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Definition

$$\mathbf{H}_{m}^{(n)}(x) = \sum_{k=1}^\infty \frac{H_k^{(n)}}{k^m} x^k\tag{1}$$

We define $$\mathbf{H}_{m}^{(1)}(x) = \mathbf{H}_{m}(x)=\sum_{k=1}^\infty \frac{H_k}{k^m} x^k \tag{2}$$

Note the alternating general formula $$\mathbf{H}_{m}(-1) = \sum_{k=1}^\infty (-1)^k \frac{H_k}{k^m} \tag{3}$$


Motivation

(1) seems to be impossible to track so we focus on (2) and (3). It has been proven in [5] and [6] that the form $\mathbf{H}_{2m}(-1)$ has a general formula in terms of zeta functions $$\begin{align*} \mathbf{H}_{2m}(-1) &=\frac{2m+1}{2}\left(1-2^{-2m}\right)\zeta(2m+1)-\frac{1}{2}\zeta(2m+1)\\ &\qquad-\sum_{k=1}^{m-1}\left(1-2^{1-2k}\right)\zeta(2k)\zeta(2m+1-2k) \end{align*}$$

Up to my knowledge the literature lacks any general formula for $\mathbf{H}_{2m+1}(-1)$. The odd formula seems to contain a finite combination of zeta and polylogs and their multiplication.

Examples

In [1] we see different evaluations for

$$\mathbf{H}_{1}(-1) = \frac{1}{2} \log^2 (2)-\frac{1}{2} \zeta(2)$$

In [2] we have

$$\mathbf{H}_{3}(-1)=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3)$$

In [3] we have some impressive calculations leading to

$$\begin{align} \color{blue}{\mathbf{H}_{3}(x)}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\color{blue}{\mathbf{H}_{2}(x)}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+\frac{\pi^4}{60}. \end{align}$$

Also in [8]

\begin{align} \color{blue}{\mathbf{H}_{4}(x)} =&\ \frac1{10}\zeta(3)\ln^2 x+\frac{\pi^4}{150}\ln x-\frac{\pi^2}{30}\operatorname{Li}_3(x)-\frac1{60}\ln^3x\ln^2(1-x)+\frac65\operatorname{Li}_5(x)\\&-\frac15\left[\operatorname{Li}_3(x)-\operatorname{Li}_2(x)\ln x-\frac12\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)-\frac15\operatorname{Li}_4(x)\\&-\frac35\operatorname{Li}_4(x)\ln x+\frac15\operatorname{Li}_3(x)\ln x+\frac15\operatorname{Li}_3(x)\ln^2x-\frac1{10}\operatorname{Li}_3(1-x)\ln^2 x\\&-\frac1{15}\operatorname{Li}_2(x)\ln^3x-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(x)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(x)} +\frac15\color{blue}{\mathbf{H}_{1}^{(3)}(x)}\ln x\\&-\frac15\color{blue}{\mathbf{H}_{1}^{(2)}(x)}\ln x+\frac25\color{blue}{\mathbf{H}_{3}(x)}\ln x-\frac15\color{blue}{\mathbf{H}_{2}(x)}\ln^2x+\frac1{15}\color{blue}{\mathbf{H}_{1}(x)}\ln^3x\\&+\frac{\pi^4}{450}+\frac{\pi^2}{5}\zeta(3)-\frac35\zeta(3)+3\zeta(5)\ \end{align}

In [4] I showed

$$\int\limits_0^1 \dfrac{\log^2 (1+x)\log^n x}{x}\; dx =2 (-1)^n(n!) \left[ \mathbf{H}_{n+2}(-1) + \left(1-2^{-n-2} \right) \zeta(n+3) \right]$$

Questions

  • Can we evaluate

$$\mathbf{H}_{5}(x) , \mathbf{H}_{5}(-1)$$

  • Can we show the following has no simple general formula ? $$\mathbf{H}_{2n+1}(x),\mathbf{H}_{2n+1}(-1)$$

Conjectures

  1. Interestingly the evaluations of $\mathbf{H}_{m}^{(n)}(-1)$ are related to $\mathbf{H}_{m}^{(n)}\left(\frac{1}{2}\right)$ with the same complexity.
  2. The form $\mathbf{H}_{m}^{(n)}(x)$ seem to involve a finite sum of products of logs,polylogs and zeta values.
  3. There can exist a recursive formula that connects

$$\mathbf{H}_{m}^{(n)}(x) = \sum_{1\leq s,t < m} (a_{s,t})\,\mathbf{H}_{s}^{(t)}(x)$$

References

[1] Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$

[2] Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$

[3] Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^32^n}$

[4] Evaluating $\int_0^1 \frac{\ln^m (1+x)\ln^n x}{x}\; dx$ for $m,n\in\mathbb{N}$

[5] https://arxiv.org/pdf/1301.7662.pdf

[6] http://algo.inria.fr/flajolet/Publications/FlSa98.pdf

[7] Alternating Euler sums

Related

[8] How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$

[9] Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^32^n}$

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  • $\begingroup$ Some of the $H$'s have different fonts. Is this on purpose? $\endgroup$ – Simply Beautiful Art Mar 3 '17 at 1:10
  • $\begingroup$ Also, why do you mention your first equation if it is never used in the question? $\endgroup$ – Simply Beautiful Art Mar 3 '17 at 1:14
  • $\begingroup$ @SimplyBeautifulArt , $\mathbf{H}_{m}(x)$ is different than $H_m$, the bold one takes an extra variable $x \in \mathbb{R}$. The first formula just to generalize the notation. $\endgroup$ – Zaid Alyafeai Mar 3 '17 at 1:21
  • $\begingroup$ Glad I favorite posts like these. $\endgroup$ – Simply Beautiful Art May 5 '17 at 22:50
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Please allow me the privilege of using a different notation so that I can make use of the results that I have derived myself without falling into a danger of making mistakes.We have: \begin{eqnarray} {\bf H}_n^{(1)}(t):= \sum\limits_{m=1}^\infty H_m^{(1)} \frac{t^m}{m^n} &=& PolyLog[n-1,2,t] + Li_{n+1}(t)\\ &=&\frac{(-1)^n}{2!(n-2)!} \int\limits_0^1 \frac{[\log(\eta)]^{n-2}}{\eta} [\log(1-t \eta)]^2d \eta + Li_{n+1}(t) \end{eqnarray} Here PolyLog[,,] is the Nielsen generalised poly-logarithm. In the first line above we used the results from the answer to Closed form expressions for harmonic sums and in the second line we wrote down explicitely the definition of the Nielsen poly-logarithm. Now are are going to set $t=-1$ and then we use the identity $\log(1+\eta) = \log(1-\eta^2) - \log(1-\eta)$. Therefore we have: \begin{eqnarray} &&\left({\bf H}_n^{(1)}(t)- Li_{n+1}(-1)\right) \cdot \frac{2!(n-2)!}{(-1)^n} = \\ && \int\limits_0^1 \frac{[\log(\eta)]^{n-2}\left(\log(1-\eta^2)^2 - 2 \log(1+\eta)\log(1-\eta) - \log(1-\eta)^2\right)}{\eta} d \eta=\\ && \left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(\eta)^{n-2} \frac{\log(1-\eta)^2}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(1-\eta)^{n-1} \frac{\log(\eta)^{1}}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}})(-1)^n (n-2)!\left(n \zeta(n+1) - \sum\limits_{j=1}^{n-2} \zeta(1+j) \zeta(n-j)\right) - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)= \end{eqnarray} In the top line we just expanded the square of the difference of the two logs. In the next line changed variables accordingly and regrouped the whole expression into two different types of integrals. In the subsequent line we integrated by parts in the first integrals and left the second integral unchanged and finally in the bottom line we used Compute an integral containing a product of powers of logarithms. . Now, the only thing that remains is to compute the remaining integral on the rhs. We have: \begin{eqnarray} {\mathcal I}^{(n-2)} &:=& \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \\ &=& \sum\limits_{r=1}^\infty \frac{(-1)^{r-1}}{r} \cdot \frac{\partial^{n-2}}{\partial \theta^{n-2}} \left.\left( \frac{\Psi^{(0)}(1) - \Psi^{(0)}(1+r+\theta)}{\theta+r}\right)\right|_{\theta=0}\\ &=& \sum\limits_{r=1}^\infty\sum\limits_{l=0}^{n-2} \frac{(-1)^{r-1}}{r} \cdot \binom{n-2}{l} \frac{(n-2-l)! (-1)^{n-2-l}}{r^{n-2-l+1}} (\Psi^{(0)}(1) \delta_{l,0} - \Psi^{(l)}(1+r)) \\ &=&(-1)^n (n-2)! \left(- \sum\limits_{l=1}^{n-2} Li_{n-l}(-1) \zeta(l+1)+ \sum\limits_{l=0}^{n-2} \sum\limits_{r=1}^\infty \frac{(-1)^r}{r^{n-l}} H_r^{(l+1)}\right)\\ &=&(-1)^n (n-2)! \left(+ \sum\limits_{l=1}^{n-2} \frac{2^{n-l-1}-1}{2^{n-l-1}}\zeta(n-l)\zeta(l+1)+ \sum\limits_{l=0}^{n-2} {\bf H}_{n-l}^{(l+1)}(-1) \right) \end{eqnarray} In the top line we expanded the term $\log(1+\eta)$ in a series and then integrated term by term by using Compute an integral containing a product of powers of logarithms. .In the following line we computed the partial derivative using the chain rule and in the last line we simplified the result. Bringing everything together we get the following: \begin{eqnarray} {\bf H}_n^{(1)}(-1) = \left(\frac{n+1}{2^n} - \frac{n}{2} - 1\right) \zeta(n+1) - \sum\limits_{j=1}^{n-2} \left(\frac{1}{2^n} + \frac{1}{2} - \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n-j) - \sum\limits_{l=0}^{n-2} {\bf H}^{(l+1)}_{n-l}(-1) \end{eqnarray} for $n\ge 2$. Otherwise by going back to the original integral representation we have: \begin{equation} {\bf H}^{(1)}_1(-1) = -\frac{\pi^2}{12} + \frac{1}{2} \log(2)^2 \end{equation} It is clearly seen that calculating our sum in question requires knowledge of sums that involve generalized harmonic numbers. It is quite likely that there exists a whole hierarchy of recurrence relations that interwine the generating functions ${\bf H}_n^{(q)}(t)$. In order to get some insight into this topic we at least write down a following identity for another sum in the hierarchy. We have: \begin{eqnarray} &&{\bf H}_n^{(q)}(t) := \sum\limits_{m=1}^\infty H_m^{(q)} \cdot \frac{t^m}{m^n} =\\ &&Li_{n+q}(t) + Li_n(t) Li_q(t) - \sum\limits_{l=1}^n \frac{1}{(n-l)!}\int\limits_0^t \frac{[\log(t/\eta)]^{n-l}}{\eta} \cdot Li_{q-1}(\eta) Li_l(\eta) d \eta =\\ &&Li_{n+q}(t) + Li_{n}(t) Li_{q}(t) - \sum\limits_{l=1}^n Li_{n-l+q}(t) Li_{l}(t) + \\ &&\sum\limits_{l=0}^{n-1} \sum\limits_{l_1=q}^{n-l-1+q} \frac{1}{(n-l-l_1+q-1)!} \int\limits_0^t \frac{[\log(t/\eta)]^{n-l-l_1+q-1}}{\eta} Li_{l_1}(\eta) Li_l (\eta) d \eta \end{eqnarray} Here $n\ge 2$ and $q \ge 1$. The above expression follows from the general integral representation given in the answer to Closed form expressions for harmonic sums and from integration by parts once (middle) and twice (bottom). We believe that the last integral on the rhs above will be easy to calculate since because it satisfies certain recurrence relations.

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  • $\begingroup$ Good job. Were you able to find some specific values using the recurrence formula ? $\endgroup$ – Zaid Alyafeai Apr 26 '17 at 9:34
  • $\begingroup$ @Zaid Alyafeai: Thanks for the nice words. It is great fun to work on this stuff. Unfortunately I haven't worked out any particular values yet since we need additional recurrence relations for generalized harmonic sums. I think that the last expression above will be quite useful for that purpose. Can I ask you why are you interested in this stuff ? $\endgroup$ – Przemo Apr 26 '17 at 11:53
  • $\begingroup$ I worked on Euler sums around 3 yrs ago and I am trying to refresh my mind on them. What I am trying to understand is whether we can find a general approach of Euler sums and their relations together. Furthermore, Can we prove if a certain family is reducible to some known closed forms? Why alternating Euler sums of even weight are harder to evaluate and there is no general formula? $\endgroup$ – Zaid Alyafeai Apr 26 '17 at 12:12
  • $\begingroup$ Congrats you earned the bounty. $\endgroup$ – Zaid Alyafeai Apr 30 '17 at 16:29
  • $\begingroup$ @Zaid Alyafeair: Thank you very much for the bounty. I have been struggling with deriving similar relations for higher order sums. In principle all one needs to do is use integration by parts. However it is important which part of the integrand one takes an anti-derivative of. On last Friday I used the identities on the bottom of the answer above and after long and tedious but straightforward transformations I arrived at a trivial identity, i.e. the generating functions that we seek cancelled from both sides of the equation. My 2nd answer below has salvaged that. $\endgroup$ – Przemo May 2 '17 at 12:53
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Let us now consider the case of an odd order of harmonic numbers. As usual we start from the integral representation of our sums . We have: \begin{eqnarray} &&{\bf H}^{(2q+1)}_n(t) - Li_{n+2q+1}(t) =\int\limits_0^t \frac{[\log(t/\xi)]^{n-1}}{(n-1)!} \cdot \frac{Li_{2q+1}(\xi)}{1-\xi} d\xi\\ &&= \sum\limits_{j=0}^q (-1)^{q+j} \left[\binom{q+j}{2 j}\frac{1}{2} + \binom{q+j}{2 j+1}\right]\cdot \int\limits_0^1 \frac{[\log(1/\xi)]^{n-2(j+1)}}{(n-2(j+1))!} \cdot \frac{[Li_{q+j+1}(t \xi)]^2}{\xi}d\xi\\ &&=\sum\limits_{l_1=0}^{2q+1}\left\{\sum\limits_{j=0 \vee (l_1-q-1)}^q (\binom{q+j}{2j+0}\frac{1}{2} + \binom{q+j}{2j+1})\binom{q+n-j-1-l_1}{n-2 j-2}\right\}(-1)^{1-l_1} \cdot \cdot (Li_{l_1}(t) 1_{l_1\ge 0} - \delta_{l_1,0}) \cdot Li_{2q+n+1-l_1}(t)+\sum\limits_{l_1=1}^{n-1} \left\{\sum\limits_{j=0 }^{q \wedge \lfloor \frac{n-1-l_1}{2}\rfloor } (\binom{q+j}{2j+0}\frac{1}{2} + \binom{q+j}{2j+1})\binom{q+n-j-1-l_1}{q+j}\right\}(-1)^1 \cdot {\bf H}^{(n+2q+1-l_1)}_{l_1}(t) \end{eqnarray} In the second line from the top we integrated by parts $(2q+2)$-times each time using the well known properties of the poly-logarithm. What we essentially did at each step was that we found the anti-derivatives of $Li_{\theta_1}(\xi) Li_{\theta_2}(\xi)/\xi$ for some integer values of $\theta_1$ and $\theta_2$. The result is a linear combination of products of pairs of poly-logs and a residual term which is either a half of a square of a poly-log or something else depending on whether $(n-p)$ is odd or even in the first and in the second case respectively.Since integration by parts produces surface terms we have to assume that $n\ge 2q+2$ for all those terms to disappear. In the subsequent line we just used Generalized definite dilogarithm integral. and we simplified the result. The result constitutes a set of recurrence relations that entwine the harmonic sums. Here $q=0,1,2,\cdots$ and $n\ge 2q+2$ and $t\in (-1,1)$. In case $n=1,\cdots,2q+1$ we have to go back to the original integral representation and take into account the surface terms. We have: \begin{eqnarray} {\bf H}^{(2q+1)}_{2 n+1}(t) &=& \sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} \cdot Li_l(t) Li_{2q+2n+2-l}(t) (-1)^{l-(2n+1)} +\\ &&(-1)^{q+n} \frac{1}{2} \binom{q+n}{2 n} \cdot [Li_{q+n+1}(t)]^2+\\ &&\sum\limits_{j=0}^{n-1} (-1)^{q+j} \cdot \left(\frac{1}{2} \binom{q+j}{2 j} + \binom{q+j}{2j+1} \right)\cdot \int\limits_0^t \frac{[\log(t/\xi)]^{2n-1-2 j}}{(2n-1-2 j)!} \cdot \frac{[Li_{q+j+1}(\xi)]^2}{\xi}d\xi+Li_{2q+2n+2}(t)\\ {\bf H}^{(2q+1)}_{2 n}(t) &=& \sum\limits_{l=2n}^{q+n} \binom{l-1}{2 n-1} \cdot Li_l(t) Li_{2q+2n+1-l}(t) (-1)^{l-(2n)} +\\ &&\sum\limits_{j=0}^{n-1} (-1)^{q+j} \cdot \left(\frac{1}{2} \binom{q+j}{2 j} + \binom{q+j}{2j+1} \right)\cdot \int\limits_0^t \frac{[\log(t/\xi)]^{2n-2-2 j}}{(2n-2-2 j)!} \cdot \frac{[Li_{q+j+1}(\xi)]^2}{\xi}d\xi+Li_{2q+2n+1}(t)\\ % {\bf H}^{(2q)}_{2 n}(t) &=& \sum\limits_{l=2n}^{q+n-1} \binom{l-1}{2 n-1} \cdot Li_l(t) Li_{2q+2n-l}(t) (-1)^{l-(2n)} +\\ &&(-1)^{q+n} \frac{1}{2} \binom{q+n-1}{2 n-1} \cdot [Li_{q+n}(t)]^2+\\ &&\sum\limits_{j=0}^{n-1} (-1)^{q+j} \cdot \left(\frac{1}{2} \binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j} \right)\cdot \int\limits_0^t \frac{[\log(t/\xi)]^{2n-1-2 j}}{(2n-1-2 j)!} \cdot \frac{[Li_{q+j}(\xi)]^2}{\xi}d\xi+Li_{2q+2n}(t)\\ % {\bf H}^{(2q)}_{2 n+1}(t) &=& \sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} \cdot Li_l(t) Li_{2q+2n+1-l}(t) (-1)^{l-(2n+1)} +\\ &&\sum\limits_{j=0}^{n} (-1)^{q+j} \cdot \left(\frac{1}{2} \binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j} \right)\cdot \int\limits_0^t \frac{[\log(t/\xi)]^{2n-2 j}}{(2n-2 j)!} \cdot \frac{[Li_{q+j}(\xi)]^2}{\xi}d\xi+Li_{2q+2n+1}(t) \end{eqnarray} both for $n\ge 0$ and for $q\ge 0$ in the two top cases above and for $n\ge 0$ and $q\ge 1$ in the two bottom cases above. The integrals on the right hand side are evaluated in Generalized definite dilogarithm integral..

Bringing everything together we have: \begin{eqnarray} &&{\bf H}^{(2q+1)}_{2n+1}(t)=\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} Li_l(t) Li_{2q+2n+2-l}(t) (-1)^{l-(2n+1)}+ (-1)^{q+n} \frac{1}{2} \binom{q+n}{2 n} [Li_{q+n+1}(t)]^2+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=0}^{q+n} \left\{\sum\limits_{j=0\vee (l-q-1)}^{n-1}(\frac{1}{2} \binom{q+j}{2 j} + \binom{q+j}{2j+1})\binom{q+2n-j-l}{2n-1-2j}\right\}(-1)^{1-l} (Li_{l}(t) 1_{l\ge 1} - \delta_{l,0})Li_{2q+2n+2-l}(t)+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=1}^{2 n} \left\{\sum\limits_{j=0}^{\lfloor n-\frac{l}{2}\rfloor}(\frac{1}{2}\binom{q+j}{2 j} + \binom{q+j}{2j+1})\binom{q+2n- j-l}{q+j}\right\} (-1)^1 {\bf H}^{(2q+2n+2-l)}_l(t)+Li_{2q+2n+2}(t)\\ % &&{\bf H}^{(2q+1)}_{2n}(t)=\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=2n}^{q+n} \binom{l-1}{2 n-1} Li_l(t) Li_{2q+2n+1-l}(t) (-1)^{l-(2n)}+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=0}^{q+n} \left\{\sum\limits_{j=0\vee (l-q-1)}^{n-1}(\frac{1}{2} \binom{q+j}{2 j} + \binom{q+j}{2j+1})\binom{q+2n-1-j-l}{2n-2-2j}\right\}(-1)^{1-l} (Li_{l}(t) 1_{l\ge 1} - \delta_{l,0})Li_{2q+2n+1-l}(t)+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=1}^{2 n-1} \left\{\sum\limits_{j=0}^{\lfloor n-\frac{1+l}{2}\rfloor}(\frac{1}{2}\binom{q+j}{2 j} + \binom{q+j}{2j+1})\binom{q+2n-1- j-l}{q+j}\right\} (-1)^1 {\bf H}^{(2q+2n+1-l)}_l(t)+Li_{2q+2n+1}(t)\\ % &&{\bf H}^{(2q)}_{2n}(t)=\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=2n}^{q+n-1} \binom{l-1}{2 n-1} Li_l(t) Li_{2q+2n-l}(t) (-1)^{l-(2n)}+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!(-1)^{q+n} \frac{1}{2} \binom{q+n-1}{2n-1} [Li_{q+n}(t)]^2+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=0}^{q+n-1} \left\{\sum\limits_{j=0\vee (l-q)}^{n-1}(\frac{1}{2} \binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j})\binom{q+2n-1-j-l}{2n-1-2j}\right\}(-1)^{l} (Li_{l}(t) 1_{l\ge 1} - \delta_{l,0})Li_{2q+2n-l}(t)+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=1}^{2 n} \left\{\sum\limits_{j=0}^{\lfloor n-\frac{l}{2}\rfloor}(\frac{1}{2}\binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j})\binom{q+2n-1- j-l}{q+j-1}\right\} (-1)^0 {\bf H}^{(2q+2n-l)}_l(t)+Li_{2q+2n}(t)\\ % &&{\bf H}^{(2q)}_{2n+1}(t)=\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} Li_l(t) Li_{2q+2n+1-l}(t) (-1)^{l-(2n+1)}+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=0}^{q+n} \left\{\sum\limits_{j=0\vee (l-q)}^{n}(\frac{1}{2} \binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j})\binom{q+2n-j-l}{2n-2j}\right\}(-1)^{l} (Li_{l}(t) 1_{l\ge 1} - \delta_{l,0})Li_{2q+2n+1-l}(t)+\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=1}^{2 n+1} \left\{\sum\limits_{j=0}^{\lfloor n+\frac{1-l}{2}\rfloor}(\frac{1}{2}\binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j})\binom{q+2n- j-l}{q+j-1}\right\} (-1)^0 {\bf H}^{(2q+2n+1-l)}_l(t)+Li_{2q+2n+1}(t) \end{eqnarray} Note that as long as the first two equations are useful because the quantity in question only appears on the left hand side the last two equations are less useful because the quantity being searched for actually cancels on both sides of the equation. In general it turns out that the odd-odd quantities always reduce to even-even quantities and poly-logs. On the other hand the odd-even quantities always reduce to even-odd quantities and poly-logs. We have \begin{eqnarray} {\bf H}^{(1)}_1(t) &=& \frac{1}{2}\left( [\log(1-t)]^2 + 2 Li_2(t)\right)\\ {\bf H}^{(1)}_2(t) &=& \frac{1}{2}\left(-{\bf H}^{(2)}_1(t) - \log(1-t) Li_2(t) + 3 Li_3(t)\right)\\ {\bf H}^{(1)}_3(t) &=& \frac{1}{4} \left( -2 {\bf H}^{(2)}_2(t) + [Li_2(t)]^2 + 6 Li_4(t)\right)\\ {\bf H}^{(1)}_4(t) &=& \frac{1}{4}\left(-2 {\bf H}^{(2)}_3(t)+{\bf H}^{(4)}_1(t) + Li_2(t) Li_3(t) + \log(1-t) Li_4(t) + 5 Li_5(t) \right) \\ {\bf H}^{(1)}_5(t) &=& \frac{1}{4}\left( -2 {\bf H}^{(2)}_4(t) + {\bf H}^{(4)}_2(t) + [Li_3(t)]^2 - Li_2(t) Li_4(t) + 5Li_6(t)\right)\\ {\bf H}^{(1)}_6(t) &=&\frac{1}{4}\left(-2{\bf H}^{(2)}_5(t)+{\bf H}^{(4)}_3(t)-2{\bf H}^{(6)}_1(t) + Li_3(t) Li_4(t) - 2 Li_2(t)Li_5(t) - 2\log(1-t)Li_6(t) + 7 Li_7(t)\right)\\ {\bf H}^{(1)}_7(t) &=& \frac{1}{8} \left( -4 {\bf H}^{(2)}_6(t)+ 2 {\bf H}^{(4)}_4(t) - 4 {\bf H}^{(6)}_2(t) + 5 [Li_4(t)]^2 - 8 Li_3(t) Li_5(t) + 4 Li_2(t) Li_6(t) + 14 Li_8(t)\right)\\ {\bf H}^{(1)}_8(t) &=& \frac{1}{8} (-4 {\bf H}^{(2)}_7(t)+2 {\bf H}^{(4)}_5(t)-4 {\bf H}^{(6)}_3(t)+17 {\bf H}^{(8)}_1(t)+5 \text{Li}_4(t) \text{Li}_5(t)-13 \text{Li}_3(t) \text{Li}_6(t)+17 \text{Li}_2(t) \text{Li}_7(t)-3 \text{Li}_9(t)+17 \text{Li}_8(t) \log (1-t))\\ {\bf H}^{(1)}_9(t) &=& \frac{1}{8} \left(-4 {\bf H}^{(2)}_8(t)+2 {\bf H}^{(4)}_6(t)-4 {\bf H}^{(6)}_4(t)+17 {\bf H}^{(8)}_2(t)+26 \text{Li}_5(t){}^2-47 \text{Li}_4(t) \text{Li}_6(t)+34 \text{Li}_3(t) \text{Li}_7(t)-17 \text{Li}_2(t) \text{Li}_8(t)-3 \text{Li}_{10}(t)\right)\\ {\bf H}^{(1)}_{10}(t) &=& \frac{1}{8} (-4 {\bf H}^{(2)}_9(t)+2 {\bf H}^{(4)}_7(t)-4 {\bf H}^{(6)}_5(t)+17 {\bf H}^{(8)}_3(t)-124 {\bf H}^{(10)}_1(t)+26 \text{Li}_5(t) \text{Li}_6(t)-73 \text{Li}_4(t) \text{Li}_7(t)+107 \text{Li}_3(t) \text{Li}_8(t)-124 \text{Li}_2(t) \text{Li}_9(t)+121 \text{Li}_{11}(t)-124 \text{Li}_{10}(t) \log (1-t))\\ \end{eqnarray} Likewise we have: \begin{eqnarray} {\bf H}^{(3)}_1(t) &=&\frac{1}{2} \left(-\text{Li}_2(t){}^2+2 \text{Li}_4(t)-2 \text{Li}_3(t) \log (1-t)\right)\\ {\bf H}^{(3)}_2(t) &=& \frac{1}{2} (-3 {\bf H}^{(4)}_1(t)-\text{Li}_2(t) \text{Li}_3(t)+5 \text{Li}_5(t)-3 \text{Li}_4(t) \log (1-t))\\ {\bf H}^{(3)}_3(t) &=& \frac{1}{2} \left(-3 {\bf H}^{(4)}_2-2 \text{Li}_3(t){}^2+3 \text{Li}_2(t) \text{Li}_4(t)+5 \text{Li}_6(t)\right)\\ {\bf H}^{(3)}_4(t) &=& \frac{1}{2} (-3 {\bf H}^{(4)}_3(t)+5 {\bf H}^{(6)}_1(t)-2 \text{Li}_3(t) \text{Li}_4(t)+5 \text{Li}_2(t) \text{Li}_5(t)+5 \text{Li}_6(t) \log (1-t))\\ {\bf H}^{(3)}_5(t) &=& \frac{1}{2} \left(-3 {\bf H}^{(4)}_4(t)+5 {\bf H}^{(6)}_2(t)-6 \text{Li}_4(t){}^2+10 \text{Li}_3(t) \text{Li}_5(t)-5 \text{Li}_2(t) \text{Li}_6(t)\right)\\ {\bf H}^{(3)}_6(t) &=& \frac{1}{2} (-3 {\bf H}^{(4)}_5(t)+5 {\bf H}^{(6)}_3(t)-21 {\bf H}^{(8)}_1(t)-6 \text{Li}_4(t) \text{Li}_5(t)+16 \text{Li}_3(t) \text{Li}_6(t)-21 \text{Li}_2(t) \text{Li}_7(t)+21 \text{Li}_9(t)-21 \text{Li}_8(t) \log (1-t))\\ {\bf H}^{(3)}_7(t) &=& \frac{1}{2} \left(-3 {\bf H}^{(4)}_6(t)+5 {\bf H}^{(6)}_4(t)-21 {\bf H}^{(8)}_2(t)-32 \text{Li}_5(t){}^2+58 \text{Li}_4(t) \text{Li}_6(t)-42 \text{Li}_3(t) \text{Li}_7(t)+21 \text{Li}_2(t) \text{Li}_8(t)+21 \text{Li}_{10}(t)\right)\\ \hline \\ {\bf H}^{(5)}_1(t)&=&\frac{1}{2} \left(\text{Li}_3(t){}^2-2 \text{Li}_2(t) \text{Li}_4(t)+2 \text{Li}_6(t)-2 \text{Li}_5(t) \log (1-t)\right)\\ {\bf H}^{(5)}_2(t)&=&\frac{1}{2} (-5 {\bf H}^{(6)}_1(t)+\text{Li}_3(t) \text{Li}_4(t)-3 \text{Li}_2(t) \text{Li}_5(t)+7 \text{Li}_7(t)-5 \text{Li}_6(t) \log (1-t))\\ {\bf H}^{(5)}_3(t)&=&\frac{1}{4} \left(-10 {\bf H}^{(6)}_2(t)+9\text{Li}_4(t){}^2-16 \text{Li}_3(t) \text{Li}_5(t)+10 \text{Li}_2(t) \text{Li}_6(t)+14 \text{Li}_8(t)\right)\\ {\bf H}^{(5)}_4(t)&=&\frac{1}{4} (-10 {\bf H}^{(6)}_3(t)+35 {\bf H}^{(8)}_1(t)+9 \text{Li}_4(t) \text{Li}_5(t)-25 \text{Li}_3(t) \text{Li}_6(t)+35 \text{Li}_2(t) \text{Li}_7(t)-21 \text{Li}_9(t)+35 \text{Li}_8(t) \log (1-t))\\ {\bf H}^{(5)}_5(t)&=& \frac{1}{4} \left(-10 {\bf H}^{(6)}_4(t)+35 {\bf H}^{(8)}_2(t)+52 \text{Li}_5(t){}^2-95 \text{Li}_4(t) \text{Li}_6(t)+70 \text{Li}_3(t) \text{Li}_7(t)-35 \text{Li}_2(t) \text{Li}_8(t)-21 \text{Li}_{10}(t)\right)\\ \hline \\ {\bf H}^{(7)}_1(t)&=&\frac{1}{2} \left(-\text{Li}_4(t){}^2+2 \text{Li}_3(t) \text{Li}_5(t)-2 \text{Li}_2(t) \text{Li}_6(t)+2 \text{Li}_8(t)-2 \text{Li}_7(t) \log (1-t)\right)\\ {\bf H}^{(7)}_2(t)&=&\frac{1}{2} (-7 {\bf H}^{(8)}_1(t)-\text{Li}_4(t) \text{Li}_5(t)+3 \text{Li}_3(t) \text{Li}_6(t)-5 \text{Li}_2(t) \text{Li}_7(t)+9 \text{Li}_9(t)-7 \text{Li}_8(t) \log (1-t))\\ {\bf H}^{(7)}_3(t)&=&\frac{1}{2} \left(-7 {\bf H}^{(8)}_2(t)-8 \text{Li}_5(t){}^2+15 \text{Li}_4(t) \text{Li}_6(t)-12 \text{Li}_3(t) \text{Li}_7(t)+7 \text{Li}_2(t) \text{Li}_8(t)+9 \text{Li}_{10}(t)\right)\\ \hline \\ {\bf H}^{(9)}_1(t)&=&\frac{1}{2} \left(\text{Li}_5(t){}^2-2 \text{Li}_4(t) \text{Li}_6(t)+2 \text{Li}_3(t) \text{Li}_7(t)-2 \text{Li}_2(t) \text{Li}_8(t)+2 \text{Li}_{10}(t)-2 \text{Li}_9(t) \log (1-t)\right) \end{eqnarray} Unfortunately both the even-even and the even-odd quantities cannot be worked out using the formalism above since the respective recurrence equations reduce to tautologies.

Update: Below we demonstrate that it is possible to get additional recurrence relations for both the even-odd and the even-even quantities provided $t=-1$. Let us start with the simplest possible example. Let us assume that $q\ge 1$ then we have: \begin{eqnarray} &&{\bf H}^{(2q)}_1(-1)= \sum\limits_{l=1}^q Li_l(-1) Li_{2q+1-l}(-1) (-1)^{l-1} + (-1)^q \underbrace{\int\limits_0^1 \frac{[Li_q(-\xi)]^2}{\xi} d\xi}_{{\mathcal A}^{(0,2)}_q(-1)} + Li_{2q+1}(-1)=\\ &&\frac{1}{4^q}\left(-1+(-2+4^q) q\right) \zeta(2q+1) - \log(2) \left(-1+\frac{1}{2^{2q-1}}\right) \zeta(2q)+\\ &&\sum\limits_{l=2}^q (-\frac{1}{2})^l \left(-2+2^{l-q}\right) \zeta(l) \zeta(2q+1-l)+\\ &&\sum\limits_{l=2}^{2q-1} \left(-\frac{1}{2} -2 (-1)^l +(-1)^l 2^{2-l} + \frac{1}{4^q} \right)\zeta(l) \zeta(2q+1-l)+\\ &&2{\bf H}^{(1)}_{2q}(-1) \end{eqnarray} In the top line we started from the integral representation which we integrated by parts $q$-times. In the bottom line we used the second answer to Generalized definite dilogarithm integral. to compute the integral on the right hand side. As a result we obtained a quite useful relation. Note that the harmonic sum on the left hand side is converging very slowly whereas the other sum on the right hand side converges quite fast. It is clear that this approach can be extended to more complicated cases. We have: \begin{eqnarray} &&{\bf H}^{(2q)}_{2n+1}(-1)=\\ &&\sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} Li_l(-1) Li_{2q+2n+1-l}(-1) (-1)^{l-(2n+1)}+\\ &&\sum\limits_{l=2}^{q+n} {\mathcal A}_0^{(n,l,q)} (-1)^l \frac{1-2^{1+l}+2^{2(n+q)}}{2^{2(n+q)}}\cdot \zeta(2n+2q+1-l) \zeta(l)+\\ &&\sum\limits_{l=2}^{2n+1} {\mathcal A}_1^{(n,l,q)} \frac{1-2^{1+l}+2^{2(n+q)}}{2^{2(n+q)}}\cdot \zeta(2n+2q+1-l) \zeta(l)+\\ &&\sum\limits_{l=1}^{2n+1} {\mathcal A}_1^{(n,l,q)}\left[(1-2^{1-2 n-2 q}) {\bf H}^{(l)}_{2n+2q+1-l}(+1) + 2 {\bf H}^{(l)}_{2n+2q+1-l}(-1)\right]+\\ &&Li_{2n+2 q+1}(-1) \end{eqnarray} for $n\ge 0$ and $q\ge 1$. Here the coefficients read: \begin{eqnarray} {\mathcal A}_0^{(n,l,q)}&:=& \sum\limits_{j=(l-q)\vee 0}^n (\frac{1}{2}\binom{q+j-1}{2j-1} + \binom{q+j-1}{2j}) \binom{-j-l+2n+q}{2(n-j)}\\ {\mathcal A}_1^{(n,l,q)}&:=& \sum\limits_{j= 0}^{n-\lfloor \frac{l-1}{2}\rfloor} (\frac{1}{2}\binom{q+j-1}{2j-1} + \binom{q+j-1}{2j}) \binom{-j-l+2n+q}{q+j-1} \end{eqnarray} In the even-even case we have: \begin{eqnarray} &&{\bf H}^{(2 q)}_{2 n}(-1)=\\ &&\sum\limits_{l=2n}^{q+n-1} \binom{l-1}{2n-1} Li_l(-1) Li_{2q+2n-l}(-1)(-1)^{l-2 n}+\\ &&(-1)^{q+n} \frac{1}{2} \binom{q+n-1}{2n-1} [Li_{q+n}(-1)]^2 +\\ &&\sum\limits_{l=2}^{q+n-1} {\mathcal A}_2^{(n,l,q)} \left( \frac{4-2^{2+l}+2^{2(n+q)}}{2^{2(n+q)}}\right)\cdot \zeta(2(n+q)-l)\zeta(l) (-1)^l+\\ &&\sum\limits_{l=2}^{2 n} {\mathcal A}_3^{(n,l,q)} \left( \frac{4-2^{2+l}+2^{2(n+q)}}{2^{2(n+q)}}\right)\cdot \zeta(2(n+q)-l)\zeta(l) +\\ &&\sum\limits_{l=1}^{2 n}{\mathcal A}_3^{(n,l,q)} \left((1-\frac{1}{2^{2(n+q-1)}}) {\bf H}^{(l)}_{2(n+q)-l}(+1) + 2 {\bf H}^{(l)}_{2(n+q)-l}(-1)\right)+\\ &&Li_{2n+2q}(-1) \end{eqnarray} where the coefficients read: \begin{eqnarray} {\mathcal A}_2^{(n,l,q)}&:=& \sum\limits_{j=(l-q)\vee 0}^n (\frac{1}{2}\binom{q+j-1}{2j-1} + \binom{q+j-1}{2j}) \binom{-1-j-l+2n+q}{2(n-j)-1}\\ {\mathcal A}_3^{(n,l,q)}&:=& \sum\limits_{j= 0}^{n-\lfloor \frac{l}{2}\rfloor} (\frac{1}{2}\binom{q+j-1}{2j-1} + \binom{q+j-1}{2j}) \binom{-1-j-l+2n+q}{q+j-1} \end{eqnarray} where $n\ge 1$ and $q\ge 1$. As we can see from the above we also need the results for plus unity. They read: \begin{eqnarray} &&{\bf H}^{(2q)}_{2n+1}(+1)=\\ &&\sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} Li_l(1) Li_{2q+2n+1-l}(1) (-1)^{l-(2n+1)}+\\ &&\sum\limits_{l=2}^{q+n} {\mathcal A}_4^{(n,l,q)} \zeta(l) \zeta(1-l+2 n+2 q) (-1)^l+\\ &&\sum\limits_{l=2}^{2n+1} {\mathcal A}_5^{(n,l,q)} \zeta(l) \zeta(1-l+2 n+2 q) +\\ &&\sum\limits_{l=1}^{2n+1} {\mathcal A}_5^{(n,l,q)}(-1)^1 {\bf H}^{(l)}_{1+2n+2q-l}(+1)+\\ &&Li_{2n+2q+1}(+1) \end{eqnarray} where \begin{eqnarray} {\mathcal A}_4^{(n,l,q)}&:=& \sum\limits_{j=(l-q) \vee 0}^n \left(\frac{1}{2} \binom{q+j-1}{2j-1}+\binom{q+j-1}{2 j}\right)\binom{q-j+2n-l}{2n-2j}\\ {\mathcal A}_5^{(n,l,q)}&:=& \sum\limits_{j=0}^{n+\lfloor \frac{1-l}{2} \rfloor}\left(\frac{1}{2} \binom{q+j-1}{2j-1}+\binom{q+j-1}{2 j}\right)\binom{q-j+2n-l}{q+j-1} \end{eqnarray} It is clear that an analogous formula exists for the remaining even-even case at plus unity. We will write it down later on. Now I am going to argue that the last two formulae above along with the relations that combine the odd-odd and the odd-even cases with the even-even and the even-odd cases-- the relations that hold for arbitrary value of $t$ -- that those relations are sufficient in order to work out closed form solutions for all the harmonic sums at plus unity. Indeed using this approach we found the following: \begin{eqnarray} {\bf H}^{(1)}_2(+1) &=& 2 \zeta(3)\\ \lim_{t\rightarrow 1} \left({\bf H}^{(2)}_1(t) + \log(1-t) Li_2(t)\right) &=& - \zeta(3)\\ \hline {\bf H}^{(1)}_3(+1) &=& -\frac{1}{6} \zeta(2)^2+ \frac{5}{3} \zeta(4)\\ {\bf H}^{(2)}_2(+1) &=& +\frac{5}{6} \zeta(2)^2 - \frac{1}{3} \zeta(4) \\ \lim_{t \rightarrow 1} \left({\bf H}^{(3)}_1(t) + \log(1-t) Li_3(t)\right) &=& -\frac{1}{2} \zeta(2)^2 + \zeta(4) \\ \hline {\bf H}^{(1)}_4(+1) &=& -\zeta(2)\zeta(3) + 3 \zeta(5)\\ {\bf H}^{(2)}_3(+1) &=&+3 \zeta(2) \zeta(3)-\frac{9}{2} \zeta(5)\\ {\bf H}^{(3)}_2(+1)&=& -2 \zeta(2) \zeta(3)+\frac{11}{2} \zeta(5)\\ \lim_{t\rightarrow 1} \left({\bf H}^{(4)}_1(t) + \log(1-t) Li_4(t)\right) &=&+1 \zeta(2) \zeta(3)-2 \zeta(5) \\ \hline {\bf H}^{(1)}_5(+1) &=& -\frac{1}{2}\zeta(3)^2-\frac{1}{3}\zeta(2)\zeta(4) + \frac{7}{3} \zeta(6)\\ {\bf H}^{(2)}_4(+1) &=&+1 \zeta(3)^2+\frac{4}{3} \zeta(2) \zeta(4)-\frac{8}{3} \zeta(6)\\ {\bf H}^{(3)}_3(+1)&=& +\frac{1}{2} \zeta(3)^2-2 \zeta(2) \zeta(4)+4\zeta(6)\\ {\bf H}^{(4)}_2(+1)&=& -1 \zeta(3)^2+\frac{7}{3} \zeta(2) \zeta(4)-1\zeta(6)\\ \lim_{t\rightarrow 1} \left({\bf H}^{(5)}_1(t) + \log(1-t) Li_5(t)\right) &=&+\frac{1}{2} \zeta(3)^2- \zeta(2) \zeta(4)+ \zeta(6) \\ \hline {\bf H}^{(1)}_6(+1) &=& -\zeta(3) \zeta(4)-\zeta(2) \zeta(5)+4 \zeta(7) \\ {\bf H}^{(2)}_5(+1) &=& +2\zeta(3) \zeta(4)+5\zeta(2) \zeta(5)-10 \zeta(7) \\ {\bf H}^{(3)}_4(+1) &=& +0\zeta(3) \zeta(4)-10\zeta(2) \zeta(5)+18 \zeta(7) \\ {\bf H}^{(4)}_3(+1) &=& +1\zeta(3) \zeta(4)+10\zeta(2) \zeta(5)-17 \zeta(7) \\ {\bf H}^{(5)}_2(+1) &=& -2\zeta(3) \zeta(4)-4\zeta(2) \zeta(5)+11 \zeta(7) \\ \lim_{t\rightarrow 1} \left({\bf H}^{(6)}_1(t) + \log(1-t) Li_6(t)\right) &=& +1 \zeta(3) \zeta(4)+ \zeta(2) \zeta(5) - 3 \zeta(7) \\ \hline \\ {\bf H}^{(1)}_7(+1) &=& \frac{9 \zeta(8)}{4}-\zeta (3) \zeta (5) \\ {\bf H}^{(2)}_6(+1) &=& \\ {\bf H}^{(3)}_5(+1) &=& -\frac{5}{2} {\bf H}^{(2)}_6(+1)-\frac{21 \zeta(8)}{8}+5 \zeta (3) \zeta (5) \\ {\bf H}^{(4)}_4(+1) &=& \frac{13 \zeta(8)}{12} \\ {\bf H}^{(5)}_3(+1) &=& \frac{5}{2} {\bf H}^{(2)}_6(+1)+\frac{29 \zeta(8)}{8}-4 \zeta (3) \zeta (5) \\ {\bf H}^{(6)}_2(+1) &=& \frac{8 \zeta(8)}{3}-{\bf H}^{(2)}_6(+1) \\ \lim_{t\rightarrow 1} {\bf H}^{(7)}_1(t) + \log(1-t) Li_7(t) &=& \zeta (3) \zeta (5)-\frac{5 \zeta(8)}{4}\\ \hline\\ {\bf H}^{(1)}_8(+1)&=&\frac{-2 \pi ^6 \zeta(3)-21 \pi ^4 \zeta(5)-315 \pi ^2 \zeta(7)+9450 \zeta(9)}{1890} \\ {\bf H}^{(2)}_7(+1)&=& \frac{2}{945} \pi ^6 \zeta(3)+\frac{2}{45} \pi ^4 \zeta(5)+\frac{7}{6} \pi ^2 \zeta(7)-\frac{35 \zeta(9)}{2} \\ {\bf H}^{(3)}_6(+1)&=& -\frac{1}{15} \pi ^4 \zeta(5)-\frac{7}{2} \pi ^2 \zeta(7)+\frac{85 \zeta(9)}{2} \\ {\bf H}^{(4)}_5(+1)&=& \frac{1}{18} \pi ^4 \zeta(5)+\frac{35}{6} \pi ^2 \zeta(7)-\frac{125 \zeta(9)}{2} \\ {\bf H}^{(5)}_4(+1)&=& -\frac{2}{45} \pi ^4 \zeta(5)-\frac{35}{6} \pi ^2 \zeta(7)+\frac{127 \zeta(9)}{2} \\ {\bf H}^{(6)}_3(+1)&=& \frac{1}{945} \pi ^6 \zeta(3)+\frac{1}{15} \pi ^4 \zeta(5)+\frac{7}{2} \pi ^2 \zeta(7)-\frac{83 \zeta(9)}{2} \\ {\bf H}^{(7)}_2(+1)&=& -\frac{2}{945} \pi ^6 \zeta(3)-\frac{2}{45} \pi ^4 \zeta(5)-\pi ^2 \zeta(7)+\frac{37 \zeta(9)}{2} \\ \lim_{t \rightarrow 1}\left( {\bf H}^{(8)}_1(t) + \log(1-t) Li_8(t) \right)&=& \frac{1}{945} \pi ^6 \zeta(3)+\frac{1}{90} \pi ^4 \zeta(5)+\frac{1}{6} \pi ^2 \zeta(7)-4 \zeta(9) \\ \hline \\ {\bf H}^{(1)}_9(+1)&=& \frac{\pi ^{10}}{34020}-\frac{\zeta (5)^2}{2}-\zeta (3) \zeta (7) \\ {\bf H}^{(2)}_8(+1)&=& {\bf H}^{(2)}_8(+1) \\ {\bf H}^{(3)}_7(+1)&=& -\frac{7}{2} {\bf H}^{(2)}_8(+1)+7 \zeta (3) \zeta (7)+4 \zeta (5)^2-\frac{\pi ^{10}}{11340} \\ {\bf H}^{(4)}_6(+1)&=& \frac{7}{2} {\bf H}^{(2)}_8(+1)-7 \zeta (3) \zeta (7)-5 \zeta (5)^2+\frac{227 \pi ^{10}}{1871100} \\ {\bf H}^{(5)}_5(+1)&=& \frac{\pi ^{10}}{187110}+\frac{\zeta (5)^2}{2} \\ {\bf H}^{(6)}_4(+1)&=& -\frac{7}{2} {\bf H}^{(2)}_8(+1)+7 \zeta (3) \zeta (7)+5 \zeta (5)^2-\frac{37 \pi ^{10}}{374220} \\ {\bf H}^{(7)}_3(+1)&=& \frac{7}{2} {\bf H}^{(2)}_8(+1)-6 \zeta (3) \zeta (7)-4 \zeta (5)^2+\frac{37 \pi ^{10}}{374220} \\ {\bf H}^{(8)}_2(+1)&=& \frac{53 \pi ^{10}}{1871100}-{\bf H}^{(2)}_8(+1) \\ \lim_{t \rightarrow 1} \left( {\bf H}^{(9)}_1(t) + \log(1-t) Li_9(t) \right) &=& -\frac{\pi ^{10}}{53460}+\frac{\zeta (5)^2}{2}+\zeta (3) \zeta (7) \\ \hline\\ \vdots\\ \hline\\ {\bf H}^{(1)}_{11}(+1)&=& -\zeta (5) \zeta (7)-\zeta (3) \zeta (9)+\frac{691 \pi ^{12}}{196465500}\\ {\bf H}^{(3)}_9(+1)&=& \frac{428652000 \zeta (5) \zeta (7)+321489000 \zeta (3) \zeta (9)-691 \pi ^{12}}{35721000}-\frac{9}{2} {\bf H}^{(2)}_{10}(+1)\\ {\bf H}^{(4)}_8(+1)&=& 8 {\bf H}^{(2)}_{10}(+1)-16 \zeta (3) \zeta (9)-28 \zeta (5) \zeta (7)+\frac{86096 \pi ^{12}}{1915538625}\\ {\bf H}^{(5)}_7(+1)&=& -7 {\bf H}^{(2)}_{10}(+1)+14 \zeta (3) \zeta (9)+28 \zeta (5) \zeta (7)-\frac{316027 \pi ^{12}}{7662154500}\\ {\bf H}^{(6)}_6(+1)&=& \frac{703 \pi ^{12}}{638512875}\\ {\bf H}^{(7)}_5(+1)&=& 7 {\bf H}^{(2)}_{10}(+1)-14 \zeta (3) \zeta (9)-27 \zeta (5) \zeta (7)+\frac{324319 \pi ^{12}}{7662154500}\\ {\bf H}^{(8)}_4(+1)&=& -8 {\bf H}^{(2)}_{10}(+1)+16 \zeta (3) \zeta (9)+28 \zeta (5) \zeta (7)-\frac{327083 \pi ^{12}}{7662154500}\\ {\bf H}^{(9)}_3(+1)&=& \frac{9}{2} {\bf H}^{(2)}_{10}(+1)-8 \zeta (3) \zeta (9)-12 \zeta (5) \zeta (7)+\frac{104341 \pi ^{12}}{5108103000}\\ {\bf H}^{(10)}_2(+1)&=& \frac{1219 \pi ^{12}}{425675250}-{\bf H}^{(2)}_{10}(+1)\\ \lim_{t\rightarrow 1} \left({\bf H}^{(11)}_1(t)+\log(1-t) Li_{11}(t) \right)&=& \frac{283783500 \zeta (5) \zeta (7)+283783500 \zeta (3) \zeta (9)-691 \pi ^{12}}{283783500} \end{eqnarray} On the face of it is seemed that all harmonic sums at plus unity are functions of zeta values at positive integers only. However when the weight became strictly bigger than seven something new happened. One of the equations appeared to be linearly dependent on the others which rendered it impossible to evaluate one of the sums. Now to the case of minus unity. In the even-even and the even-odd cases we will be using the relations for minus unity whereas in the odd-odd and in the odd-even cases we will be using the relations that are valid for arbitrary $t$. \begin{eqnarray} {\bf H}^{(1)}_1(-1) &=& \frac{1}{2} [\log(2)]^2 - \frac{1}{2} \zeta(2)\\ \hline\\ {\bf H}^{(1)}_2(-1) &=& - \frac{5}{8} \zeta(3)\\ {\bf H}^{(2)}_1(-1) &=& \frac{1}{2} [\log(2)] \zeta(2) - \zeta(3)\\ \hline\\ {\bf H}^{(1)}_3(-1) &=& \frac{1}{360} \left(30 \left(24 \text{Li}_4\left(\frac{1}{2}\right)+21 \zeta (3) \log (2)+\log ^4(2)\right)-11 \pi ^4-30 \pi ^2 \log ^2(2)\right)\\ {\bf H}^{(2)}_2(-1) &=& -4\text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{2} \zeta (3) \log (2)+\frac{17 \pi ^4}{480}-\frac{\log ^4(2)}{6}+\frac{1}{6} \pi ^2 \log ^2(2)\\ {\bf H}^{(3)}_1(-1) &=& \frac{1080 \zeta (3) \log (2)-19 \pi ^4}{1440}\\ \hline\\ {\bf H}^{(1)}_4(-1) &=& \frac{1}{96} \left(8 \pi ^2 \zeta (3)-177 \zeta (5)\right)\\ {\bf H}^{(2)}_3(-1) &=& \frac{11 \zeta (5)}{32}-\frac{5 \pi ^2 \zeta (3)}{48} \\ {\bf H}^{(3)}_2(-1) &=& \frac{21 \zeta (5)}{32}-\frac{\pi ^2 \zeta (3)}{8} \\ {\bf H}^{(4)}_1(-1) &=& \frac{\pi ^2 \zeta (3)}{16}-2 \zeta (5)+\frac{7}{720} \pi ^4 \log (2)\\ \hline \\ {\bf H}^{(1)}_5(-1) &=& \frac{1}{2} \int\limits_0^1 \frac{[log(1/\xi)]^3}{3!} \cdot \frac{[\log(1+\xi)]^2}{\xi} d\xi + Li_6(-1) \\ {\bf H}^{(2)}_4(-1) &=& \frac{1}{2} \int\limits_0^1 \frac{[log(1/\xi)]^1}{1!} \cdot \frac{Li_2(-\xi)^2}{\xi} d\xi - \int\limits_0^1 \frac{[log(1/\xi)]^3}{3!} \cdot \frac{[\log(1+\xi)]^2}{\xi} d\xi + Li_6(-1) \\ {\bf H}^{(3)}_3(-1) &=& -6 {\bf H}^{(1)}_5(-1)-3 {\bf H}^{(2)}_4(-1)+\frac{1701 \zeta (3)^2-62 \pi ^6}{6048} \\ {\bf H}^{(4)}_2(-1) &=& 4 {\bf H}^{(1)}_5(-1)+2 {\bf H}^{(2)}_4(-1)-\frac{9 \zeta (3)^2}{16}+\frac{359 \pi ^6}{60480} \\ {\bf H}^{(5)}_1(-1) &=& \frac{5670 \zeta (3)^2+18900 \zeta (5) \log (2)-37 \pi^6}{20160}\\ \hline\\ {\bf H}^{(1)}_6(-1)&=&+\frac{56 \pi ^4 \zeta (3)+480 \pi ^2 \zeta (5)-16965 \zeta (7)}{5760}\\ {\bf H}^{(2)}_5(-1)&=&+\frac{249 \zeta (7)}{64}-\frac{49 \pi ^2 \zeta (5)}{192}-\frac{7 \pi ^4 \zeta (3)}{360}\\ {\bf H}^{(3)}_4(-1)&=&-\frac{363 \zeta(7)}{128}+\frac{3 \pi ^2 \zeta (5)}{16}\\ {\bf H}^{(4)}_3(-1)&=&-\frac{199 \zeta (7)}{64}+\frac{13 \pi ^2 \zeta (5)}{96}+\frac{7 \pi ^4 \zeta (3)}{960}\\ {\bf H}^{(5)}_2(-1)&=&+\frac{519 \zeta (7)}{128}-\frac{5 \pi ^2 \zeta (5)}{16}-\frac{7 \pi ^4\zeta (3)}{480}\\ {\bf H}^{(6)}_1(-1)&=&-3 \zeta (7)+\frac{5 \pi ^2 \zeta (5)}{64}+\frac{7 \pi ^4 \zeta (3)}{960}+\frac{31 \pi ^6 \log (2)}{30240} \\ \hline\\ {\bf H}^{(1)}_7(-1) &=& {\bf H}^{(1)}_7(-1)\\ {\bf H}^{(2)}_6(-1) &=& {\bf H}^{(2)}_6(-1)\\ {\bf H}^{(3)}_5(-1) &=& -9 {\bf H}^{(1)}_7(-1)-4 {\bf H}^{(2)}_6(-1)-\frac{63}{128} {\bf H}^{(2)}_6(+1)+\frac{123 \zeta (3) \zeta (5)}{64}-\frac{127 \pi ^8}{76800}\\ {\bf H}^{(4)}_4(-1) &=& 16 {\bf H}^{(1)}_7(-1)+6 {\bf H}^{(2)}_6(-1)+\frac{63}{32} {\bf H}^{(2)}_6(+1)-\frac{123 \zeta (3) \zeta (5)}{16}+\frac{3097 \pi ^8}{1036800}\\ {\bf H}^{(5)}_3(-1) &=& -15 {\bf H}^{(1)}_7(-1)-5 {\bf H}^{(2)}_6(-1)-\frac{315}{128} {\bf H}^{(2)}_6(+1)+\frac{165 \zeta (3) \zeta (5)}{16}-\frac{2257 \pi ^8}{691200}\\ {\bf H}^{(6)}_2(-1) &=& 6 {\bf H}^{(1)}_7(-1)+2 {\bf H}^{(2)}_6(-1)+\frac{63}{64} {\bf H}^{(2)}_6(+1)-\frac{21 \zeta (3) \zeta (5)}{4}+\frac{193 \pi ^8}{145152}\\ {\bf H}^{(7)}_1(-1) &=& \frac{45 \zeta (3) \zeta (5)}{64}+\frac{63}{64} \zeta (7) \log (2)-\frac{23 \pi ^8}{96768}\\ \hline\\ {\bf H}^{(1)}_8(-1)&=&+\frac{496 \pi ^6 \zeta (3)+4704 \pi ^4 \zeta (5)+40320 \pi ^2 \zeta (7)-1926855 \zeta (9)}{483840}\\ {\bf H}^{(2)}_7(-1)&=&+\frac{4837 \zeta (9)}{512}-\frac{107 \pi ^2 \zeta (7)}{256}-\frac{7 \pi ^4 \zeta (5)}{180}-\frac{31 \pi ^6 \zeta (3)}{15120}\\ {\bf H}^{(3)}_6(-1)&=&-\frac{7367 \zeta (9)}{512}+\frac{97 \pi ^2 \zeta (7)}{128}+\frac{7 \pi ^4 \zeta (5)}{120}\\ {\bf H}^{(4)}_5(-1)&=&+\frac{3259 \zeta (9)}{512}-\frac{335 \pi ^2 \zeta (7)}{768}-\frac{343 \pi ^4 \zeta (5)}{11520}\\ {\bf H}^{(5)}_4(-1)&=&+\frac{3385 \zeta (9)}{512}-\frac{25 \pi ^2 \zeta (7)}{64}-\frac{7 \pi ^4 \zeta (5)}{192}\\ {\bf H}^{(6)}_3(-1)&=&-\frac{7451 \zeta (9)}{512}+\frac{187 \pi ^2 \zeta (7)}{256}+\frac{7 \pi ^4 \zeta (5)}{128}+\frac{31 \pi ^6 \zeta (3)}{40320}\\ {\bf H}^{(7)}_2(-1)&=&+\frac{4873 \zeta (9)}{512}-\frac{63 \pi ^2 \zeta (7)}{128}-\frac{7 \pi ^4 \zeta (5)}{192}-\frac{31 \pi ^6 \zeta (3)}{20160}\\ {\bf H}^{(8)}_1(-1)&=&-4 \zeta (9)+\frac{21 \pi ^2 \zeta (7)}{256}+\frac{7 \pi ^4 \zeta (5)}{768}+\frac{31 \pi ^6 \zeta (3)}{40320}+\frac{127 \pi ^8 \log (2)}{1209600}\\ \hline \\ {\bf H}^{(9)}_1(-1)&=&+\frac{189 \zeta (3) \zeta (7)}{256}+\frac{225 \zeta (5)^2}{512}+\frac{255}{256} \zeta (9) \log (2)-\frac{563 \pi ^{10}}{19160064}\\ {\bf H}^{(8)}_2(-1)&=&+8 {\bf H}^{(1)}_9(-1)+2 {\bf H}^{(2)}_8(-1)+\frac{255}{256} {\bf H}^{(2)}_8(+1)-\frac{237 \zeta(3) \zeta (7)}{32}-\frac{15 \zeta (5)^2}{4}+\frac{36067 \pi ^{10}}{159667200}\\ {\bf H}^{(7)}_3(-1)&=&-28 {\bf H}^{(1)}_9(-1)-7 {\bf H}^{(2)}_8(-1)-\frac{1785}{512} {\bf H}^{(2)}_8(+1)+\frac{2751 \zeta (3) \zeta (7)}{128}+\frac{615 \zeta (5)^2}{64}-\frac{223 \pi^{10}}{304128}\\ {\bf H}^{(6)}_4(-1)&=&+24 {\bf H}^{(1)}_9(-1)+3 {\bf H}^{(2)}_8(-1)+\frac{2295}{512} {\bf H}^{(2)}_8(+1)-2 {\bf H}^{(3)}_7(-1)-\frac{6831 \zeta (3) \zeta (7)}{256}-\frac{2745 \zeta (5)^2}{256}+\frac{64811 \pi ^{10}}{95800320}\\ {\bf H}^{(5)}_5(-1)&=&+10 {\bf H}^{(1)}_9(-1)+10{\bf H}^{(2)}_8(-1)-\frac{1275}{512} {\bf H}^{(2)}_8(+1)+5 {\bf H}^{(3)}_7(-1)+\frac{3795 \zeta (3) \zeta (7)}{256}+\frac{2775 \zeta (5)^2}{512}+\frac{893 \pi ^{10}}{31933440}\\ {\bf H}^{(4)}_6(-1)&=&-16 {\bf H}^{(1)}_9(-1)-9 {\bf H}^{(2)}_8(-1)+\frac{255}{512} {\bf H}^{(2)}_8(+1)-4{\bf H}^{(3)}_7(-1)-\frac{759 \zeta (3) \zeta (7)}{256}-\frac{255 \zeta (5)^2}{256}-\frac{43817 \pi ^{10}}{159667200} \end{eqnarray}

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    $\begingroup$ Wow, this is really impressive. $\endgroup$ – Zaid Alyafeai May 5 '17 at 22:58
  • $\begingroup$ @Zaid Alyafeai: I think I will be able to find all possible Euler sums at both plus and minus unity. The algorithm for this is given in the most recent update of the post above. Since you seem to be an expert on those things do you know if these quantities have been computed and compiled in a concise way somewhere else? I know that there is a review article on this which you sent to me before but it doesn't really contain all the results. Thanks for replying in advance. $\endgroup$ – Przemo May 11 '17 at 11:43
  • $\begingroup$ To be honest it will be difficult for me to follow. However, I do encourage you to publish these results. You will have to do a literature review to make sure they are new. $\endgroup$ – Zaid Alyafeai May 11 '17 at 12:01
  • $\begingroup$ Holy freaking crap. Just saw this. I'm tempted to hand you a +50 bounty and I haven't even read through this yet. Give me a bit and I just might :D $\endgroup$ – Brevan Ellefsen May 23 '17 at 15:14
  • $\begingroup$ @Brevan Ellefsen: Thanks for the nice words. I checked all the result using wayback.cecm.sfu.ca/cgi-bin/EZFace/zetaform.cgi . What really puzzles me why ${\bf H}^{(1)}_5(-1)$ (and their higher analogues) cannot be reduced to one dimensional zeta functions.I have tried to compute them in different ways but it was always a roundabout. Maybe it is simply the case that as complexity arises there will always be some "new" numbers, i.e. such that they cannot be reduced to the numbers we were dealing with in the first place? $\endgroup$ – Przemo May 23 '17 at 17:32
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Let us now write down the recurrence relations for ${\bf H}^{(2)}_n(t)$. As usual we start from the integral representation. We have: \begin{eqnarray} &&{\bf H}^{(2)}_n(t) =\\ && \int\limits_0^t \frac{[\log(t/\xi)]^{n-1}}{(n-1)!} \cdot \underbrace{\frac{Li_2(\xi)}{1-\xi}}_{[Li_1(\xi) Li_2(\xi) - \int \frac{Li_1(\xi)^2}{\xi} d\xi]^{'}} d \xi +Li_{n+2}(t)=\\ && \frac{1_{n\ge 3}}{2 (n-3)!} \int\limits_0^t [\log(t/\xi)]^{n-3} \cdot \frac{Li_2(\xi)^2}{\xi} d\xi - \frac{1_{n\ge 2}}{(n-1)!} \int\limits_0^t [\log(t/\xi)]^{n-1} \cdot \frac{Li_1(\xi)^2}{\xi} d\xi + Li_{n+2}(t) +\\ && \left[Li_1(t) Li_2(t) - \int\limits_0^t \frac{Li_1(\xi)^2}{\xi} d \xi\right] 1_{n=1} + \frac{1}{2} Li_2(t)^2 1_{n=2} \end{eqnarray} In the bottom line we integrated by parts and expressed the result through integrals that contain squares of a poly-log only rather than products of two different poly-logs. Now we set $t=-1$. We can immediately say that from the two integrals on the right hand side the middle one has already been computed in my previous answer to this question. The first integral is harder but it has been calculated in here Generalized definite dilogarithm integral. . Then all we need to do is to write down the results , add then up and simplify which, despite appearances, is a tedious and mundane task. We have: \begin{eqnarray} &&\frac{1}{(n-1)!} \int\limits_0^1 [\log(1/\xi)]^{n-1} \cdot \frac{\log(1+\xi)^2}{\xi} d\xi = \\ &&\left(\frac{1}{2^n} - 1\right)(n+1) \zeta(n+2) + \sum\limits_{j=1}^{n-1} \left(-1-\frac{1}{2^n} + \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n+1-j)\\ && - 2 \sum\limits_{l=0}^{n-1} {\bf H}^{(l+1)}_{n+1-l} (-1) \end{eqnarray} Likewise: \begin{eqnarray} &&\frac{1}{2(n-3)!} \int\limits_0^1 [\log(1/\xi)]^{n-3} \cdot \frac{Li_2(-\xi)^2}{\xi} d\xi =\\ && \frac{1}{4} \left(1-\frac{1}{2^{n+1}}\right)(n-1)_{(2)} \zeta(n+2) + (n-2)\left(-\frac{1}{2} + \frac{1}{2^{n+1}}\right) \log(2) \zeta(n+1) + \\ &&\frac{1}{4} \left(1-\frac{1}{2^{n-1}}\right)\zeta(2) \zeta(n)+\frac{1}{2} \sum\limits_{l=3}^n (l-2) {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} Now since \begin{equation} Li_{n+2}(-1) = \left(-1+\frac{1}{2^{n+1}}\right) \zeta(n+2) \end{equation} for $n=0,1,\cdots$ this concludes the calculation. All we need to do is to collect all the different terms. Bringing everything together we get: \begin{eqnarray} &&{\bf H}^{(2)}_n(-1)=\\ &&\frac{1}{3\cdot 2^{n+3}} \left(\right.\\ &&3 \left(\left(2^{n+1}-1\right) n^2+\left(2^{n+1}-5\right) n+2^{n+2}-6\right)\zeta (n+2)\\ &&-4 \left(2^n-1\right) (n-2) \log (8) \zeta (n+1)\\ &&+\pi ^2 \left(2^n-2\right) \zeta (n)\\ &&\left.\right) +\\ &&\sum\limits_{j=1}^{n-1} (1+\frac{1}{2^n}-\frac{1}{2^{n-j-1}}) \zeta(1+j) \zeta(n+1-j) +\\ &&\sum\limits_{l=0}^n \left(2 \cdot 1_{0\le l \le 2} + (\frac{l}{2}+1)\cdot 1_{3 \le l \le n-1} + (\frac{l}{2}-1) 1_{l=n}\right)\cdot {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} for $n\ge 3$. In the case $n=1,2$ surface terms need to be taken into account. To be specific we have: \begin{eqnarray} {\bf H}^{(2)}_1(-1) &=& \frac{1}{12} \pi ^2 \log (2)-\zeta (3)\\ {\bf H}^{(2)}_2(-1) &=& -4 \text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{2} \zeta (3) \log (2)+\frac{17 \pi ^4}{480}-\frac{\log ^4(2)}{6}+\frac{1}{6} \pi ^2 \log ^2(2) \end{eqnarray}

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  • $\begingroup$ Very nice. This thread will be of great interest to me. Thanks. $\endgroup$ – Zaid Alyafeai May 3 '17 at 6:48
  • $\begingroup$ Impressive work! I was happy to see here the closed form ${\bf H}^{(2)}_2(-1) = -4 \text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{2} \zeta (3) \log (2)+\frac{17 \pi ^4}{480}-\frac{\log ^4(2)}{6}+\frac{1}{6} \pi ^2 \log ^2(2)$. Do you have more of the type ${\bf H}^{(m)}_m(-1)$ for $m\gt 2$? $\endgroup$ – Dr. Wolfgang Hintze Jan 3 '18 at 22:34
  • $\begingroup$ @ Przemo Please have a look on my question and, especially, my self-answer math.stackexchange.com/questions/2587838/… which, in its conclusions, would rely much on your work here. I would appreciate your comments. $\endgroup$ – Dr. Wolfgang Hintze Jan 3 '18 at 23:14
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This is not an answer but too long for a comment.

Referring to the impressive work of Przemo here, I have a specific question:

First of all we adopt the definition

$$\mathbf{H}_{m}^{(n)}(x) = \sum_{k=1}^\infty \frac{H_k^{(n)}}{k^m} x^k\tag{1}$$

My question concerns the case $n=m$ and $x=-1$, i.e. the alternating series with equal indices.

Question

For which $m = 1, 2, 3, ...$ the quantity

$$S^{+-}_{m,m}(-1) = \mathbf{H}_{m}^{(m)}(-1) = \sum_{k=1}^\infty (-1)^k \frac{H_k^{(m)}}{k^m}\tag{2}$$

has a closed form?

List of cases gathered

I have collected what I have found here up to now

$m=1$ $$ {\bf H}^{(1)}_1(-1) =\frac{1}{2} [\log(2)]^2 - \frac{1}{2} \zeta(2)$$

$m=2$ $${\bf H}^{(2)}_2(-1) = -4 \text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{2} \zeta (3) \log (2)+\frac{17 \pi ^4}{480}-\frac{\log ^4(2)}{6}+\frac{1}{6} \pi ^2 \log ^2(2)$$

$m=3$ $${\bf H}^{(3)}_3(t) = \frac{1}{2} \left(-3 {\bf H}^{(4)}_2-2 \text{Li}_3(t){}^2+3 \text{Li}_2(t) \text{Li}_4(t)+5 \text{Li}_6(t)\right)$$

where, however,

$${\bf H}^{(2)}_4(-1) = \frac{1}{2} \int\limits_0^1 \frac{[log(1/\xi)]^1}{1!} \cdot \frac{Li_2(-\xi)^2}{\xi} d\xi-\int\limits_0^1 \frac{[log(1/\xi)]^3}{3!} \cdot \frac{[\log(1+\xi)]^2}{\xi} d\xi + Li_6(-1) $$

is not "closed"

$m=4$ $${\bf H}^{(4)}_4(-1) = 16 {\bf H}^{(1)}_7(-1)+6 {\bf H}^{(2)}_6(-1)+\frac{63}{32} {\bf H}^{(2)}_6(+1)-\frac{123 \zeta (3) \zeta (5)}{16}+\frac{3097 \pi ^8}{1036800}$$

where only these "explanations" are given.

$${\bf H}^{(1)}_7(-1) = {\bf H}^{(1)}_7(-1)$$ $${\bf H}^{(2)}_6(-1) = {\bf H}^{(2)}_6(-1)$$

Conclusion

I conclude from this list based on the results of Przemo that closed forms of the alternating series with equal indices (2) exist for $m=1$ and $m=2$. If closed forms for $m\ge3$ exist, and if so in what terms, is an open question.

Clarifying comments, especially from Przemo, are greatly appreciated.

$\endgroup$
  • $\begingroup$ @ Przemo I'd like to draw you attention to my answer above which collects your results obtained so far on $\mathbf{H}_{m}^{(m)}(-1)$. Could you please comment if I understood correctly that there is no closed expression for m>=3. $\endgroup$ – Dr. Wolfgang Hintze Jan 7 '18 at 12:07

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