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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.

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Set $z_0=0$, then $$\begin{split} 0&=\frac 1 {2\pi} \int_0^{2\pi}\ln |1+re^{i\theta}|d\theta\\ &\rightarrow \frac 1 {2\pi} \int_0^{2\pi}\ln |1+e^{i\theta}|d\theta\\ &= \frac 1 {2\pi} \int_0^{2\pi}\ln\left |2\cos \frac \theta 2\right|d\theta\\ &= \ln 2 +\frac 1 {\pi} \int_0^{\pi}\ln |\cos \theta |d\theta\\ &= \ln 2 +\frac 1 {\pi} \int_0^{\pi}\ln (\sin \theta )d\theta\\ \end{split} $$

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  • $\begingroup$ How did you changed$\rightarrow \frac 1 {2\pi} \int_0^{2\pi}\ln |1+e^{i\theta}|d\theta\\ $ to $= \frac 1 {2\pi} \int_0^{2\pi}\ln\left |2\cos \frac \theta 2\right|d\theta\\$? Can you please elaborate? I am geting $1+e^{i\theta} = 2 cos (\theta /2) \times( cos(\theta /2) + i sin(\theta /2) )$ . But unable to comprehend how did you wrote $1/2 \pi \int_{0^{2\pi}} ln | cos(\theta /2 + i sin(\theta /2)| d{\theta} $ =0? $\endgroup$ – James Jan 6 at 13:51
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    $\begingroup$ Sure:$|1+e^{i\theta} |= 2| cos (\theta /2) | \times |cos(\theta /2) + i sin(\theta /2) | = 2| cos (\theta /2) | $ $\endgroup$ – Stefan Lafon Jan 6 at 21:14
  • $\begingroup$ Can you please also tell how $ 1/\pi \int_{)}^{\pi} ln | cos (\theta) | d(\theta) = 1/\pi \int_{0}^{\pi} ln(sin\theta) d(\theta) $? $\endgroup$ – James Jan 7 at 14:38
  • $\begingroup$ $\int_0^\pi \ln |\cos\theta|d\theta=2\int_0^{\frac \pi 2} \ln\cos(\theta) d\theta = 2\int_{\frac \pi 2}^0 \ln \cos(\frac \pi 2 -\theta)d(\frac \pi 2 - \theta) = 2\int_0^{\frac \pi 2} \ln\sin(\theta) d\theta = \int_0^\pi \ln |\sin\theta|d\theta$ $\endgroup$ – Stefan Lafon Jan 7 at 21:00

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