Integral and Gamma Function (1)

$\begin{aligned} I & = \int_0^1 \ln \Gamma (x) \ dx & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^1 \left(\ln \Gamma (x) + \ln \Gamma(1-x) \right) dx \\ & = \frac 12 \int_0^1 \ln \left(\Gamma (x) \Gamma(1-x) \right) dx \\ & = \frac 12 \int_0^1 \ln \left(\frac \pi{\sin (\pi x)} \right) dx \\ & = \frac 12 \left(\int_0^1 \ln \pi \ dx - {\color{#3D99F6} \int_0^1 \ln (\sin (\pi x))\ dx} \right) & \small \color{#3D99F6} \text{Let }u = \pi x \implies du = \pi \ dx \\ & = \frac 12 \left(x \ln \pi \ \bigg|_0^1 - {\color{#3D99F6} \frac 1\pi \int_0^\pi \ln (\sin u)\ du} \right) \\ & = \frac {\ln \pi}2 - \color{#3D99F6} \frac 1{2\pi} \int_0^\pi \ln (\sin u)\ du & \small \color{#3D99F6} \text{See note.} \\ & = \frac {\ln \pi}2 - \color{#3D99F6} \left(-\frac {\ln 2}2\right) \\ & = \frac {\ln (2\pi)}2 \end{aligned}$

Therefore, $A = \boxed{2}$ .

Note:

$\begin{aligned} I_1 & = \int_0^\pi \ln (\sin u) \ du & \small \color{#3D99F6} \sin u \text{ is symmetrical about }u = \frac \pi 2 \\ & = 2 \int_0^\frac \pi 2 \ln (\sin u) \ du & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \int_0^\frac \pi 2 (\ln (\sin u) + \ln (\cos u))\ du \\ & = \int_0^\frac \pi 2 \ln (\sin u\cos u)\ du \\ & = \int_0^\frac \pi 2 \ln \left(\frac {\sin (2u)}2 \right) du & \small \color{#3D99F6} \text{Let }\theta = 2u \implies d\theta = 2 \ du \\ & = \frac 12 \int_0^\pi \ln (\sin \theta) \ d\theta - \int_0^\frac \pi 2 \ln 2 \ du \\ & = \frac 12 I_1 - \frac {\pi \ln 2}2 \\ & = - \pi \ln 2 \end{aligned}$