Closed form of $\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$ Today I discussed the following integral in the chat room

$$\int_0^\infty  \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$$
  where $0\leq a, b\leq \pi$ and $k>0$.

Some users suggested me that I can use Frullani's theorem:
$$\int_0^\infty \frac{f(ax) - f(bx)}{x} = \big[f(0) - f(\infty)\big]\ln \left(\frac ab\right)$$
So I tried to work with that way.
\begin{align}   I&=\int_0^\infty  \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}\\   &=\int_0^\infty \frac{\ln \left( x^2+2kx\cos b+k^2\right)-\ln \left( x^2+2kx\cos a+k^2\right)}{x}\mathrm dx\tag{1}\\   &=\int_0^\infty \frac{\ln \left( 1+\dfrac{2k\cos b}{x}+\dfrac{k^2}{x^2}\right)-\ln \left( 1+\dfrac{2k\cos a}{x}+\dfrac{k^2}{x^2}\right)}{x}\mathrm dx\tag{2}\\   \end{align}
The issue arose from $(1)$ because $f(\infty)$ diverges and the same issue arose from $(2)$ because $f(0)$ diverges. I then tried to use D.U.I.S. by differentiating w.r.t. $k$, but it seemed no hope because WolframAlpha gave me this horrible form. Any idea? Any help would be appreciated. Thanks in advance.

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$\ds{\int_{0}^{\infty}\ln\pars{x^{2} + 2kx\cos\pars{b} + k^{2}\over
     x^{2} + 2kx\cos\pars{a} + k^2}\,{\dd x \over x}:\ {\large ?}
     \,,\qquad 0 <\ a\,,\ b\ <\pi\,,\quad k > 0.}$

\begin{align}&\overbrace{\color{#66f}{\large\int_{0}^{\infty}
\ln\pars{x^{2} + 2kx\cos\pars{b} + k^{2}\over
         x^{2} + 2kx\cos\pars{a} + k^2}\,{\dd x \over x}}}
^{\dsc{x \over k}\ \ds{\mapsto}\ \dsc{x}}
\\[5mm]&=\int_{0}^{\infty}
\ln\pars{x^{2} + 2x\cos\pars{b} + 1\over x^{2} + 2x\cos\pars{a} + 1}
\,{\dd x \over x}
=\lim_{R\ \to\ \infty}\bracks{\fermi\pars{a,R} - \fermi\pars{b,R}}
\\[5mm]&\mbox{where}\
\begin{array}{|c|}\hline\\
\ \fermi\pars{\mu,R}\equiv
\int_{0}^{R}\ln\pars{x}
{2x + 2\cos\pars{\mu}\over x^{2} + 2x\cos\pars{\mu} + 1}\,\dd x\,,\quad 0 < \mu < \pi\
\\ \\ \hline
\end{array}\qquad\qquad\quad\pars{1}
\end{align}

Note that $\ds{r_{-} \equiv \exp\pars{\bracks{\pi - \mu}\ic}}$ and
$\ds{r_{+} \equiv \exp\pars{\bracks{\pi + \mu}\ic}}$ are the roots of
$\ds{x^{2} + 2x\cos\pars{\mu} + 1 = 0}$ such that $\ds{\pars{~\mbox{with}\ R > 1~}}$:
\begin{align}&\dsc{\int_{0}^{R}\ln^{2}\pars{x}
{x + \cos\pars{\mu}\over x^{2} + 2x\cos\pars{\mu} + 1}\,\dd x}
\\[5mm]&=2\pi\ic\bracks{\half\,\ln^{2}\pars{r_{-}} + \half\,\ln^{2}\pars{r_{+}}}
-\int_{R}^{0}\bracks{\ln\pars{x} + 2\pi\ic}^{2}
{x + \cos\pars{\mu}\over x^{2} + 2x\cos\pars{\mu} + 1}\,\dd x
\\[5mm]&=\pi\ic\bracks{-\pars{\pi - \mu}^{2} - \pars{\pi + \mu}^{2}}
+\dsc{\int_{0}^{R}\ln^{2}\pars{x}
{x + \cos\pars{\mu}\over x^{2} + 2x\cos\pars{\mu} + 1}\,\dd x}
\\[5mm]&+2\pi\ic\ \overbrace{\int_{0}^{R}\ln\pars{x}
{2x + 2\cos\pars{\mu}\over x^{2} + 2x\cos\pars{\mu} + 1}\,\dd x}
^{\dsc{\fermi\pars{\mu,R}}}\ +\
\pars{2\pi\ic}^{2}\int_{0}^{R}
{x + \cos\pars{\mu} \over x^{2} + 2x\cos\pars{\mu} + 1}\,\dd x
\\[5mm]&-{\mathfrak C}\pars{R,\mu}
\end{align}
where $\ds{\left.{\mathfrak C}\pars{R,\mu}
\equiv\oint\ln^{2}\pars{z}
{z + \cos\pars{\mu}\over z^{2} + 2z\cos\pars{\mu} + 1}\,\dd z\,\right\vert
_{z\ \equiv\ R\expo{\ic\theta}\,,\ 0\ <\ \theta\ <\ 2\pi}}$

This expression leads to:
  \begin{align}
0&=-2\pi\ic\pars{\pi^{2} + \mu^{2}} + 2\pi\ic\int_{0}^{R}\ln\pars{x}
{2x + 2\cos\pars{\mu}\over x^{2} + 2x\cos\pars{\mu} + 1}\,\dd x
\\[5mm]&+\pars{2\pi\ic}^{2}\int_{\cos\pars{\mu}}^{R + \cos\pars{\mu}}
{x \over x^{2} + \sin^{2}\pars{\mu}}\,\dd x - {\mathfrak C}\pars{R,\mu}
\end{align}

and
\begin{align}
\fermi\pars{\mu,R}&=\int_{0}^{R}\ln\pars{x}
{2x + 2\cos\pars{\mu}\over x^{2} + 2x\cos\pars{\mu} + 1}\,\dd x
\\[5mm]&=\pi^{2} + \mu^{2}
-\pi\ic\ln\pars{\bracks{R + \cos\pars{\mu}}^{2} + \sin^{2}\pars{\mu}}
+ {{\mathfrak C}\pars{R,\mu} \over 2\pi\ic}
\end{align}

In the $\ds{R \to \infty}$ we'll have:
  \begin{align}
\lim_{R \to\ \infty}\bracks{\fermi\pars{a,R} - \fermi\pars{b,R}}
&=a^{2} - b^{2}
\end{align}

such that expression $\pars{1}$ is reduced to:
\begin{align}&\color{#66f}{\large\int_{0}^{\infty}
\ln\pars{x^{2} + 2kx\cos\pars{b} + k^{2}\over
         x^{2} + 2kx\cos\pars{a} + k^2}\,{\dd x \over x}}
=\color{#66f}{\large a^{2} - b^{2}}
\end{align}
