Basta con hacer estas integrales:
$$
\begin{align}
\int_0^1 \log(\Gamma(s))\;ds &= \frac{\log(2\pi)}{2}
\tag{1a}\\
\int_0^1 \log(\Gamma(s))\;\cos(2k \pi s)\;ds &= \frac{1}{4k},\qquad k \ge 1
\tag{1b}\\
\int_0^1 \log(\Gamma(s))\;\sin(2k \pi s)\;ds &= \frac{\gamma+\log(2k\pi)}{2k\pi},\qquad k \ge 1
\tag{1c}
\\
\int_0^1 \frac{\log(\sin(\pi s))}{2}\;ds &= \frac{-\log 2}{2}
\tag{2a}
\\
\int_0^1 \frac{\log(\sin(\pi s))}{2}\;\cos(2k \pi s)\;ds &= \frac{-1}{4k},\qquad k \ge 1
\tag{2b}
\\
\int_0^1 \frac{\log(\sin(\pi s))}{2}\;\sin(2k \pi s)\;ds &= 0,\qquad k \ge 1
\tag{2c}
\\
\int_0^1 1 \;ds &= 1
\tag{3a}
\\
\int_0^1 1 \cdot \cos(2k \pi s)\;ds &= 0,\qquad k \ge 1
\tag{3b}
\\
\int_0^1 1 \cdot \sin(2k \pi s)\;ds &= 0,\qquad k \ge 1
\tag{3c}
\\
\int_0^1 s \;ds &= \frac{1}{2}
\tag{4a}
\\
\int_0^1 s \cdot \cos(2k \pi s)\;ds &= 0,\qquad k \ge 1
\tag{4b}
\\
\int_0^1 s \cdot \sin(2k \pi s)\;ds &= \frac{-1}{2k\pi},\qquad k \ge 1
\tag{4c}
\end{align}
$$
A continuación, para $f(s) = \pi \left(\log(\Gamma(s)) +\frac{1}{2}\log \sin(\pi s)-(1-s)\log(\pi)- \left(\frac{1}{2}-s\right)(\gamma+\log 2)\right)$, obtenemos
$$
\begin{align}
\int_0^1 f(s)\;ds &= 0
\\
2\int_0^1f(s) \cos(2k\pi s)\;\;ds &= 0,\qquad k \ge 1
\\
2\int_0^1f(s) \sin(2k\pi s)\;\;ds &= \frac{\log k}{k},\qquad k \ge 1
\end{align}
$$
y la fórmula sigue como una serie de Fourier:
$$
f(s) = \sum_{k=1}^\infty \frac{\log k}{k}\;\sin(2 k\pi s),\qquad 0 < s < 1.
$$
referencia
Gradshteyn & Ryzhik, Tabla de Integrales de la Serie y de los Productos
(1a) 6.441.2
(1b) 6.443.3
(1c) 6.443.1
(2a) 4.384.3
(2b) 4.384.3
(2c) 4.384.1