Show that $\int_0^\infty \frac{x^2+1}{x^4+x^2+1}\frac{\log(1-x+x^2)}{\log(x)}dx=\frac{\pi}{\sqrt{3}}$

Question

$$\displaystyle \int_0^\infty \frac{x^2+1}{x^4+x^2+1}\frac{\log(1-x+x^2)}{\log(x)}dx=\frac{\pi}{\sqrt{3}}$$

It involves Beta.

Start with $$\displaystyle \int_{0}^{\infty}\frac{x^{2n-2}}{(1+x^{2})^{2n}}dx=\frac{\sqrt{\pi}\Gamma(n-1/2)}{2^{2n}\Gamma(n)}$$

Multiply by $\displaystyle \frac{x^{2}+1}{x^{a}+1}$

$$\displaystyle \int_{0}^{\infty}\frac{x^{2n-2}}{(x^{2}+1)^{2n}}\frac{x^{2}+1}{x^{a}+1}dx$$

integrate w.r.t a from 1 to 3:

$$\displaystyle \int_{0}^{\infty}\frac{(\ln(x^{3}+1)-\ln(x+1)-\ln(x^{3})+\ln(x))}{\ln(x)}\cdot \frac{x^{2n-2}}{(1+x^{2})^{2n-1}}da$$

and note the identity:

Now, note the identity $$\displaystyle -\int_{1}^{3}\frac{1}{x^{a}+1}da=\frac{\ln(x^{3}+1)-\ln(x+1)-\ln(x^{3})+\ln(x)}{\ln(x)}$$

$$\displaystyle =\frac{\ln\left(\frac{x^{2}-x+1}{x^{2}}\right)}{\ln(x)}=\frac{\ln(x^{2}-x+1)}{\ln(x)}-2$$

So, what we have is

$$\displaystyle \int_{0}^{\infty}\frac{x^{2n-2}}{(x^{2}+1)^{2n-1}}\left(\frac{\ln(x^{2}-x+1)}{\ln(x)}-2\right)dx=\frac{-1}{2^{2n-1}}\beta(n-1/2,1/2)$$

Answer 1

$$ \begin{align} &\int_0^\infty \frac{x^2+1}{x^4+x^2+1}\frac{\log(1-x+x^2)}{\log(x)}dx\\ &=\int_0^1 \frac{x^2+1}{x^4+x^2+1}\frac{\log(1-x+x^2)}{\log(x)}dx+\int_0^1 \frac{x^2+1}{x^4+x^2+1}\frac{\log(1-x+x^2)-2\log(x)}{-\log(x)}dx\\ &=2\int_0^1 \frac{x^2+1}{x^4+x^2+1}dx=\frac{\pi}{\sqrt 3} \end{align} $$

Answer 2

Letting $x\mapsto \frac{1}{x} $ yields \begin{aligned} I & =\int_0^{\infty} \frac{\frac{1}{x^2}+1}{\frac{1}{x^4}+\frac{1}{x^2}+1} \frac{\ln \left(1-\frac{1}{x}+\frac{1}{x^2}\right)}{\ln \frac{1}{x}} \frac{d x}{x^2} \\ & =\int_0^{\infty} \frac{1+x^2}{1+x^2+x^4} \frac{\ln \left(1-x+x^2\right)-2 \ln x}{-\ln x} d x \end{aligned} Re-arranging gives \begin{aligned} 2 I & =2 \int_0^{\infty} \frac{1+x^2}{1+x^2+x^4} d x \\ I & =\int_0^{\infty} \frac{\frac{1}{x^2}+1}{\frac{1}{x^2}+1+x^2} d x \\ & =\int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+3} \\ & =\frac{1}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)\right]_0^\frac{\pi}2 \\ & =\frac{\pi}{\sqrt{3}} \end{aligned}