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!


$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{I \equiv {1 \over \pi}\int_{0}^{\pi}\theta^{2}\ln^{2}\pars{2\cos\pars{\theta \over 2}} \,\dd\theta = {11\pi^{4} \over 180} = {11 \over 2}\,\zeta\pars{4}}$

\begin{align} I&={1 \over 2\pi}\int_{-\pi}^{\pi}\theta^{2} \ln^{2}\pars{2\root{1 + \cos\pars{\theta} \over 2}}\,\dd\theta \\[3mm]&={1 \over 2\pi} \int_{\verts{z} = 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}} \bracks{-\ln^{2}\pars{z}}\ln^{2}\pars{\root{2}\root{1 + {z^{2} + 1 \over 2z}}}\, {\dd z \over \ic z} \\[3mm]&={\ic \over 2\pi} \int_{\verts{z} = 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}} \ln^{2}\pars{z}\ln^{2}\pars{z + 1 \over z^{1/2}}\,{\dd z \over z} \\[3mm]&={\ic \over 2\pi}\lim_{\mu \to -1 \atop \nu \to 0} \partiald[2]{}{\mu}\partiald[2]{}{\nu} \int_{\verts{z} = 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}} z^{\mu}\pars{z + 1 \over z^{1/2}}^{\nu}\,\dd z \\[3mm]&={\ic \over 2\pi}\lim_{\mu \to -1 \atop \nu \to 0} \partiald[2]{}{\mu}\partiald[2]{}{\nu} \int_{\verts{z} = 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}}z^{\mu - \nu/2}\pars{z + 1}^{\nu}\,\dd z \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\pars{1} \end{align}

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.

With $\pars{1}$ and $\pars{2}$, $\ds{I}$ is reduced to: $$ I=-\,{1 \over \pi}\lim_{\mu \to -1 \atop \nu \to 0} \partiald[2]{}{\mu}\partiald[2]{}{\nu}\bracks{% \sin\pars{\pi\bracks{{\nu \over 2} - \mu}} {\rm B}\pars{\mu - {\nu \over 2} + 1,\nu + 1}} $$

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 $\ds{\Gamma\pars{z}\Gamma\pars{1 - z} = {\pi \over \sin\pars{\pi z}}}$

Then, $$ I \equiv \color{#00f}{\large% {1 \over \pi}\int_{0}^{\pi}\theta^{2}\ln^{2}\pars{2\cos\pars{\theta \over 2}} \,\dd\theta} = \color{#00f}{\large{11\pi^{4} \over 180} = {11 \over 2}\,\zeta\pars{4}} $$

  • $\begingroup$ This is an excellent solution. Thanks a lot Felix Marin. I had thought I had commented earlier on this. Sorry for the delay, this answer is GREAT. Nice Integral $\endgroup$ – Jeff Faraci Mar 20 '14 at 0:00
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    $\begingroup$ @Jeff Thanks a lot. You're welcome. $\endgroup$ – Felix Marin Mar 20 '14 at 0:33
  • $\begingroup$ you may appreciate this: math.uwo.ca/~dborwein/cv/zeta4.pdf $\endgroup$ – Jeff Faraci Apr 27 '14 at 19:05
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    $\begingroup$ @Integrals That link is very helpful. Manipulation of Harmonic Numbers are always somehow tricky and that paper has a lot about them. Thanks. $\endgroup$ – Felix Marin Apr 27 '14 at 21:49

One way to go about this is to use the identity:

$$\int_{0}^{\frac{\pi}{2}}\cos^{p-1}(x)\cos(ax)dx=\frac{\pi}{2^{p}}\cdot \frac{\Gamma(p)}{\Gamma\left(\frac{a+p+1}{2}\right)\Gamma\left(\frac{p-a+1}{2}\right)}....(1)$$

Then, diff this twice w.r.t 'a', and let a=0.

Then, diff twice w.r.t p and let p=1.

The diffing on the right side may be a little tedious, but get tech to do it.

diffing once w.r.t p will give you $$\int_{0}^{\frac{\pi}{2}}x^{2}\ln(\cos(x))dx=\frac{-\pi^{3}}{24}\ln(2)-\frac{\pi}{3}\zeta(3)$$.

Then, diff again to get the integral in question.

a fun way to go about evaluating this integral is to use contours.

consider $$f(z)=zlog^{3}(1+e^{2iz})$$ over a rectangular contour with vertices $$-\frac{\pi}{2}, \;\ \frac{\pi}{2}, \;\ \frac{\pi}{2}+Ri, \;\ \frac{-\pi}{2}+Ri$$, with quarter-circle indents around $\pm \frac{\pi}{2}$.

A good while back, Nick Strehle wrote up a nice post on this method of evaluating log-trig integrals via residues. It is on the site somewhere if you nose around.


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