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

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!

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 $$(0,1)$$.
• +1. Can I ask you if you've chosen arbitrarily the main branch of the complex logarithm or if is there a specific reason you must choose that? And why you're considering the open set $(0,1)$ for the sign of arctangent and not $[0,1]$? Thanks. Commented Apr 15, 2020 at 17:21
• @Dunkelheit The OP’s integral identity is only true if the log in the integrand uses the main branch (otherwise the imaginary part would equal $3\pi$ or $-\pi$ or some other odd multiple of pi). My use of the open interval $(0,1)$ doesn’t really make any difference. Commented Apr 15, 2020 at 20:38