The integral ${I=\int_0^\pi \log \sin x dx}$ is normally computed (e.g. in Ahlfors’ book) to be ${-\pi \log 2}$ using complex integration over a suitable almost-rectangular contour. There is also a simple and direct way to get the value of this integral by a substitution and elementary calculus.

First, by the substitution ${x=2t}$ and the identity ${\sin(2x)=2\sin x \cos x}$,

$\displaystyle I = 2 \int_0^{\pi/2} \log \sin t dt + 2 \int_0^{\pi/2} \log \cos t dt + \pi \log 2;$

then using the symmetry of ${\sin}$ and ${\cos}$ gives:

$\displaystyle I = I + I + \pi \log 2,$

whence the result. There are slight technicalities regarding the improperness of these integrals, but they can be directly justified (or one may use the Lebesgue integral).

[Edit (7/25)- Todd Trimble posted solutions to similar integrals, which use the result of this post as a lemma, here.  AM]