The Fourier transform of $e^{2 \pi i / x}$ makes sense as a distribution, I believe. Does it have a nice expression in terms of functions and well-known distributions (e.g. Dirac delta)?

  • 2
    $\begingroup$ Since the limit as $x\to\infty$ is $1$, I suggest putting the constant $1$ aside (which contributes Dirac delta), and working with $e^{2\pi i/x}-1$, which is a (discontinuous) $L^2$ function. So its transform is also in $L^2$, although it's not in $L^1$... I've little hope for an explicit form of it. $\endgroup$ – user147263 Jan 6 '16 at 8:28

We want to compute

$$\int_{-\infty}^{\infty} dx \, e^{i 2 \pi/x} e^{i 2 \pi \nu x} $$

Let's consider the following integral in the complex plane:

$$\oint_C dz \, e^{-\left (2 \pi \nu z + \frac{2 \pi}{z} \right )} $$

where $C$ is a circular wedge of radius $R$ and angle $90$ degrees in the upper right quadrant. By Cauchy's theorem, the integral is zero. However, the integral is also equal to

$$\int_0^{R} dx \, e^{-\left (2 \pi \nu x + \frac{2 \pi}{x} \right )} + i R \int_0^{\pi/2} d\theta \, e^{i \theta} \, e^{- 2 \pi \nu R e^{i \theta} - 2 \pi e^{-i \theta}/R}+ i \int_R^0 dy \, e^{i 2 \pi (-\nu y + 1/y)} = 0$$

We take the limit as $R \to \infty$. The second integral may be evaluated as follows:

$$R \int_0^{i} dz \, e^{-2 \pi \nu R z} = \frac1{2 \pi \nu} \left (1-e^{-i 2 \pi \nu R} \right )$$

which is a distribution in the limit as $R \to \infty$. In this limit, the first integral has been done in these pages, for example here, and is equal to

$$\int_0^{\infty} dx \, e^{-\left (2 \pi \nu x + \frac{2 \pi}{x} \right )} = \frac{2}{\sqrt{\nu}} \operatorname{K_1}{\left (4 \pi \sqrt{\nu}\right )}$$

Thus, we have

$$i \int_0^{\infty} dy \, e^{i 2 \pi (-\nu y + 1/y)} = \frac{2}{\sqrt{\nu}} \operatorname{K_1}{\left (4 \pi \sqrt{\nu}\right )} + \lim_{R \to \infty} \frac1{2 \pi \nu} \left (1-e^{-i 2 \pi \nu R} \right )$$

Now a little bit of a dirty trick. Let's reverse the sign of $\nu$ and use the fact that

$$\operatorname{K_1}{(i x)} = \frac{\pi}{2} \left [Y_1(x) - i J_1(x) \right ] $$

We thus have

$$-i \int_0^{\infty} dy \, e^{i 2 \pi (\nu y + 1/y)} = \frac{\pi}{\sqrt{\nu}} \left [-Y_1 \left (4 \pi \sqrt{\nu}\right ) + i J_1 \left (4 \pi \sqrt{\nu}\right ) \right ] + \lim_{R \to \infty} \frac1{2 \pi \nu} \left (1-e^{i 2 \pi \nu R} \right )$$

Now take the imaginary part of both sides and get

$$\int_0^{\infty} dy \, \cos{\left [2 \pi (\nu y + 1/y)\right ]} = -\frac{\pi}{\sqrt{\nu}} J_1 \left (4 \pi \sqrt{\nu}\right )+ \lim_{R \to \infty} \frac1{2 \pi \nu} \sin{2 \pi \nu R}$$

Noting that

$$\delta(t) = \lim_{R \to \infty} \frac{\sin{R t}}{\pi t} $$

The FT we seek is twice the LHS, or

$$\int_{-\infty}^{\infty} dx \, e^{i 2 \pi/x} e^{i 2 \pi \nu x} = \delta(\nu) - 2 \pi \frac{J_1 \left (4 \pi \sqrt{\nu}\right )}{\sqrt{\nu}}$$

As a reminder, $J_1$ is the Bessel function of the first kind of first order.


The contour $C$ I described technically should avoid the essential singularity at the origin. However, a quick computation verifies that the deformation of $C$ to a small quarter-circle of radius $\epsilon$ about the origin vanishes in the limit as $\epsilon \to 0$, so we are OK.


The original answer had an error in that I had misplaced a factor of $i$ in the middle of the calculation. The result here is now correct and verified to be so.

  • $\begingroup$ And last time I checked, $2 \pi \delta(2 \pi \nu) = \delta(\nu)$, simplifying it just a bit more. $\endgroup$ – Marty Jan 6 '16 at 11:35
  • $\begingroup$ Thanks! Exactly the sort of clean answer I was hoping for. $\endgroup$ – Marty Jan 6 '16 at 11:36
  • $\begingroup$ @Ineptigrator: Thanks for pointing that out. $\endgroup$ – Ron Gordon Jan 6 '16 at 12:35

A partial solution :

As Normal say, we can looks at $f(x) = e^{2i\pi/x} -1$ instead

Your function verify the following ODE

$$f'(x) - 1= -\frac{2i\pi}{x^2} f(x) $$

By "multiplying by $x^2$", you get that

$$f(x) + \frac{x^2}{2i\pi} f'(x) = x^2$$

Using Fourier transform, we get

$$\widehat{f}(x) + \widehat{\frac{x^2}{2i\pi} f'(x)} = \delta''$$

$$\widehat{f}(x) + \widehat{\frac{x^2}{2i\pi}} \ast \widehat{ f'(x)} = \delta''$$

$$\widehat{f}(x) + \frac{1}{i\sqrt{2\pi}} \delta'' \ast ( -i\sqrt{2\pi} x \widehat{ f}(x) ) = \delta''$$

$$\widehat{f}(x) - ( x \widehat{ f}(x) )'' = \delta''$$

$$\widehat{f}(x) - 2 \widehat{f}'(x)- x\widehat{ f}''(x) = \delta''$$

So, for $x \neq 0$, $\widehat{f}$ verify

$$\widehat{f}(x) - 2 \widehat{f}'(x)- x\widehat{ f}''(x) = 0$$

And my numerical slave gives me a solution in term of Modified Bessel functions :

$$g(x) = \frac{c_1 I_1(2 \sqrt{x})}{\sqrt{x}}+ \frac{c_2 K_1(2 \sqrt{x})}{\sqrt{x}}$$

with $c_1$ and $c_2$ to find

But I'm too tired to get the complete solution for now (we're missing a distribution with support $\{0\}$).


$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{-\infty}^{\infty}\expo{2\pi\ic/x} \expo{2\pi\nu x\ic}\,\dd x} = \int_{0}^{\infty}\cos\pars{2\pi\root{\nu}\bracks{{1 \over \root{\nu}x} + \root{\nu}x}} \end{align} With $\ds{x = {\expo{\theta} \over \root{\nu}}}$: \begin{align} &\bbox[5px,#ffd]{\int_{-\infty}^{\infty}\expo{2\pi\ic/x} \expo{2\pi\nu x\ic}\,\dd x} = \int_{-\infty}^{\infty} \cos\pars{4\pi\root{\nu}\cosh\pars{\theta}} {\expo{\theta} \over \root{\nu}}\,\dd\theta \\[5mm] = &\ {1 \over \root{\nu}}\int_{0}^{\infty} \cos\pars{4\pi\root{\nu}\cosh\pars{\theta}} \cosh{\theta}\,\dd\theta \\[5mm] = &\ \left.{1 \over \root{\nu}}\,\partiald{}{\alpha}\ \underbrace{\int_{0}^{\infty} \sin\pars{\alpha\cosh\pars{\theta}} \,\dd\theta}_{\ds{{\pi \over 2}\on{J}_{0}\pars{\alpha}}}\,\right\vert_{\,\alpha\ =\ 4\pi\root{\nu}}\label{1}\tag{1} \\[5mm] = &\ \left.{1 \over \root{\nu}}\,{\pi \over 2}\, \partiald{\on{J}_{0}\pars{\alpha}}{\alpha}\ \,\right\vert_{\,\alpha\ =\ 4\pi\root{\nu}} = \bbx{-\,{\pi \over 2\root{\nu}}\,\on{J}_{1}\pars{4\pi\root{\nu}}} \label{2}\tag{2} \\ & \end{align}

(\ref{2}) as a function of $\ds{\nu}$:

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