## Floor functions using complex logarithms

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):