A curious formula related to the Euler Gamma function

Abstract

In this note, we prove that for all $x\in (0,1)$, we have: $$\log \Gamma (x)=\frac{1}{2}\log \pi +\pi \mathbf{\eta }\left(\frac{1}{2}-x\right)-\frac{1}{2}\log \sin (\pi x)+\frac{1}{\pi }\sum _{n=1}^{\infty }\frac{\log n}{n}\sin (2\pi nx),$$ where $\Gamma$ denotes the Euler Gamma function and $$\mathbf{\eta }:=2{\int }_0^1\log \Gamma (x)\cdot \sin (2\pi x)dx=0.7687478924\dots$$

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