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$$\int\limits_{0}^{\frac{\pi }{2}}{x\log \left( 1-\cos x \right)dx}=\frac{35}{16}\zeta \left( 3 \right)-\frac{\pi ^{2}}{8}\log 2-\pi G$$

whitexlotus wrote:$$\int\limits_{0}^{\frac{\pi }{2}}{x\log \left( 1-\cos x \right)dx}=\frac{35}{16}\zeta \left( 3 \right)-\frac{\pi ^{2}}{8}\log 2-\pi G$$

Hi pprime. Here is something!


\begin{align*}
\int_{0}^{\pi/2} x \log \left ( 1-\cos x \right ) &= \int_{0}^{\pi/2} x \log \left ( 2 \sin^2 \frac{x}{2} \right )\, {\rm d}x \\
&= \int_{0}^{\pi/2} x \log 2 \, {\rm d}x + 2 \int_{0}^{\pi/2} x \log \sin \frac{x}{2} \, {\rm d}x\\
&\!\!\!\!\!\!\overset{u=x/2}{=\! =\! =\! =\! =\!} \frac{\pi^2 \log 2}{8} + 8 \int_{0}^{\pi/4} u \log \sin u \, {\rm d}u \\
&= \frac{\pi^2 \log 2}{8} + 8 \int_{0}^{\pi/4} u \left ( - \log 2 - \sum_{m=1}^{\infty} \frac{\cos 2mu}{m} \right ) \, {\rm d}u \\
&= \frac{\pi^2 \log 2}{8} - \frac{2\pi^2 \log 2}{8} - 8 \int_{0}^{\pi/4} u \sum_{m=1}^{\infty} \frac{\cos 2mu}{m} \, {\rm d}u\\
&= - \frac{\pi^2 \log 2}{8} - 8 \sum_{m=1}^{\infty} \frac{1}{m} \int_{0}^{\pi/4} u \cos 2mu \, {\rm d}u \\
&= -\frac{\pi^2 \log 2}{8} - 8 \sum_{m=1}^{\infty} \frac{1}{m}\left [ \frac{\pi \sin \left ( \frac{m \pi}{2} \right )}{8m} + \frac{2 \cos \left ( \frac{m \pi}{2} \right )}{8m^2} - \frac{2}{8m^2} \right ] \\
&= -\frac{\pi^2 \log 2}{8} - \sum_{m=1}^{\infty} \frac{1}{m}\left [ \frac{\pi \sin \left ( \frac{m \pi}{2} \right )}{m} + \frac{2 \cos \left ( \frac{m \pi}{2} \right )}{m^2} - \frac{2}{m^2} \right ] \\
&=-\frac{\pi^2 \log 2}{8} - \pi \mathcal{G} + \frac{35}{16} \zeta(3)
\end{align*}

since we have seen in the tut. topic that

$${\rm Cl}_2 \left ( \frac{\pi}{2} \right )= \sum_{n=1}^{\infty} \frac{\sin \frac{n \pi}{2}}{n^2}=\mathcal{G}$$

On the other hand the series $\displaystyle \sum_{n=1}^{\infty} \frac{\cos \frac{n \pi}{2}}{n^3}$ can be dealt either with Fourier series or again by splitting it up to its even and odd components . Finally we get that:

$$\sum_{n=1}^{\infty} \frac{\cos \frac{n \pi}{2}}{n^3} = - \frac{3\zeta(3)}{32}$$

That was nice!

Well in general invoking Fourier series one can see that:

$$\sum_{n=1}^{\infty} \frac{\cos nx}{n^3}= \frac{1}{2}\left [ {\rm Li}_3 \left ( e^{-ix} \right ) + {\rm Li}_3 \left ( e^{ix} \right ) \right ]$$

Hence for $x=\frac{\pi}{2}$ we have that:

$$\sum_{n=1}^{\infty} \frac{\cos \frac{n \pi}{2}}{n^3} = \frac{1}{2}\left [ {\rm Li}_3(-i) + {\rm Li}_3(i) \right ]$$

However we have that $\displaystyle {\rm Li}_3(i)= - \frac{3\zeta(3)}{32} + i \frac{\pi^3}{32}$. Also $${\rm Li}_3(-i)= -\frac{3\zeta(3)}{32} - i \frac{\pi^3}{32}$$

Other interesting results are:

1. $\displaystyle \int_0^{\pi/2} x \log \sin x \, {\rm d}x = \frac{7 \zeta(3)}{16} - \frac{\pi^2 \log 2}{8}$
2. $\displaystyle \int_0^{\pi/2} x \log \cos x \, {\rm d}x = - \frac{7\zeta(3)}{16} - \frac{\pi^2 \log 2}{8}$
3. $\displaystyle \int_0^{\pi/2} x \log \tan x \, {\rm d}x = \frac{7 \zeta(3)}{8}$.

All of them are dealt using Fourier series of $\log \sin $, $\log \cos $ and $\log \tan $ respectively. Now, giving an extension of the original integral one can prove that:

$$\int_{0}^{\pi/2} x \log \left ( 1+\cos x \right ) \, {\rm d}x = \pi \mathcal{G} - \frac{21 \zeta(3)}{16} - \frac{\pi^2 \log 2}{8}$$

as well as many other interesting facts , mainly:

• $\displaystyle \int_{0}^{\pi/2} x \log \left ( 1 - \sin x \right ) \, {\rm d}x = - \frac{35 \zeta(3)}{16} - \frac{\pi^2 \log 2}{8}$.
• $\displaystyle \int_{0}^{\pi/2} x \log \left ( 1 + \sin x \right ) \, {\rm d}x = \frac{21 \zeta(3)}{16} - \frac{\pi^2 \log 2}{8}$.