# Closed-form of $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx$

Does the following series or integral have a closed-form

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

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

Here is my attempt. Using equation (11) from Mathworld Wolfram: $$\Psi_n(z)=(-1)^{n+1} n!\left(\zeta(n+1)-H_{z-1}^{(n+1)}\right)$$ I got $$\Psi_3(n+1)=6\left(\zeta(4)-H_{n}^{(4)}\right)$$ 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 $$\Psi_m(z)=(-1)^m\int_0^1\frac{x^{z-1}}{1-x}\ln^m x\,dx$$ 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.

• Downvote for no reason again! Why the heck this OP got downvoted?? May The Lord forgive your sin (҂⌣̀_⌣́)ᕤ Commented Oct 11, 2014 at 9:06

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.

• Interesting alternate approach together with nice reference! +1 Commented Oct 10, 2014 at 9:52
• @MarkusScheuer Thank you for your compliment. Commented Oct 10, 2014 at 10:05
• 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) Commented Oct 10, 2014 at 12:56
• @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. Commented Oct 11, 2014 at 1:08
• 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. Commented Oct 11, 2014 at 9:09

\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.

• 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) Commented Sep 18, 2014 at 13:29
• @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 $\displaystyle\sum^\infty_{n=1}\frac{(-1)^nH_n^{(4)}}{n^2}$ Commented Sep 18, 2014 at 13:44
• 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? Commented Sep 18, 2014 at 13:46
• @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. Commented Sep 18, 2014 at 13:50
• No doubts! :-) Nice answer, good reference! +1 Commented Oct 7, 2014 at 19:57

\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 $$\int_0^1\frac{\ln(1+x)\ln^3x}{1-x}\ dx=12\zeta(5)-\frac{45}{4}\ln2\zeta(4)-\frac94\zeta(2)\zeta(3)$$

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.

• 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 $m\geq1$. Commented Jun 9, 2019 at 5:46
• @TitoPiezasIII Thanks for sharing. Commented Jun 10, 2019 at 13:16

We can have a nice generalization,

From

$$\frac{\ln(1+x)}{1-x}=\sum_{n=1}^\infty \overline{H}_n x^n$$

We have

$$I_m=\int_0^1\frac{\ln(1+x)\ln^{m-1}x}{1-x}\ dx=\sum_{n=1}^\infty \overline{H}_n \int_0^1 x^n\ln^{m-1}x\ dx$$

$$=(-1)^{m-1}(m-1)!\sum_{n=1}^\infty \frac{\overline{H}_n}{(n+1)^m}$$

$$=(-1)^{m-1}(m-1)!\sum_{n=1}^\infty \frac{\overline{H}_{n-1}}{n^m}$$

$$=(-1)^{m-1}(m-1)!\sum_{n=1}^\infty \frac{\overline{H}_n+\frac{(-1)^n}{n}}{n^m}$$

$$=(-1)^{m-1}(m-1)!\left[\sum_{n=1}^\infty \frac{\overline{H}_n}{n^m}-\eta(m+1)\right]$$

Substitute

$$\sum_{n = 1}^\infty \frac{\overline H_n}{n^m} = \ln 2\zeta (m) - \frac{1}{2} m \zeta (m + 1) + \eta (m + 1) + \frac{1}{2} \sum_{i = 1}^m \eta (i) \eta (m - i + 1)$$

We get

$$I_m=(-1)^{m}(m-1)!\left[\frac{1}{2} m \zeta (m + 1)-\ln 2\zeta (m) - \frac{1}{2} \sum_{i = 1}^m \eta (i) \eta (m - i + 1)\right]$$

The generalization $$\displaystyle \small \sum_{n = 1}^\infty \frac{\overline H_n}{n^m}$$ can be found here (see Theorem 3.5 on page 9).

Computation of $$\displaystyle U=\int_0^1 \frac{\ln(1+x)\ln^3 x}{1-x}\,dx$$

\begin{align*} U&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln^3 t}{1-t}\,dt\right)\ln(1+x)\right]_0^1-\int_0^1 \frac{1}{1+x}\left(\int_0^x\frac{\ln^3 t}{1-t}\,dt\right)\,dx\\ &=-6\zeta(4)\ln 2+\int_0^1\int_0^1 \left(\frac{\ln^3(tx)}{(1+t)(1+x)}-\frac{\ln^3(tx)}{(1+t)(1-tx)}\right)\,dt\,dx\\ &=-6\zeta(4)\ln 2+6\left(\int_0^1\frac{\ln^2 t}{1+t}\,dt\right)\left(\int_0^1\frac{\ln x}{1+x}\,dx\right)+\\ &2\left(\int_0^1\frac{\ln^3 t}{1+t}\,dt\right)\left(\int_0^1\frac{1}{1+x}\,dx\right)-\int_0^1 \frac{1}{t(1+t)}\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\,dt\\ &=-\frac{33}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)-\int_0^1 \frac{1}{t(1+t)}\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\,dt\\ &\overset{\text{IBP}}=-\frac{33}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)-\left[\ln\left(\frac{t}{1+t}\right)\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\right]_0^1+\int_0^1 \frac{\ln\left(\frac{t}{1+t}\right)\ln^3 t}{1-t}\,dt\\ &=-\frac{45}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)+\int_0^1 \frac{\ln\left(\frac{t}{1+t}\right)\ln^3 t}{1-t}\,dt\\ &=-\frac{45}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)+24\zeta(5)-U\\ U&=\boxed{-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{4}\zeta(2)\zeta(3)+12\zeta(5)} \end{align*}

• Impressive and clever. (+1) Commented Apr 2, 2023 at 19:26

Different approach using only series manipulations.

By using the identity

$$\frac{\ln(1+x)}{1-x}=\sum_{n=1}^\infty \overline{H}_n x^n$$ which can be easily proved by series-expanding the numerator and denominator.

Multiply both sides by $$\ln^3x$$ then $$\int_0^1$$ we get

$$I=\int_0^1\frac{\ln(1+x)\ln^3x}{1-x}\ dx=\sum_{n=1}^\infty \overline{H}_n\int_0^1 x^n \ln^3x\ dx=-6\sum_{n=1}^\infty\frac{\overline{H}_n}{(n+1)^4}=-6\sum_{n=1}^\infty\frac{\overline{H}_{n-1}}{n^4}$$

Now use $$\overline{H}_{n-1}=\overline{H}_n+\frac{(-1)^n}{n}$$

$$\Longrightarrow I=-6\sum_{n=1}^\infty\frac{\overline{H}_n}{n^4}-6\sum_{n=1}^\infty\frac{(-1)^n}{n^5}=\frac{45}{8}\zeta(5)-6\sum_{n=1}^\infty\frac{\overline{H}_n}{n^4}\tag1$$

$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^4}=1+\sum_{n=2}^\infty\frac{\overline{H}_n}{n^4}=1+\sum_{n=1}^\infty\frac{\overline{H}_{2n}}{(2n)^4}+\sum_{n=1}^\infty\frac{\overline{H}_{2n+1}}{(2n+1)^4}$$

By writing $$\overline{H}_{2n}=H_{2n}-H_n$$ and $$\overline{H}_{2n+1}=H_{2n+1}-H_n$$ we have

$$\sum_{n=1}^\infty\frac{\overline{H}_{2n}}{(2n)^4}=\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}-\sum_{n=1}^\infty\frac{H_{n}}{(2n)^4}=\frac12\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{n^4}+\frac7{16}\sum_{n=1}^\infty\frac{H_{n}}{n^4}$$

and

$$\sum_{n=1}^\infty\frac{\overline{H}_{2n+1}}{(2n+1)^4}=\color{blue}{\sum_{n=1}^\infty\frac{H_{2n+1}}{(2n+1)^4}}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}$$

$$=\color{blue}{-1+\sum_{n=0}^\infty\frac{H_{2n+1}}{(2n+1)^4}}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}$$

$$=\color{blue}{-1+\frac12\sum_{n=0}^\infty\frac{(-1)^nH_{n+1}}{(n+1)^4}+\frac12\sum_{n=0}^\infty\frac{H_{n+1}}{(n+1)^4}}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}$$

$$=\color{blue}{-1-\frac12\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{n^4}+\frac12\sum_{n=1}^\infty\frac{H_{n}}{n^4}}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}$$

Combine the two sums,

$$\Longrightarrow \sum_{n=1}^\infty\frac{\overline{H}_n}{n^4}=\frac{15}{16}\sum_{n=1}^\infty\frac{H_n}{n^4}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}$$

From here we have

$$\sum_{n=1}^{\infty} \frac{H_{n}}{(n+a)^{2}}=\left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2}$$

Differentiate with respect to $$a$$ twice then set $$a=1/2$$ we get

$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}=\frac{31}{8}\zeta(5)-\frac{15}{8}\ln2\zeta(4)-\frac78\zeta(2)\zeta(3)$$

Substituting this result along with $$\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$ gives

$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^4}=-\frac{17}{16}\zeta(5)+\frac{15}{8}\ln2\zeta(4)+\frac38\zeta(2)\zeta(3)\tag2$$

Finally plug $$(2)$$ in $$(1)$$ we get

$$I=12\zeta(5)-\frac{45}{4}\ln2\zeta(4)-\frac94\zeta(2)\zeta(3)$$

Edit

Another way to calculate $$\displaystyle \sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}$$ is to use the generalization

$$\sum_{k = 1}^\infty \frac{\overline H_k}{k^m} = \zeta (m) \log 2 - \frac{1}{2} m \zeta (m + 1) + \eta (m + 1) + \frac{1}{2} \sum_{i = 1}^m \eta (i) \eta (m - i + 1).$$

where $$\eta (s) = \sum_{n = 1}^\infty \frac{(-1)^{n - 1}}{n^s} = (1 - 2^{1 - s}) \zeta (s)$$ is the Dirichlet eta function and $$\zeta (s) = \sum_{n = 1}^\infty \frac{1}{n^s}$$ is the Riemann zeta function.

With $$m=4$$ we have

$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^4}=-\frac{17}{16}\zeta(5)+\frac{15}{8}\ln2\zeta(4)+\frac38\zeta(2)\zeta(3)$$

The generalization can be found here (see Theorem 3.5 on page 9).

• The skew-harmonic number $\overline H_n$ is a beautiful new tool. Commented Feb 16, 2020 at 23:12
• @omegadot I agree and I recently started using it. it simplifies some tough problems. Commented Feb 16, 2020 at 23:17

The following proof is very simple and short but it works for only the odd powers of $$\ln(x):$$

By integrating

$$\operatorname{Li}_2(-x)+\operatorname{Li}_2\left(-\frac1x\right)=-\frac{\ln^2(x)}{2}-2\eta(2)$$

repeatedly, we get

$$\operatorname{Li}_{2a}(-x)+\operatorname{Li}_{2a}\left(-\frac1x\right)=-2\sum_{k=0}^a \frac{\eta(2a-2k)}{(2k)!}\ln^{2k}(x)$$

Using the integral form of the polylogarithm function, we have

$$\begin{gather*} \operatorname{Li}_{2a}(-x)+\operatorname{Li}_{2a}\left(-\frac1x\right)\\ =\frac{-1}{(2a-1)!}\int_0^1\frac{-x\ln^{2a-1}(y)}{1+xy}\mathrm{d}y-\frac1{(2a-1)!}\int_0^1\frac{-\frac1{x}\ln^{2a-1}(y)}{1+\frac{y}{x}}\mathrm{d}y\\ =\frac{1}{(2a-1)!}\int_0^1\ln^{2a-1}(y)\left(\frac{x}{1+xy}+\frac{1}{x+y}\right)\mathrm{d}y\\ =\frac{1}{(2a-1)!}\int_0^1\frac{(1+2xy+x^2)\ln^{2a-1}(y)}{(1+xy)(x+y)}\mathrm{d}y. \end{gather*}$$ and so $$$$\int_0^1\frac{(1+2xy+x^2)\ln^{2a-1}(y)}{(1+xy)(x+y)}\mathrm{d}y=-2(2a-1)!\sum_{k=0}^a \frac{\eta(2a-2k)}{(2k)!}\ln^{2k}(x).$$$$

Divide the latter identity by $$1+x$$ using $$\int_0^1\frac{\ln^s(x)}{1+x}dx=(-1)^s s!\eta(s+1),$$ we get

$$-2(2a-1)!\sum_{k=0}^a \eta(2a-2k)\eta(2k+1)=\int_0^1\ln^{2a-1}(y)\left(\int_0^1\frac{1+2xy+x^2}{(1+xy)(x+y)(1+x)}\mathrm{d}x\right)\mathrm{d}y$$

$$=\int_0^1\ln^{2a-1}(y)\left(\frac{2\ln(1+y)}{1-y}+\frac{\ln(1+y)}{y}-\frac{\ln(y)}{1-y}-\frac{2\ln(2)}{1-y}\right)\mathrm{d}y$$

$$=2\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{1-x}\mathrm{d}x-(2a-1)!\eta(2a+1)-(2a)!\zeta(2a+1)$$ $$+2(2a-1)!\ln(2)\zeta(2a)$$

$$\Longrightarrow \int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{1-x}\mathrm{d}x$$ $$=(2a-1)!\left(\frac12(2a-4^{-a}+1)\zeta(2a+1)-\ln(2)\zeta(2a)-\sum_{k=0}^a \eta(2a-2k)\eta(2k+1)\right).$$