A contour integration approach for $\int_{0}^{\infty} \frac{\ln \left(x^{4}+2 x^{2} \cos 2 a+1\right)}{1+x^{2}} \, d x$

Question

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!

Answer 1

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 $$ \begin{align} \int_{0}^{\infty} \frac{\ln(x^{4}+2x^{2}\cos(2a)+1)}{1+x^{2}} \, \mathrm dx &= \Re \, 2 \pi i \left( \operatorname{Res}[f(z), i] - \operatorname{Res}[g(z), -i] \right) \\ &= \Re \, 2 \pi i\left(\frac{\ln(i+ie^{ia})}{2i} - \frac{\ln(-i-ie^{ia})}{-2i} \right) \\ &= \pi \left(\frac{1}{2} \, \ln \left(2+2 \cos(a) \right) + \frac{1}{2} \, \ln \left(2+2 \cos(a) \right)\right) \\ &= \pi \ln \left(4 \cos^{2} \left(\frac{a}{2} \right) \right) \\ &= 2 \pi \ln \left(2 \cos \left(\frac{a}{2} \right) \right). \end{align}$$

Answer 2

There are four ugly branch cuts so lets try Feynman's trick. Let $I(a)$ be the given integral. Then, $$I'(a)=\int_0^\infty\frac{-4x^2\sin(2a)}{(1+x^2)(x^4+2x^2\cos2a+1)}dx.$$ By means of the contour integral around the closed counter-clockwise semicircular contour and the residue theorem we get $$\begin{align} I'(a)&=\pi i(Res_{z=\Large i}f(z)+Res_{\Large z=ie^{ai}}f(z)+Res_{\Large z=ie^{-ai}}f(z))\\ &=\pi i(-i\cot a+\frac{e^{ai}}{1-e^{2ai}}-\frac{e^{-ai}}{1-e^{-2ai}})\\ &=-\pi\tan\frac a2. \end{align}$$ Hence, $I(a)=2\pi\ln(\cos\frac a2)+c.$ But, $c=I(0)=\int_0^\infty\frac{2\ln(x^2+1)}{x^2+1}dx=2\pi\ln2.$ Hence, $I(a)=2\pi\ln(2\cos\frac a2).$