Out of curiosity (and in relation to this MSE question), I computed numerically (an approximation to) the Fourier transform of $\mu(n)/n$ where $\mu$ is the Möbius function, viꝫ. $f\colon t \mapsto \sum_{n=1}^{+\infty} \frac{\mu(n)}{n}\,\exp(2i\pi nt)$ (in whatever sense it converges; it obviously makes sense in $L^2$, but this answer shows that convergence is uniform so that $f$ is, in fact, continuous).
The graph of $f$ looks like this (real part in blue, imaginary part in red, plotted over two periods):
(This has been computed using $403\,200$ Fourier coefficients, a number round enough and large enough that I hope would give a decent approximation of $f$; actual Sage code used is here. Numerically, with this many coefficients, $|f(0)|$ turns out to be $<7.3\times 10^{-5}$, which gives some indication that the overall error must not be too large — keeping in mind that $f(0)=0$ is an equivalent of the Prime Number Theorem.)
Now the above looks more or less what I expected. As mentioned above, the value at $0$ is $0$, the value at $\frac{1}{2}$ is also $0$, as explained here on MSE, and otherwise it looks like a random walk, as I expected from the Fourier transform of a sequence whose Fourier coefficients decrease in $1/n$ but are otherwise “random”.
On the other hand, the graph of the image of $f$ inside $\mathbb{C}$ does not look like what I expected to see:
To me this does not look like a typical Brownian motion: most notably, there are places where the curve looks “thin” and others where it looks “thick”, and there are repeated patterns where the curve thins and appears to “bounce” on a vertical axis.
I don't know what to think of this, and I don't know whether the explanation is to be sought in the realm of analytic number theory or harmonic analysis.
Can someone shed some light on this? (Not necessarily with proofs, but at least give some heuristic explanations.)
