Special Gamma function integral

Solution 1:

Let $u=\cos(x)$, then $$ \begin{align} \int_0^\pi\sin(x)\log(\sin(x))\,\mathrm{d}x &=\frac12\int_{-1}^1\log\left(1-u^2\right)\,\mathrm{d}u\\ &=\frac12\left(\int_{-1}^1\log(1-u)\,\mathrm{d}u+\int_{-1}^1\log(1+u)\,\mathrm{d}u\right)\\ &=\int_0^2\log(v)\,\mathrm{d}v\\[3pt] &=\left.v\log(v)-v\right]_0^2\\[9pt] &=2\log(2)-2\tag{1} \end{align} $$

Using the comment by Jack D'Aurizio $$ \begin{align} &\int_0^1\sin(\pi x)\log(\Gamma(x))\,\mathrm{d}x\tag{2}\\ &=\int_0^1\sin(\pi x)\log(\Gamma(1-x))\,\mathrm{d}x\tag{3}\\ &=\frac12\int_0^1\sin(\pi x)\log\left(\frac\pi{\sin(\pi x)}\right)\,\mathrm{d}x\tag{4}\\ &=\frac1{2\pi}\int_0^\pi\sin(x)\log(\pi)\,\mathrm{d}x -\frac1{2\pi}\int_0^\pi\sin(x)\log(\sin(x))\,\mathrm{d}x\tag{5}\\ &=\frac{\log(\pi)}\pi-\frac{\log(2)-1}\pi\tag{6}\\[3pt] &=\frac{\log(e\pi/2)}\pi\tag{7} \end{align} $$ Explanation:
$(3)$: substitute $x\mapsto1-x$
$(4)$: average $(2)$ and $(3)$ and use Euler's Reflection Formula
$(5)$: substitute $x\mapsto x/\pi$
$(6)$: apply $(1)$
$(7)$: algebra

Solution 2:

Let we put everything together. $$ I = \int_{0}^{1}\sin(\pi x)\log\Gamma(x)\,dx = \int_{0}^{1}\sin(\pi z)\log\Gamma(1-z)\,dz \tag{1}$$ leads to: $$ I = \frac{1}{2}\int_{0}^{1}\sin(\pi x)\log\left(\Gamma(x)\,\Gamma(1-x)\right)\,dx \tag{2}$$ but $\Gamma(x)\,\Gamma(1-x) = \frac{\pi}{\sin(\pi x)}$, hence: $$ I = \frac{\log \pi}{\pi}-\frac{1}{\pi}\int_{0}^{\pi/2}\sin(x)\log\sin(x)\,dx\tag{3} $$ or, with a change of variable and integration by parts: $$ I = \frac{\log \pi}{\pi}-\frac{1}{\pi}\int_{0}^{1}\frac{x\log x}{\sqrt{1-x^2}}\,dx = \frac{\log \pi}{\pi}+\frac{1}{\pi}\int_{0}^{1}\frac{1-\sqrt{1-x^2}}{x}\,dx\tag{4}$$ so:

$$ \int_{0}^{1}\sin(\pi x)\log\Gamma(x)\,dx = \color{red}{\frac{1}{\pi}\left(1+\log\frac{\pi}{2}\right)}\tag{5}$$

since a primitive for $\frac{1-\sqrt{1-x^2}}{x}=\frac{x}{1+\sqrt{1-x^2}}$ is given by $\log(1+\sqrt{1-x^2})-\sqrt{1-x^2}$.

By using the reflection formula, $(5)$ can be seen as a consequence of Raabe's formula, too.

It also follows from Kummer's Fourier series expansion.

Solution 3:

$$ \eqalign{ & \int_{x\, = \,0}^\pi {\sin (x)\ln (\sin (x))dx} = - {1 \over 2}\int_{x\, = \,0}^\pi {\ln (1 - \cos ^{\,2} (x))d\cos (x)} = \cr & = - {1 \over 2}\int_{t\, = \,1}^{ - 1} {\left( {\ln (1 - t) + \ln (1 + t)} \right)dt} = \; \cdots \cr} $$