Question

Prove $\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\log \left(\sin x\right)dx=-\frac{\pi }{2}\log 2$

Answer 1

$I={\int }_0^{\frac{\pi }{2}}\log \left(\sin x\right)dx$

$I={\int }_0^{\frac{\pi }{2}}\log \left(\cos x\right)dx$

$2I={\int }_0^{\frac{\pi }{2}}\left[\log \right(\sin x)+\log (\cos x\left)\right]dx$

$2I={\int }_0^{\frac{\pi }{2}}\log \left(\sin x\cos x\right)dx$

$2I={\int }_0^{\frac{\pi }{2}}\log \left(\frac{\sin 2x}{2}\right)dx$

$2I={\int }_0^{\frac{\pi }{2}}\log \left(\sin 2x\right)dx-{\int }_0^{\frac{\pi }{2}}\log 2dx$

$2I=I_1-\log 2[x{]}_0^{\frac{\pi }{2}}$

$2I=I_1-\frac{\pi }{2}\log 2$...(1)

$I_1={\int }_0^{\frac{\pi }{2}}\log \left(\sin 2x\right)dx$

substitute $2x=t\to dt=2dx\Rightarrow x=0-\frac{\pi }{2}\to t=0-\pi$

$\therefore I_1=\frac{1}{2}{\int }_0^{\pi }\log \left(\sin t\right)dt$

$\Rightarrow I_1=I$...(2)

Placing (2) in (1)

$\Rightarrow 2I=I-\frac{\pi }{2}\log 2$

$\therefore I=-\frac{\pi }{2}\log 2$