# 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 – Anastasiya-Romanova 秀 Oct 11 '14 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 – Markus Scheuer Oct 10 '14 at 9:52
• @MarkusScheuer Thank you for your compliment. – M.N.C.E. Oct 10 '14 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) – Anastasiya-Romanova 秀 Oct 10 '14 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. – M.N.C.E. Oct 11 '14 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. – Anastasiya-Romanova 秀 Oct 11 '14 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) – Anastasiya-Romanova 秀 Sep 18 '14 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 $\displaystyle\sum^\infty_{n=1}\frac{(-1)^nH_n^{(4)}}{n^2}$ – SuperAbound Sep 18 '14 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? – Anastasiya-Romanova 秀 Sep 18 '14 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. – SuperAbound Sep 18 '14 at 13:50
• No doubts! :-) Nice answer, good reference! +1 – Markus Scheuer Oct 7 '14 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 $m\geq1$. – Tito Piezas III Jun 9 '19 at 5:46
• @TitoPiezasIII Thanks for sharing. – user97357329 Jun 10 '19 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).

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. – omegadot Feb 16 '20 at 23:12
• @omegadot I agree and I recently started using it. it simplifies some tough problems. – Ali Shadhar Feb 16 '20 at 23:17

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*}