A log-sine and log-gamma integral

[polygamma]

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

 

[Aaron Bex]

We will use the following integrals:

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

Then,

\begin{align*}
\int_{0}^{1} \log (\sin \pi x) \log \Gamma(x) \ dx &= \int_{0}^{1} \log (\sin \pi x) \left(- \frac{\zeta(2)}{2}\right) \ dx \\
&= - \frac{\zeta(2)}{2} \int_{0}^{1} \log (\sin \pi x) \ dx \\
&= - \frac{\zeta(2)}{2} (- \log (2)) \\
&= \frac{\log 2 \pi}{2} \int_{0}^{1} \log (\sin \pi x) \ dx - \frac{\zeta(2)}{4} \\
&= \frac{\log 2 \pi}{2} (- \log (2)) - \frac{\zeta(2)}{4} \\
&= - \frac{\log (2 \pi