Integral arising from Fourier series

Evaluate:

$$\int_0^{2\pi} \ln^2 \left( 2 \sin \frac{x}{2} \right)\, {\rm d}x$$

Solution



 Invoking the Taylor series of $\ln(1-e^{ix})$ we have that:

$$\begin{aligned}
\ln \left ( 1-e^{ix} \right ) &= - \sum_{n=1}^{\infty}\frac{e^{inx}}{n}\\
 &= - \sum_{n=1}^{\infty}\frac{\cos nx}{n}- i \sum_{n=1}^{\infty}\frac{\sin nx}{n}
\end{aligned}$$

However,

$$\begin{aligned}
1-e^{ix} &=\left ( 1-\cos x \right )-i\sin x \\
 &=2\sin^2 \frac{x}{2}- 2i \sin \frac{x}{2}\cos \frac{x}{2} \\
 &= 2\sin \frac{x}{2}\left ( \sin \frac{x}{2} - i
\cos \frac{x}{2} \right )\\
 &=2\sin \frac{x}{2}e^{i \frac{x-\pi}{2}}
\end{aligned}$$

Therefore
$$\ln \left ( 1-e^{ix} \right )= \ln \left ( 2\sin \frac{x}{2}  \right )+i \frac{\pi-x}{2}$$

Hence:

$$\sum_{n=1}^{\infty}\frac{\cos nx}{n}= - \ln \left ( 2 \sin \frac{x}{2} \right )$$

Using the Parseval identity we get:

$$\frac{1}{\pi}\int_{0}^{2\pi}\ln^2 \left ( 2 \sin \frac{x}{2} \right )\, {\rm d}x=\sum_{n=1}^{\infty}\frac{1}{n^2}= \frac{\pi^2}{6}\Leftrightarrow \int_{0}^{2\pi}\ln^2 \left ( 2\sin \frac{x}{2} \right )\, {\rm d}x = \frac{\pi^3}{6}$$

and this completes the exercise.