What is the Fourier transform of $e^{2 \pi i / x}$? 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)? 
 A: 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.
ADDENDUM
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.
ADDENDUM II
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.
A: 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 $\{0\}$).
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\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{1}): See a $\ds{\on{J}_{0}}$ Integral Representation.

*(\ref{2}): See the $\ds{\on{J}_{0}}$ Recurrence.

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

