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By generalizing the approach in Integral involving a dilogarithm versus an Euler sum. meaning by using the integral representation of the harmonic numbers and by computing a three dimensional integral over a unit cube analytically we have found the generating function of cubes of harmonic numbers. We have: \begin{eqnarray} &&S^{(3)}(x) := \sum\limits_{n=1}^\infty H_n^3 x^n = \frac{-18 \text{Li}_3\left(1-\frac{1}{x}\right)+6 \text{Li}_3\left(\frac{1}{x}\right)-18 \text{Li}_3(x)}{6(1-x)}+ \frac{6 \log ^3(1-x)-9 \log (x) \log ^2(1-x)+3 \left(3 \log ^2(x)+\pi ^2\right) \log (1-x)}{6(x-1)}+\frac{-\log (x) \left(2 \log ^2(x)+ 3 i \pi \log (x)+5 \pi ^2\right)}{6 (x-1)} \end{eqnarray} Clearly some of the terms on the right hand side are complex even though the whole expression is of course real. The first two terms in the first fraction on the rhs are complex and the middle term in the last fraction is complex. My question is how do I simplify the right hand side to get rid of the complex terms?

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By using the functional equations for the trilogarithm we simplified the result as follows: \begin{eqnarray} &&S^{(3)}(x)= \\ &&\frac{ \text{Li}_3(x)}{(1-x)}+ 3\frac{\text{Li}_3(1-x)-\zeta (3)}{(1-x)}+ \log(1-x)\frac{ \left(-2 \log ^2(1-x)+3 \log (x) \log(1-x)-\pi ^2 \right)}{2 (1-x)} \end{eqnarray}

For a sanity check we expand each of the terms in the formula in a Taylor series about zero we have: \begin{eqnarray} &&\frac{Li_3(x)}{1-x} =\\ && x+\frac{9 x^2}{8}+\frac{251 x^3}{216} + O(x^4) \\ &&3\frac{\text{Li}_3(1-x)-\zeta (3)}{(1-x)} =\\ &&-\frac{\pi ^2 x}{2}+\frac{9 x^2}{4}-\frac{3}{2} x^2 \log (x)-\frac{3 \pi ^2 x^2}{4}-\frac{11 \pi ^2 x^3}{12}+4 x^3-3 x^3 \log (x) + O(x^4) \\ &&\log(1-x)\frac{ \left(-2 \log ^2(1-x)+3 \log (x) \log(1-x)-\pi ^2 \right)}{2 (1-x)} =\\ && \frac{\pi ^2 x}{2}+\frac{3 \pi ^2 x^2}{4}+\frac{3}{2} x^2 \log (x) +\frac{11 \pi ^2 x^3}{12}+x^3+3 x^3 \log (x)+ O(x^4) \end{eqnarray}

As we can see the terms proportional to $\log(x)$being present in the second and the third term exactly cancel each other. The formula is correct.

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  • $\begingroup$ @ Przemo : I doubt that this result is correct. The power series about $x = 0$ should start by definition like this $x+\frac{27 x^2}{8}+\frac{1331 x^3}{216}+\frac{15625 x^4}{1728}+\frac{2571353 x^5}{216000}$ but your expression contains $\log x$ $\endgroup$ – Dr. Wolfgang Hintze Nov 22 '17 at 13:09
  • $\begingroup$ @ Przemo : I tried to find the error. We can make te the $\log$ terms vanish if we replace the factor 3 in your second term by 2, but the formula still is not correct. $\endgroup$ – Dr. Wolfgang Hintze Nov 22 '17 at 14:13
  • $\begingroup$ @Dr. Wolfgang Hintze: Thank you for this remark. Unfortunately I haven't kept record of my old notes. Now I have checked if the original (longish) formula is correct . It is. As far as a I remember the only thing I did to simplify it was to use the reflection formulae for the trilogarithm from Wolfram's website mathworld.wolfram.com/Trilogarithm.html . I will try to redo it now and fix the formula. Btw, what do you want to use this formula for if I may ask ? $\endgroup$ – Przemo Nov 22 '17 at 15:34
  • $\begingroup$ @Dr. Wolfgang Hintze: I have double checked the formula. It is correct. See above. $\endgroup$ – Przemo Nov 22 '17 at 16:20
  • $\begingroup$ @ Przemo The second formula is now ok. So we agree on the g.f. for the third power. I need to type into Mathematica very carefully the second one with which I still have problems. $\endgroup$ – Dr. Wolfgang Hintze Nov 22 '17 at 17:06
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Here we provide a closed form for another related sum. We have: \begin{eqnarray} &&\sum\limits_{n=1}^\infty \frac{H_n^3}{n} \cdot x^n =\\ &&3 \zeta(4)-3 \text{Li}_4(1-x)+3 \text{Li}_3(1-x) \log (1-x)+\log (x) \log ^3(1-x)+\\ &&\text{Li}_2(x){}^2-2 \text{Li}_4(x)-3 \text{Li}_4\left(\frac{x}{x-1}\right)+\frac{3}{2} \left(\text{Li}_2(x)-\frac{\pi ^2}{6}\right) \log ^2(1-x)+2 \text{Li}_3(x) \log (1-x)+\frac{1}{8} \log ^4(1-x) \end{eqnarray} We obtained this formula by dividing the right hand side in the question above by $x$ and then integrating.

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