What does the Fourier transform of $1/x^2$ mean?

Question

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).

Answer 1

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

Answer 2

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