# Find the result of $\sum_{x = 1}^{\infty}\frac{1}{x}\,\log\left(\frac{kx}{\left\lfloor kx\right\rfloor}_{o}\right)$

I would like to calculate the sum

$$\sum_{x = 1}^{\infty}\frac{1}{x}\, \log\left(\frac{kx}{\left\lfloor kx \right\rfloor}_{o}\right)$$

where $k = \sqrt{\, 2\,} + 1$, $x$ is an odd integer and $\left\lfloor z \right\rfloor_o$ indicates the greatest odd integer $\leq z$. Considering that the odd greatest integer function satisfies $\left\lfloor z \right\rfloor_{o} = 2\left\lfloor\frac{z + 1}{2}\right\rfloor - 1$, and setting $x = 2n - 1$ to obtain a summation over all positive integers, the sum can also be written as

$$\sum_{n = 1}^{\infty}\frac{1}{2n - 1}\, \log\left(\frac{k\left[2n - 1\right]} {2\left\lfloor kn - k/2 + 1/2\right\rfloor - 1}\right)$$

The numerical value of the sum converges to $0.952842\ldots$ ( these first six decimal digits are stable after summing the first $10^{8}$ terms ). I tried several approaches to determine an explicit expression for this limit, including transformations of floor function by Fourier series or Laplace transforms, but I was not able to get it. Even if a closed expression could maybe not exist, I would be very interested in obtaining a simpler expression for this infinite sum.

For first, we may use: $$\lfloor x\rfloor_{o} = 2\left\lfloor\frac{x-1}{2}\right\rfloor+1= x-2\left\{\frac{x-1}{2}\right\}\tag{1}$$ where $\{x\}$ stands for the fractional part of $x$. Then we may recall Frullani's integral: $$(a>b>0)\qquad \log\frac{a}{b}=\int_{0}^{+\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx \tag{2}$$ to state:
$$\sum_{n\geq 1}\frac{1}{n}\cdot\log\frac{kn}{\lfloor kn\rfloor_{o}}=\int_{0}^{+\infty}\sum_{n\geq 1}\frac{e^{-knx}}{nx}\left(\exp\left(2\left\{\frac{kn-1}{2}\right\}\right)-1\right).\tag{3}$$ Now the path is the following: $\exp\left(2\left\{\frac{x-1}{2}\right\}\right)$ is a $2$-periodic function, hence we may decompose it as: $$\exp\left(2\left\{\frac{x-1}{2}\right\}\right) = 1+\sum_{m\geq 1}s_m \sin(\pi m x)+\sum_{m\geq 1}c_m\left(1-\cos(\pi m x)\right) \tag{4}$$ and plug the single terms back into $(3)$. The contribute given by $s_m$ depends on: $$s_m\int_{0}^{+\infty}\frac{e^{-knx}}{nx}\sin(\pi k m n x)\,dx=s_m\cdot\frac{\arctan(m\pi)}{n}\tag{5}$$ while the contribute given by $c_m$ depends on: $$c_m\int_{0}^{+\infty}\frac{e^{-knx}}{nx}\left(1-\cos(\pi k m n x)\right)\,dx=c_m\cdot\frac{\log(1+m^2\pi^2)}{2n}\tag{6}$$ hence your initial series ultimately depends on the logarithm of an infinite product and a Dedekind eta function value.