TITLE: How do I evaluate this integral $\int_0^\pi{\frac{{{x^2}}}{{\sqrt 5-2\cos x}}}\operatorname d\!x$?
QUESTION [20 upvotes]: Show that $$\int\limits_0^\pi{\frac{{{x^2}}}{{\sqrt 5-2\cos x}}}\operatorname d\!x =\frac{{{\pi^3}}}{{15}}+2\pi \ln^2 \left({\frac{{1+\sqrt 5 }}{2}}\right).$$

I don't have any idea how to start, but maybe I could use the Polylogarithm.

REPLY [6 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{\color{#c00000}{\int_{0}^{\pi}{x^{2} \over \sqrt 5 - 2\cos\pars{x}}\,\dd x}
     ={\pi^{3} \over 15} + 2\pi\,\ln^{2}\pars{1 + \sqrt 5 \over 2}:\ {\large ?}}$

$$
\color{#c00000}{\int_{0}^{\pi}{x^{2} \over \sqrt 5 - 2\cos\pars{x}}\,\dd x}
=\half\int_{-\pi}^{\pi}{x^{2} \over 2\varphi - 1 - 2\cos\pars{x}}\,\dd x\tag{1}
$$
  where $\ds{\varphi \equiv {1 + \root{5} \over 2} \approx 1.6180}$ is the
  Golden Ratio.

\begin{align}
&\color{#c00000}{\int_{0}^{\pi}{x^{2} \over \sqrt 5 - 2\cos\pars{x}}\,\dd x}
=\half
\int_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}}
{-\ln^{2}\pars{z} \over 2\varphi - 1 -2\,\pars{z^{2} + 1}/\pars{2z}}
\,{\dd z \over \ic z}
\\[3mm]&=-\,\half\,\ic
\int_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}}
{\ln^{2}\pars{z} \over z^{2} - \pars{2\varphi - 1}z + 1}
\,\dd z\tag{2}
\end{align}
Zeros $\ds{z_{\pm}}$ of $\ds{z^{2} - \pars{2\varphi - 1}z + 1 = 0}$ are given by:
$$
z_{+} = \varphi > 1\,;\qquad\qquad z_{-} = {1 \over \varphi}\,,\quad 0 < z_{-} < 1
$$

Expression $\pars{2}$ is reduced to:
  \begin{align}
&\color{#c00000}{\int_{0}^{\pi}{x^{2} \over \sqrt 5 - 2\cos\pars{x}}\,\dd x}
=-\,\half\,\ic\braces{2\pi\ic\lim_{z \to z_{-}}\bracks{%
{\pars{z - z_{-}}\ln^{2}\pars{z} \over z^{2} - \pars{2\varphi - 1}z + 1}}}
\\[3mm]&\mbox{}+ \half\,\ic\int_{-1}^{0}
{\bracks{\ln\pars{-x} + \ic\pi}^{2} \over x^{2} - \pars{2\varphi - 1}x + 1}\,\dd x
+ \half\,\ic\int_{0}^{-1}
{\bracks{\ln\pars{-x} - \ic\pi}^{2} \over x^{2} - \pars{2\varphi - 1}x + 1}\,\dd x
\\[3mm]&=\pi\,{\ln^{2}\pars{1/\varphi} \over 2\pars{1/\varphi} - \pars{2\varphi - 1}}
+ \half\,\ic\int_{0}^{-1}
{-4\pi\ic\ln\pars{-x} \over x^{2} - \pars{2\varphi - 1}x + 1}\,\dd x\tag{3}
\end{align}

Since $\ds{2\,{1 \over \varphi} - \pars{2\varphi - 1} = -1}$, expression $\pars{3}$
becomes:
\begin{align}
&\color{#c00000}{\int_{0}^{\pi}{x^{2} \over \sqrt 5 - 2\cos\pars{x}}\,\dd x}
=-\pi\ln^{2}\pars{\varphi}
-2\pi\int_{0}^{1}{\ln\pars{x} \over x^{2} + \pars{2\varphi - 1}x + 1}\,\dd x
\tag{4}
\end{align}
Roots of $\ds{x^{2} + \pars{2\varphi - 1}x + 1 = 0}$ are $\ds{-\varphi}$
and $\ds{-\,{1 \over \varphi}}$ such that $\pars{4}$ is written as:

\begin{align}
&\color{#c00000}{\int_{0}^{\pi}{x^{2} \over \sqrt 5 - 2\cos\pars{x}}\,\dd x}
\\[3mm]&=-\pi\ln^{2}\pars{\varphi} - 2\pi\int_{0}^{1}\ln\pars{x}\bracks{%
\pars{{1 \over x + \varphi} - {1 \over x + 1/\varphi}}\
\overbrace{\pars{{1 \over \varphi} - \varphi}^{-1}}^{\ds{=\ -1}}}\,\dd x
\\[3mm]&=-\pi\ln^{2}\pars{\varphi}
+ 2\pi\bracks{%
\int_{0}^{1}{\ln\pars{x} \over \varphi + x}\,\dd x
-\int_{0}^{1}{\ln\pars{x} \over 1/\varphi + x}\,\dd x}\tag{5}
\end{align}

With $\ds{a > 0}$:
\begin{align}
\int_{0}^{1}{\ln\pars{x} \over a + x}\,\dd x&=
-\int_{0}^{1}{\ln\pars{-a\bracks{-x/a}} \over 1 - \pars{-x/a}}\,
\pars{-\,{\dd x \over a}}
=-\int_{0}^{-1/a}{\ln\pars{-ax} \over 1 - x}\,\dd x
\\[3mm]&=\left.\vphantom{\LARGE A}
\ln\pars{1 - x}\ln\pars{-ax}\,\right\vert_{\,x\ =\ 0}^{\,x\ =\ -1/a}
-\int_{0}^{-1/a}\ln\pars{1 - x}\,{-a \over -ax}\,\dd x
\\[3mm]&=\int_{0}^{-1/a}{-\ln\pars{1 - x} \over x}\,\dd x
=\int_{0}^{-1/a}{{\rm Li}_{1}\pars{x} \over x}\,\dd x\tag{6}
\end{align}
where $\ds{{\rm Li}_{s}\pars{z}}$ is the
Polylogarithm Function which satisfies the recurrence relation
$${\rm Li}_{s + 1}\pars{z} =
\int_{0}^{z}{{\rm Li}_{s}\pars{t} \over t}\,\dd t\qquad\mbox{and}
\qquad -\ln\pars{1 - z} = {\rm Li}_{1}\pars{z}
$$
$$
\mbox{Expression}\ \pars{6}\ \mbox{becomes}\quad
\begin{array}{|c|}\hline\\ 
\quad\int_{0}^{1}{\ln\pars{x} \over x + a}\,\dd x
={\rm Li}_{2}\pars{-\,{1 \over a}}\quad
\\ \\ \hline
\end{array}
\,,\qquad\qquad{\rm Li}_{2}\pars{0} = 0
$$
such that $\ds{\pars{~\mbox{see expression}\ \pars{5}~}}$
$$
\int_{0}^{1}{\ln\pars{x} \over x + \varphi}\,\dd x
-\int_{0}^{1}{\ln\pars{x} \over x + 1/\varphi}\,\dd x
={\rm Li}_{2}\pars{-\,{1 \over \varphi}}
-{\rm Li}_{2}\pars{-\varphi}
$$

Expression $\pars{5}$ is reduced to:
  $$
\color{#c00000}{\int_{0}^{\pi}{x^{2} \over \sqrt 5 - 2\cos\pars{x}}\,\dd x}
=-\pi\ln^{2}\pars{\varphi}
+ 2\pi\bracks{%
{\rm Li}_{2}\pars{-\,{1 \over \varphi}} - {\rm Li}_{2}\pars{-\varphi}}\tag{7}
$$
  With the identities
  \begin{align}&\left.\begin{array}{rcl}
{\rm L_{i}}_{2}\pars{-\,{1 \over \varphi}}
& = & -\,{\pi^{2} \over 15} + \half\,\ln^{2}\pars{\varphi}
\\[1mm]
{\rm L_{i}}_{2}\pars{-\varphi}
& = & -\,{\pi^{2} \over 10} - \ln^{2}\pars{\varphi}
\\[5mm]
\mbox{we find}\quad
{\rm Li}_{2}\pars{-\,{1 \over \varphi}} - {\rm Li}_{2}\pars{-\varphi}
& = & {\pi^{2} \over 30} + {3 \over 2}\,\ln^{2}\pars{\varphi}
\end{array}\right\rbrace
\end{align}

the final result is given by:
$$
\color{#00f}{\large\int_{0}^{\pi}{x^{2} \over \sqrt 5 - 2\cos\pars{x}}\,\dd x}
={\pi^{3} \over 15} + 2\pi^{2}\ln\pars{\varphi}
=\color{#00f}{\large{\pi^{3} \over 15} + 2\pi\ln^{2}\pars{\root{5} + 1 \over 2}}
$$