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

Question

We consider a following class of Euler sums: \begin{equation} {\bf H}^{(1)}_p(\frac{1}{2}) := \sum\limits_{m=1}^\infty \frac{H_m}{m^p} \cdot \frac{1}{2^m} \end{equation} Now by using the following integral representation: \begin{equation} {\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 \end{equation} 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.

Answer 1

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 \begin{equation} {\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} \end{equation} where $\Psi$ is the polygamma function. On the other hand from Calculating alternating Euler sums of odd powers we know that : \begin{equation} {\bf H}^{(1)}_6(-1)=\frac{56 \pi ^4 \zeta (3)+480 \pi ^2 \zeta (5)-16965 \zeta (7)}{5760} \end{equation} 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}