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Is there a way to evaluate the following integral involving three logarithmic functions

$$ I(y)=\int \frac{\log(1+y)\log(y)\log(1-y)}{y}\mathrm{d}y $$

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user12588
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  • Where'd you come across the problem? What have you tried so far on it? – Steven Stadnicki Nov 16 '21 at 23:28
  • Is there a reason to believe that there is a closed form in terms of elementary functions? Do you care if it is in closed form of non-elementary functions? – Bonnaduck Nov 16 '21 at 23:41
  • Wolfram Alpha outputs "no result found in terms of standard mathematical functions". In other words, it seems that no closed form is possible. – Crostul Nov 16 '21 at 23:45
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    https://math.stackexchange.com/questions/465444/evaluating-int1-0-frac-log1x-log1-x-logxx-mathrm-dx?noredirect=1 obtained using searching tool approach0.xyz – Jean Marie Nov 16 '21 at 23:49

1 Answers1

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According to the formula section of sequence $A049281$ in $OEIS$ $$\frac{\log (1-y) \log (1+y)}{y^2}=-\sum_{n=0}^\infty\Bigg[ \int_0^1x^{n} \log \left(1+\frac{1}{\sqrt{x}}\right)\,dx\Bigg] y^{2n}$$ $$a_n=\int_0^1x^{n} \log \left(1+\frac{1}{\sqrt{x}}\right)\,dx=\frac{1}{2 (n+1)^2}+\frac{\log (2)-\Phi (-1,1,2 n+3)}{n+1}$$ where appears s the Lerch transcendent function.

So $$\frac{\log(1+y)\log(y)\log(1-y)}{y}=-\sum_{n=0}^\infty a_n\,y^{2n+1}\log(y)$$ $$\int y^{2n+1}\log(y)\,dy=\frac{y^{2 n+2} (2 (n+1) \log (y)-1)}{4 (n+1)^2}$$

So, here is the formula.

Integrating between $0$ and $1$ and using $30$ terms, what is obtained is

$$\frac{129331573539024596038309003920483687963710634649045728750361929932743}{444999142480969921593398700100589348217215915374013669213491200000000}$$ which is $0.29063$ while the exact solution given in this question is, numerically, $0.29072$

Claude Leibovici
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