A difficult logarithmic integral and its relation to alternating Euler Sums

closed-formeuler-sumsharmonic-numbersintegrationsequences-and-series

The following integral was recently brought up in this thread on AoPS.

$$\mathfrak I~=~\int_0^1\frac{\log(1-x)\log^2(x)\log(1+x)}{1+x}\mathrm dx\tag1$$

It is reasonable to ask for a closed-form of $(1)$ as similar (namely taking $x$ instead of $1+x$ as numerator) have known closed-form representations. The crux here appears to be the inherent alternating structure induced by both, the $1+x$ within the numerator as well as in the logarithm. Let me elaborate on this by converting this integral into a sum. Using the generating function for the harmonic numbers combined with various well-known results we may obtain

$$\small\begin{align*}
\int_0^1\frac{\log(1-x)\log^2(x)\log(1+x)}{1+x}\mathrm dx&=\sum_{n\geqslant1}(-1)^{n+1}H_n\int_0^1x^n\log^2(x)\log(1-x)\mathrm dx\\
&=\sum_{n\geqslant1}(-1)^{n+1}H_n\left(\frac{\mathrm d^2}{\mathrm dn^2}\left[-\frac{\psi^{(0)}(n+2)+\gamma}{n+1}\right]\right)\\
&=\sum_{n\geqslant1}(-1)^nH_n\left(2\frac{\psi^{(0)}(n+2)+\gamma}{(n+1)^3}-2\frac{\psi^{(1)}(n+2)}{(n+1)^2}+\frac{\psi^{(2)}(n+2)}{n+1}\right)\\
&=2\sum_{n\geqslant1}(-1)^nH_n\left(\frac{H_{n+1}}{(n+1)^3}-\frac{\zeta(2)-H_{n+1}^{(2)}}{(n+1)^2}-\frac{\zeta(3)-H_{n+1}^{(3)}}{n+1}\right)\\
&=2\sum_{n\geqslant1}(-1)^{n+1}\left(H_n-\frac1n\right)\left(\frac{H_n}{n^3}-\frac{\zeta(2)-H_n^{(2)}}{n^2}-\frac{\zeta(3)-H_n^{(3)}}n\right)
\end{align*}$$

So, basically we are left with alternating sums of the form $\sum\limits_{n\geqslant1}(-1)^n a_n$ where $a_n$ is a coefficient up to a weight of $5$ (by the usual definition of weight). Expanding the parenthesis (!) we are left with the following (ordered by weight and complexity)

$$\small\frac12\mathfrak I-\frac54\zeta(2)\zeta(3)=\zeta(3)\sum_{n\geqslant1}(-1)^n\frac{H_n}n+\zeta(2)\sum_{n\geqslant1}(-1)^n\frac{H_n}{n^2}+\sum_{n\geqslant1}(-1)^n\frac{H_n}{n^4}-\sum_{n\geqslant1}(-1)^n\frac{H_n^2}{n^3}\\\small-\sum_{n\geqslant1}(-1)^n\frac{H_nH_n^{(2)}}{n^2}-\sum_{n\geqslant1}(-1)^n\frac{H_nH_n^{(3)}}n+\sum_{n\geqslant1}(-1)^n\frac{H_n^{(2)}}{n^3}+\sum_{n\geqslant1}(-1)^n\frac{H_n^{(3)}}{n^2}$$

I am not entirely sure about splitting the sum as the first series is only conditionally convergent instead of absolutely convergent as the rest. However, the first two series fall rather easily by using the generating function again and integrating once and twice, respectively, giving us the following results.

\begin{align*}
\sum_{n\geqslant1}(-1)^n\frac{H_n}n&=\frac12\zeta(2)-\frac12\log^2(2)\tag2\\
\sum_{n\geqslant1}(-1)^n\frac{H_n}{n^2}&=-\frac58\zeta(3)\tag3
\end{align*}

While this approach essentially works for the third sum too the calculations are nearly impossible by hand and WolframAlpha already returns this monstrosity for a denominator of only $n^3$. But, let's put this aside as "in-fact-doable" (even though the result may not admit a closed-form in terms of known constants alone).

For the remaining series – except the sixth – I have a vague idea. Making use of more generating functions, namely the following, one can get these sums. Anyway, I do not know if the so-occurring integrals are easier than $(1)$ or in the worst case not even harder.

$$\small\begin{align*}
\sum_{n\geqslant1}H_n^{(p)}x^n&=\frac{\operatorname{Li}_p(x)}{1-x}\tag4\\
\sum_{n\geqslant1}H_n^2x^n&=\frac1{1-x}(\log^2(1-x)+\operatorname{Li}_2(x))\tag5\\
\sum_{n\geqslant1}H_nH_n^{(2)}x^n&=\frac1{1-x}\left(\frac12\log(x)\log^2(1-x)+\operatorname{Li}_3(x)+\operatorname{Li}_3(1-x)\tag6\\
-\zeta(2)\log(1-x)-\zeta(3)\right)
\end{align*}$$

However, there is no similar formula I am aware of to generate $H_nH_n^{(3)}$, so this part remains unknown. It might be possible to shorten things up by considering linear combinations of some of the occuring series but honestly speaking I am not longer able to keep track of all what is going on.

I have a few questions, but I would be glad for answers referring to only one of them too.

$\textbf{Q. 1}$ What is the current state of art concerning regular alternating Euler Sums? Is there a similar formula available as there is for non-alternating (I recall to have seen a post dealing with this issue on MSE, but I am unable to find it again)?

$\textbf{Q. 2}$ Are the given generating functions $(4)$–$(6)$ of any use? To put it in other words: are the occurring integrals easier to handle than $(1)$ (e.g. avoiding alternating sums at all)? I played a little bit with them but soon I ran into problems concerning convergence and I was not able to resolve them.

$\textbf{Q. 3}$ How can we deal with $H_nH_n^{(3)}$ at all? Is there a generating function known for solely this coefficient or is it necessary to use harmonic series containing this coefficient among other (actually I know some of them)?

$\textbf{Q. 4}$ Is there a closed-form for $(1)$, possibly including non-expressible constants immanent to the field of polylogarithms?

Thanks in advance!

EDIT: As pointed out by user97357329 the series having the $H_nH_n^{(3)}$ can be found in Cornel I. Valean's (Almost) Impossible Integrals, Sums, and Series, derived on page $528-529$. Searching through the book I found all of the remaining series presented as problems $4.53,$ $4.54$, $4.55$ and $4.57$ (Thanks to Ali Shather, who noticed a crucial typo).

Best Answer

Different approach to compute our main sum $\displaystyle\sum_{n=1}^\infty(-1)^n\frac{H_nH_n^{(3)}}{n}$.

From here we have

$$\int_0^1\frac{\ln^ax\ln\left(\frac{1+x}{2}\right)}{1-x}=(-1)^aa!\sum_{n=1}^\infty\frac{(-1)^nH_n^{a+1}}{n}\tag{1}$$ Using the identity

$$\ln^2(1+x)=2\sum_{n=1}^\infty\frac{H_n}{n+1}(-x)^{n+1}=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)x^n\tag{2}$$

Multiply both sides of (2) by $\frac{\ln^2x}{1-x}$ then integrate from $x=0$ to $1$ we have

\begin{align} I&=\int_0^1\frac{\ln^2x\ln^2(1+x)}{1-x}\ dx=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\int_0^1\frac{x^n\ln^2x}{1-x}\ dx\\ &=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\left(2\zeta(3)-2H_n^{(3)}\right)\\ &=4\zeta(3)\underbrace{\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)}_{\text{use (2) where}\ x=1}+4\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n^2}-4\sum_{n=1}^\infty(-1)^n\frac{H_nH_n^{(3)}}{n}\\ &=2\ln^22\zeta(3)+4\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n^2}-4\sum_{n=1}^\infty(-1)^n\frac{H_nH_n^{(3)}}{n}\tag{3} \end{align}

On the other hand:

\begin{align} I&=\small{\int_0^1\frac{\ln^2x\ln^2(1+x)}{1-x}\ dx\overset{x\mapsto 1-x}=\int_0^1\frac{\ln^2(1-x)\ln^2(2-x)}{x}=\int_0^1\frac{\ln^2(1-x)}{x}\left(\ln2+\ln\left(1-\frac x2\right)\right)^2\ dx}\\ &=\small{\ln^22\int_0^1\frac{\ln^2(1-x)}{x}\ dx+2\ln2\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}\ln\left(1-\frac x2\right)\ dx}_{x\mapsto 1-x}+\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}\ln^2\left(1-\frac x2\right)\ dx}_{\text{use (2)}}}\\ &=\small{2\ln^22\zeta(3)+2\ln2\underbrace{\int_0^1\frac{\ln^2x}{1-x}\ln\left(\frac{1+x}{2}\right)\ dx}_{\text{use (1)}}+2\sum_{n=1}^\infty\frac1{2^n}\left(\frac{H_n}{n}-\frac1{n^2}\right)\int_0^1x^{n-1}\ln^2(1-x)\ dx}\\ &=2\ln^22\zeta(3)+4\ln2\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n}+2\sum_{n=1}^\infty\frac1{2^n}\left(\frac{H_n}{n}-\frac1{n^2}\right)\left(\frac{H_n^2+H_n^{(2)}}{n}\right)\\ &=\small{2\ln^22\zeta(3)+4\ln2\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n}+2\sum_{n=1}^\infty\frac{H_n^3}{n^22^n}+2\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}-2\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}}\quad \quad \quad \quad \text{(4)} \end{align}

From (3) and (4) we conclude that

$$\sum_{n=1}^\infty(-1)^n\frac{H_nH_n^{(3)}}{n}=\\ \small{\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n^2}-\ln2\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n}-\frac12\sum_{n=1}^\infty\frac{H_n^3}{n^22^n}-\frac12\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n} +\frac12\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}+\frac12\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}}\tag{5}$$

We have the following results:

$$S_1=\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n^2}=\frac{21}{32}\zeta(5)-\frac34\zeta(2)\zeta(3)$$

$$S_2=\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n}=\frac34\ln2\zeta(3)-\frac{19}{16}\zeta(4)$$

$$S_3=\sum_{n=1}^\infty\frac{H_n^3}{n^22^n}=-14\operatorname{Li}_5\left(\frac12\right)-9\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{279}{16}\zeta(5)-\frac{25}{4}\ln2\zeta(4)-\frac78\zeta(2)\zeta(3)\\-\frac74\ln^22\zeta(3)+\frac{13}{12}\ln^32\zeta(2)-\frac{31}{120}\ln^52$$

$$S_4=\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}=2\operatorname{Li}_5\left(\frac12\right)+\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{31}{32}\zeta(5)+\frac{1}{8}\ln2\zeta(4)+\frac18\zeta(2)\zeta(3)\\-\frac{1}{12}\ln^32\zeta(2)+\frac{1}{40}\ln^52$$

$$S_5=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}=-2\operatorname{Li}_5\left(\frac12\right)-3\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{23}{64}\zeta(5)-\frac1{16}\ln2\zeta(4)+\frac{23}{16}\zeta(2)\zeta(3)\\-\frac{23}{16}\ln^22\zeta(3)+\frac7{12}\ln^32\zeta(2)-\frac{13}{120}\ln^52$$

$$S_6=\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}=-2\operatorname{Li}_5\left(\frac12\right)-\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{279}{64}\zeta(5)-\frac{37}{16}\ln2\zeta(4)-\frac{9}{16}\zeta(2)\zeta(3)\\+\frac{7}{16}\ln^22\zeta(3)+\frac1{12}\ln^32\zeta(2)-\frac{1}{40}\ln^52$$

By substituting these results in (5) we get

$$\sum_{n=1}^\infty(-1)^n\frac{H_nH_n^{(3)}}{n}=4 \operatorname{Li}_5\left(\frac{1}{2}\right)+2\ln2\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{167}{32}\zeta(5)+\frac{49}{16}\ln2\zeta(4)-\frac{3}{8}\ln^22\zeta(3)\\-\frac{1}{6}\ln^32\zeta(2)+\frac{1}{16}\zeta(2)\zeta(3)+\frac{1}{20}\ln^52$$

NOTE:

$S_1$ and $S_2$ can be found here, $S_3$ and $S_4$ can be found here and $S_5$ and $S_6$ can be found here.

Related Solutions

Euler Sums – Calculating Alternating Sums of Odd Powers

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}

Two powerful alternating sums $\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}$

To calculate these two sums, we are going to establish two relations and solve them by elimination.

To establish the first relation, we use $\displaystyle I=\int_0^1\frac{\ln^4(1+x)+6\ln^2(1-x)\ln^2(1+x)}{x}\ dx=\frac{21}4\zeta(5)\tag{1}$

which was proved by Khalef Ruhemi ( unfortunately he is not an MSE user).

The proof as follows: using the algebraic identity $\ b^4+6a^2b^2=\frac12(a-b)^4+\frac12(a+b)^4-a^4$

with $\ a=\ln(1-x)$ and $\ b=\ln(1+x)$ , divide both sides by $x$ then integrate, we get

$$I=\frac12\underbrace{\int_0^1\frac1x{\ln^4\left(\frac{1-x}{1+x}\right)}\ dx}_{\frac{1-x}{1+x}=y}+\underbrace{\frac12\int_0^1\frac{\ln^4(1-x^2)}{x}\ dx}_{x^2=y}-\int_0^1\frac{\ln^4(1-x)}{x}\ dx$$

$$=\int_0^1\frac{\ln^4x}{1-x^2}+\frac14\int_0^1\frac{\ln^4(1-x)}{x}\ dx-\int_0^1\frac{\ln^4(1-x)}{x}\ dx$$ $$=\frac12\int_0^1\frac{\ln^4x}{1-x}+\frac12\int_0^1\frac{\ln^4x}{1+x}-\frac34\underbrace{\int_0^1\frac{\ln^4(1-x)}{x}\ dx}_{1-x=y}$$ $$=\frac12\int_0^1\frac{\ln^4x}{1+x}\ dx+\frac14\int_0^1\frac{\ln^4x}{1-x}\ dx=\frac12\left(\frac{45}{2}\zeta(5)\right)+\frac14(24\zeta(5))=\frac{21}4\zeta(5)$$

On the other hand, $\quad\displaystyle I=\underbrace{\int_0^1\frac{\ln^4(1+x)}{x}\ dx}_{I_1}+6\int_0^1\frac{\ln^2(1-x)\ln^2(1+x)}{x}\ dx$

Using $\ln^2(1+x)=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)x^n\ $ for the second integral, we get

\begin{align} I&=I_1+12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\int_0^1x^{n-1}\ln^2(1-x)\ dx\\ I&=I_1+12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\left(\frac{H_n^2+H_n^{(2)}}{n}\right)\\ I&=I_1+12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n^3+H_nH_n^{(2)}}{n^2}\right)-12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n^2+H_n^{(2)}}{n^3}\right)\tag{2} \end{align} From $(1)$ and $(2)$, we get

$$\boxed{\small{R_1=\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}+\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}=\frac{7}{16}\zeta(5)+\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}+\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}-\frac{1}{12}I_1}}$$

and the first relation is established.

To get the second relation, we need to use the sterling number formula ( check here) $$ \frac{\ln^k(1-x)}{k!}=\sum_{n=k}^\infty(-1)^k \begin{bmatrix} n \\ k \end{bmatrix}\frac{x^n}{n!}$$ letting $k=4$ and using $\displaystyle\begin{bmatrix} n \\ 4 \end{bmatrix}=\frac{1}{3!}(n-1)!\left[\left(H_{n-1}\right)^3-3H_{n-1}H_{n-1}^{(2)}+2H_{n-1}^{(3)}\right],$ we get $$\frac14\ln^4(1-x)=\sum_{n=1}^\infty\frac{x^{n+1}}{n+1}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$

differentiate both sides with respect to $x$, we get $$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$

Now replace $x$ with $-x$ then multiply both sides by $\frac{\ln x}{x}$ and integrate, we get $$-\sum_{n=1}^\infty(-1)^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)\int_0^1x^{n-1}\ln x\ dx=\int_0^1\frac{\ln^3(1+x)\ln x}{x(1+x)}\ dx$$ $$\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)=\int_0^1\frac{\ln^3(1+x)\ln x}{x}\ dx-\underbrace{\int_0^1\frac{\ln^3(1+x)\ln x}{1+x}\ dx}_{IBP}$$ $$\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)=\int_0^1\frac{\ln^3(1+x)\ln x}{x}\ dx+\frac14I_1$$ Rearranging the terms, we get $$\boxed{R_2=\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}-3\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}=\int_0^1\frac{\ln^3(1+x)\ln x}{x}-2\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}+\frac14I_1}$$ and the second relation is established.

Now we are ready to calculate the first sum. \begin{align} \sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}&=\frac{3R_1+R_2}{4}\\ &=\frac34\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}+\frac34\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}-\frac12\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}\\ &\quad+\frac14\int_0^1\frac{\ln x\ln^3(1+x)}{x}\ dx+\frac{21}{64}\zeta(5) \end{align} the closed form of the first and second sum can be found here and the closed form of the third sum is evaluated here. as for the integral, I evaluated it here. by combining these results, we get our closed form.

and the second sum. $$\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}=\frac{R_1-R_2}{4}$$ $$\small{=\frac14\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}+\frac14\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}+\frac12\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}-\frac14\int_0^1\frac{\ln x\ln^3(1+x)}{x}\ dx-\frac1{12}I_1+\frac{7}{64}\zeta(5)}$$ lets calculate $I_1$ and by setting $\frac1{1+x}=y$, we get \begin{align} I_1&=\int_0^1\frac{\ln^4(1+x)}{x}=\int_{1/2}^1\frac{\ln^4x}{x}\ dx+\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx\\ &=\frac15\ln^52+\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln^4x\ dx\\ &=\frac15\ln^52+\sum_{n=1}^\infty\left(\frac{24}{n^5}-\frac{24}{n^52^n}-\frac{24\ln2}{n^42^n}-\frac{12\ln^22}{n^32^n}-\frac{4\ln^32}{n^22^n}-\frac{\ln^42}{n2^n}\right)\\ &=4\ln^32\zeta(2)-\frac{21}2\ln^22\zeta(3)+24\zeta(5)-\frac45\ln^52-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right) \end{align} by combining the result of $I_1$ along with the results we used in our first sum, we get the closed form of the second sum.

UPDATE:

The identity used above:

$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$

can also be proved this way.

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Euler Sums – Calculating Alternating Sums of Odd Powers

Two powerful alternating sums $\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}$

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