica to compute the Fourier transform of $\frac{1}{x^2}$ using the FourierTransform function, it gives me a result of

$$\mathcal{F}\biggl[\frac{1}{x^2}\biggr] = -\sqrt{\frac{\pi}{2}}k\operatorname{sign}k\tag{1}$$

Similarly if I ask for the Fourier transform of $\frac{1}{x^2 + y^2}$, I get

$$\mathcal{F}\biggl[\frac{1}{x^2 + y^2}\biggr] = -\gamma_E + \ln\frac{1}{\sqrt{k_x^2 + k_y^2}}\tag{2}$$

(Mathematica claims this is true as long as $k_x > 0$; I'm not convinced that's necessary, but I'm happy to accept that restriction for the scope of this question.)

But the integrals I would normally expect to do to obtain these results, namely

$$\begin{align} &\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} \frac{e^{ikx}}{x^2}\mathrm{d}x & &\frac{1}{2\pi}\iint_{\mathbb{R}^2} \frac{e^{i(k_x x + k_y y)}}{x^2 + y^2}\mathrm{d}x\,\mathrm{d}y \end{align}$$

don't converge. So when Mathematica gives me the results (1) and (2), what is it really telling me? In what sense are those functions the Fourier transforms of $1/x^2$ and $1/(x^2 + y^2)$? How would I derive these results?

I'm assuming that there is some mathematical meaning to the results, and it's not just something that Mathematica "decided" to do arbitrarily (but of course if that's not the case I'd like to know).


The Fourier transform is defined as a certain integral on absolutely integrable functions $L^1(\mathbb{R}^n)$, as you may know. $L^1(\mathbb{R}^n)$ fits inside a space of "generalized functions" called tempered distributions. Examples of tempered distributions are the dirac delta function and locally integrable functions with algebraic growth at infinity. It turns out that the Fourier transform extends continuously to this bigger space.

I probably should have just directed you here: http://en.wikipedia.org/wiki/Fourier_transform#Tempered_distributions

| cite | improve this answer | |
  • 1
    $\begingroup$ Yes, but $1/x^2$ is not a tempered distribution. You need to very careful with the interpretation here. $\endgroup$ – mrf Jun 17 '14 at 9:36
  • $\begingroup$ Good point. Then again I never said anything about how you identify functions with distributions, just that it can be done. $\endgroup$ – mathematician Jun 17 '14 at 20:22

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align}&\partiald[2]{}{k}\int_{-\infty}^{\infty}{\expo{\ic kx} \over x^{2}} \,{\dd x \over \root{2\pi}} =-\int_{-\infty}^{\infty}\expo{\ic kx}\,{\dd x \over \root{2\pi}} =-\root{2\pi}\delta\pars{k} \end{align}

\begin{align}&\partiald{}{k}\int_{-\infty}^{\infty}{\expo{\ic kx} \over x^{2}} \,{\dd x \over \root{2\pi}} =-\root{2\pi}\Theta\pars{k} \end{align}

\begin{align}&\int_{-\infty}^{\infty}{\expo{\ic kx} \over x^{2}} \,{\dd x \over \root{2\pi}} =-\root{2\pi}k\ \overbrace{\Theta\pars{k}}^{\ds{1 + \sgn{k} \over 2}} \end{align}

| cite | improve this answer | |
  • 2
    $\begingroup$ How do you justify the first steps? Note that $1/x^2$ is not even a distribution without some careful interpretation. $\endgroup$ – mrf Jun 17 '14 at 9:37
  • $\begingroup$ This does actually help, since it provides a decent justification for the results in physicist-math. Though I am also interested in the other part of the question, about the proper mathematical interpretation of the result. $\endgroup$ – David Z Jun 17 '14 at 16:56
  • $\begingroup$ I just put that to "illustrate" the $\tt Mathematica$ approach. $\tt Mathematica$ doesn't say too much about its integration which is a quite bad signal. Instead, we should study the equation $\large\displaystyle{\partial^{2}{\rm G}\left(k\right) \over \partial k^{2}} = -\sqrt{2\pi}\ \delta\left(k\right)$ in order "to guess" what conditions $\tt Mathematica$ imposes in its calculation. $\endgroup$ – Felix Marin Jun 17 '14 at 17:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.