A beautiful function

We have $I=\displaystyle\int \limits^{1}_{0}ln\left( \Gamma \left( x\right) \right) dx=\displaystyle\int \limits^{1}_{0}ln\left( \Gamma \left( 1-x\right) \right) dx$ So $I=\frac{1}{2} \displaystyle\int \limits^{1}_{0}ln\left( \Gamma \left( x\right) \Gamma \left( 1-x\right) \right) dx= \frac{1}{2} \displaystyle\int \limits^{1}_{0}ln\left( \frac{\pi }{\sin \left( \pi x\right) } \right) dx$ $I= \frac{1}{2} \displaystyle\int \limits^{1}_{0}ln\left( \pi \right) dx-\frac{1}{2} \displaystyle\int \limits^{1}_{0}ln\left( sin\pi x\right) dx$ $I=\frac{ln\left( \pi \right) }{2} -\frac{1}{2\pi } \displaystyle\int \limits^{\pi }_{0}ln\left( \sin \left( x\right) \right) dx$ $I =\frac{ln\left( \pi \right) }{2} -\frac{1}{\pi } \displaystyle\int \limits^{\frac{\pi }{2} }_{0}ln\left( \sin \left( x\right) \right) dx$ Finally $I=\frac{ln\left( \pi \right) }{2} - \frac{1}{\pi } \left( \frac{-\pi ln\left( 2\right) }{2} \right) =\frac{ln\left( \pi \right) }{2} +\frac{ln\left( 2\right) }{2} =\frac{ln\left( 2\pi \right) }{2}$ So $A=\boxed {2}$ For $\displaystyle\int \limits^{\frac{\pi }{2} }_{0}ln\left( \sin \left( x\right) \right) dx$ Let $I_{1}=\displaystyle\int \limits^{\frac{\pi }{2} }_{0}ln\left( \sin \left( x\right) \right) =\displaystyle\int \limits^{\frac{\pi }{2} }_{0}ln\left( \cos \left( x\right) \right)$ So $2I_{1}=\displaystyle\int \limits^{\frac{\pi }{2} }_{0}ln\sin \left( 2x\right) -\displaystyle\int \limits^{\frac{\pi }{2} }_{0}ln\left( 2\right) dx$ Whitch is equal to 2I_{1} =2\int \limits^{\frac{\pi }{4} }_{0}ln\left( \sin \left( 2x\right) \right) dx-\frac{\pi ln\left( 2\right) }{2} = \underbrace{\int \limits^{\frac{\pi }{2} }_{0}ln\left( \sin \left( x\right) \right) dx} \limits_{I_{1}}-\frac{\pi ln\left( 2\right) }{2} Finally $2I_{1}=I_{1}-\frac{\pi ln\left( 2\right) }{2}$ $\boxed { I_{1}=\frac{-\pi ln\left( 2\right) }{2} }$

Raabe's Formula ;)