What is $\int_0^1{\ln\Gamma(x)\sin(\pi x)}\mathop{\mathrm{d}x}$? Hello I am stuck with this integral:
$$\int_0^1{\ln\Gamma(x)\sin(\pi x)}\mathop{\mathrm{d}x}$$
My questions are:


*

*What is the integrand, $\ln(\Gamma(x)\sin(x\pi))$ or $\ln(\Gamma(x))\:\sin(x\pi)$?

*Are there any other suggestions?


Thank you for your help.
 A: Let $I_1$ and $I_2$ be the integrals
$$
I_1 \stackrel{def}{=} \int_0^1 \log(\Gamma(x)\sin(\pi x)) dx
\quad\text{ and }\quad
I_2 \stackrel{def}{=} \int_0^1 \log(\Gamma(x))\sin(\pi x) dx
$$
I believe $I_2$ is the integral intended. Notice 
$$\sin\pi x = \sin \pi(1-x)\quad\text{ and }\quad\Gamma(x)\Gamma(1-x) = \frac{\pi}{\sin \pi x}$$ 
We have
$$
I_2
= \frac12\int_0^1 \left(\log\Gamma(x) + \log\Gamma(1-x)\right)\sin\pi x dx
= \frac12\int_0^1 \log\left(\frac{\pi}{\sin\pi x}\right)\sin \pi x dx
$$
Change variable first to $y = \pi x$ and then to $z = \cos y$, we have
$$\begin{align}
I_2 &= \frac{1}{2\pi}\int_0^{\pi}\left(\log\pi - \log\sin y\right)\sin y dy
   = \frac{1}{\pi}\int_0^{\pi/2}\left(\log\pi - \log\sin y\right)\sin y dy\\
  &= \frac{1}{\pi}\left(\log \pi - \int_0^1 \log \sqrt{1-z^2} dz\right)\\
  &= \frac{1}{\pi}\left(\log\pi - \frac12\bigg[(1+z)\log(1+z)-(1-z)\log(1-z)-2z\bigg]_0^1\right)\\
  &= \frac{1+\log(\pi/2)}{\pi}
\end{align}
$$
For completeness, let us evaluate $I_1$ too. By a similar argument like above, we have
$$I_1 = \frac{1}{\pi}\int_0^{\pi/2}(\log \pi + \log\sin y)dy$$
Notice
$$\int_0^{\pi/2}\log\sin(2y) dy
= \frac12 \int_0^{\pi}\log\sin y dy = \int_0^{\pi/2}\log\sin y dy$$
We have
$$\int_0^{\pi/2}\log\sin y dy
= \int_0^{\pi/2}\log\cos y dy
= \int_0^{\pi/2}(\log\sin(2y) - \log\sin y - \log 2) dy
= -\log 2\frac{\pi}{2}$$
This implies
$$I_1 = \frac{1}{\pi}\left(\log\pi\cdot\frac{\pi}{2} - \log 2\cdot\frac{\pi}{2}\right) = \frac12\log\frac{\pi}{2}$$
the result first pointed by @J.M.
A: 
$$\begin{eqnarray*} \int_{0}^{1}\log\Gamma(x)\,\sin(\pi x)\,dx &=& \log(\pi)\int_{0}^{1}(1-x)\sin(\pi x)\,dx-\frac{1}{2}\int_{0}^{1}\sin(\pi x)\log\sin(\pi x)\,dx\\&=&\frac{\log \pi}{\pi}-\frac{\log(2)-1}{\pi}=\color{red}{\frac{1}{\pi}\left(1+\log\frac{\pi}{2}\right)}\end{eqnarray*}$$

can be directly computed from the Kummer's Fourier series expansion for $\log\Gamma$.
A simple alternative is given by integration by parts and the reflection formula for the $\psi$ function:
$$\begin{eqnarray*} \pi I = -\int_{0}^{1}\psi(x)(1-\cos(\pi x))\,dx.\end{eqnarray*}$$
