Background
In this week, I am tackling the integral
$$\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}} d x$$
and found that a general formula below in my post, $$
\begin{aligned}
J(a)= \int_{0}^{\infty} \frac{\ln \left(1+2 x \sin a+x^{2}\right)}{1+x^{2}} d x=& \pi\left(\ln \left(2 \cos \frac{a}{2}\right)\right)+|a| \ln \left(\tan \frac{|a|}{2}\right)-2 \operatorname{sgn} (a) \int_{0}^{\frac{|a|}{2}} \ln (\tan x) d x
\end{aligned}
$$
by which, I found a decent formula for the quartic one, $$
\boxed{I(a)=\int_{0}^{\infty} \frac{\ln \left(x^{4}+2 x^{2} \cos 2 a+1\right)}{1+x^{2}} d x=2 \pi \ln \left(2 \cos \frac{a}{2}\right)}
$$
Proof:
$$
\begin{aligned}
I(a)&=\int_{0}^{\infty} \frac{\ln \left(x^{4}+2 x^{2} \cos 2 a+1\right)}{1+x^{2}} d x\\
&=\int_{0}^{\infty} \frac{\ln \left[\left(x^{2}+1\right)^{2}+2 x^{2}(\cos 2 a-1)+1\right]}{1+x^{2}} d x\\
&=\int_{0}^{\infty} \frac{\ln \left[\left(x^{2}+1\right)^{2}-4 x^{2} \sin ^{2} a\right]}{1+x^{2}} d x\\
&=\int_{0}^{\infty} \frac{\ln \left(x^{2}+2 x \sin a+1\right)}{1+x^{2}}+\int_{0}^{\infty} \frac{\ln \left(x^{2}-2 x \sin a+1\right)}{1+x^{2}} d x\\& =J\left(\frac{a}{2}\right)+J\left(-\frac{a}{2}\right)\\& =2 \pi \ln \left(2 \cos \frac{a}{2}\right)
\end{aligned}
$$
For examples:
$$
K=\int_{0}^{\infty} \frac{\ln \left(x^{4}+1\right)}{1+x^{2}} d x= I\left(\frac{\pi}{4}\right)=2 \pi \ln \left(2 \cos \frac{\pi}{8}\right)= \pi \ln (2+\sqrt{2})
$$
$$
\begin{aligned}
L=\int_{0}^{\infty} \frac{\ln \left(x^{4}+x^{2}+1\right)}{1+x^{2}} d x&=I\left(\frac{\pi}{6}\right)=2 \pi \ln \left(2 \cos \frac{\pi}{12}\right)=\pi\ln (2+\sqrt{3})
\end{aligned}
$$
$$
\begin{aligned}
M=\int_{0}^{\infty} \frac{\ln \left(x^{4}-x^{2}+1\right)}{1+x^{2}} d x&=I\left(\frac{\pi}{3}\right)=2 \pi \ln \left(2 \cos \frac{\pi}{6}\right)=\pi\ln 3
\end{aligned}
$$
$$
\begin{aligned}
N=&\int_{0}^{\infty} \frac{\ln \left(x^{8}+x^{4}+1\right)}{1+x^{2}} dx = L+M=\pi \ln (6+3 \sqrt{3})
\end{aligned}
$$
Last but not least,
$$
\boxed{\int_{0}^{\infty} \frac{\ln \left(x^{4}+b x^{2}+1\right)}{1+x^{2}} d x = 2 \pi \ln \left[2 \cos \left(\frac{1}{4} \cos ^{-1}\left(\frac{b}{2}\right)\right)\right]}
$$
$$
\begin{aligned}
\int_{0}^{\infty} \frac{\ln \left(x^{8}+1\right)}{1+x^{2}}=& \int_{0}^{\infty} \frac{\ln \left(x^{4}+\sqrt{2} x+1\right)}{1+x^{2}} d x +\int_{0}^{\infty} \frac{\ln \left(x^{4}-\sqrt{2} x+1\right)}{1+x^{2}} d x \\
=& 2 \pi \ln \left[2 \cos \frac{1}{4} \cos ^{-1}\left(\frac{\sqrt{2}}{2}\right)\right] +2 \pi \ln \left[2 \cos \frac{1}{4} \cos ^{-1}\left(-\frac{\sqrt{2}}{2}\right)\right] \\
=& 2 \pi \ln \left(2 \cos \frac{\pi}{16}\right)+2 \pi \ln \left(2 \cos \frac{3 \pi}{16}\right)\\
=& 2 \pi \ln \left(2^2 \cos \frac{\pi}{16} \cos \frac{3 \pi}{16}\right)\\=& 2 \pi \ln (\sqrt{2}+\sqrt{2+\sqrt{2}})
\end{aligned}
$$
Eager to know whether it can be proved by contour integration.
Your help and solutions are highly appreciated!
Best Answer
Assume that $0 \le a < \frac{\pi}{2}$.
Using the principal branch of the logarithm, we have $$ \begin{align}\int_{0}^{\infty} \frac{\ln(x^{4}+2x^{2}\cos(2a)+1)}{1+x^{2}} \, \mathrm dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\ln(x^{4}+2x^{2}\cos(2a)+1)}{1+x^{2}} \, \mathrm dx \\ &= \Re\int_{-\infty}^{\infty} \frac{\ln(x^{2}+e^{2ia})}{1+x^{2}} \, \mathrm dx \\ &= \Re \int_{-\infty}^{\infty} \frac{\ln \left((x+ie^{ia})(x-ie^{ia}) \right)}{1+x^{2}} \, \mathrm dx \\ &= \Re \left( \int_{-\infty}^{\infty} \frac{\ln(x+ie^{ia})}{1+x^{2}} \, \mathrm dx + \int_{-\infty}^{\infty}\frac{\ln(x-ie^{ia})}{1+x^{2}} \, \mathrm dx \right) \\ &= \Re \left( \int_{-\infty}^{\infty} f(x) \, \mathrm dx + \int_{-\infty}^{\infty}\ g(x) \, \mathrm dx \right). \end{align} $$
(The identity $\ln(z_{1}z_{2}) = \ln(z_{1}) + \ln(z_{2})$ holds if $-\pi < \arg(z_{1}) +\arg (z_{2}) \le \pi$).
The function $f(z)$ has has a branch cut in the lower half-plane, while the function $g(z)$ has a branch cut in the upper half-plane.
Therefore, we'll integrate $f(z)$ counterclockwise around a closed semicircular contour in the upper half-plane, and we'll integrate $g(z)$ clockwise around a closed semicircular contour in the lower half-plane.
We get $$ \begin{align} \int_{0}^{\infty} \frac{\ln(x^{4}+2x^{2}\cos(2a)+1)}{1+x^{2}} \, \mathrm dx &= \Re \, 2 \pi i \left( \operatorname{Res}[f(z), i] - \operatorname{Res}[g(z), -i] \right) \\ &= \Re \, 2 \pi i\left(\frac{\ln(i+ie^{ia})}{2i} - \frac{\ln(-i-ie^{ia})}{-2i} \right) \\ &= \pi \left(\frac{1}{2} \, \ln \left(2+2 \cos(a) \right) + \frac{1}{2} \, \ln \left(2+2 \cos(a) \right)\right) \\ &= \pi \ln \left(4 \cos^{2} \left(\frac{a}{2} \right) \right) \\ &= 2 \pi \ln \left(2 \cos \left(\frac{a}{2} \right) \right). \end{align}$$