Integral and Gamma Function (2)

$\begin{aligned} I&=\int_0^1\left(\ln\Gamma(x)\right)\sin(\pi x)dx&\small\color{#3D99F6}\text{Using }\int_a^bf(x)dx=\int_a^bf(a+b-x)dx\\ &=\frac12\int_0^1\left(\ln\Gamma(x)+\ln\Gamma(1-x)\right)\sin(\pi x)dx\\ &=\frac12\int_0^1\left(\ln\Gamma(x)\Gamma(1-x)\right)\sin(\pi x)dx&\small\color{#3D99F6}\text{Using Euler's reflection formula}\\ &=\frac12\int_0^1\left(\ln\frac\pi{\sin(\pi x)}\right)\sin(\pi x)dx\\ &=\frac{\ln\pi}2\int_0^1\sin(\pi x)dx-\color{#3D99F6}\frac12\int_0^1\left(\ln\sin(\pi x)\right)\sin(\pi x)dx&\small\color{#3D99F6}\text{Let }u=\pi x\implies du=\pi dx\\ &=-\frac{\ln\pi}{2\pi}\cos(\pi x)\bigg|_0^1-\color{#3D99F6}\frac1{2\pi}\int_0^\pi\left(\ln\sin u\right)\sin udu&\small\color{#3D99F6}\text{See note}\\ &=\frac{\ln\pi}\pi-\color{#3D99F6}\frac{\ln 2-1}\pi\\ &=\frac1\pi\left(1+\ln{\frac\pi 2}\right) \end{aligned}$

Therefore, $A=\boxed2$ .

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Note:

$\begin{aligned} I_1&=\int_0^\pi\left(\ln\sin u\right)\sin udu&\small\color{#3D99F6}\sin u\text{ is symmetrical about }u=\frac\pi 2\\ &=2\int_0^{\frac\pi2}\left(\ln\sin u\right)\sin udu\\ &=2\int_0^{\frac\pi2}\left(\ln\sin u\right)d(1-\cos u)\\ &=2\left(\ln\sin u\right)(1-\cos u)\bigg|_{0^+}^{\frac\pi2}-2\int_0^{\frac\pi2}\frac{(1-\cos u)\cos u}{\sin u}du&\small\color{#3D99F6}\text{Clearly, }\lim\limits_{u\rightarrow 0^+}\left(\ln\sin u\right)(1-\cos u)=0\\ &=2\int_0^{\frac\pi2}\frac{(1-\cos u)\cos u}{\sin^2 u}d\cos u\\ &=2\int_0^{\frac\pi2}\left(1-\frac1{1+\cos u}\right)d\cos u\\ &=2\left(\cos u-\ln(1+\cos u)\right)\bigg|_0^{\frac\pi2}\\ &=2(\ln 2-1) \end{aligned}$