Integration of $ \int_{0}^{\frac{\pi}{2}} x \log(1-\cos x) \,dx $

Question

Question:

What is the closed form of this following integral? $$ \int_{0}^{\frac{\pi}{2}} x \log(1-\cos x) \,dx.$$

Here is my solution

we know that $$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{\cos nx}}{n}} = \frac{1}{2}\left( {\sum\limits_{n = 1}^\infty {\frac{{{e^{inx}}}}{n}} + \sum\limits_{n = 1}^\infty {\frac{{{e^{ - inx}}}}{n}} } \right) = - \frac{1}{2}\log \left( {1 - {e^{ix}}} \right)\left( {1 - {e^{ - ix}}} \right) = - \frac{1}{2}\log 2\left( {1 - \cos x} \right) = }$$

$$\displaystyle{ = - \frac{{\log 2}}{2} - \frac{1}{2}\log \left( {1 - \cos x} \right) \Rightarrow \boxed{\log \left( {1 - \cos x} \right) = - \log 2 - 2\sum\limits_{n = 1}^\infty {\frac{{\cos nx}}{n}} }}$$, so

$$\displaystyle{\int\limits_0^{\pi /2} {x\log \left( {1 - \cos x} \right)dx} = - \log 2\int\limits_0^{\pi /2} {x\,dx} - 2\sum\limits_{n = 1}^\infty {\frac{1}{n}\int\limits_0^{\pi /2} {x\cos nx\,dx} } = - \frac{{{\pi ^2}\log 2}}{8} - 2\sum\limits_{n = 1}^\infty {\frac{1}{n}\left( { - \frac{1}{{{n^2}}} + \frac{{\cos \dfrac{{n\pi }}{2}}}{{{n^2}}} + \pi \frac{{\sin \dfrac{{n\pi }}{2}}}{{2n}}} \right)} = }$$

$$\displaystyle{ = - \frac{{{\pi ^2}\log 2}}{8} + 2\zeta \left( 3 \right) - 2\sum\limits_{n = 1}^\infty {\frac{{\cos \dfrac{{n\pi }}{2}}}{{{n^3}}}} - \pi \sum\limits_{n = 1}^\infty {\frac{{\sin \dfrac{{n\pi }}{2}}}{{{n^2}}}} = - \frac{{{\pi ^2}\log 2}}{8} + 2\zeta \left( 3 \right) - \sum\limits_{n = 1}^\infty {\frac{{\left( {{i^n} + {{\left( { - i} \right)}^n}} \right)}}{{{n^3}}}} - \frac{\pi }{{2i}}\sum\limits_{n = 1}^\infty {\frac{{\left( {{i^n} - {{\left( { - i} \right)}^n}} \right)}}{{{n^2}}}} \Rightarrow }$$

$$\displaystyle{ \Rightarrow \boxed{\int\limits_0^{\pi /2} {x\log \left( {1 - \cos x} \right)dx} = - \frac{{{\pi ^2}\log 2}}{8} + 2\zeta \left( 3 \right) - \left( {L{i_3}\left( i \right) + L{i_3}\left( { - i} \right)} \right) - \frac{\pi }{{2i}}\left( {L{i_2}\left( i \right) - L{i_2}\left( { - i} \right)} \right)}}$$

However, it holds that $$\displaystyle{L{i_3}\left( z \right) + L{i_3}\left( { - z} \right) = \frac{1}{4}L{i_3}\left( {{z^2}} \right)}$$ , so $$\displaystyle{L{i_3}\left( i \right) + L{i_3}\left( { - i} \right) = \frac{1}{4}L{i_3}\left( { - 1} \right) = \frac{1}{4}\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^3}}}} = - \frac{3}{{16}}\zeta \left( 3 \right)}$$

and $$\displaystyle{L{i_2}\left( i \right) - L{i_2}\left( { - i} \right) = \sum\limits_{n = 1}^\infty {\frac{{{i^n}}}{{{n^2}}}} - \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - i} \right)}^n}}}{{{n^2}}}} = \sum\limits_{n = 1}^\infty {\frac{{{i^{2n - 1}}}}{{{{\left( {2n - 1} \right)}^2}}}} + \sum\limits_{n = 1}^\infty {\frac{{{i^{2n}}}}{{{{\left( {2n} \right)}^2}}}} - \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - i} \right)}^{2n - 1}}}}{{{{\left( {2n - 1} \right)}^2}}}} - \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - i} \right)}^{2n}}}}{{{{\left( {2n} \right)}^2}}}} = }$$

$$\displaystyle{ = 2\sum\limits_{n = 1}^\infty {\frac{{{i^{2n - 1}}}}{{{{\left( {2n - 1} \right)}^2}}}} = \frac{2}{i}\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{{\left( {2n - 1} \right)}^2}}}} = - 2i\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{{\left( {2n - 1} \right)}^2}}}} = 2i\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{{\left( {2n + 1} \right)}^2}}}} = 2i \cdot G}$$

Therefore, $$\displaystyle{\int\limits_0^{\pi /2} {x\log \left( {1 - \cos x} \right)dx} = - \frac{{{\pi ^2}\log 2}}{8} + 2\zeta \left( 3 \right) - \left( {L{i_3}\left( i \right) + L{i_3}\left( { - i} \right)} \right) - \frac{\pi }{{2i}}\left( {L{i_2}\left( i \right) - L{i_2}\left( { - i} \right)} \right) = }$$

$$\displaystyle{ = - \frac{{{\pi ^2}\log 2}}{8} + 2\zeta \left( 3 \right) + \frac{3}{{16}}\zeta \left( 3 \right) - \frac{\pi }{{2i}}\left( {2i \cdot G} \right) = - \frac{{{\pi ^2}\log 2}}{8} + \frac{{35}}{{16}}\zeta \left( 3 \right) - \pi \cdot G}.$$

Sorry to the community for the misunderstanding. When DecarbonatedOdessa asked, "Is there a question in there somewhere?" I misunderstood what they were asking and answered, "Yes, on AOPS, but I’m trying a new way to find the answer." It was my mistake; they asked me something else, and I answered something else.

What should I add to my question to get it reopened?

Answer 1

\begin{align} &\int_{0}^{\frac{\pi}{2}} x \ln(1-\cos x) \,dx \\ = & \int_{0}^{\frac{\pi}{2}} x \ln(\sin x) \overset{x\to\frac \pi2-x}{dx } + \int_{0}^{\frac{\pi}{2}} x\ln(\tan\frac x2)\,dx \\ = &\ \frac12\int_{0}^{\frac{\pi}{2}} \left[x \ln(\sin x) +(\frac\pi2-x)\ln(\cos x)\right]dx\\ & + \int_{0}^{{\pi}} x\ln(\tan\frac x2) \overset{x\to 2x}{dx } -\int_{\frac\pi2}^{{\pi}} x\ln(\tan\frac x2) \overset{x\to\pi-x}{dx } \\ =& \ \frac52\int_{0}^{\frac{\pi}{2}} x\ln(\tan x)dx + \frac\pi4\int_{0}^{\frac{\pi}{2}} \ln(\cos x) dx + \frac\pi2 \int_0^{\frac\pi2}\ln(\tan \frac x2)dx\\ =& \ \frac{35}{16}\zeta(3) -\frac{\pi^2}8\ln2-\pi G \end{align}

where $ \int_{0}^{\frac{\pi}{2}} \ln(\cos x) dx=-\frac\pi2\ln2$, $ \int_0^{\frac\pi2}\ln(\tan \frac x2)dx =-2G$ and $\int_{0}^{\frac{\pi}{2}} x\ln(\tan x)= \frac78\zeta(3)$.

Answer 2

Using $$ \Re\log(1-e^{ix})=\ln|1-e^{ix}|=\frac12(\log2+\log(1-\cos x)) $$ one has $$\begin{eqnarray} I&=&\int_{0}^{\frac{\pi}{2}} x \log(1-\cos x) \,dx\\ &=&-\frac18\pi^2\log2+2\Re\int_{0}^{\frac{\pi}{2}} x \log(1-e^{ix}) \,dx\\ &=&-\frac18\pi^2\log2+2\Re i\int_{0}^{\frac{\pi}{2}} x d\text{Li}_2\left(e^{i x}\right)\\ &=&-\frac18\pi^2\log2+2\Re \bigg[i x \text{Li}_2\left(e^{i x}\right)-\text{Li}_3\left(e^{i x}\right)\bigg]\bigg|_{0}^{\frac{\pi}{2}}\\ &=&-\frac18\pi^2\log2+2\Re \bigg[\frac\pi2 i\text{Li}_2(i)-\text{Li}_3(i)+\zeta(3)\bigg]. \end{eqnarray}$$ Update: (1) It is easy to see $$ i\text{Li}_2(i)=-C-i\frac{\pi^2}{48}. $$ (2). From https://en.wikipedia.org/wiki/Polylogarithm, one has $$ \text{Li}_3(i)=-2^{-3}\eta(3)+i\beta(3)=-\frac18(1-\frac1{2^2})\zeta(3)+i\beta(3)=-\frac3{32}\zeta(3)+i\beta(3). $$ So $$ I=-\frac{\pi^2}8\ln2+\frac{35}{16}\zeta(3)-\pi C $$

Answer 3

First, we have

$$\begin{align} \log(1-\cos(x))&=\log\left(2\sin^2(x/2)\right)\\\\ &=\log(2)+2\log(\sin(x/2))\\\\ &=\log(2)+2\left(-\log(2)-\sum_{n=1}^\infty \frac{\cos(nx)}{n}\right)\\\\ &=-\log(2)-2\sum_{n=1}^\infty \frac{\cos(nx)}{n} \end{align}$$

where we used the well-known Fourier Series (See Here for example) of $\log(\sin(x/2))$.

Then, we have

$$\begin{align} \int_0^{\pi/2}x\log(1-\cos(x))\,dx&=-\frac{\pi^2}8 \log(2)-2\sum_{n=1}^\infty \frac1n\int_0^{\pi/2}x\cos(nx)\,dx\\\\ &=-\frac{\pi^2}8 \log(2)+2\sum_{n=1}^\infty \left(\frac1{n^3}-\frac\pi2 \frac{\sin(n\pi/2)}{n^2}-\frac{\cos(n\pi/2)}{n^3}\right)\\\\ &=-\frac{\pi^2}8 \log(2)+2\sum_{n=1}^\infty \frac1{n^3}+\pi \sum_{n=1}^\infty \frac{(-1)^{n}}{(2n-1)^2}+2\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n)^3}\\\\ &=-\frac{\pi^2}8 \log(2)+2\zeta(3)-\pi G+\frac3{16}\zeta(3)\\\\ &=-\frac{\pi^2}8 \log(2)+\frac{35}{16}\zeta(3)-\pi G \end{align}$$

where we used $\sum_{n=1}^\infty \frac{1}{n^3}=\zeta(3)$, $\sum_{n=1}^\infty \frac{(-1)^n}{(2n-1)^2}=-G$, and $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n)^3}=\frac{3}{4}\zeta(3)$.