The following question was asked in my complex analysis assignment and I am confused about how this should be done.
Show that $\int_{0}^{\pi} \ln( \sin {\theta}) d{\theta} = -\pi \ln 2$ by applying the mean value theorem to $ \ln|1+z|$ for $|z|\leq r <1$ and then letting $r\to 1$.
If I use mean value theorem to $\ln(1+z)$,I get $ln|(1+z_0)| = \frac{1}{2\pi} \int_{0}^{2\pi}\ln|1+z_0+ r e^{i\theta}|d {\theta}$.
But how to change the $1+z_0 + re^{i \theta}$ in RHS into $\sin\theta$ if $r$ approaches $1$?
I am not able to manipulate into that.
Set $z_0=0$, then $$\begin{split} 0&=\frac 1 {2\pi} \int_0^{2\pi}\ln |1+re^{i\theta}|d\theta\\ &\rightarrow \frac 1 {2\pi} \int_0^{2\pi}\ln |1+e^{i\theta}|d\theta\\ &= \frac 1 {2\pi} \int_0^{2\pi}\ln\left |2\cos \frac \theta 2\right|d\theta\\ &= \ln 2 +\frac 1 {\pi} \int_0^{\pi}\ln |\cos \theta |d\theta\\ &= \ln 2 +\frac 1 {\pi} \int_0^{\pi}\ln (\sin \theta )d\theta\\ \end{split} $$