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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!

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  • $\begingroup$ Wolfram Alpha can do it. Let $u=\ln\left(1 + \frac{x^2}{2-2x}\right)$ and then do integration by parts. $\endgroup$ Aug 21, 2015 at 22:40
  • $\begingroup$ Observing $\log\left(\log\left(1+\frac{x^2}{2-2x}\right)\right) = \log(\log(2-2x+x^2) - \log(2-2x))$ could be useful $\endgroup$
    – Blex
    Aug 21, 2015 at 22:43
  • $\begingroup$ @Blex: how could it be useful, can you explain?i don't understand please $\endgroup$ Sep 21, 2016 at 10:39

2 Answers 2

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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

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

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    $\begingroup$ Finally, someone ended their answer with flavour! $\endgroup$
    – Mr Pie
    Feb 18, 2018 at 13:52
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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.

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    $\begingroup$ What/ who is Namagiri? $\endgroup$ Aug 21, 2015 at 23:49
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    $\begingroup$ How did you obtain this? I would recommend posting a development of the solution rather than the end result, which apparently is available via Maple. $\endgroup$
    – Mark Viola
    Aug 22, 2015 at 6:15
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    $\begingroup$ @Dr.MV Compare the score of this and this answer and then recommend me something again. $\endgroup$ Aug 22, 2015 at 8:37
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    $\begingroup$ @user109899 Compute and simplify the derivative of $\ln\left(1+\frac{x^2}{2-2x}\right)$. $\endgroup$ Aug 22, 2015 at 10:04
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    $\begingroup$ @L.G. So, you believe that receiving a high score is the goal here? What do thise points do for you if I may ask? $\endgroup$
    – Mark Viola
    Aug 22, 2015 at 16:22

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