A curious formula related to the Euler Gamma function

Bakir Farhi

In this note, we prove that for all $x\in (0,1)$, we have: $$ \logΓ(x) = \frac{1}{2} \logπ+ π\boldsymbolη \left(\frac{1}{2} - x\right) - \frac{1}{2} \log\sin(πx) + \frac{1}π \su…