How to evaluate $ \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}$

Question

I am having a difficult time evaulating $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}.$$ I have tried the following relation: $$\frac{1}{2}\int_0^1 \frac{\mathrm{Li}_2(x)}{x(1-x)}\log^2{x}\ \mathrm{d}x.$$ Or, alternatively, we have $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3} = \sum_{n=1}^\infty \frac{1}{n^3}\sum_{m=1}^n \frac{1}{m^2};$$ from which I wrote \begin{align}-\sum_{n=1}^\infty \frac{1}{n^3}\sum_{m=1}^n \int_0^1 x^{m-1}\log{x}\ \mathrm{d}x &= -\sum_{n=1}^\infty \frac{1}{n^3}\int_0^1 \sum_{m=1}^n x^{m-1}\log{x}\ \mathrm{d}x,\\ &=-\sum_{n=1}^\infty \frac{1}{n^3}\int_0^1 \frac{x^n-1}{x-1}\log{x}\ \mathrm{d}x \end{align} followed by distributing the latter sum into the integral and integrating parts. Using this technique I obtained $$-\zeta(2)\zeta(3)-\int_0^1\frac{\mathrm{Li}_2(x)\mathrm{Li}_2(1-x)}{x}\ \mathrm{d}x.$$ However, I do not know where to go. I know the value of the sum, but I am looking to evaulate the sum by the means of integrals. A hint/comment would be much appreciated. Thank you!

Answer 1

Calculating your First integral:

$$\int_0^1 \frac{\mathrm{Li}_2(x)\ln^2(x)}{x(1-x)}dx=\int_0^1 \frac{\mathrm{Li}_2(x)\ln^2(x)}{x}dx+\int_0^1 \frac{\mathrm{Li}_2(x)\ln^2(x)}{1-x}dx=A+B$$

where

$$A=\sum_{n=1}^\infty\frac{1}{n^2}\int_0^1 x^{n-1}\ln^2(x)dx=2\sum_{n=1}^\infty\frac{1}{n^5}=2\zeta(5)$$

and

$$B\overset{IBP}{=}-\ln(1-x)\text{Li}_2(x)\ln^2(x)|_0^1+\int_0^1 \ln(1-x)\left[\frac{2\text{Li}_2(x)\ln(x)}{x}-\frac{\ln(1-x)\ln^2(x)}{x}\right]dx$$

$$=2\int_0^1\frac{\ln(x)\ln(1-x)\text{Li}_2(x)}{x}dx-\int_0^1\frac{\ln^2(1-x)\ln^2(x)}{x}dx$$

$$=B_1-B_2$$

$$B_1\overset{IBP}{=}-\text{Li}_2^2(x)\ln(x)|_0^1+\int_0^1\frac{\text{Li}_2^2(x)}{x}dx$$

$$=\sum_{n=1}^\infty\frac{1}{n^2}\int_0^1 x^{n-1}\text{Li}_2(x)dx=\sum_{n=1}^\infty\frac{1}{n^2}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)$$

$$=\zeta(2)\zeta(3)-\sum_{n=1}^\infty\frac{H_n}{n^4}=2\zeta(2)\zeta(3)-3\zeta(5)$$

$$B_2=2\sum_{n=1}^\infty \frac{H_{n-1}}{n}\int_0^1 x^{n-1}\ln^2(x)dx=4\sum_{n=1}^\infty \frac{H_{n-1}}{n^4}=4\sum_{n=1}^\infty \frac{H_n}{n^4}-4\zeta(5)$$

$$=8\zeta(5)-4\zeta(2)\zeta(3)$$

Combine all the pieces we get

$$\begin{equation} \int_0^1 \frac{\mathrm{Li}_2(x)\ln^2(x)}{x(1-x)}dx=6\zeta(2)\zeta(3)-9\zeta(5) \end{equation}$$

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Calculating your second integral:

Make use of the dilog reflection identity $\text{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)-\text{Li}_2(x)$

$$\int_0^1\frac{\text{Li}_2(x)\text{Li}_2(1-x)}{x}dx$$

$$=\zeta(2)\int_0^1\frac{\text{Li}_2(x)}{x}dx-\underbrace{\int_0^1\frac{\ln x\ln(1-x)\text{Li}_2(x)}{x}dx}_{IBP}-\int_0^1\frac{\text{Li}_2^2(x)}{x}dx$$

$$=\zeta(2)\zeta(3)-\frac32\int_0^1\frac{\text{Li}_2^2(x)}{x}dx$$

$$=\zeta(2)\zeta(3)-\frac32\left(2\zeta(2)\zeta(3)-3\zeta(5)\right)$$

$$=\frac92\zeta(5)-2\zeta(2)\zeta(3)$$

Answer 2

The closed forms are\begin{equation} \sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)\label{H_n^(2)/n^3} \end{equation} \begin{equation} \sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\frac{11}{2}\zeta(5)-2\zeta(2)\zeta(3)\label{H_n^(3)/n^2} \end{equation} Proof: We are going to establish two relations and solve for the two sums by elimination.

The first relation: By Cauchy product we have

$$\text{Li}_2(x)\text{Li}_3(x)=6\sum_{n=1}^\infty \frac{H_n}{n^4}x^n+3\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}x^n+\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^2}-10\text{Li}_5(x)$$

and set $x=1$ and rearrange the terms we have

$$3\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}+\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^2}=\zeta(2)\zeta(3)+10\zeta(5)-6\sum_{n=1}^\infty \frac{H_n}{n^4}$$

$$=7\zeta(2)\zeta(3)-8\zeta(5)\tag1$$

where the last result follows from substituting $\sum_{n=1}^\infty \frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$

The second relation: By Abels' summation we have $$\sum_{k=1}^\infty \frac{H_k^{(p)}}{k^q}+\sum_{k=1}^\infty \frac{H_k^{(q)}}{k^p}=\zeta(p)\zeta(q)+\zeta(p+q)$$

and set $p=2$ and $q=3$ we get \begin{equation} \sum_{k=1}^\infty \frac{H_k^{(3)}}{k^2}+\sum_{k=1}^\infty \frac{H_k^{(2)}}{k^3}=\zeta(2)\zeta(3)+\zeta(5)\tag2 \end{equation}

Solving $(1)$ and $(2)$ systematically, the closed forms follows.

Answer 3

\begin{align}S&=\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}\\ &=-\sum_{n=1}^\infty\frac1{n^3}\left(\int_0^1\frac{(1-x^n)\ln x}{1-x}dx\right)\\ &=-\int_0^1\frac{\Big(\text{Li}_3(1)-\text{Li}_3(x)\Big) \ln x}{1-x}dx\\ &=\underbrace{\int_0^1\frac{\ln x\text{Li}_3(x)}{1-x}dx}_{\color{red}{=A}}+\zeta(2)\zeta(3)\\ A&=\frac12\int_0^1\int_0^1\frac{x\ln x\ln^2t}{(1-x)(1-tx)}dtdx\\ &=\frac12\int_0^1\int_0^1\frac{x\ln^2(tx)\ln x-2x\ln(tx)\ln^2x+x\ln^3x }{(1-x)(1-tx)}dtdx\\ &=\frac12\underbrace{\int_0^1\frac{\ln x}{1-x}\left(\int_0^x\frac{\ln^2t}{1-t}dt\right)dx}_{\text{IBP}}-\int_0^1\frac{\ln^2 x}{1-x}\left(\int_0^x\frac{\ln t}{1-t}dt\right)dx-\\&\frac12\underbrace{\int_0^1\frac{\ln(1-x)\ln^3x}{1-x}dx}_{\color{red}{=B}}\\ &=-\frac32\underbrace{\int_0^1\frac{\ln^2 x}{1-x}\left(\int_0^x\frac{\ln t}{1-t}dt\right)dx}_{\color{red}{=C}}-\frac12B-\zeta(2)\zeta(3)\\ C&=\int_0^1\int_0^1\frac{x\ln^2 x\ln(tx)}{(1-x)(1-tx)}dtdx\\ &=\int_0^1\int_0^1\frac{2x\ln^2(tx)\ln x-x\ln^3(tx)+x\ln(tx)\ln^2 t}{(1-x)(1-tx)}dtdx\\ &=2\underbrace{\int_0^1\frac{\ln x}{1-x}\left(\int_0^x\frac{\ln^2 t}{1-t}dt\right)dx}_{\text{IBP}}-\underbrace{\int_0^1\frac1{1-x}\left(\int_0^x\frac{\ln^3t}{1-t}dt-\int_0^1\frac{\ln^3t}{1-t}dt\right)dx}_{\text{IBP}}+\\ &\int_0^1\int_0^1\left(\underbrace{\frac{x\ln(tx)\ln^2t}{(1-x)(1-tx)}}_{\text{Partial fraction décomposition}}-\frac{\ln^3t}{(1-t)(1-x)}\right)dtdx\\ &=-4\zeta(2)\zeta(3)-2C-B+\underbrace{\int_0^1\int_0^1\frac{\ln^2 t\ln x}{(1-t)(1-x)}dtdx}_{\text{Fubini}}-\\&\int_0^1\frac{\ln^2t}t\left(\int_0^t\frac{\ln x}{1-x}dx\right)dt-C\\ &\color{blue}{\boxed{C=2\zeta(5)-\frac32\zeta(2)\zeta(3)-\frac14B}} \color{blue}{\boxed{A=\frac54\zeta(2)\zeta(3)-3\zeta(5)-\frac18B}} \end{align} \begin{align}B&\overset{\text{IBP}}=\int_0^1\frac1{1-x}\left(\int_0^x\frac{\ln^3t}{1-t}dt-\int_0^1\frac{\ln^3t}{1-t}dt\right)dx\\ &=\int_0^1\int_0^1\left(\underbrace{\frac{x\ln^3(tx)}{(1-x)(1-tx)}}_{\text{Partial fraction décomposition}}-\frac{\ln^3t}{(1-t)(1-x)}\right)dtdx\\ &=\underbrace{\int_0^1\int_0^1\frac{\ln^3(tx)-\ln^3t-\ln^3x}{(1-t)(1-x)}dtdx}_{\text{Fubini}}-\\&\int_0^1\int_0^1\left(\frac{\ln^3(tx)}{(1-t)(1-tx)}-\frac{\ln^3 x}{(1-t)(1-x)}\right)dtdx\\ &=-12\zeta(2)\zeta(3)-\int_0^1\left(\frac1{t(1-t)}\int_0^t\frac{\ln^3x}{1-x}dx-\frac1{1-t}\int_0^1\frac{\ln^3x}{1-x}dx\right)dt\\ &=-12\zeta(2)\zeta(3)-\underbrace{\int_0^1\frac1{t}\left(\int_0^t\frac{\ln^3x}{1-x}dx\right)dt}_{\text{IBP}}-\\&\underbrace{\int_0^1\frac1{1-t}\left(\int_0^t\frac{\ln^3x}{1-x}dx-\int_0^1\frac{\ln^3x}{1-x}dx\right)dt}_{\text{IBP}}\\ &=-12\zeta(2)\zeta(3)+24\zeta(5)-B\\ &\color{blue}{\boxed{B=12\zeta(5)-6\zeta(2)\zeta(3)}}\color{green}{\boxed{A=\int_0^1\frac{\text{Li}_3(x)\ln x}{1-x}dx=2\zeta(2)\zeta(3)-\frac92\zeta(5)}} \end{align} Therefore, \begin{align}\color{green}{\boxed{S=\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)}}\end{align}

NB: For $n\geq 1$, integer, \begin{align}H_n^{(2)}=-\int_0^1\frac{(1-x^n)\ln x}{1-x}dx\end{align}