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This question is a continuation of the topic in Closed form expressions for harmonic sums . By using the integral representation of harmonic numbers and by induction we have derived the following integral representation of an infinite harmonic sum. We have: \begin{equation} {\mathfrak S}^{(p)}_n(t):=\sum\limits_{m=0}^\infty H_m^p \cdot \frac{t^{m+1}}{(m+1)^n} = \int\limits_{[0,1]^p} \prod\limits_{\eta=0}^{p-1} \log\left(1-\xi_\eta\right) \cdot \frac{\left(\sum\limits_{l=0}^{p}(-1)^l \binom{p}{l} Li_{n-l}(t\prod\limits_{\eta=0}^{p-1} \xi_\eta )\right)}{\prod\limits_{\eta=0}^{p-1} \xi_\eta^2} \prod\limits_{\eta=0}^{p-1} d \xi_\eta \end{equation} Here $Li_n()$ are poly-logarithms, $p$ and $n$ are strictly positive integers and $t\in (-1,1)$. Now the question is how can we use the result above to calculate those sums at unity,i.e. how do we calculate ${\mathfrak S}^{(n)}_p(1)$ ? Does the result always depend only on certain values of the zeta function (as it does in case $p=1$ as shown in the question quoted above) or instead does it also depend on something else?

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  • $\begingroup$ For $p+m $ is odd it is reducible to zeta values and there is a general formula. For the other case I have never seen one. $\endgroup$ – Zaid Alyafeai Apr 24 '17 at 17:08
  • $\begingroup$ See google.com.sa/url?sa=t&source=web&rct=j&url=http://… $\endgroup$ – Zaid Alyafeai Apr 24 '17 at 17:12
  • $\begingroup$ @Zaid Alyafeai: Thank you for this document and for the comments. I am interested in this topic because those Euler sums appear to be related to solutions to certain recurrence relations with time dependent coefficients. Again, ${\mathfrak S}^{(1)}_n(1)$ is known -- you posted it yourself some time ago. I was thinking that the integral I gave above could help us to derive the result for $p=2$ or maybe $p=3$. I will be working on this and I will post results if I achieve any. $\endgroup$ – Przemo Apr 24 '17 at 17:20
  • $\begingroup$ It seems difficult to work with $\mathrm{Li}_q(xy)$. Moreover in the link I provided they claim that Euler sums with large even weight should be considered as 'new' constants. Perhaps the best someone can do is consider ${\mathfrak S}^{(2)}_4(1)$. I guess you already know that ${\mathfrak S}^{(p)}_p(1)$ is trivial. $\endgroup$ – Zaid Alyafeai Apr 25 '17 at 6:45
  • $\begingroup$ If you like working with Euler sums. You can take a look at my latest question math.stackexchange.com/questions/2169507/… $\endgroup$ – Zaid Alyafeai Apr 25 '17 at 6:50

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