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.

First note that by substituting $x\mapsto kx$, we get $$ \int_0^\infty\log\left(\frac{x^2+2kx\cos(a)+k^2}{x^2+2kx\cos(b)+k^2}\right)\frac{\mathrm{d}x}{x} =\int_0^\infty\log\left(\frac{x^2+2x\cos(a)+1}{x^2+2x\cos(b)+1}\right)\frac{\mathrm{d}x}{x} $$ Let $u=\frac{x+\cos(a)}{\sin(a)}$. Then $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}a}\int_0^\infty\log\left(\frac{x^2+2x\cos(a)+1}{x^2+2x+1}\right)\frac{\mathrm{d}x}{x} &=-2\int_0^\infty\frac{\sin(a)}{x^2+2x\cos(a)+1}\,\mathrm{d}x\\ &=-2\int_0^\infty\frac{\sin(a)}{(x+\cos(a))^2+\sin^2(a)}\,\mathrm{d}x\\ &=-\frac2{\sin(a)}\int_0^\infty\frac1{\frac{(x+\cos(a))^2}{\sin^2(a)}+1}\,\mathrm{d}x\\ &=-2\int_{\cot(a)}^\infty\frac1{u^2+1}\,\mathrm{d}u\\[9pt] &=-2a \end{align} $$ Integrating in $a$ gives $$ \int_0^\infty\log\left(\frac{x^2+2x\cos(a)+1}{x^2+2x+1}\right)\frac{\mathrm{d}x}{x} =-a^2 $$ Therefore, by subtraction, $$ \int_0^\infty\log\left(\frac{x^2+2kx\cos(a)+k^2}{x^2+2kx\cos(b)+k^2}\right)\frac{\mathrm{d}x}{x} =b^2-a^2 $$

\begin{align} \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{1}{x}\int_a^b \frac{\mathrm d}{\mathrm dy}\ln \left(x^2+2kx\cos y+k^2\right) \;\mathrm dy\,\mathrm dx\\[10pt] &=-\int_0^\infty\frac{1}{x}\int_a^b \frac{2kx\sin y}{x^2+2kx\cos y+k^2} \;\mathrm dy\,\mathrm dx\\[10pt] &=2k\int_b^a\sin y\int_0^\infty \frac{\mathrm dx}{x^2+2kx\cos y+k^2} \;\mathrm dy\\[10pt] &=2k\int_b^a\sin y\int_0^\infty \frac{\mathrm dx}{\left(x+k\cos y\right)^2+k^2-k^2\cos^2y} \;\mathrm dy\\[10pt] &=2k\int_b^a\sin y\underbrace{\int_0^\infty \frac{\mathrm dx}{\left(x+k\cos y\right)^2+k^2\sin^2y}}_{\large\color{blue}{(k\sin y)\, u\,=\,x+k\cos y}} \;\mathrm dy\\[10pt] &=2\int_b^a\int_{\cot y}^\infty \frac{\mathrm du}{u^2+1} \;\mathrm dy\\[10pt] &=2\int_b^a\left(\frac{\pi}{2}-\arctan\left(\cot y\right)\right)\;\mathrm dy\\[10pt] &=2\int_b^ay\;\mathrm dy\\[10pt] &=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large a^2-b^2}} \end{align}

Here is a complex-analytic method: Notice that

$$ \int_{0}^{\infty} \log\left( \frac{x^{2} + 2x\cos b + 1}{x^{2} + 2x\cos a + 1} \right) \frac{dx}{x} = 2 \int_{0}^{\infty} \Re \frac{\log(1 + e^{ib}x) - \log(1 + e^{ia}x)}{x} \, dx. \tag{1} $$

Let $R$ be a positive large number. Then the function $z \mapsto \log(1+z)/z$ is analytic on $\Bbb{C} \setminus (-\infty, 1]$ with the standard branch cut and we get

\begin{align*} \int_{0}^{R} \frac{\log(1 + e^{ib}x)}{x} \, dx &=\int_{0}^{Re^{ib}} \frac{\log(1 + z)}{z} \, dz \qquad (z = e^{ib}x) \\ &= \int_{0}^{R} \frac{\log(1 + z)}{z} \, dz + \int_{R}^{Re^{ib}} \frac{\log(1 + z)}{z} \, dz \\ &= \int_{0}^{R} \frac{\log(1 + z)}{z} \, dz + i \int_{0}^{b} \log(1 + Re^{i\theta}) \, d\theta, \quad (z=Re^{i\theta}) \end{align*}

and likewise for the integral with $b$ replaced by $a$. This shows that

$$ \int_{0}^{R} \frac{\log(1 + e^{ib}x) - \log(1 + e^{ia}x)}{x} \, dx = i \int_{a}^{b} \log(1 + Re^{i\theta}) \, d\theta. $$

Multiplying by 2 and taking real part, we get

$$ 2 \int_{0}^{R} \Re \frac{\log(1 + e^{ib}x) - \log(1 + e^{ia}x)}{x} \, dx = - 2 \Im \int_{a}^{b} \log(R^{-1} + e^{i\theta}) \, d\theta. $$

(Here we exploited the fact that $\log R$ is real.) Taking $R \to \infty$, in view of the identity $\text{(1)}$ it follows that

$$ \int_{0}^{\infty} \log\left( \frac{x^{2} + 2x\cos b + 1}{x^{2} + 2x\cos a + 1} \right) \frac{dx}{x} = - \int_{a}^{b} 2\theta \, d\theta = a^{2} - b^{2}. $$

Another approach is to use the Fourier series $$\sum_{k=1}^{\infty}\frac{x^{k} \cos(ka)}{k} = - \frac{1}{2} \log \left(x^{2} - 2 x \cos(a) +1 \right) \ , \ |x| <1 $$ which can be derived from the Maclaurin series of $\log(1-z)$ by replacing $z$ with $xe^{ia}$ and equating the real parts on both sides.

$$ \begin{align} \int_0^\infty\log\left(\frac{x^2+2kx\cos(b)+k^2}{x^2+2kx\cos(a)+k^2}\right)\frac{dx}{x} &=\int_0^\infty\log\left(\frac{u^2+2u\cos(b)+1}{u^2+2u\cos(a)+1}\right)\frac{du}{u} \tag{1} \\ &= 2 \int_{0}^{1} \tag{2} \log\left(\frac{u^2+2u\cos(b)+1}{u^2+2u\cos(a)+1}\right)\frac{du}{u} \\ &= 4 \int_{0}^{1} \frac{1}{u} \sum_{k=1}^{\infty} (-u)^{k} \frac{\cos(ka) - \cos(kb)}{k} \\ &= 4 \sum_{k=1}^{\infty} (-1)^{k} \frac{\cos(ka) -\cos(kb)}{k} \int_{0}^{1} u^{k-1} \ du \\ &= 4 \sum_{k=1}^{\infty} (-1)^{k} \frac{\cos(ka) - \cos(kb)}{k^{2}} \\ &= 4 \left(\frac{a^{2}}{4} - \frac{b^{2}}{4} \right) \tag{3} \\ &= a^{2}- b^{2} \end{align} $$

$ $

$(1)$ Let $ \displaystyle u= \frac{x}{k}.$

$(2)$ Separate the integral into two integrals, namely one over the interval $(0,1)$ and one over the interval $(1, \infty)$, and replace $u$ with $\frac{1}{u}$ in the second integral.

$(3)$ By integrating the Fourier series $ \displaystyle \sum_{k=1}^{\infty} (-1)^{k} \frac{\sin(ka)}{k} = - \frac{a}{2} \ , \ |a| < \pi$, we get $$ \sum_{k=1}^{\infty} (-1)^{k} \frac{\cos(ka)}{k^{2}} = \frac{a^{2}}{4} - \frac{\pi^{2}}{12} \ , \ |a| < \pi .$$