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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 similiar (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 aswell 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 defintion of weight). Expanding the parathesis (!) 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-occuring 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 similiar 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 similiar 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 occuring 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-expressable 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).

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    $\begingroup$ In (Almost) Impossible Integrals, Sums, and Series you may find the series $$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_n H_n^{(3)}}{n}$$ $$=\frac{1}{6}\log^3(2)\zeta(2)+\frac{3}{8}\log^2(2)\zeta(3)-\frac{49}{16}\log(2)\zeta(4)+\frac{167}{32}\zeta(5)-\frac{1}{16}\zeta(2)\zeta(3)$$ $$-\frac{\log^5(2)}{20}-2\log(2)\operatorname{Li}_4\left(\frac{1}{2}\right)-4 \operatorname{Li}_5\left(\frac{1}{2}\right),$$ together with a derivation. See Section $6.58$, particularly page $529$. Also, another three important tough alternating series are given on pages $528-529$. $\endgroup$ – user97357329 Aug 26 at 20:09
  • $\begingroup$ @user97357329 Thank you kindly. I've totally overlooked this derivation as it was not explicitely stated as problem in Section $4.58$. $\endgroup$ – mrtaurho Aug 26 at 20:14
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    $\begingroup$ For the generalization, $\sum_{k=1}^{\infty} (-1)^{k-1}\frac{H_k}{k^{2m}}=\left(m+\frac{1}{2}\right)\eta(2m+1)-\frac{1}{2}\zeta(2m+1)-\sum_{i=1}^{m-1}\eta(2i)\zeta(2m-2i+1)$, you may find a simple proof here researchgate.net/publication/…, which is based in general on series manipulations. $\endgroup$ – user97357329 Aug 26 at 20:22
  • $\begingroup$ @user97357329 Ah, I see. Thank you yet again! $\endgroup$ – mrtaurho Aug 26 at 20:26
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    $\begingroup$ Other way to deal with meant sum is by using $$\sum_{n=1}^\infty H_n^{(3)}x^n=\frac{\operatorname{Li}_3(x)}{1-x}$$ Replace $x$ with $-x$ $$\sum_{n=1}^\infty (-1)^n H_n^{(3)}x^n=\frac{\operatorname{Li}_3(-x)}{1+x}$$ Now multiply both sides by $-\frac{\ln(1-x)}{x}$ and use the fact that $\int_0^1-x^{n-1}\ln(1-x)\ dx=\frac{H_n}{n}$ we get $$\sum_{n=1}^\infty(-1)^n \frac{H_nH_n^{(3)}}{n}=-\int_0^1\frac{\ln(1-x)\operatorname{Li}_3(-x)}{x(1+x)}\ dx$$ which can done using IBP $\endgroup$ – Ali Shather Aug 26 at 21:08
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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.

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    $\begingroup$ $$I = -8\text{Li}_5(1/2) - 4\ln 2\text{Li}_4(1/2) + \frac {213}{16}\zeta(5) -\frac {5\pi^2}{12}\zeta(3)+\frac {7\ln^2 2}{4} \zeta(3)+\frac {\pi^2\ln^3 2}{18}-\frac{\ln^5 2}{10}-\frac{49}{720}{\pi^4}\log2$$ $\endgroup$ – user178256 Sep 3 at 16:25

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