5
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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 ?

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  • $\begingroup$ If I'm not mistaken this can be written as $$I(x) = \int_{1}^{\infty} \ln^2 x \ln^2(1-1/x) \ln^2(1+1/x)$$ $\endgroup$ – Brevan Ellefsen Aug 11 '16 at 19:45
  • $\begingroup$ 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. $\endgroup$ – Aditya Narayan Sharma Aug 11 '16 at 19:49
  • $\begingroup$ I just used $u = \frac 1 x$, noting that some symmetry occurs. Most importantly, it gets rid of the denominator $\endgroup$ – Brevan Ellefsen Aug 11 '16 at 19:51
  • $\begingroup$ 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 $\endgroup$ – Brevan Ellefsen Aug 11 '16 at 19:52
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    $\begingroup$ @JackD'Aurizio Nice observation. If a closed form does exist I don't imagine it will be pretty. $\endgroup$ – Brevan Ellefsen Aug 11 '16 at 20:07

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