Complex Analysis – Evaluate $\int_0^{\pi} \ln \left( \sin \theta \right) d\theta$

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Evaluate: $ \displaystyle \int_0^{\pi} \ln \left( \sin \theta \right) d\theta$ using Gauss Mean Value theorem.

Given hint: consider $f(z) = \ln ( 1 +z)$.

EDIT:: I know how to evaluate it, but I am looking if I can evaluate it using Gauss MVT.

ADDED:: Here is what I have got so far!!

$$\ln 2 = \frac{1}{2 \pi } \int_0^{2\pi } \log(2+e^{i \theta}) d\theta = \frac{1}{2 \pi } \int_0^{2\pi } \log(2+e^{-i \theta}) d\theta$$

Hence, $ \displaystyle 2 \ln 2 = \frac{1}{2 \pi } \int_{0}^{2 \pi} \log(5 + 4 \cos \theta )d \theta = \frac{1}{\pi} \int_0^{\pi} \log(1 + 8 \cos^2 \theta) d \theta$, now to problem is how to reduce it to the above form?

Best Answer

I got this as the first part of this answer:

Start with $$ \begin{align} \int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x &=\frac12\int_0^\pi\log(\sin(x))\,\mathrm{d}x\\ &=\int_0^{\pi/2}\log(\sin(2x))\,\mathrm{d}x\\ &=\int_0^{\pi/2}\Big(\log(2)+\log(\sin(x))+\log(\cos(x))\Big)\,\mathrm{d}x\\ &=\frac\pi2\log(2)+2\int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x\tag{1} \end{align} $$ Therefore, $$ \int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x=-\frac\pi2\log(2)\tag{2} $$

Thus, $$ \int_0^\pi\log(\sin(x))\,\mathrm{d}x=-\pi\log(2) $$

Using Gauss Mean Value

$\mathrm{Re}(\log(z))=\log(|z|)=\log\left(\sqrt{2-2\cos(x)}\right)$

$\hspace{4.5cm}$enter image description here $$ \begin{align} \int_0^\pi\log(\sin(x))\,\mathrm{d}x &=\int_0^\pi\log\left(\color{#C00000}{\frac12}\sqrt{2-2\cos(x)}\right)\,\mathrm{d}x\\ &=\pi\color{#00A000}{\frac1{2\pi}\int_0^{2\pi}\log\left(\sqrt{2-2\cos(x)}\right)\,\mathrm{d}x}\color{#C00000}{-\pi\log(2)}\\[6pt] &=\pi\color{#00A000}{\log(1)}-\pi\log(2)\\[12pt] &=-\pi\log(2) \end{align} $$

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