Another Series $\sum\limits_{k=2}^\infty \frac{\log(k)}{k}\sin(2k \mu \pi)$

It suffices to do these integrals: $$ \begin{align} \int_0^1 \log(\Gamma(s))\;ds &= \frac{\log(2\pi)}{2} \tag{1a}\\ \int_0^1 \log(\Gamma(s))\;\cos(2k \pi s)\;ds &= \frac{1}{4k},\qquad k \ge 1 \tag{1b}\\ \int_0^1 \log(\Gamma(s))\;\sin(2k \pi s)\;ds &= \frac{\gamma+\log(2k\pi)}{2k\pi},\qquad k \ge 1 \tag{1c} \\ \int_0^1 \frac{\log(\sin(\pi s))}{2}\;ds &= \frac{-\log 2}{2} \tag{2a} \\ \int_0^1 \frac{\log(\sin(\pi s))}{2}\;\cos(2k \pi s)\;ds &= \frac{-1}{4k},\qquad k \ge 1 \tag{2b} \\ \int_0^1 \frac{\log(\sin(\pi s))}{2}\;\sin(2k \pi s)\;ds &= 0,\qquad k \ge 1 \tag{2c} \\ \int_0^1 1 \;ds &= 1 \tag{3a} \\ \int_0^1 1 \cdot \cos(2k \pi s)\;ds &= 0,\qquad k \ge 1 \tag{3b} \\ \int_0^1 1 \cdot \sin(2k \pi s)\;ds &= 0,\qquad k \ge 1 \tag{3c} \\ \int_0^1 s \;ds &= \frac{1}{2} \tag{4a} \\ \int_0^1 s \cdot \cos(2k \pi s)\;ds &= 0,\qquad k \ge 1 \tag{4b} \\ \int_0^1 s \cdot \sin(2k \pi s)\;ds &= \frac{-1}{2k\pi},\qquad k \ge 1 \tag{4c} \end{align} $$ Then for $f(s) = \pi \left(\log(\Gamma(s)) +\frac{1}{2}\log \sin(\pi s)-(1-s)\log(\pi)- \left(\frac{1}{2}-s\right)(\gamma+\log 2)\right)$, we get $$ \begin{align} \int_0^1 f(s)\;ds &= 0 \\ 2\int_0^1f(s) \cos(2k\pi s)\;\;ds &= 0,\qquad k \ge 1 \\ 2\int_0^1f(s) \sin(2k\pi s)\;\;ds &= \frac{\log k}{k},\qquad k \ge 1 \end{align} $$ and the formula follows as a Fourier series: $$ f(s) = \sum_{k=1}^\infty \frac{\log k}{k}\;\sin(2 k\pi s),\qquad 0 < s < 1. $$

reference

Gradshteyn & Ryzhik, Table of Integrals Series and Products

(1a) 6.441.2
(1b) 6.443.3
(1c) 6.443.1
(2a) 4.384.3
(2b) 4.384.3
(2c) 4.384.1