# How to use mean value theorem For harmonic functions to prove this

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 $$\begin{split} 0&=\frac 1 {2\pi} \int_0^{2\pi}\ln |1+re^{i\theta}|d\theta\\ &\rightarrow \frac 1 {2\pi} \int_0^{2\pi}\ln |1+e^{i\theta}|d\theta\\ &= \frac 1 {2\pi} \int_0^{2\pi}\ln\left |2\cos \frac \theta 2\right|d\theta\\ &= \ln 2 +\frac 1 {\pi} \int_0^{\pi}\ln |\cos \theta |d\theta\\ &= \ln 2 +\frac 1 {\pi} \int_0^{\pi}\ln (\sin \theta )d\theta\\ \end{split}$$
• 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? – James Jan 6 at 13:51
• Sure:$|1+e^{i\theta} |= 2| cos (\theta /2) | \times |cos(\theta /2) + i sin(\theta /2) | = 2| cos (\theta /2) |$ – Stefan Lafon Jan 6 at 21:14
• Can you please also tell how $1/\pi \int_{)}^{\pi} ln | cos (\theta) | d(\theta) = 1/\pi \int_{0}^{\pi} ln(sin\theta) d(\theta)$? – James Jan 7 at 14:38
• $\int_0^\pi \ln |\cos\theta|d\theta=2\int_0^{\frac \pi 2} \ln\cos(\theta) d\theta = 2\int_{\frac \pi 2}^0 \ln \cos(\frac \pi 2 -\theta)d(\frac \pi 2 - \theta) = 2\int_0^{\frac \pi 2} \ln\sin(\theta) d\theta = \int_0^\pi \ln |\sin\theta|d\theta$ – Stefan Lafon Jan 7 at 21:00