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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Harish Chandra Rajpoot
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Redundant Aunt
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Wolfram Alpha can do it. Let $u=\ln\left(1 + \frac{x^2}{2-2x}\right)$ and then do integration by parts. – Christopher Carl Heckman Aug 21 '15 at 22:40
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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 – Blex Aug 21 '15 at 22:43
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@Blex: how could it be useful, can you explain?i don't understand please – Bhaskara-III Sep 21 '16 at 10:39
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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
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\int_{0}^{\frac{e-1}{e}}\frac{x(2-x)}{1-x}\frac{\log\left(\log\left(1+\frac{x^2}{2-2x}\right)\right)}{2-2x+x^2}dx}=\color{blue}{\log\left(\frac{e^2+1}{2e}\right)\cdot \log\left(\frac{1}{e}\log\left(\frac{e^2+1}{2e}\right)\right)}}$$
Harish Chandra Rajpoot
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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.
Bhaskara-III
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Start wearing purple
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Beside Namagiri, both Mathematica and Maple could evaluate the integral in this form. – user153012 Aug 21 '15 at 23:06
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1How did you obtain this? I would recommend posting a development of the solution rather than the end result, which apparently is available via Maple. – Mark Viola Aug 22 '15 at 06:15
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1@Dr.MV Compare the score of this and this answer and then recommend me something again. – Start wearing purple Aug 22 '15 at 08:37
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1@user109899 Compute and simplify the derivative of $\ln\left(1+\frac{x^2}{2-2x}\right)$. – Start wearing purple Aug 22 '15 at 10:04
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1@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? – Mark Viola Aug 22 '15 at 16:22
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1@Dr.MV I am not asking myself such questions, although it might be that game aspects excite me more than helping people and stuff like that. This is especially true for "could-we-find-a-closed-form" questions where the askers only pretend that they don't know the answer. You might have noticed that the number of people who pressed the button "This answer is useful" in my example relate as 45:1 in the two cases. The incentive is thus very clear. – Start wearing purple Aug 22 '15 at 16:48
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1@L.G. Two examples make up a very small sample on which to base a general conclusion ... especially when that person is so highly mathematically inclined as you are. – Mark Viola Aug 22 '15 at 16:55
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@Lucian I was not successful with that problem in spite of considerable effort. – Start wearing purple Aug 23 '15 at 15:40