I am trying to calculate
$$
I=\frac{1}{\pi}\int_0^\pi \theta^2 \ln^2\big(2\cos\frac{\theta}{2}\big)d \theta=\frac{11\pi^4}{180}=\frac{11\zeta(4)}{2}.
$$
Note, we can expand the log in the integral to obtain three interals, one trivial, the other 2 are not so easy, any ideas? We will use
$$
\left( \ln 2 +\ln \cos \frac{\theta}{2} \right)^2=\ln^2(2)+\ln^2\cos\frac{\theta}{2}+2\ln (2)\ln \cos\big(\frac{\theta}{2}\big)
$$ and re-write I as
$$
\pi I=\ln^2(2)\int_0^\pi \theta^2d\theta +\int_0^\pi\theta^2 \ln^2 \cos \frac{\theta}{2}d\theta+2\ln 2 \int_0^\pi\theta^2 \ln \cos{\frac{\theta}{2}}d\theta.
$$
Simplfying this further by using $y=\theta/2$ we obtain
$$
\pi I=\frac{\pi^3\ln^2(2)}{3}+16\ln(2)\int_0^{\pi/2} y^2 \ln \cos (y) dy+8\int_0^{\pi/2} y^2 \ln^2 \cos (y) dy
$$
Any Idea how to approach these two integrals? I know that
$$
\int_0^{\pi/2} \ln \cos y dy= \frac{-\pi\ln(2)}{2}\approx -1.08879
$$
but I am unsure how to use that here. I do not think partial integration will work, Also the Riemann Zeta function is given by
$$
\zeta(4)=\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90}.
$$
Thanks!
Best Answer
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{I \equiv {1 \over \pi}\int_{0}^{\pi}\theta^{2}\ln^{2}\pars{2\cos\pars{\theta \over 2}} \,\dd\theta = {11\pi^{4} \over 180} = {11 \over 2}\,\zeta\pars{4}}$
The integration in $\pars{1}$ is given by: \begin{align} &\int_{\verts{z} = 1} z^{\mu - \nu/2}\pars{z + 1}^{\nu}\,\dd z \\[3mm]&=-\int_{-1}^{0}\pars{-x}^{\mu - \nu/2}\expo{\ic\pi\pars{\mu - \nu/2}} \pars{x + 1}^{\nu}\,\dd x -\int_{0}^{-1}\pars{-x}^{\mu - \nu/2}\expo{-\ic\pi\pars{\mu - \nu/2}} \pars{x + 1}^{\nu}\,\dd x \\[3mm]&=-\expo{\ic\pi\pars{\mu - \nu/2}}\int_{0}^{1}x^{\mu - \nu/2} \pars{-x + 1}^{\nu}\,\dd x +\expo{-\ic\pi\pars{\mu - \nu/2}}\int_{0}^{1}x^{\mu - \nu/2} \pars{-x + 1}^{\nu}\,\dd x \\[3mm]&=2\ic\sin\pars{\pi\bracks{{\nu \over 2} - \mu}} {\rm B}\pars{\mu - {\nu \over 2} + 1,\nu + 1}\tag{2} \end{align} where $\ds{{\rm B}\pars{x,y} = \int_{0}^{1}t^{x - 1}\pars{1 - t}^{y - 1}\,\dd t}$, $\ds{\pars{~\mbox{with}\ \Re\pars{x} > 0,\ \Re\pars{y} > 0~}}$ is the Beta Function.
Since $\ds{{\rm B}\pars{x,y}= {\Gamma\pars{x}\Gamma\pars{y} \over \Gamma\pars{x + y}}}$ ( $\ds{\Gamma\pars{z}}$ is the GammaFunction ): \begin{align} I&=-\,{1 \over \pi}\ \lim_{\mu \to -1 \atop \nu \to 0} \partiald[2]{}{\mu}\partiald[2]{}{\nu}\bracks{% \sin\pars{\pi\bracks{{\nu \over 2} - \mu}} {\Gamma\pars{\mu - \nu/2 + 1}\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu/2 + 2}}} \\[3mm]&=-\ \overbrace{\lim_{\mu \to -1 \atop \nu \to 0} \partiald[2]{}{\mu}\partiald[2]{}{\nu}\bracks{% {\Gamma\pars{\nu + 1} \over \Gamma\pars{\nu/2 - \mu}\Gamma\pars{\mu + \nu/2 + 2}}}} ^{\ds{=\ -\,{11\pi^{4} \over 180}}} \end{align} where we used the identity $\ds{\Gamma\pars{z}\Gamma\pars{1 - z} = {\pi \over \sin\pars{\pi z}}}$