Nice integral $\Im\Big(\int_{0}^{1}\ln\Big(\arctan\Big(\frac{x^2-x-1}{x^2+x+1}\Big)\Big)dx\Big)=\pi$

Question

It's a little result that I found interesting :

$$\Im\Big(\int_{0}^{1}\ln\Big(\arctan\Big(\frac{x^2-x-1}{x^2+x+1}\Big)\Big)dx\Big)=\pi$$

I have spend two hours to extract the imaginary part without success .I have tried some obvious things as factorize the numerator to appear the golden ratio .I have learned somethings about the residue calculus on Wikipedia but I would be happy if there exists a "real" proof . Moreover I have tried the following substitution : $$\frac{x^2-x-1}{x^2+x+1}=t$$

But I don't know if we can use integration by parts in the case of complex numbers. Finally I have spend one hour to find an antiderivative without success .

Even I found this beautiful I think it's a little bit hard (for me).

If you have nice ideas...

Thanks a lot for your conrtibutions!

Answer 1

This is easier than it might seem at first. If $x$ is apositive real number, then we have that $$\ln(-x)=\pi i+x$$ on the main branch of the complex log function. This means that the imaginary part of your integral is just $$\int_0^1 \pi dx=\pi$$ since your integrand is the logarithm of an arctangent that is always negative on the interval $(0,1)$.