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Due to curiosity and also since I evaluated lower degree sums like these but this one is too hard to manipulate I am eager to know does this have a closed form ?

I broke it into the series $\displaystyle \sum_{m,n\ge 1}(-1)^{m+n}\frac{{\rm H}_m{\rm H}_n}{(m+1)(n+1)} \frac{2}{(2n-1)^3} $ , but does this help ?

Aditya Narayan Sharma
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  • If I'm not mistaken this can be written as $$I(x) = \int_{1}^{\infty} \ln^2 x \ln^2(1-1/x) \ln^2(1+1/x)$$ – Brevan Ellefsen Aug 11 '16 at 19:45
  • Using the substitution ? And to proceed from here we would have a double harmonic sum and the limits are 1 to infinity which won't give a proper one maybe. I'm not sure. – Aditya Narayan Sharma Aug 11 '16 at 19:49
  • I just used $u = \frac 1 x$, noting that some symmetry occurs. Most importantly, it gets rid of the denominator – Brevan Ellefsen Aug 11 '16 at 19:51
  • From here I think the squared logarithms can be expanded into a series with $9$ terms in it, that I bet can each be integrated. Now, I doubt that will be pretty. Hopefully I did make an error and the integrand is easier than I wrote it XD – Brevan Ellefsen Aug 11 '16 at 19:52
  • Scratch that doubt. Numerically the integrals are equivalent, so I will try breaking up the integral. – Brevan Ellefsen Aug 11 '16 at 19:53
  • Yes I guess breaking the integrals would be helpful. In my approach If you can break that series in a simpler form using MZV values , then also we might reach a conclusion – Aditya Narayan Sharma Aug 11 '16 at 19:55
  • Unfortunately, that still left some squared logarithms in there, so a solution along this route is unlikely at best. I'm not too experienced with MZV identities myself, but hopefully someone can pick up my slack! – Brevan Ellefsen Aug 11 '16 at 20:02
  • Even worse, it appears some of those integrals don't converge, so a more careful approach will be necessary than blindy splitting. Back to the drawing board! – Brevan Ellefsen Aug 11 '16 at 20:05
  • Yet $$\int_{0}^{1}\log^2(x)\log^2(1-x),dx = 24-\frac{1}{90} \pi ^2 \left(120+\pi ^2\right)-8\zeta(3) $$ is not entirely trivial, and the term depending on $x+1$ makes everything much worse in our case... – Jack D'Aurizio Aug 11 '16 at 20:06
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    @JackD'Aurizio Nice observation. If a closed form does exist I don't imagine it will be pretty. – Brevan Ellefsen Aug 11 '16 at 20:07
  • $\displaystyle{\approx 0.025039435836443526297487364729844045358794142742411}$ – Felix Marin Aug 11 '16 at 23:30

1 Answers1

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See this paper. One have

  • $\scriptsize \int_{0}^{1} \frac{\ln^2 x \ln^2 (1+x)\ln^2(1-x)}{x^2}dx=8 \zeta(\bar5,1)-\frac{8}{3} \pi ^2 \text{Li}_4\left(\frac{1}{2}\right)+8 \text{Li}_4\left(\frac{1}{2}\right)+24 \text{Li}_5\left(\frac{1}{2}\right)+16 \text{Li}_6\left(\frac{1}{2}\right)+8 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+24 \text{Li}_4\left(\frac{1}{2}\right) \log (2)+16 \text{Li}_5\left(\frac{1}{2}\right) \log (2)+\frac{19 \pi ^2 \zeta (3)}{6}-62 \zeta (5)-\frac{227 \zeta (3)^2}{8}-7 \zeta (3) \log ^3(2)-\frac{35}{2} \zeta (3) \log ^2(2)+\frac{35}{6} \pi ^2 \zeta (3) \log (2)-9 \zeta (3) \log (2)-\frac{217}{2} \zeta (5) \log (2)+\frac{73 \pi ^6}{1260}+\frac{\pi ^4}{90}+\frac{2 \log ^6(2)}{9}+\frac{4 \log ^5(2)}{5}-\frac{5}{18} \pi ^2 \log ^4(2)-\frac{5 \log ^4(2)}{3}+\frac{2}{3} \pi ^2 \log ^3(2)+\frac{13}{36} \pi ^4 \log ^2(2)+\pi ^2 \log ^2(2)+\frac{1}{6} \pi ^4 \log (2)$

Where $\zeta(\bar5,1)$ is a Multiple Zeta Value. Bonus:

  • $\scriptsize \int_0^1 \frac{\log ^3(1-x) \log ^2(x) \log ^2(x+1)}{x^2} \, dx=-12 \zeta(\bar5,1)+36 \zeta(\bar5,1,1)-6 \zeta(5,\bar1,1)+54 \log (2) \zeta(\bar5,1)+156 \text{Li}_4\left(\frac{1}{2}\right) \zeta (3)-8 \pi ^2 \text{Li}_4\left(\frac{1}{2}\right)-34 \pi ^2 \text{Li}_5\left(\frac{1}{2}\right)-48 \text{Li}_5\left(\frac{1}{2}\right)+144 \text{Li}_6\left(\frac{1}{2}\right)-96 \text{Li}_7\left(\frac{1}{2}\right)-22 \pi ^2 \text{Li}_4\left(\frac{1}{2}\right) \log (2)-51 \zeta (3)^2-\frac{71 \pi ^4 \zeta (3)}{320}+\frac{9 \pi ^2 \zeta (3)}{4}+\frac{4461 \pi ^2 \zeta (5)}{64}-3 \zeta (5)-\frac{1755 \zeta (7)}{4}-\frac{1}{2} \zeta (3) \log ^4(2)-28 \zeta (3) \log ^3(2)+\frac{95}{8} \pi ^2 \zeta (3) \log ^2(2)-24 \zeta (3) \log ^2(2)-\frac{3999}{16} \zeta (5) \log ^2(2)-\frac{1563}{16} \zeta (3)^2 \log (2)+\frac{31}{2} \pi ^2 \zeta (3) \log (2)-\frac{279}{2} \zeta (5) \log (2)+\frac{\pi ^6}{112}+\frac{2 \log ^7(2)}{105}+\frac{\log ^6(2)}{5}-\frac{7}{10} \pi ^2 \log ^5(2)-\frac{6 \log ^5(2)}{5}+\frac{2}{3} \pi ^2 \log ^4(2)+\frac{121}{180} \pi ^4 \log ^3(2)+\frac{4}{3} \pi ^2 \log ^3(2)+\frac{29}{60} \pi ^4 \log ^2(2)+\frac{1321 \pi ^6 \log (2)}{10080}+\frac{1}{15} \pi ^4 \log (2)$

The latter will be extremely difficult via ordinary means. See the second link of my answer here for algorithmic reference. One may verify their correctness by numerical methods.

Infiniticism
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