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- #1

- Jan 31, 2012

- 253

Show that $$\int_{0}^{1} \log (\sin \pi x) \log \Gamma(x) \ dx = \frac{\log 2 \pi}{2} \int_{0}^{1} \log (\sin \pi x) \ dx - \frac{\zeta(2)}{4} = - \frac{\log (2 \pi ) \log (2)}{2} - \frac{\pi^{2}}{24}$$

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