Let $x \in [0,1]$. By using the integral representation of a generalized harmonic number and then by integrating by parts we have shown that: \begin{eqnarray} \tiny &&\sum\limits_{n=1}^\infty [H^{(2)}_n]^2 x^n =\\ &&\frac{360 \zeta (3) \log (x)+60 \pi ^2 \log (1-x) \log (x)+4 \pi ^4}{180 (x-1)}+\\ &&\frac{45 \log ^4(1-x)-90 \log ^2(x) \log ^2(1-x)-30 \pi ^2 \log ^2(1-x)}{180 (x-1)}+\\ &&\frac{-90 \text{Li}_2(x){}^2+180 \text{Li}_2(x) \log ^2\left(\frac{1}{x}-1\right)+180 \text{Li}_2\left(\frac{x}{x-1}\right) \log ^2\left(\frac{1}{x}-1\right)}{180 (x-1)}+\\ &&\frac{360 \text{Li}_3(1-x) \log \left(\frac{1}{x}-1\right)-360 \text{Li}_3(x) \log (x)+360 \text{Li}_3\left(\frac{x}{x-1}\right) \log \left(\frac{1}{x}-1\right)}{180 (x-1)}+\\ &&\frac{-360 \text{Li}_4(1-x)+360 \text{Li}_4(x)+360 \text{Li}_4\left(\frac{x}{x-1}\right)}{180 (x-1)}+\\ &&\frac{1}{1-x}\left( \zeta\left(\begin{array}{r}2 & 2\\x&\frac{1}{x}\end{array}\right)+ 2\zeta\left(\begin{array}{r}3 & 1\\1&x\end{array}\right)+ 2\zeta\left(\begin{array}{r}3 & 1\\x&\frac{1}{x}\end{array}\right) \right) \end{eqnarray} where $\zeta\left(\right)$ are multivariate zeta functions (see http://mathworld.wolfram.com/MultivariateZetaFunction.html for the definition). Now, the interesting question would be is it possible to reduce the later functions to polylogarithms and to elementary functions only?

  • $\begingroup$ Thanks for bringing this to my attention. I fixed the typo. $\endgroup$
    – Przemo
    Sep 5, 2016 at 14:14
  • $\begingroup$ There is a $+$ sign just before the last parenthesis. Probably it should be deleted, or perhaps a term is missing. $\endgroup$
    – ajotatxe
    Sep 5, 2016 at 14:22
  • $\begingroup$ Again, thank you for the review. I again double checked everything and there is no term missing. The plus sign at the end of the line appeared because I just copied and pasted the previous line and then modified it slightly. $\endgroup$
    – Przemo
    Sep 5, 2016 at 14:28

1 Answer 1


In here we show that the middle term can indeed be reduced to polylogarithms. We have: \begin{eqnarray} \zeta\left(\begin{array}{rr}3 & 1 \\ 1 & x \end{array} \right) &:=& \sum\limits_{\infty > k > k_1 \ge 1} \frac{1}{k^3} \cdot \frac{x^{k_1}}{k_1}\\ &=& \sum\limits_{k_1=1}^\infty \frac{x^{k_1}}{k_1} \cdot\left(\zeta(3) -H^{(3)}_{k_1} \right)\\ &=&-\log(1-x) \cdot \zeta(3) - \int\limits_0^1 \frac{Li_3(\xi)}{\xi(1-\xi)} d\xi \\ &=& -\log(1-x) \cdot \zeta(3) - \left(Li_4(x) - \frac{1}{2} Li_2(x) - \log(1-x) Li_3(x)\right) \\ \end{eqnarray} Now we tackle the other term: \begin{eqnarray} \zeta\left(\begin{array}{r} 2& 2\\x& \frac{1}{x} \end{array}\right) &:=& \sum\limits_{k_1=1}^\infty \sum\limits_{k=k_1+1}^\infty \frac{x^{k-k_1}}{k^2 k_1^2}\\ &=& \sum\limits_{k=1}^\infty x^k \sum\limits_{k_1=1}^\infty \frac{1}{(k+k_1)^2 k_1^2} \\ &=& \sum\limits_{k=1}^\infty x^k \left(2\zeta(2) \frac{1}{k^2} -2 \frac{H_k}{k^3} - \frac{H^{(2)}_k}{k^2} \right)\\ &=& 2\zeta(2) Li_2(x) - 2 \sum\limits_{k=1}^\infty \frac{x^k}{k^3} H_k -\sum\limits_{k=1}^\infty \frac{x^k}{k^2} H^{(2)}_k \end{eqnarray} Clearly the first sum on the right hand side has been almost computed here in MSE (see Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$ for example). In fact we have: \begin{eqnarray} &&\sum\limits_{k=1}^\infty \frac{x^k}{k^3} H_k =\\ &&\left\{ \begin{array}{rr} -\text{Li}_4(1-x)-\text{Li}_4\left(\frac{x-1}{x}\right)+2 \text{Li}_4(x)-\text{Li}_3(x) \log (1-x)+\zeta (3) \log (1-x)-\frac{1}{24} \log ^4(x)+\frac{1}{6} \log (1-x) \log ^3(x)-\frac{1}{4} \log ^2(1-x) \log ^2(x)-\frac{1}{12} \pi ^2 \log ^2(x)+\frac{1}{6} \pi ^2 \log (1-x) \log (x)-\frac{\pi ^4}{120} & \mbox{if $x>0$}\\ \text{Li}_4\left(\frac{1}{1-x}\right)+2 \text{Li}_4(x)+\text{Li}_4\left(\frac{x}{x-1}\right)-\text{Li}_3(x) \log (1-x)+\zeta (3) \log (1-x)+\frac{1}{12} \log ^4(1-x)-\frac{1}{6} \log (-x) \log ^3(1-x)-\frac{1}{12} \pi ^2 \log ^2(1-x)-\frac{\pi ^4}{90} & \mbox{if $x<0$} \end{array} \right. \end{eqnarray} The remaining sum reads: \begin{eqnarray} &&\sum\limits_{k=1}^\infty \frac{x^k}{k^2} H^{(2)}_k = \int\limits_0^x \log(\frac{x}{\xi})\cdot \frac{1}{\xi}\cdot \frac{Li_2(\xi)}{1-\xi} d\xi = \int\limits_0^1 \log(\frac{1}{\xi}) \cdot \frac{1}{\xi} \cdot \frac{Li_2(x \xi)}{1-x \xi} d \xi \\ &&-\int\limits_0^1 \log(\xi) \cdot Li_2(x \xi) \cdot \left[\frac{1}{\xi} + \frac{x}{1-x \xi}\right] d\xi = Li_4(x) + \int\limits_0^1 \log(\xi) \cdot Li_2(x \xi) \left(\frac{-x}{1-x \xi}\right) d\xi =\\ &&Li_4(x) + Li_2^2(x) - \int\limits_0^1 \left[\log(\xi) \cdot \log(1-x \xi)+Li_2(x \xi)\right] \cdot \frac{(-)\log(1- x \xi)}{\xi} d\xi =\\ && Li_4(x) + \frac{1}{2} Li_2(x)^2 + \int\limits_0^1 \frac{\log(\xi)}{\xi} \cdot \log(1-x \xi)^2 d \xi \end{eqnarray} Now, the integral in the last equation on the right hand side is proportional to the generalized Nielsen polylogarithms. Therefore it can be expressed in terms of polylogarithms as well. The final result reads: \begin{eqnarray} &&\sum\limits_{k=1}^\infty \frac{x^k}{k^2} H^{(2)}_k =\\ &&-2 \zeta (3) \log (x)-\frac{1}{4} \log ^4(1-x)+\frac{1}{2} \log ^2(x) \log ^2(1-x)+\frac{1}{6} \pi ^2 \log ^2(1-x)-\frac{1}{3} \pi ^2 \log (x) \log (1-x)-\frac{\pi ^4}{45}\frac{\text{Li}_2(x){}^2}{2}-\text{Li}_2(x) \log ^2\left(\frac{1}{x}-1\right)-\text{Li}_2\left(\frac{x}{x-1}\right) \log ^2\left(\frac{1}{x}-1\right)-2 \text{Li}_3(1-x) \log (1-x)+2 \text{Li}_3(1-x) \log (x)+2 \text{Li}_3(x) \log (x)-2 \text{Li}_3\left(\frac{x}{x-1}\right) \log (1-x)+2 \text{Li}_3\left(\frac{x}{x-1}\right) \log (x)+2 \text{Li}_4(1-x)-\text{Li}_4(x)-2 \text{Li}_4\left(\frac{x}{x-1}\right) \end{eqnarray} Now we proceed to compute the remaining term. We carry out the same steps as before, meaning changing the order of summation, collecting all terms corresponding to a given power of $x$ and then decomposing the coefficient at the power of $x$ into simple fractions and then resumming those simple fractions using the definition of harmonic numbers and the zeta function. We have: \begin{eqnarray} \zeta\left(\begin{array}{rr} 3 & 1 \\ x & \frac{1}{x}\end{array} \right)= \log(1-x) \cdot \zeta(3) - Li_2(x) \cdot \zeta(2) + \sum\limits_{k=1}^\infty x^k \frac{H_k}{k^3} +\sum\limits_{k=1}^\infty x^k \frac{H^{(3)}_k}{k} +\sum\limits_{k=1}^\infty x^k \frac{H^{(2)}_k}{k^2} \end{eqnarray} As before the result has been reduced to harmonic sums (those sums are always simpler than the original harmonic sum that we are attempting to compute). Out of the sums on the right hand side only the middle one has not been computed yet. We compute it now: \begin{eqnarray} &&\sum\limits_{n=1}^\infty x^n \frac{H^{(3)}_n}{n} = \int\limits_0^x \frac{Li_3(\xi)}{\xi(1-\xi)} d\xi = \int\limits_0^x Li_3(\xi) \left[\frac{1}{\xi} + \frac{1}{1-\xi}\right] d\xi =\\ &&Li_4(x) + \int\limits_0^x Li_3(\xi) \frac{1}{1-\xi} d\xi = Li_4(x) - \log(1-x) Li_3(x) - \int\limits_0^x Li_2^{'}(\xi) Li_2(\xi) d \xi =\\&& Li_4(x) - \log(1-x) Li_3(x) - \frac{1}{2} Li_2(x)^2 \end{eqnarray} This finishes the calculation.

  • $\begingroup$ @ Przemo I have calculated $\sum _ {} x^k H_{k} /k^3$ independently, and found a different expression. I confirm that your expression leads to the correct power series at $x=0$ in the range $0\le x \le 1$. But the value at $x= -1$ is not correct as it does not coincide with the numerical calculation of the alternating sum (-0.8527...). My expression gives a real part which coincides with the sum. I believe it is worthwhile to request an expression which represents the analytic function of $x$ in the unit circle (the region of absolute convergence). $\endgroup$ Nov 26, 2017 at 20:12
  • $\begingroup$ @Dr Wolfgang Hintze: Thank you for this remark. It is likely that I made an error related to the argument of $x$. I will check the calculations and get back to you. What is frustrating about this stuff is that I am not aware of any reflection formula for the polylogarithm of order four and as such going beyond weight four is hard because one needs to deal with a plethora of terms yet the calculations are simple. Do you know any functional identities of the polylogarithms of order four or higher? $\endgroup$
    – Przemo
    Nov 27, 2017 at 11:21
  • $\begingroup$ @ Przemo My strategy is to look for expressions which first of all are valid in the whole region -1<x<1 Then I try to split real and imaginary parts. I have the impression that reflexion formulae for $Li_{m}(x)$ exist only for m=2. But we have inversion formulas for any $m$ which ca be used to shift the region of the argument of $Li_{m}(x)$ to $x \lt 1$ where the function is real. $\endgroup$ Nov 27, 2017 at 16:48
  • $\begingroup$ @Dr Wolfgang Hintze: I have got two expressions for the sum in question , one is valid for positive and one for negative values of $x$. It is simply that I do not know how to add here a link to the Mathematica generated text file. If you want I can send you the file via email. $\endgroup$
    – Przemo
    Nov 27, 2017 at 17:06
  • $\begingroup$ @ Przemo What are the limits for $x\to+1$ and $x\to-1$ ? To continue our discussion we could go to a chat room. There one can also upload files like Mathematica code. $\endgroup$ Nov 27, 2017 at 22:23

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