# If $I = \int _0 ^1 \frac{\ln (- \ln x)}{x^2 + x + 1} \, dx$ and $J = \frac{1}{2} \int _0 ^1 \frac{\ln (- \ln x)}{x^2 - x + 1} \,dx$ then what's $I-J$?

I'm trying to calculate a difficult integral that reduces to the following question:

If $\displaystyle I = \int _0 ^1 \frac{\ln (- \ln x)}{x^2 + x + 1} \mathrm{d}x$ and and $\displaystyle J = \frac{1}{2} \int _0 ^1 \frac{\ln (- \ln x)}{x^2 - x + 1} \mathrm{d}x$ then what's $I-J$?

This comes from while trying to calcululate the integral $\displaystyle \int_{0}^{1}{t-3t^3+t^5\over 1+t^4+t^8}\cdot \ln(-\ln t) \mathrm dt.$

I've been told that this integral evaluates to $\displaystyle {{\pi\over 3\sqrt{3}}}\cdot {\ln 2\over 2}$ (probably by symbolic calculator).

It's can be shown it's equal to $I-J$. Now trying to find $I$ and $J$ separately seems to be harder than finding the difference, but still I can't find any obvious symmetry to exploit betweeen them. I have made the following observation - which appears to be rather useless unless I'm missing something.

Define $\displaystyle \mathcal{I}(\lambda)= \int_0^1 \frac{\log(-\lambda \log(x))}{1+x+x^2}\mathrm{d}x, ~~ \mathcal{J}(\lambda)= -\frac{1}{2}\int_0^1 \frac{\log(-\lambda \log(x))}{1-x+x^2}\mathrm{d}x$,

Then we have $\displaystyle \mathcal{I}(\lambda) = \frac{\pi \log(\lambda)}{3\sqrt{3}}+\mathcal{I}(1)$ and $\displaystyle \mathcal{J}(\lambda) = -\frac{\pi \log(\lambda)}{3\sqrt{3}}+\mathcal{J}(1)$

But it seems all it's saying is that $\mathcal{I}(\lambda)+\mathcal{J}(\lambda) =\mathcal{I}(1)+\mathcal{J}(1)$ (i.e. $\mathcal{I}+\mathcal{J}$ is independent of $\lambda$). At this point it feels like there's some revelation that could do the trick, but I can't see it.

• Inserting a parameter $\lambda$ as above in $\mathcal I(\lambda)$, $\mathcal J(\lambda)$,does not help much, since inside the $\log$ numerators we can split terms $\log\lambda$, and $$\int_0^1\frac {dx}{1+x+x^2}+\int_0^1\frac {dx}{1-x+x^2}=\frac\pi{\sqrt3}$$ and $$\int_0^1\frac {dx}{1+x+x^2}-\frac 12\int_0^1\frac {dx}{1-x+x^2}=0$$ – dan_fulea May 25 '18 at 10:04
• @dan_fulea Yes, that's how found the expressions for $\mathcal I(\lambda), ~ \mathcal J(\lambda)$. And indeed it does feel like going back to square one. But the form $\displaystyle \frac{\pi \log(\lambda)}{3\sqrt{3}}\cdots$ resembles the answer for the integral a lot which made me suspicious! – Symposium May 25 '18 at 10:11

It is enough to play a bit with simple substitutions. For instance, through $x\mapsto x^2$ we have
$$\begin{eqnarray*} I = \int_{0}^{1}\frac{\log(-\log x)}{x^2+x+1}\,dx &=& \int_{0}^{1}\frac{2x\left[\log 2+\log(-\log x)\right]}{x^4+x^2+1}\,dx\\&=&\frac{\pi\log 2}{3\sqrt{3}}+\int_{0}^{1}\log(-\log x)\left[\frac{1}{x^2-x+1}-\frac{1}{x^2+x+1}\right]\,dx\tag{1}\end{eqnarray*}$$ and $I-J=\frac{\pi\log 2}{6\sqrt{3}}$ simply follows by rearranging.