便利さんの積分・級数botを解く②

\begin{align} I&=\int_{0}^{1}\log\Gamma(x)dx \\&=\int_{0}^{1}\log\Gamma(1-x)dx \\&=\frac{1}{2}\int_{0}^{1}\log\Gamma(x)\Gamma(1-x)dx \\&=\frac{1}{2}\int_{0}^{1}\log\frac{\pi}{\sin\pi x}dx \\&=\frac{1}{2}\int_{0}^{1}\log\pi dx- \frac{1}{2}\int_{0}^{1}\log\sin\pi xdx \\&=\frac{1}{2}\log\pi-\frac{1}{2\pi}\int_{0}^{\pi}\log\sin xdx \\&=\frac{1}{2}\log\pi+\frac{1}{2}\log2 \\&=\frac{1}{2}\log2\pi \end{align}