I read in a paper about Frullani integrals the following claim
$$ 
\begin{align\*}
I & :=\int\_0^\infty \frac{1}{x}\log\left(\frac{1 + 2\alpha \cos px + \alpha^2}{1 + 2\alpha \cos qx + \alpha^2}\right)\,\mathrm{d}x 
 =
\left\{ 
 \begin{array}{lrr} 
 \log \frac{p}{q} \log\left( 1 + \alpha^2\right) & \text{if} & \alpha^2>1\\
 \log \frac{p}{q} \log\left( 1 + \frac{1}{\alpha}\right) & \text{if} & \alpha^2<1
 \end{array} 
\right.
\end{align\*}
$$
Where the author points at the dangers of using this method. The problem
lies in the fact that the function $f(n,r) := \log(1+2n \cos rx + n^2)$ does not have a limit
as $x\to \infty$. And hence the [Frullani identity](https://math.stackexchange.com/questions/280329/frullani-identity-justifying-the-application-of-fubini-theorem) can not be used. 

A very similar integral $f(n,-r)$ has been evaluated several times before. 
See for an example [A question in Complex Analysis $\int\_0^{2\pi}\log(1-2r\cos x +r^2)\,dx$](https://math.stackexchange.com/questions/352046/a-question-in-complex-analysis-int-02-pi-log1-2r-cos-x-r2-dx?lq=1)
and can also be solved by [differentiating under the integral sign](http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign#Example_3) . Even $f(n,-r)^2$ has been evaluated. 

My question is: Can any of this be used to prove the claim above?