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Does the following series or integral have a closed-form

\begin{equation} \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx \end{equation}

where $\Psi_3(x)$ is the polygamma function of order $3$.


Here is my attempt. Using equation (11) from Mathworld Wolfram: \begin{equation} \Psi_n(z)=(-1)^{n+1} n!\left(\zeta(n+1)-H_{z-1}^{(n+1)}\right) \end{equation} I got \begin{equation} \Psi_3(n+1)=6\left(\zeta(4)-H_{n}^{(4)}\right) \end{equation} then \begin{align} \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)&=6\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\left(\zeta(4)-H_{n}^{(4)}\right)\\ &=6\zeta(4)\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}-6\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}\\ &=\frac{\pi^4}{15}\ln2-6\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}\\ \end{align} From the answers of this OP, the integral representation of the latter Euler sum is \begin{align} \sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}&=\int_0^1\int_0^1\int_0^1\int_0^1\int_0^1\frac{dx_1\,dx_2\,dx_3\,dx_4\,dx_5}{(1-x_1)(1+x_1x_2x_3x_4x_5)} \end{align} or another simpler form \begin{align} \sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}&=-\int_0^1\frac{\text{Li}_4(-x)}{x(1+x)}dx\\ &=-\int_0^1\frac{\text{Li}_4(-x)}{x}dx+\int_0^1\frac{\text{Li}_4(-x)}{1+x}dx\\ &=\text{Li}_5(-1)-\int_0^{-1}\frac{\text{Li}_4(x)}{1-x}dx\\ \end{align} I don't know how to continue it, I am stuck. Could anyone here please help me to find the closed-form of the series preferably with elementary ways? Any help would be greatly appreciated. Thank you.


Edit :

Using the integral representation of polygamma function \begin{equation} \Psi_m(z)=(-1)^m\int_0^1\frac{x^{z-1}}{1-x}\ln^m x\,dx \end{equation} then we have \begin{align} \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)&=-\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\int_0^1\frac{x^{n}}{1-x}\ln^3 x\,dx\\ &=-\int_0^1\sum_{n=1}^\infty\frac{(-1)^{n+1}x^{n}}{n}\cdot\frac{\ln^3 x}{1-x}\,dx\\ &=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx\\ \end{align} I am looking for an approach to evaluate the above integral without using residue method or double summation.

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    $\begingroup$ Downvote for no reason again! Why the heck this OP got downvoted?? May The Lord forgive your sin (҂⌣̀_⌣́)ᕤ $\endgroup$ – Anastasiya-Romanova 秀 Oct 11 '14 at 9:06
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Edited: I have changed the approach as I realised that the use of summation is quite redundant (since the resulting sums have to be converted back to integrals). I feel that this new method is slightly cleaner and more systematic.


We can break up the integral into \begin{align} -&\int^1_0\frac{\ln^3{x}\ln(1+x)}{1-x}{\rm d}x\\ =&\int^1_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x-\int^1_0\frac{(1+x)\ln^3{x}\ln(1-x^2)}{(1+x)(1-x)}{\rm d}x\\ =&\int^1_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x-\int^1_0\frac{\ln^3{x}\ln(1-x^2)}{1-x^2}{\rm d}x-\int^1_0\frac{x\ln^3{x}\ln(1-x^2)}{1-x^2}{\rm d}x\\ =&\frac{15}{16}\int^1_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x-\frac{1}{16}\int^1_0\frac{x^{-1/2}\ln^3{x}\ln(1-x)}{1-x}{\rm d}x\\ =&\frac{15}{16}\frac{\partial^4\beta}{\partial a^3 \partial b}(1,0^{+})-\frac{1}{16}\frac{\partial^4\beta}{\partial a^3 \partial b}(0.5,0^{+}) \end{align} After differentiating and expanding at $b=0$ (with the help of Mathematica), \begin{align} &\frac{\partial^4\beta}{\partial a^3 \partial b}(a,0^{+})\\ =&\left[\frac{\Gamma(a)}{\Gamma(a+b)}\left(\frac{1}{b}+\mathcal{O}(1)\right)\left(\left(-\frac{\psi_4(a)}{2}+(\gamma+\psi_0(a))\psi_3(a)+3\psi_1(a)\psi_2(a)\right)b+\mathcal{O}(b^2)\right)\right]_{b=0}\\ =&-\frac{1}{2}\psi_4(a)+(\gamma+\psi_0(a))\psi_3(a)+3\psi_1(a)\psi_2(a) \end{align} Therefore, \begin{align} -&\int^1_0\frac{\ln^3{x}\ln(1+x)}{1-x}{\rm d}x\\ =&-\frac{15}{32}\psi_4(1)+\frac{45}{16}\psi_1(1)\psi_2(1)+\frac{1}{32}\psi_4(0.5)+\frac{1}{8}\psi_3(0.5)\ln{2}-\frac{3}{16}\psi_1(0.5)\psi_2(0.5)\\ =&-12\zeta(5)+\frac{3\pi^2}{8}\zeta(3)+\frac{\pi^4}{8}\ln{2} \end{align} The relation between $\psi_{m}(1)$, $\psi_m(0.5)$ and $\zeta(m+1)$ is established easily using the series representation of the polygamma function.

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  • $\begingroup$ Interesting alternate approach together with nice reference! +1 $\endgroup$ – Markus Scheuer Oct 10 '14 at 9:52
  • $\begingroup$ @MarkusScheuer Thank you for your compliment. $\endgroup$ – M.N.C.E. Oct 10 '14 at 10:05
  • $\begingroup$ Very nice answer. This kind of solution that I'm looking for. I know that if we use the residue approach the solution will be much cleaner and more efficient, but I don't know anything about it. My knowledge in residue is almost zero that's why I love Feynman's style answers to evaluate integral problems. +1 for your answer. I hope I get other interesting approach (ô‿ô) $\endgroup$ – Anastasiya-Romanova 秀 Oct 10 '14 at 12:56
  • $\begingroup$ @Anastasiya-Romanova If you are interested, I have edited my answer and removed the (redundant) use of summation altogether. The method of evaluation is slightly shorter and more systematic now, if I may say so myself. Thanks. $\endgroup$ – M.N.C.E. Oct 11 '14 at 1:08
  • $\begingroup$ Very clever! I love this approach. No double summation's involved. Another way beside using multiple derivative of beta function, we can use generating function $$\sum_{n=1}^\infty H_{n}x^n=\frac{\ln(1-x)}{1-x}$$to solve the integral. I wish I could give 1000 upvotes for this answer. $\endgroup$ – Anastasiya-Romanova 秀 Oct 11 '14 at 9:09
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\begin{align} \sum^\infty_{n=1}\frac{(-1)^{n-1}\psi_3(n+1)}{n} &=-12\zeta(5)+\frac{45}{4}\zeta(4)\ln{2}+\frac{9}{4}\zeta(2)\zeta(3) \end{align}


Let $\displaystyle f(z)=\frac{\pi\csc(\pi z)\psi_3(-z)}{z}$. Then at the positive integers, \begin{align} \sum^\infty_{n=1}{\rm Res}(f,n) &=\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\left[\frac{6(-1)^n}{z(z-n)^5}+\frac{6(-1)^n\zeta(2)}{z(z-n)^3}+(-1)^n\frac{(33/2)\zeta(4)+6H_n^{(4)}}{z(z-n)}\right]\\ &=6\sum^\infty_{n=1}\frac{(-1)^n}{n^5}+6\zeta(2)\sum^\infty_{n=1}\frac{(-1)^n}{n^3}+\frac{33}{2}\zeta(4)\sum^\infty_{n=1}\frac{(-1)^n}{n}+6\sum^\infty_{n=1}\frac{(-1)^nH_n^{(4)}}{n}\\ &=-\frac{45}{8}\zeta(5)-\frac{9}{2}\zeta(2)\zeta(3)-\frac{33}{2}\zeta(4)\ln{2}+6\sum^\infty_{n=1}\frac{(-1)^nH_n^{(4)}}{n} \end{align} At zero, $${\rm Res}(f,0)=24\zeta(5)$$ At the negative integers, \begin{align} \sum^\infty_{n=1}{\rm Res}(f,-n) &=\sum^\infty_{n=1}\frac{(-1)^{n-1}\psi_3(n)}{n}\\ &=6\zeta(4)\ln{2}-6\sum^\infty_{n=1}\frac{(-1)^{n-1}H_{n-1}^{(4)}}{n}\\ &=\frac{45}{8}\zeta(5)+6\zeta(4)\ln{2}+6\sum^\infty_{n=1}\frac{(-1)^{n}H_{n}^{(4)}}{n}\\ \end{align} Since the sum of residues is zero, \begin{align} 12\sum^\infty_{n=1}\frac{(-1)^{n}H_{n}^{(4)}}{n}=-24\zeta(5)+\frac{21}{2}\zeta(4)\ln{2}+\frac{9}{2}\zeta(2)\zeta(3)\\ \end{align} This implies that \begin{align} \sum^\infty_{n=1}\frac{(-1)^{n-1}\psi_3(n+1)}{n} &=-12\zeta(5)+\frac{45}{4}\zeta(4)\ln{2}+\frac{9}{4}\zeta(2)\zeta(3) \end{align} Refer to this paper if you have any doubts.

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  • $\begingroup$ I'm looking for elementary method, without using complex analysis. Still I upvote you, +1. Anyway, could you please give me references or link of papers to evaluate sum of harmonic series using residue? I mean simpler than cited link. It looks like that technique very useful & easy. Thank you (ô‿ô) $\endgroup$ – Anastasiya-Romanova 秀 Sep 18 '14 at 13:29
  • $\begingroup$ @Anastasiya-Romanova I actually picked up this method from the cited link, and it seems that this paper is the only paper I have seen online that elaborates on this method to such a great degree. Of course, if I remember correctly, users like Random Variable and Galactus have also used this method quite a few times to compute Euler sums on I&S, and you might want to refer to their work there as well. Unfortunately, this method is not without its limitations, for example, it cannot be used to evaluate $\displaystyle\sum^\infty_{n=1}\frac{(-1)^nH_n^{(4)}}{n^2}$ $\endgroup$ – SuperAbound Sep 18 '14 at 13:44
  • $\begingroup$ Okay, could you please elaborate this part: \begin{align} \sum^\infty_{n=1}{\rm Res}(f,n) &=\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\left[\frac{6(-1)^n}{z(z-n)^5}+\frac{6(-1)^n\zeta(2)}{z(z-n)^3}+(-1)^n\frac{(33/2)\zeta(4)+6H_n^{(4)}}{z(z-n)}\right]\\ &=6\sum^\infty_{n=1}\frac{(-1)^n}{n^5}+6\zeta(2)\sum^\infty_{n=1}\frac{(-1)^n}{n^3}+\frac{33}{2}\zeta(4)\sum^\infty_{n=1}\frac{(-1)^n}{n}+6\sum^\infty_{n=1}\frac{(-1)^nH_n^{(4)}}{n} \end{align} I don't understand the first & second line? $\endgroup$ – Anastasiya-Romanova 秀 Sep 18 '14 at 13:46
  • $\begingroup$ @Anastasiya-Romanova I expanded $\pi\csc(\pi z)\psi_3(-z)$ as a Laurent series about the positive integers. The Laurent series of both functions are found in the paper, and they can also be derived rather easily. $\endgroup$ – SuperAbound Sep 18 '14 at 13:50
  • $\begingroup$ No doubts! :-) Nice answer, good reference! +1 $\endgroup$ – Markus Scheuer Oct 7 '14 at 19:57
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The following generalizations with solutions are proposed by Cornel Ioan Valean, using ideas about symmetry from his book, (Almost) Impossible Integrals, Sums, and Series.

(First main result) Let $m\ge2$ be a positive integer. The following equalities hold: \begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(m)}}{n}=\frac{(-1)^m}{(m-1)!}\int_0^1\frac{\displaystyle \log^{m-1}(x)\log\left(\frac{1+x}{2}\right)}{1-x}\textrm{d}x \end{equation*} \begin{equation*} =\frac{1}{2}\biggr(m\zeta (m+1)-2\log (2) \left(1-2^{1-m}\right) \zeta (m)-\sum_{k=1}^{m-2} \left(1-2^{-k}\right)\left(1-2^{1+k-m}\right)\zeta (k+1)\zeta (m-k)\biggr), \end{equation*} where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}$ represents the $n$th generalized harmonic number of order $m$ and $\zeta$ denotes the Riemann zeta function.

Proof: \begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(m)}}{n} \end{equation*} \begin{equation*} =\sum_{n=1}^{\infty}\frac{ (-1)^{n-1}}{n}\sum_{k=1}^n \frac{1}{k^m}=\frac{(-1)^{m-1}}{(m-1)!}\sum_{n=1}^{\infty}\frac{ (-1)^{n-1}}{n}\sum_{k=1}^n\int_0^1 x^{k-1}\log^{m-1}(x)\textrm{d}x \end{equation*} \begin{equation*} =\frac{(-1)^{m-1}}{(m-1)!}\int_0^1 \log^{m-1}(x)\sum_{n=1}^{\infty}\frac{ (-1)^{n-1}}{n}\sum_{k=1}^nx^{k-1}\textrm{d}x \end{equation*} \begin{equation*} =\frac{(-1)^m}{(m-1)!}\int_0^1\frac{\displaystyle \log^{m-1}(x)\log\left(\frac{1+x}{2}\right)}{1-x}\textrm{d}x=\frac{(-1)^{m-1}}{(m-1)!}\int_0^1\left(\int_x^1 \frac{\displaystyle \log^{m-1}(x)}{(1+y)(1-x)}\textrm{d}y \right)\textrm{d}x \end{equation*} \begin{equation*} =\frac{(-1)^{m-1}}{(m-1)!}\int_0^1\left(\int_0^y \frac{\displaystyle \log^{m-1}(x)}{(1+y)(1-x)}\textrm{d}x \right)\textrm{d}y\overset{x=yz}{=}\frac{(-1)^{m-1}}{(m-1)!}\int_0^1\left(\int_0^1 \frac{y\log^{m-1}(y z)}{(1+y)(1-yz)}\textrm{d}z \right)\textrm{d}y \end{equation*} \begin{equation*} =\frac{(-1)^{m-1}}{2\cdot(m-1)!}\left(\int_0^1\left(\int_0^1 \frac{y\log^{m-1}(x y)}{(1+y)(1-x y)}\textrm{d}x \right)\textrm{d}y+\int_0^1\left(\int_0^1 \frac{x\log^{m-1}(x y)}{(1+x)(1-x y)}\textrm{d}x \right)\textrm{d}y\right) \end{equation*} \begin{equation*} =\frac{(-1)^{m-1}}{2\cdot(m-1)!}\int_0^1\left(\int_0^1 \frac{((1 + x) (1 + y) - (1 - x y))\log^{m-1}(x y)}{(1+x)(1+y)(1-x y)}\textrm{d}x \right)\textrm{d}y \end{equation*} \begin{equation*} =\frac{(-1)^{m-1}}{2\cdot(m-1)!}\left(\int_0^1\left(\int_0^1 \frac{\log^{m-1}(x y)}{1-x y}\textrm{d}x \right)\textrm{d}y-\int_0^1\left(\int_0^1 \frac{\log^{m-1}(x y)}{(1+x)(1+y)}\textrm{d}x \right)\textrm{d}y\right) \end{equation*} \begin{equation*} =\frac{(-1)^{m-1}}{2\cdot(m-1)!}\biggr(\int_0^1\left(\int_0^y \frac{\log^{m-1}(x)}{(1-x)y}\textrm{d}x \right)\textrm{d}y \end{equation*} \begin{equation*} -\int_0^1\left(\int_0^1 \sum_{k=0}^{m-1}\binom{m-1}{k} \frac{\log^k(x)\log^{m-k-1}(y)}{(1+x)(1+y)}\textrm{d}x \right)\textrm{d}y\biggr) \end{equation*} \begin{equation*} =\frac{(-1)^{m-1}}{2\cdot(m-1)!}\biggr(\int_0^1\left(\int_x^1 \frac{\log^{m-1}(x)}{(1-x)y}\textrm{d}y \right)\textrm{d}x \end{equation*} \begin{equation*} - \sum_{k=0}^{m-1}\binom{m-1}{k}\int_0^1\left(\int_0^1 \frac{\log^k(x)\log^{m-k-1}(y)}{(1+x)(1+y)}\textrm{d}x \right)\textrm{d}y\biggr) \end{equation*} \begin{equation*} =\frac{(-1)^m}{2\cdot(m-1)!}\left(\int_0^1\frac{\log^m(x)}{1-x}\textrm{d}x+\sum_{k=0}^{m-1}\binom{m-1}{k}\int_0^1\frac{\log^{m-k-1}(y)}{1+y}\left(\int_0^1 \frac{\log^k(x)}{1+x}\textrm{d}x \right)\textrm{d}y\right) \end{equation*} \begin{equation*} =\frac{(-1)^m}{2\cdot(m-1)!}\biggr((-1)^m m!\zeta (m+1) +(-1)^{m-1} 2\log (2)(1-2^{1-m}) (m-1)!\zeta (m) \end{equation*} \begin{equation*} +\sum_{k=1}^{m-2}\binom{m-1}{k}\int_0^1\frac{\log^{m-k-1}(y)}{1+y}\left(\int_0^1 \frac{\log^k(x)}{1+x}\textrm{d}x \right)\textrm{d}y\biggr) \end{equation*} \begin{equation*} =\frac{1}{2}\biggr(m\zeta (m+1)-2\log (2) \left(1-2^{1-m}\right) \zeta (m)-\sum_{k=1}^{m-2} \left(1-2^{-k}\right)\left(1-2^{1+k-m}\right)\zeta (k+1)\zeta (m-k)\biggr). \end{equation*}

A few cases of the first generalization

For $m=2$, \begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n}=\zeta(3)-\frac{1}{2}\log(2)\zeta(2); \end{equation*} For $m=3$, \begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(3)}}{n}=\frac{19}{16}\zeta(4)-\frac{3}{4}\log(2)\zeta(3); \end{equation*} For $m=4$, \begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(4)}}{n}=2\zeta(5)-\frac{3}{8}\zeta(2)\zeta(3)-\frac{7}{8}\log(2)\zeta(4); \end{equation*} For $m=5$, \begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(5)}}{n}=\frac{111}{64}\zeta(6)-\frac{9}{32}\zeta^2(3)-\frac{15}{16}\log(2)\zeta(5); \end{equation*} For $m=6$, \begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(6)}}{n}=3\zeta(7)-\frac{15}{32}\zeta(2)\zeta(5)-\frac{21}{32}\zeta(3)\zeta(4)-\frac{31}{32}\log(2)\zeta(6). \end{equation*}

(Second main result) Let $m\ge2$ be a positive integer. The following equalities hold:

\begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^{(m)}}{n}=\frac{(-1)^m}{(m-1)!}\int_0^1\frac{\displaystyle \log^{m-1}(x)\log\left(\frac{1+x^2}{2}\right)}{1-x}\textrm{d}x \end{equation*} \begin{equation*} =m\zeta (m+1)- 2^{-m} \left(1-2^{-m+1}\right) \log(2 ) \zeta (m) -\sum_{k=0}^{m-1}\beta(k+1)\beta(m-k) \end{equation*} \begin{equation*} -\sum_{k=1}^{m-2}2^{- m-1}(1-2^{-k})(1-2^{-m+k+1}) \zeta (k+1)\zeta (m-k), \end{equation*} where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}$ represents the $n$th generalized harmonic number of order $m$, $\zeta$ denotes the Riemann zeta function and $\beta$ designates the Dirichlet beta function.

Proof: \begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^{(m)}}{n} \end{equation*} \begin{equation*} =\sum_{n=1}^{\infty}\frac{ (-1)^{n-1}}{n}\sum_{k=1}^{2n} \frac{1}{k^m}=\frac{(-1)^{m-1}}{(m-1)!}\sum_{n=1}^{\infty}\frac{ (-1)^{n-1}}{n}\sum_{k=1}^{2n}\int_0^1 x^{k-1}\log^{m-1}(x)\textrm{d}x \end{equation*} \begin{equation*} =\frac{(-1)^{m-1}}{(m-1)!}\int_0^1 \sum_{n=1}^{\infty}\frac{ (-1)^{n-1}}{n}\sum_{k=1}^{2n}x^{k-1}\log^{m-1}(x)\textrm{d}x \end{equation*} \begin{equation*} =\frac{(-1)^m}{(m-1)!}\int_0^1\frac{\displaystyle \log^{m-1}(x)\log\left(\frac{1+x^2}{2}\right)}{1-x}\textrm{d}x=\frac{2(-1)^{m-1}}{(m-1)!}\int_0^1\left(\int_x^1 \frac{y\log^{m-1}(x)}{(1+y^2)(1-x)}\textrm{d}y \right)\textrm{d}x \end{equation*} \begin{equation*} =\frac{2(-1)^{m-1}}{(m-1)!}\int_0^1\left(\int_0^y \frac{y\log^{m-1}(x)}{(1+y^2)(1-x)}\textrm{d}x \right)\textrm{d}y\overset{x=y z}{=}\frac{2(-1)^{m-1}}{(m-1)!}\int_0^1\left(\int_0^1 \frac{y^2\log^{m-1}(yz)}{(1+y^2)(1-yz)}\textrm{d}z \right)\textrm{d}y \end{equation*} \begin{equation*} =\frac{(-1)^{m-1}}{(m-1)!}\left(\int_0^1\left(\int_0^1 \frac{y^2\log^{m-1}(xy)}{(1+y^2)(1-xy)}\textrm{d}x \right)\textrm{d}y+\int_0^1\left(\int_0^1 \frac{x^2\log^{m-1}(xy)}{(1+x^2)(1-xy)}\textrm{d}x \right)\textrm{d}y\right) \end{equation*} \begin{equation*} =\frac{(-1)^{m-1}}{(m-1)!}\int_0^1\left(\int_0^1 \frac{((1+x^2)(1+y^2)-(1-(x y)^2))\log^{m-1}(xy)}{(1+x^2)(1+y^2)(1-xy)}\textrm{d}x \right)\textrm{d}y \end{equation*} \begin{equation*} =\frac{(-1)^{m-1}}{(m-1)!}\biggr(\int_0^1\left(\int_0^1 \frac{\log^{m-1}(xy)}{1-xy}\textrm{d}x \right)\textrm{d}y-\int_0^1\left(\int_0^1 \frac{\log^{m-1}(xy)}{(1+x^2)(1+y^2)}\textrm{d}x \right)\textrm{d}y \end{equation*} \begin{equation*} -\int_0^1\left(\int_0^1 \frac{x y\log^{m-1}(xy)}{(1+x^2)(1+y^2)}\textrm{d}x \right)\textrm{d}y\biggr) \end{equation*} \begin{equation*} =\frac{(-1)^m}{(m-1)!}\biggr(\int_0^1 \frac{\log^{m}(x)}{1-x}\textrm{d}x+\sum_{k=0}^{m-1}\binom{m-1}{k}\int_0^1 \frac{\log^{m-k-1}(y)}{1+y^2}\left(\int_0^1 \frac{\log^k(x)}{1+x^2}\textrm{d}x \right)\textrm{d}y \end{equation*} \begin{equation*} +\sum_{k=0}^{m-1}\binom{m-1}{k}\int_0^1\frac{y \log^{m-k-1}(y)}{1+y^2} \left(\int_0^1 \frac{x\log^k(x)}{1+x^2}\textrm{d}x \right)\textrm{d}y\biggr) \end{equation*} \begin{equation*} =m\zeta (m+1)- 2^{-m} \left(1-2^{-m+1}\right) \log(2 ) \zeta (m) -\sum_{k=0}^{m-1}\beta(k+1)\beta(m-k) \end{equation*} \begin{equation*} -\sum_{k=1}^{m-2}2^{- m-1}(1-2^{-k})(1-2^{-m+k+1}) \zeta (k+1)\zeta (m-k). \end{equation*}

A few cases of the second generalization

For $m=2$, \begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^{(2)}}{n}=2\zeta(3)-\frac{1}{8}\log(2)\zeta(2)-\frac{\pi}{2}G; \end{equation*} For $m=3$, \begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^{(3)}}{n}=\frac{199}{128}\zeta (4)-\frac{3}{32} \log (2)\zeta (3)-G^2; \end{equation*} For $m=4$, \begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^{(4)}}{n} \end{equation*} \begin{equation*} =4\zeta(5)-\frac{3}{128}\zeta(2)\zeta(3)-\frac{7}{128}\log(2)\zeta(4)+\frac{\pi^5}{192}-\frac{\pi^3}{16}G-\frac{\pi}{1536}\psi^{(3)}\left(\frac{1}{4}\right); \end{equation*} For $m=5$, \begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^{(5)}}{n} \end{equation*} \begin{equation*} =\frac{5151}{2048}\zeta(6)-\frac{15}{512}\log(2)\zeta(5)-\frac{9}{1024}\zeta^2(3)+\frac{15}{8}\zeta(4)G-\frac{1}{384}G\psi^{(3)}\left(\frac{1}{4}\right); \end{equation*} For $m=6$, \begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^{(6)}}{n} \end{equation*} \begin{equation*} =6\zeta(7)-\frac{15}{2048}\zeta (2) \zeta (5)-\frac{21}{2048}\zeta (3) \zeta (4)-\frac{31}{2048}\log(2)\zeta(6)+\frac{3}{2560}\pi^7-\frac{5}{768}\pi^5 G \end{equation*} \begin{equation*} -\frac{\pi^3}{12288}\psi^{(3)}\left(\frac{1}{4}\right)-\frac{\pi}{491520}\psi^{(5)}\left(\frac{1}{4}\right). \end{equation*}

The following equalities have been necessary during calculations: \begin{equation*} i) \ \int_0^1\frac{\log^m(x)}{1-x}\textrm{d}x=(-1)^m m!\zeta(m+1); \end{equation*} \begin{equation*} ii) \ \int_0^1\frac{\log^m(x)}{1+x}\textrm{d}x=(-1)^m (1-2^{-m})m!\zeta(m+1); \end{equation*} \begin{equation*} iii) \ \int_0^1\frac{\log^m(x)}{1+x^2}\textrm{d}x=(-1)^m m!\beta(m+1); \end{equation*} \begin{equation*} iv) \ \int_0^1\frac{x\log^m(x)}{1+x^2}\textrm{d}x=(-1)^m 2^{-(m+1)} (1-2^{-m})m!\zeta(m+1), \end{equation*} where $\zeta$ denotes the Riemann zeta function and $\beta$ represents the Dirichlet beta function.

Proof: The results are obtained immediately if we use geometric series.

An important observation: the strategy presented above works for the more general case \begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{kn}^{(m)}}{n}=\frac{(-1)^m}{(m-1)!}\int_0^1\frac{\displaystyle \log^{m-1}(x)\log\left(\frac{1+x^k}{2}\right)}{1-x}\textrm{d}x, \end{equation*} where $k\ge1$, $m\ge2$ are positive integers.

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    $\begingroup$ Note that a similar general equality, but where $m$ is now in the denominator of the binomial sum, is, $$\sum_{n=1}^{\infty} \frac{H_{n}^{(1)}}{(n+1)^m}z^{n}=\frac{(-1)^m}{(m-1)!}\int_0^1\frac{\displaystyle \log^{m-1}(x)\log\left(1-z\,x\right)}{1-z\,x}\textrm{d}x$$ for $-1\leq z\leq 1$ and integer $m\geq1$. $\endgroup$ – Tito Piezas III Jun 9 at 5:46
  • $\begingroup$ @TitoPiezasIII Thanks for sharing. $\endgroup$ – user97357329 Jun 10 at 13:16
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\begin{align} \int_0^1\frac{\ln(1+x)\ln^3x}{1-x}\ dx&=-\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1\frac{x^{n}\ln^3x}{1-x}\ dx=6\sum_{n=1}^\infty\frac{(-1)^n}{n}\left(\zeta(4)-H_n^{(4)}\right)\\ &=-6\ln2\zeta(4)-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}\tag{1} \end{align} evaluating the sum: \begin{align} \sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}&=\int_0^1\frac{\operatorname{Li}_4(-x)}{x(1+x)}\ dx=\int_0^1\frac{\operatorname{Li}_4(-x)}{x}\ dx-\underbrace{\int_0^1\frac{\operatorname{Li}_4(-x)}{1+x}\ dx}_{\text{IBP}}\\ &=\operatorname{Li}_5(-1)-\ln2\operatorname{Li}_4(-1)+\underbrace{\int_0^1\frac{\ln(1+x)\operatorname{Li}_3(-x)}{x}\ dx}_{\text{IBP}}\\ &=\operatorname{Li}_5(-1)-\ln2\operatorname{Li}_4(-1)-\operatorname{Li}_2(-1)\operatorname{Li}_3(-1)+\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx\\ &=-\frac{15}{16}\zeta(5)+\frac78\ln2\zeta(4)-\frac38\zeta(2)\zeta(3)-\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx \tag{2} \end{align} and the last integral: \begin{align} \int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx&=\int_0^1\frac1x\left(\frac12\operatorname{Li}_2(x^2)-\operatorname{Li}_2(x)\right)^2\ dx\\ &=\underbrace{\frac14\int_0^1\frac{\operatorname{Li}_2^2(x^2)}{x}\ dx}_{x^2=y}-\int_0^1\frac{\operatorname{Li}_2(x^2)\operatorname{Li}_2(x)}{x}\ dx+\int_0^1\frac{\operatorname{Li}_2^2(x)}{x}\ dx\\ &=\frac98\int_0^1\frac{\operatorname{Li}_2^2(x)}{x}\ dx-\int_0^1\frac{\operatorname{Li}_2(x^2)\operatorname{Li}_2(x)}{x}\ dx\\ &=\frac98\sum_{n=1}^\infty\frac1{n^2}\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx-\sum_{n=1}^\infty\frac1{n^2}\int_0^1x^{2n-1}\operatorname{Li}_2(x)\ dx\\ &=\frac98\sum_{n=1}^\infty\frac1{n^2}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)-\sum_{n=1}^\infty\frac1{n^2}\left(\frac{\zeta(2)}{2n}-\frac{H_{2n}}{(2n)^2}\right)\\ &=\frac98\zeta(2)\zeta(3)-\frac98\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac12\zeta(2)\zeta(3)+4\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}\\ &=\frac58\zeta(2\zeta(3)+\frac78\sum_{n=1}^\infty\frac{H_n}{n^4}+2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}\\ &=\frac58\zeta(2\zeta(3)+\frac78\left(3\zeta(5)-\zeta(2)\zeta(3)\right)+2\left(\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)\right)\\ &=\frac34\zeta(2)\zeta(3)-\frac{17}{16}\zeta(5)\tag{3} \end{align}

plugging $(3)$ in $(2)$ we have $$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}=\frac78\ln2\zeta(4)+\frac38\zeta(2)\zeta(3)-2\zeta(5)$$ plugging this result in $(1)$ we have $$\int_0^1\frac{\ln(1+x)\ln^3x}{1-x}\ dx=12\zeta(5)-\frac{45}{4}\ln2\zeta(4)-\frac94\zeta(2)\zeta(3)$$

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