# What is $\sum\limits_{m=1}^\infty \frac{H_m}{m^6} \cdot (\frac{1}{2})^m$?

We consider a following class of Euler sums: $${\bf H}^{(1)}_p(\frac{1}{2}) := \sum\limits_{m=1}^\infty \frac{H_m}{m^p} \cdot \frac{1}{2^m}$$ Now by using the following integral representation: $${\bf H}^{(1)}_p(\frac{1}{2}) = \int\limits_0^{1/2} \frac{[\log(\frac{1/2}{x})]^{p-1}}{(p-1)!} \cdot \frac{Li_1(x)}{x(1-x)} dx$$ and then by integration by parts we computed those sums for $p\le 5$. We have: \begin{eqnarray} {\bf H}^{(1)}_1(1/2) &=& \frac{\pi ^2}{12}\\ {\bf H}^{(1)}_2(1/2) &=& \zeta (3)-\frac{1}{12} \pi ^2 \log (2)\\ {\bf H}^{(1)}_3(1/2) &=& \text{Li}_4\left(\frac{1}{2}\right)-\frac{1}{8} \zeta (3) \log (2)+\frac{\pi ^4}{720}+\frac{\log ^4(2)}{24}\\ {\bf H}^{(1)}_4(1/2) &=& 2 \text{Li}_5\left(\frac{1}{2}\right)+\text{Li}_4\left(\frac{1}{2}\right) \log (2)-\frac{\pi ^2 \zeta (3)}{12}+\frac{\zeta (5)}{32}+\frac{1}{2} \zeta (3) \log ^2(2)+\frac{\log ^5(2)}{40}-\frac{1}{36} \pi ^2 \log ^3(2)-\frac{1}{720} \pi ^4 \log (2)\\ {\bf H}^{(1)}_5(1/2) &=& 3 \text{Li}_6\left(\frac{1}{2}\right)+\text{Li}_5\left(\frac{1}{2}\right) \log (2)-\frac{\zeta (3)^2}{4}-\frac{1}{6} \zeta (3) \log ^3(2)+\frac{1}{12} \pi ^2 \zeta (3) \log (2)-\frac{1}{32} \zeta (5) \log (2)-\frac{19 \pi ^6}{8640}-\frac{\log ^6(2)}{240}+\frac{1}{144} \pi ^2 \log ^4(2)+\frac{\pi ^4 \log ^2(2)}{1440}-\frac{1}{2} {\bf H}^{(1)}_5(-1) \end{eqnarray} Note that the last case above involves a new quantity ${\bf H}^{(1)}_5(-1) = \zeta(-5,1)+Li_6(-1)$ a quantity which is not expressible via poly-logarithms. Now, my question is here quite humble. Can we push this thread up one step further and compute the result for $p=6$? Is the quantity in question also "new" or can it be reduced to the univariate zeta functions only?

I have used the following code to check in http://wayback.cecm.sfu.ca/cgi-bin/EZFace/zetaform.cgi for possible linear dependencies between the quantity in question and zeta functions.

lindep([zp(2,6,1)+zp(2,7), z(7), z(2)^3*log(2), z(3)^2*log(2), z(5)*log(2)^2, z(5)*z(2), z(2)^2*z(3), z(2)^2*log(2)^3, z(3)*log(2)^4, z(2)*log(2)^5, log(2)^7, zp(2,4)*log(2)^3, zp(2,5)*log(2)^2, zp(2,6)*log(2), zp(2,7), zp(2,5,1)*log(2)])


Unfortunately I couldn't find any results.

• Just out of curiosity : which CAS did you use to produce these beautiful results ? – Claude Leibovici Dec 21 '17 at 13:09
• @Claude Leibovici: I was using Mathematica but I wasn't really relying on it entirely but instead only verifying if each step in the derivation was correct. The derivation itself ran roughly according to the lines given in the answer below. – Przemo Dec 21 '17 at 19:19
• We are already familiar with the generalization $\sum_{n=1}^\infty \frac{(-1)^n H_n}{n^{2p}}$ , so if we can relate $\sum_{n=1}^\infty \frac{ H_n}{n^62^n}$ to $\sum_{n=1}^\infty \frac{(-1)^nH_n}{n^6}$ just like how M.N.C.E related $\sum_{n=1}^\infty \frac{ H_n}{n^{4}2^n}$ to $\sum_{n=1}^\infty \frac{(-1)^n H_n}{n^{4}}$ here math.stackexchange.com/q/973408 , then the mission will be easier. – Ali Shather Aug 10 at 22:02

This is not going to be a full answer but I think that this approach is quite viable and will eventually lead to a full answer and therefore is worth presenting. Let us start with a different quantity ${\bf H}^{(1)}_6(-1)$. We have: \begin{eqnarray} &&{\bf H}^{(1)}_6(-1) = \int\limits_0^{-1} \frac{[\log(-1/t)]^5}{5!} \cdot \frac{Li_1(t)}{t(1-t)} dt \\ &&\underbrace{=}_{u=t/(t-1)} \frac{1}{5!} \sum\limits_{p=0}^5 \binom{5}{p} (-1)^p \int\limits_0^{1/2} \frac{\log(u)^p \log(1-u)^{6-p}}{u} du \end{eqnarray} Now, all we need to do is to evaluate the integrals above.
Take $p=0$ and $p=5$ first: \begin{eqnarray} \int\limits_0^{1/2} \frac{\log(1-u)^6}{u} du &=& \left.\sum\limits_{q=1}^7 (-1)^q 6_{(q-1)} Li_q(1-u) \log(1-u)^{7-q}\right|_0^{1/2}\\ \int\limits_0^{1/2} \frac{\log(u)^5 \log(1-u)^1}{u} du &=& \left.\sum\limits_{q=1}^6 (-1)^q 5_{(q-1)} Li_{1+q}(u) \log(u)^{6-q}\right|_0^{1/2} \end{eqnarray} Now we move on to the case $p=1,2$. We have: \begin{eqnarray} \int\limits_0^{1/2} \frac{\log(u)^p \log(1-u)^{6-p}}{u} du &=& \frac{1}{p+1} \log(u)^{p+1} \log(1-u)^{6-p} + \frac{(6-p)}{p+1} \int\limits_{1/2}^1 \frac{\log(u)^{5-p}}{u} \log(1-u)^{p+1} du\\ &=& \frac{1}{p+1} \log(u)^{p+1} \log(1-u)^{6-p} + \frac{(6-p)}{p+1} ({\mathcal I}^{(p+1,5-p)} - \int\limits_{0}^{1/2} \frac{\log(u)^{5-p}}{u} \log(1-u)^{p+1} du) \end{eqnarray} Here we integrated by parts and then changed variables $u \rightarrow 1-u$. Then we used Compute an integral containing a product of powers of logarithms. . Here $${\mathcal I}^{(q,p)} = \left.\sum\limits_{l=0}^{p+1} \frac{1}{(l+1)!} \sum\limits_{\begin{array}{r}j_0+j_1+\cdots+j_l=q\\j_0\ge1,\cdots,j_l\ge 1\end{array}} \binom{q}{j_0,\cdots,j_l} \cdot \frac{\partial^p}{\partial \theta^p} \left(\frac{\prod\limits_{\xi=0}^l [ \Psi^{(j_\xi-1)}(1) - \Psi^{(j_\xi-1)}(1+\theta)]}{\theta}\right) \right|_{\theta=0}$$ where $\Psi$ is the polygamma function. On the other hand from Calculating alternating Euler sums of odd powers we know that : $${\bf H}^{(1)}_6(-1)=\frac{56 \pi ^4 \zeta (3)+480 \pi ^2 \zeta (5)-16965 \zeta (7)}{5760}$$ Bringing everything together we have: \begin{eqnarray} &&\int\limits_0^{1/2} \frac{\log(u)^4 \log(1-u)^2}{u} du=\\ &&48 \text{Li}_7\left(\frac{1}{2}\right)+8 \text{Li}_4\left(\frac{1}{2}\right) \log ^3(2)+24 \text{Li}_5\left(\frac{1}{2}\right) \log ^2(2)+48 \text{Li}_6\left(\frac{1}{2}\right) \log (2)+\frac{\pi ^4 \zeta (3)}{15}-\frac{765 \zeta (7)}{8}+\pi ^2 \left(4 \zeta (5)-\frac{2}{15} \log ^5(2)\right)+\frac{7}{4} \zeta (3) \log ^4(2)+\frac{5 \log ^7(2)}{21}+\\ &&\frac{4}{3} \int\limits_0^{1/2} \frac{\log(u)^3 \log(1-u)^3}{u} du \end{eqnarray} Finally we have: \begin{eqnarray} \int\limits_0^{1/2} \frac{\log(u)^4 \log(1-u)^2}{u} du&=& 2 \sum\limits_{j=0}^4 \binom{4}{j} \log(2)^{4-j} j! \sum\limits_{m=1}^\infty \frac{H_{m-1}}{m^{j+2}} \frac{1}{2^m}\\ \int\limits_0^{1/2} \frac{\log(u)^3 \log(1-u)^3}{u} du &=&3 \sum\limits_{j=0}^3 \binom{3}{j} \log(2)^{3-j} j! \sum\limits_{m=1}^\infty \frac{H_{m-1}^2-H_{m-1}^{(2)}}{m^{j+2}} \cdot \frac{1}{2^m} \end{eqnarray} Now we use the following results for the non-linear Euler sums: \begin{eqnarray} &&-3 \sum\limits_{m=1}^\infty \frac{H_{m-1}^2-H_{m-1}^{(2)}}{m^p}\cdot \frac{1}{2^m} =\left\{\right.\\ &&\left. \begin{array}{rr} 6 \text{Li}_4\left(\frac{1}{2}\right)+\frac{21}{4} \zeta (3) \log (2)-\frac{\pi ^4}{15}+\frac{\log ^4(2)}{2}-\frac{1}{4} \pi ^2 \log ^2(2) & \mbox{if $p=2$}\\ 6 \text{Li}_5\left(\frac{1}{2}\right)+\frac{\pi ^2 \zeta (3)}{2}-\frac{189 \zeta (5)}{16}-\frac{21}{8} \zeta (3) \log ^2(2)-\frac{\log ^5(2)}{10}+\frac{1}{12} \pi ^2 \log ^3(2)+\frac{1}{15} \pi ^4 \log (2) & \mbox{if $p=3$}\\ -3 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)-6 \text{Li}_5\left(\frac{1}{2}\right) \log (2)+3 \zeta (3)^2-\frac{1}{2} \pi ^2 \zeta (3) \log (2)+\frac{189}{16} \zeta (5) \log (2)-\frac{23 \pi ^6}{5040}-\frac{\log ^6(2)}{15}+\frac{1}{24} \pi ^2 \log ^4(2)-\frac{1}{30} \pi ^4 \log ^2(2) & \mbox{if $p=4$}\\ 12 \text{Li}_7\left(\frac{1}{2}\right)+3 \text{Li}_5\left(\frac{1}{2}\right) \log ^2(2)+12 \text{Li}_6\left(\frac{1}{2}\right) \log (2)+\frac{\pi ^4 \zeta (3)}{120}+\frac{\pi ^2 \zeta (5)}{2}-\frac{765 \zeta (7)}{64}-\frac{3}{4} \zeta (3) \log ^4(2)+\frac{1}{2} \pi ^2 \zeta (3) \log ^2(2)-6 \zeta (5) \log ^2(2)-\zeta (3)^2 \log (8)-\frac{\log ^7(2)}{84}+\frac{1}{40} \pi ^2 \log ^5(2)+\frac{1}{72} \pi ^4 \log ^3(2)+\frac{23 \pi ^6 \log (2)}{5040} -6 {\bf H}^{(1)}_6(1/2)-6 \log(2) {\bf H}^{(1)}_5(1/2) & \mbox{if $p=5$} \end{array} \right. \end{eqnarray} We solve the last equation above for our quantity in question and we get: \begin{eqnarray} &&{\bf H}^{(1)}_6(\frac{1}{2})=\\ && 2 \text{Li}_7\left(\frac{1}{2}\right)-\frac{1}{2} \text{Li}_5\left(\frac{1}{2}\right) \log ^2(2)-\text{Li}_6\left(\frac{1}{2}\right) \log (2)+\frac{\zeta (2)^2 \zeta (3)}{20}+\frac{\zeta (2) \zeta (5)}{2}-\frac{255 \zeta (7)}{128}-\frac{1}{60} \zeta (2) \log ^5(2)+\frac{1}{24} \zeta (3) \log ^4(2)+\frac{7}{120} \zeta (2)^2 \log ^3(2)-\frac{31}{32} \zeta (5) \log ^2(2)+\frac{179}{280} \zeta (2)^3 \log (2)-\frac{1}{4} \zeta (3)^2 \log (2)+\frac{11 \log ^7(2)}{5040} +\frac{1}{2} \log(2) {\bf H}^{(1)}_5(-1) + \frac{1}{2} \sum\limits_{m=1}^\infty \frac{H_{m-1}^2-H_{m-1}^{(2)}}{m^5} \cdot \frac{1}{2^m} \end{eqnarray}