Is it hard to evaluate $\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}} d x$?

Question

Latest Edit

As suggested by @Quanto, $I(a)$ can be utilised to give more examples as below.

$$ \boxed{\begin{aligned} I(a)&= \int_{0}^{\infty} \frac{\ln \left(1+2 x \sin a+x^{2}\right)}{1+x^{2}} d x \\&= \pi\ln \left|2 \cos \frac{a}{2}\right|+a\ln \left|\tan \frac{a}{2}\right|-2 \operatorname{sgn} (a) \int_{0}^{\frac{|a|}{2}} \ln (\tan x) d x \end{aligned}} $$ For examples:

Example 1 $$\begin{aligned}\quad \int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}} d x&= I\left(-\frac{\pi}{6}\right)\\&= \frac{\pi}{2}[\ln (2+\sqrt{3})]+\frac{\pi}{6} \ln (2+\sqrt{3})+2\left(-\frac{2}{3} G\right)\\& =\frac{2 \pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G \end{aligned}$$

Example 2

$$ \begin{aligned} &\int_{0}^{\infty} \frac{\ln \left(1+\sqrt{2} x+x^{2}\right)}{1+x^{2}}\\=&I\left(\frac{\pi}{4}\right) \\ =& \pi \ln \left(2 \cos \frac{\pi}{8}\right)-\frac{\pi}{4} \ln \left(\tan \frac{\pi}{8}\right)-2 \int_{0}^{\frac{\pi}{8}} \ln (\tan x) d x \\ =& \frac{\pi}{2} \ln (2+\sqrt{2})+\frac{\pi}{4} \ln (\sqrt{2}+1) -2\left[\frac{\pi}{8} \ln (\sqrt{2}-1)-\Im\left(\operatorname{Li}_{2}(i(\sqrt{2}-1))\right]\right.\\=& \pi \ln [\sqrt[4]{2}(\sqrt{2}+1)] +2 \Im\left(\operatorname{Li}_{2}(i(\sqrt{2}-1))\right. \end{aligned} $$ where the last integral see post

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We are going to prove that $$\boxed{J=\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}} d x =\frac{2 \pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G }\tag*{} $$

by Feynman’s Technique Integration.

We first deal with its partner integral $$\displaystyle I=\int_{0}^{\infty} \frac{\ln \left(1+x+x^{2}\right)}{1+x^{2}} d x \tag*{} $$ which is parameterised by $\displaystyle I(a)=\int_{0}^{\infty} \frac{\ln \left(1+2 x \sin a+x^{2}\right)}{1+x^{2}} d x,\tag*{} $

where $ \displaystyle a\in [-\frac{\pi}{2}, \frac{\pi}{2}]. $ Differentiating $I(a)$ w.r.t. $a$ yields $\displaystyle \begin{aligned}I^{\prime}(a) &=\int_{0}^{\infty} \frac{2 x \cos a}{\left(1+x^{2}\right)\left(1+2 x \sin a+x^{2}\right)} d x \\&=\cot a\int_{0}^{\infty}\left(\frac{1}{1+x^{2}}-\frac{1}{1+2 x \sin a+x^{2}}\right) d x \\&=\cot a\left[\tan ^{-1} x-\frac{1}{\cos a} \tan ^{-1}\left(\frac{x+\sin a}{\cos a}\right)\right]_{0}^{\infty} \\&=\cot a\left[\frac{\pi}{2}-\frac{1}{\cos a}\left(\frac{\pi}{2}-a\right)\right]\end{aligned}\tag*{} $

Integrating $I’(a)$ back, we have $\displaystyle \begin{aligned}I\left(\frac{\pi}{6}\right)- \underbrace{I(0)}_{=\pi\ln 2} &=\int_{0}^{\frac{\pi}{6}} \cot a\left[\frac{\pi}{2}-\frac{1}{\cos a}\left(\frac{\pi}{2}-a\right)\right] d a \\&=\frac{\pi}{2} \underbrace{ \int_{0}^{\frac{\pi}{6}}\left(\cot a-\frac{1}{\sin a}\right) d a}_{=\ln \left(\frac{2+\sqrt{3}}{4}\right)} + \underbrace{\int_{0}^{\frac{\pi}{6}} \frac{a}{\sin a} d a}_{K}\end{aligned}\tag*{} $ $\displaystyle \begin{aligned}K &=\int_{0}^{\frac{\pi}{6}} \frac{a}{\sin a} d a=\int_{0}^{\frac{\pi}{6}} a\, d\left[\ln \left(\tan \frac{a}{2}\right)\right] \\&=\left[a \ln \left(\tan \frac{a}{2}\right)\right]_{0}^{\frac{\pi}{6}}-\int_{0}^{\frac{\pi}{6}} \ln \left(\tan \frac{a}{2}\right) d a \\&=\frac{\pi}{6} \ln \left(\tan \frac{\pi}{12}\right)-2 \int_{0}^{\frac{\pi}{12}} \ln (\tan a) d a \\&=-\frac{\pi}{6} \ln (2+\sqrt{3})+\frac{4}{3} G,\end{aligned}\tag*{} $ where $G$ is the Catalan’s constant and the last integral refer to the post.

Now we can conclude that

$\displaystyle \int_{0}^{\infty} \frac{\ln \left(1+x+x^{2}\right)}{1+x^{2}} d x =I\left(\frac{\pi}{6}\right)=\frac{\pi}{3} \ln (2+\sqrt{3})+\frac{4}{3} G\tag*{} $

Back to our integral $\displaystyle J=\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}} d x, \tag*{} $ using the result from my post, $\displaystyle \int_{0}^{\infty} \frac{\ln \left(x^{4}+x^{2}+1\right)}{x^{2}+1} d x=\pi \ln (2+\sqrt 3) \tag*{} $ yields immediately: $\displaystyle \begin{aligned}\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}} d x&=\int_{0}^{\infty} \frac{\ln \left(x^{4}+x^{2}+1\right)}{x^{2}+1} d x-\int_{0}^{\infty} \frac{\ln \left(1+x+x^{2}\right)}{1+x^{2}} d x\\&=\pi \ln (2+\sqrt{3})-\frac{\pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G \\&=\frac{2 \pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G\end{aligned} \tag*{} $

Is there any method other than Feynman’s Technique?

Your comments and alternative methods are highly appreciated.

Answer 1

Let $$x^2-x+1=(x-a)(x-b)\qquad \text{where} \qquad a=\frac{1+i \sqrt{3}}{2} \quad \text{and} \quad b=\frac{1-i \sqrt{3}}{2}$$ and $$\frac 1 {x^2+1}=\frac 1{(x-i)(x+i)}=\frac i 2\left(\frac{1}{x+i}-\frac{1}{x-i}\right)$$ Expanding the logarithm, we face four integrals $$I(\alpha,\beta)=\int \frac{\log (x+\alpha )}{x+\beta } \,dx$$ where $\alpha$ and $\beta$ are complex numbers

$$I(\alpha,\beta)=\text{Li}_2\left(\frac{x+\alpha }{\alpha -\beta }\right)+\log (\alpha +x) \log\left(\frac{\beta +x}{\beta -\alpha }\right)$$

Recombining all results and simplifying the polylogarithms leads to the result for the definite integral.

Answer 2

Utilize still $I’(a)$ to integrate \begin{aligned} &\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}} d x= I\left(-\frac{\pi}{6}\right)\\ &={I(0)}+\int_{0}^{-\frac{\pi}{6}} \cot a\left[\frac{\pi}{2}-\frac{1}{\cos a}\left(\frac{\pi}{2}-a\right)\right] \overset{a\to -a}{d a }\\ &= \pi\ln 2+\frac{\pi}{2} \underset{ \ln (2+\sqrt{3})-2\ln 2}{ \int_{0}^{\frac{\pi}{6}}\left(\cot a-\frac{1}{\sin a}\right) d a}-\underset{-\frac{\pi}{6} \ln (2+\sqrt{3})+\frac{4}{3} G}{\int_{0}^{\frac{\pi}{6}} \frac{a}{\sin a} d a}\\ &= \frac{2\pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G \end{aligned}

Answer 3

Define: $$x=\frac{1+t}{1-t}$$ The integral goes to

$$\begin{align} I&=\int_{-1}^1 \frac{\ln(1+3t^2)}{1+t^2}dt-2\int_{-1}^1 \frac{\ln(1-t)}{1+t^2}dt\\ \\ &=2\int_{0}^1 \frac{\ln(1+3t^2)}{1+t^2}dt-2\int_{0}^1 \frac{\ln(1+t)}{1+t^2}dt-2\int_{0}^1 \frac{\ln(1-t)}{1+t^2}dt\\ \\ &=2\int_{0}^1 \frac{\ln(1+3t^2)}{1+t^2}dt+2G-\frac{\pi}{2}\ln(2)\\ \end{align}$$

The first term can be handled by $$F(a)=\int_{0}^1 \frac{\ln(a+3t^2)}{1+t^2}dt,~~~F(0)=-2G+\frac{\pi}{4}\ln(3)$$