Rewriting the floor function using inverse hyperbolic trig functions, instead of the sine function

Question

Given that:

$$\lfloor x \rfloor = x - \frac{1}{2} + \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\sin{(2\pi kx)}}{k}$$

How can $\lfloor x \rfloor$ be rewritten using inverse hyperbolic trigonometric functions ($\sinh^{-1}$, $\cosh^{-1}$, $\tanh^{-1}$) instead of the sine function? Additionally, is it possible to do so without using any other trigonometric functions (other than the inverse hyperbolic functions)?

Answer 1

Using Maple:

restart:
func1 := x-1/2+(sum(sin(2*Pi*k*x)/k, k = 1 .. infinity, formal))/Pi;
func2 := evalc(Re(func1));
func3 := `assuming`([combine(func2, arctan, symbolic)], [x > 0]);
func4 := convert(func3, arctanh);

$$\lfloor x\rfloor =x-\frac{1}{2}-\frac{i \tanh ^{-1}\left(-\frac{i \sin (2 \pi x)}{\cos (2 \pi x)-1}\right)}{\pi }$$ Maple can't convert all for arctan.