How to compute the monstrous $ \int_0^{\frac{e-1}{e}}{\frac{x(2-x)}{(1-x)}\frac{\log\left(\log\left(1+\frac{x^2}{2-2x}\right)\right)}{2-2x+x^2}dx} $

Question

A friend told me, that he found a closed form for the following integral: $$ \int_0^{\frac{e-1}{e}}{\frac{x(2-x)}{(1-x)}\frac{\log\left(\log\left(1+\frac{x^2}{2-2x}\right)\right)}{\left(2-2x+x^2\right)}dx} $$ I don't know if he's just messing around with me, but I wonder if this integral admits a closed form. I tried to expand the $\log(\log)$ term into a power series, but things got worse. So any help will be appreciated!

Answer 1

Notice, we have $$\int_{0}^{\frac{e-1}{e}}\frac{x(2-x)}{1-x}\frac{\log\left(\log\left(1+\frac{x^2}{2-2x}\right)\right)}{2-2x+x^2}dx$$ $$=\int_{0}^{\frac{e-1}{e}}\frac{x(2-x)}{1-x}\frac{\log\left(\log\left(\frac{2-2x+x^2}{2-2x}\right)\right)}{2-2x+x^2}dx$$

Let, $$\log\left(\frac{2-2x+x^2}{2-2x}\right)=u$$ $$\implies \frac{d}{dx}\left(\log\left(\frac{2-2x+x^2}{2-2x}\right)\right)=\frac{d}{dx}(u)$$ $$\frac{1}{\left(\frac{2-2x+x^2}{2-2x}\right)}\cdot \left(\frac{(2-2x)(-2+2x)-(2-2x+x^2)(-2)}{(2-2x)^2} \right)=\frac{du}{dx}$$ $$\left(\frac{2-2x}{2-2x+x^2}\right)\cdot \left(\frac{2x(2-x)}{(2-2x)^2} \right)=\frac{du}{dx}$$ $$\frac{x(2-x)}{(1-x)}\frac{1}{(2-2x+x^2)}dx=du$$ Now, we have $$\int_{0}^{\log\left(\frac{e^2+1}{2e}\right)}\log(u)du$$

$$=\left[u\log(u)-u\right]_{0}^{\log\left(\frac{e^2+1}{2e}\right)}$$ $$=\left[u\log\left(\frac{u}{e}\right)\right]_{0}^{\log\left(\frac{e^2+1}{2e}\right)}$$

$$=\log\left(\frac{e^2+1}{2e}\right)\cdot\log\left(\frac{1}{e}\log\left(\frac{e^2+1}{2e}\right)\right)-\lim_{u\to 0}u\log\left(\frac{u}{e}\right)$$

$$=\log\left(\frac{e^2+1}{2e}\right)\cdot\log\left(\frac{1}{e}\log\left(\frac{e^2+1}{2e}\right)\right)-0$$

Hence, we get

$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\int_{0}^{\frac{e-1}{e}}\frac{x(2-x)}{1-x}\frac{\log\left(\log\left(1+\frac{x^2}{2-2x}\right)\right)}{2-2x+x^2}dx}=\color{blue}{\log\left(\frac{e^2+1}{2e}\right)\cdot \log\left(\frac{1}{e}\log\left(\frac{e^2+1}{2e}\right)\right)}}$$

Answer 2

Here comes the help! $$\mathcal{I}=(\varphi-1)\left(\ln(\varphi-1)-1\right)$$ $$\text{with}\qquad \varphi=\ln\left(\frac{1+e^{2}}{2}\right)$$ Namagiri is on fire today.