A curious formula related to the Euler Gamma function

In this note, we prove that for all $x \in (0 , 1)$, we have: $$\log\Gamma(x) = \frac{1}{2} \log\pi + \pi \boldsymbol{\eta} \left(\frac{1}{2} - x\right) - \frac{1}{2} \log\sin(\pi x) + \frac{1}{\pi} \sum_{n = 1}^{\infty} \frac{\log n}{n} \sin(2 \pi n x) ,$$ where $\Gamma$ denotes the Euler Gamma function and $$\boldsymbol{\eta} := 2 \int_{0}^{1} \log\Gamma(x) \cdot \sin(2 \pi x) \, d x = 0.7687478924\dots$$