TITLE: Yet another difficult integration question
QUESTION [10 upvotes]: For $r \in (0,1)$ and $k \in \mathbb Z^+$, prove
$$ \frac{1}{\pi} \int_{0}^\pi \ln\left(1 + r \cos(u)\right) \ln \left( 1 + r \cos(3^k u)\right) du = \left(\ln\left(\frac{2(1-\sqrt{1-r^2})}{r^2}\right)\right)^2.$$
Given that for any $n \in \mathbb Z^+$
$$ - \frac{1}{\pi} \int_{0}^\pi \ln\left(1 + r \cos(nu)\right) du = \ln\left(\frac{2(1-\sqrt{1-r^2})}{r^2}\right).$$
In my mind, the strategy should be to use the fact that $\cos(u)$ and $\cos(3^ku)$ are orthogonal. Write $[0,\pi]$ as a product space where $\cos(u)$ and $\cos(3^ku)$ are constant along the (orthogonal) fibres. Then use Fubini's theorem. But I am having difficulty in putting the pieces together. Thanks for your help.

REPLY [3 votes]: For a positive integer $s$, and $r\in(0,1)$, let $I_s(r)$ be defined by
$$
I_s(r)=\frac{1}{\pi}\int_0^\pi\ln(1+r\cos(t))\ln(1+r\cos(s t))dt.
$$
We will prove the following result:

Proposition. For every positive integer $s$, and every $r\in(0,1)$ we have
  $$
I_s(r)=\left(\ln\left(\frac{2(1-\sqrt{1-r^2})}{r^2}\right)\right)^2
+\frac{2}{s}{\rm Li}_2\left(\left(\frac{\sqrt{1-r^2}-1}{r}\right)^{s+1}\right)
$$
  where ${\rm Li}_2$ is the well-known Dilogarithm.

This result shows that the proposed formula in the question is wrong.
${\it Proof.}$ In what follows $z=\frac{1-\sqrt{1-r^2}}{r}$. $z$ is the root that belongs to $(0,1)$ of the equation $z^2-\frac{2}{r}z+1=0$.
Now
$$\eqalign{
\ln(1+z^2+2z \cos t)&=\ln(|1+z e^{it}|^2)=2{\rm Re}\,{\rm Log}(1+ze^{it})\cr
&=\sum_{n=1}^\infty\frac{2(-1)^{n-1}z^n}{n}\cos(nt)
}
$$
using the fact that $\frac{2z}{1+z^2}=r$ we conclude that
$$
\ln(1+r \cos t)=-\ln\left(\frac{2z}{r}\right)+\sum_{n=1}^\infty\frac{2(-1)^{n-1}z^n}{n}\cos(nt)\tag{1}
$$
Replacing $t$ by $st$ we get also
$$
\ln(1+r \cos (st))=-\ln\left(\frac{2z}{r}\right)+\sum_{n=1}^\infty\frac{2(-1)^{n-1}z^n}{n}\cos(sn t)\tag{1}
$$
Using Parseval's formula we conclude that
$$ 
\frac{1}{2\pi}\int_{-\pi}^{\pi}
\ln(1+r \cos (t))\ln(1+r \cos (st))dt =\left(\ln\left(\frac{2z}{r}\right)\right)^2
+\frac{2}{s}\sum_{n=1}^\infty
\frac{(-z)^{(s+1) n}}{n^2}
$$
Or, equivalently
$$
I_s(r)=\left(\ln\left(\frac{2z}{r}\right)\right)^2
+\frac{2}{s}\,{\rm Li}_2((-z)^{s+1}).
$$
which is the announced result.$\qquad\square$
Finally, note that the proposed integral corresponds to $s=3^k$.