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It may be rather tedious and I will have to delve into deeper, but I have a little something. We probably already know this one. The thing is, the first one results in yet another Euler sum. But, I think we can get it without contours. It may require some work, though.

$$ \sum_{n=1}^\infty\frac{H_n^{(2)}}{n^{4}} = \sum_{n=1}^\infty\sum_{k=1}^n \frac 1 {k^4} = \sum_{n=1}^\infty\frac 1{n^2} \sum_{k=n}^\infty\frac 1{k^4}\ . $$ This result:

$$ \sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n^4} = \zeta(2)\zeta(4)+\zeta(6) -\sum_{n=1}^{\infty}\frac{H_n^{(4)}}{n^2} $$ where that sum on the end is equal to $$ \sum_{n=1}^{\infty}\frac{H_n^{(4)}}{n^2} =\frac{37}{12}\zeta(6)-\zeta^{2}(3)\ . $$ Which matches up if we change it all to $\zeta(6)$

i.e. $\displaystyle \zeta(4)\zeta(2)=\frac{7}{4}\zeta(6)$

and $$ \sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n^4} =\zeta(2)\zeta(4) + \int_0^1\frac{\log(x)\operatorname{Li}_4(x)}{1-x}\; dx $$

My main question is how to evaluate the last integral. Any better ways to evaluate the main Euler sum are welcome.

dan_fulea
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Martin.s
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    A clear message is missing in the first few sentences. We have a question without question mark in the title, but the true question seems to be an other one, related to how to evaluate an integral. If this is the real question, then the title should reflect it, maybe also some tag. I have to guess that $H_n^{(2)}$ is the sum over $1/k^2$ for $k$ between $1$ and $n$, but then i am missing something in that $$\sum_{\substack{k,n\1\le k\le n}}\frac 1{k^4}\ .$$(And now to get more confused we have a $k^4$, not a $n^4$.) Then "This result:" seems to want to introduce a proposition, the verb is? – dan_fulea Feb 22 '24 at 16:21
  • It is unclear, what we know, please give references, and what should be shown, and if the result for the integral is already known. Please edit... – dan_fulea Feb 22 '24 at 16:23
  • The sum $\sum_{n=1}^{\infty}\frac{H_n^{(4)}}{n^2}$ is a different form of the integral you are asking about. To find a different proof for the sum $\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n^4}$, see the link https://math.stackexchange.com/a/3238703/432085 – Ali Shadhar Feb 22 '24 at 16:38
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    $$\int_0^1 \frac{\log (x) \text{Li}_4(x)}{1-x} , dx=-\frac{5 \pi ^6}{2268}+\zeta (3)^2$$ – Mariusz Iwaniuk Feb 22 '24 at 16:59
  • To evaluate the last integral, probably $\displaystyle x\in[0,1],\text{Li}_4(x)=-\frac{1}{6}\int_0^1 \frac{x\ln^3 t}{1-tx}dt$ is useful. – FDP Feb 25 '24 at 13:00

1 Answers1

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\begin{align}J&=\int_0^1\frac{\text{Li}_4(x)\ln x}{1-x}dx\\ &=-\frac{1}{6}\int_0^1\int_0^1\frac{x\ln^3 t\ln x}{(1-x)(1-tx)}dtdx\\ &=-\frac{1}{6}\int_0^1\int_0^1\left(\frac{\ln^3 t\ln x}{(1-x)(1-t)}-\frac{\ln^3 t\ln x}{(1-tx)(1-t)}\right)dtdx\\ &=-\zeta(2)\zeta(4)+\frac{1}{6}\int_0^1\int_0^1\frac{\ln^3 t\Big(\ln(t x)-\ln t\Big)}{(1-tx)(1-t)}dtdx\\ &=-\zeta(2)\zeta(4)+\frac{1}{6}\int_0^1 \frac{\ln^3 t}{t(1-t)}\left(\int_0^t\frac{\ln x}{1-x}dx\right)dt-\frac{1}{6}\int_0^1 \frac{\ln^4 t}{t(1-t)}\left(\int_0^t\frac{1}{1-x}dx\right)dt\\ &\overset{\text{IBP}}=-\zeta(2)\zeta(4)-\frac{1}{24}\int_0^1\frac{\ln^5 t}{1-t}dt+\frac{1}{6}\int_0^1\frac{\ln^3 t}{1-t}\left(\int_0^t\frac{\ln x}{1-x}dx\right)dt+\frac{1}{30}\int_0^1\frac{\ln^5 t}{1-t}dt+\\ &\frac{1}{6}\int_0^1\frac{\ln^4 t\ln(1-t)}{1-t}dt\\ &=\zeta(6)-\zeta(2)\zeta(4)+\frac{1}{6}\underbrace{\int_0^1\frac{\ln^3 t}{1-t}\left(\int_0^t\frac{\ln x}{1-x}dx \right)dt}_{=A}+\frac{1}{6}\int_0^1\frac{\ln^4 t\ln(1-t)}{1-t}dt\\ A&=\int_0^1\int_0^1\frac{t\ln(tx)\ln^3 t}{(1-t)(1-tx)}dtdx=\int_0^1\int_0^1\left(\frac{\ln(tx)\ln^3 t}{(1-t)(1-x)}-\frac{\ln(tx)\ln^3 t}{(1-x)(1-tx)}\right)dtdx\\ &\overset{\text{Fubini}}=6\zeta(2)\zeta(4)+\int_0^1\int_0^1\left(\frac{\ln^4 t}{(1-t)(1-x)}-\frac{\ln(tx)\Big(\ln(tx)-\ln x\Big)^3}{(1-x)(1-tx)}\right)dtdx\\ &=6\zeta(2)\zeta(4)+\int_0^1\int_0^1\left(\frac{\ln^4 t}{(1-t)(1-x)}-\frac{\ln^4(tx)}{(1-x)(1-tx)}\right)dtdx+\\&3\!\int_0^1\!\!\int_0^1\frac{\ln^3(tx)\ln x}{(1-x)(1-tx)}\!dtdx-3\int_0^1\int_0^1\frac{\ln^2(tx)\ln^2 x}{(1-x)(1-tx)}dtdx+\\&\int_0^1\int_0^1\frac{\ln(tx)\ln^3 x}{(1-x)(1-tx)}dtdx\\ &=6\zeta(2)\zeta(4)+\int_0^1\frac{1}{1-x}\left(\int_0^1\frac{\ln^4 t}{1-t}dt-\frac{1}{x}\int_0^x\frac{\ln^4 t}{1-t}dt\right)dx+\\&3\int_0^1\frac{\ln x}{(1-x)x}\left(\int_0^x \frac{\ln^3 t}{1-t}dt\right)dx-3\int_0^1\frac{\ln^2 x}{(1-x)x}\left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)dx+\\&\int_0^1\frac{\ln^3 x}{(1-x)x}\left(\int_0^x \frac{\ln t}{1-t}dt\right)dx\\ &\overset{\text{IBP}}=6\zeta(2)\zeta(4)-30\zeta(6)-\int_0^1\frac{\ln(1-t)\ln^4 t}{1-t}dt+3\underbrace{\int_0^1\frac{\ln x}{1-x}\left(\int_0^x \frac{\ln^3 t}{1-t}dt\right)dx}_{\text{IBP}}-\\ &3\underbrace{\int_0^1\frac{\ln^2 x}{1-x}\left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)dx}_{=2\zeta(3)^2}+A\\ &=24\zeta(2)\zeta(4)-30\zeta(6)-6\zeta(3)^2-\int_0^1\frac{\ln(1-t)\ln^4 t}{1-t}dt-2A\\ &=8\zeta(2)\zeta(4)-10\zeta(6)-2\zeta(3)^2-\frac{1}{3}\int_0^1\frac{\ln(1-t)\ln^4 t}{1-t}dt\\ J&=-\frac{2}{3}\zeta(6)+\frac{1}{3}\zeta(2)\zeta(4)-\frac{1}{3}\zeta(3)^2+\frac{1}{9}\underbrace{\int_0^1\frac{\ln(1-t)\ln^4 t}{1-t}dt}_{=B} \end{align} \begin{align} B&\overset{\text{IBP}}=\left[\left(\int_0^t\frac{\ln^4 x}{1-x}dx-\int_0^1\frac{\ln^4 x}{1-x}dx\right)\ln(1-t)\right]_0^1+\\&\int_0^1\int_0^1\frac{1}{1-t}\left(\frac{t\ln^4(tx)}{1-tx}-\frac{\ln^4 x}{1-x}\right)dtdx\\ &=\int_0^1\int_0^1\left(\frac{\ln^4(tx)}{(1-t)(1-x)}-\frac{\ln^4(tx)}{(1-x)(1-tx)}-\frac{\ln^4 x}{(1-t)(1-x)}\right)dtdx\\ &=\int_0^1\int_0^1\frac{4\ln t\ln^3 x+6\ln^2 t\ln^2 x+4\ln^3 t\ln x}{(1-t)(1-x)}dtdx+\\ &\int_0^1\int_0^1\left(\frac{\ln^4 t}{(1-t)(1-x)}-\frac{\ln^4(tx)}{(1-x)(1-tx)}\right)dtdx\\ &\overset{\text{Fubini}}=24\zeta(2)\zeta(4)+24\zeta(3)^2+24\zeta(2)\zeta(4)-\int_0^1\frac{1}{x}\int_0^x\frac{\ln^4 t}{1-t}dt+\int_0^1\frac{1}{1-x}\int_x^1\frac{\ln^4 t}{1-t}dt\\ &\overset{\text{IBP}}=48\zeta(2)\zeta(4)+24\zeta(3)^2-120\zeta(6)-B\\ &=\boxed{24\zeta(2)\zeta(4)+12\zeta(3)^2-60\zeta(6)} \end{align} Therefore, \begin{align}\boxed{J=-\frac{22}{3}\zeta(6)+3\zeta(2)\zeta(4)+\zeta(3)^2=\zeta(3)^2-\frac{5\pi^6}{2268}}\end{align}

FDP
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