An interesting discovery, thanks to a problem presented by K. Srinivasa Raghava:

A standard result for \(x\) real, where \(\lfloor .\rfloor\) is the floor function:

\(\lfloor x\rfloor =\frac{1}{\pi }\sum _{k=1}^{\infty } \frac{\sin (2 \pi k x)}{k}+x-\frac{1}{2}\).

Now it so happens that:

\(\frac{1}{\pi }\sum _{k=1}^{\infty } \frac{\sin (2 \pi k x)}{k}=\frac{i \left(\log \left(1-e^{-2 i \pi x}\right)-\log \left(1-e^{2 i \pi x}\right)\right)}{2 \pi }\)

which is of course intuitive owing to Riemann surfaces produced by the complex logarithm. I could not find the result but I am nearly certain it must be somewhere.

Now to get an idea, let us examine the compensating function \(f(x)= \frac{i \left(\log \left(1-e^{-2 i \pi x}\right)-\log \left(1-e^{2 i \pi x}\right)\right)}{2 \pi } \)

And of course, the complex logarithm (here is the standard function, just to illustrate):