Generating function of squares of generalized harmonic numbers

Question

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?

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.