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

Question

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 $3$.

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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.

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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.

Answer 1

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.

Answer 2

\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}

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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.

Answer 3

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

Answer 4

Preprint source: A simple strategy of calculating two alternating harmonic series generalizations by Cornel Ioan Valean

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.

Answer 5

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

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

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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).

Answer 6

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

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

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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).

Answer 7

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*}