Integral ${\large\int}_0^1\ln(1-x)\ln(1+x)\ln^2x\,dx$

Question

This problem was posted at I&S a week ago, and no attempts to solve it have been posted there yet. It looks very alluring, so I decided to repost it here:

Prove: $$\int_0^1\ln(1-x)\ln(1+x)\ln^2x\,dx=24-\frac{4\pi^2}3-\frac{11\pi^4}{720}-12\ln2\\+2\ln^22-\frac16\ln^42+\pi ^2\ln2+\frac{\pi^2}6\ln^22-4\operatorname{Li}_4\!\left(\tfrac12\right)-\frac{35}4\zeta(3)+\frac72\zeta(3)\ln2.$$

I found a paper where some similar integrals are evaluated: J. A. M. Vermaseren, Harmonic sums, Mellin transforms and Integrals, Int. J. Mod. Phys. A, 14 (1999), 2037-2076, DOI: 10.1142/S0217751X99001032 , but it's not quite easy to read for me. Maybe it could be of some help for this problem.

Answer 1

This answer is split into 3 main steps.

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Step 1: Expressing the integral as a sum

\begin{align} &\ \ \ \ \ \int^1_0\ln(1+x)\ln(1-x)\ln^2{x} \ {\rm d}x\\ &=\sum^\infty_{j=1}\frac{(-1)^j}{j}\sum^\infty_{k=1}\frac{1}{k}\int^1_0x^{j+k}\ln^2{x} \ {\rm d}x\\ &=2\sum^\infty_{j=1}\frac{(-1)^j}{j}\sum^\infty_{k=1}\frac{1}{k(k+j+1)^3}\\ &=\small{2\sum^\infty_{j=1}\frac{(-1)^j}{j}\sum^\infty_{k=1}\frac{1}{(j+1)^3k}-\frac{1}{(j+1)^3(k+j+1)}-\frac{1}{(j+1)^2(k+j+1)^2}-\frac{1}{(j+1)(k+j+1)^3}}\\ &=2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}}{j(j+1)^3}-2\sum^\infty_{j=1}\frac{(-1)^j\left[\zeta(2)-H_{j+1}^{(2)}\right]}{j(j+1)^2}-2\sum^\infty_{j=1}\frac{(-1)^j\left[\zeta(3)-H_{j+1}^{(3)}\right]}{j(j+1)} \end{align}

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Step 2a: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n}{n}$ \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n}{n} &=\frac{1}{2}\ln^2{2}-\frac{\pi^2}{12} \end{align} See here for the details.

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Step 2b: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^2}$ \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^2} &=-\frac{5}{8}\zeta(3) \end{align} See here for the details.

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Step 2c: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^3}$ \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^3} &=\int^{-1}_0\frac{1}{y}\left[\int^y_0\frac{1}{x}\left[\int^x_0\frac{\ln(1-t)}{t(t-1)}{\rm d}t\right]{\rm d}x\right]{\rm d}y\\ &=2{\rm Li}_4\left(\frac{1}{2}\right)-\frac{11\pi^4}{360}+\frac{1}{12}\ln^4{2}+\frac{7}{4}\zeta(3)\ln{2}-\frac{\pi^2}{12}\ln^2{2} \end{align} Tunk-Fey did a calculation of this type here.

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Step 2d: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n}$ \begin{align} \sum^\infty_{n=1}\frac{H_n^{(2)}}{n}x^n &=\int^x_0\frac{{\rm Li}_2(t)}{t(1-t)}{\rm d}t\\ &={\rm Li}_3(x)+\int^x_0\frac{{\rm Li}_2(t)}{1-t}{\rm d}t\\ &={\rm Li}_3(x)-{\rm Li}_2(x)\ln(1-x)-\int^x_0\frac{\ln^2(1-t)}{t}{\rm d}t\\ &={\rm Li}_3(x)-{\rm Li}_2(x)\ln(1-x)+\int^{1-x}_1\frac{\ln^2{t}}{1-t}{\rm d}t\\ &={\rm Li}_3(x)-{\rm Li}_2(x)\ln(1-x)-\ln^2(1-x)\ln{x}+\int^{1-x}_1\frac{2\ln(1-t)\ln{t}}{t}{\rm d}t\\ &\small{={\rm Li}_3(x)-{\rm Li}_2(x)\ln(1-x)-\ln^2(1-x)\ln{x}-2{\rm Li}_2(1-x)\ln(1-x)+\int^{1-x}_1\frac{2{\rm Li}_2(t)}{t}{\rm d}t}\\ &\small{={\rm Li}_3(x)-{\rm Li}_2(x)\ln(1-x)-\ln^2(1-x)\ln{x}-2{\rm Li}_2(1-x)\ln(1-x)+2{\rm Li}_3(1-x)-2\zeta(3)} \end{align} Therefore \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n} &={\rm Li}_3(-1)-{\rm Li}_2(-1)\ln{2}-\ln^2{2}\ln(-1)-2{\rm Li}_2(2)\ln{2}+2{\rm Li}_3(2)-2\zeta(3)\\ &=-\zeta(3)+\frac{\pi^2}{12}\ln{2} \end{align} You can use polylogarithm identities to simplify the last equation. I took the easy way out and used Wolfram Alpha. Note that contour integration is a slightly more efficient method to solve this sum, however this method is required if I want to solve $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n^2}$ as well.

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Step 2e: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n^{(3)}}{n}$

\begin{align} \sum^\infty_{n=1}\frac{H_n^{(3)}}{n}x^n &=\int^x_0\frac{{\rm Li}_3(t)}{t(1-t)}{\rm d}t\\ &={\rm Li}_4(x)+\int^x_0\frac{{\rm Li}_3(t)}{1-t}{\rm d}t\\ &={\rm Li}_4(x)-{\rm Li}_3(x)\ln(1-x)-\int^x_0\frac{-\ln(1-t){\rm Li}_2(t)}{t}{\rm d}t\\ &={\rm Li}_4(x)-{\rm Li}_3(x)\ln(1-x)-\frac{1}{2}{\rm Li}^2_2(x) \end{align} Therefore \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n^{(3)}}{n} &={\rm Li}_4(-1)-{\rm Li}_3(-1)\ln{2}-\frac{1}{2}{\rm Li}^2_2(-1)\\ &=-\frac{19\pi^4}{1440}+\frac{3}{4}\zeta(3)\ln{2} \end{align}

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Step 2f: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n^2}$

This part is rather similar to Tunk-Fey's answer, so he certainly deserves credit. \begin{align} &\ \ \ \ \ \sum^\infty_{n=1}\frac{H_n^{(2)}}{n^2}x^n\\ &=\small{{\rm Li}_4(x)-2\zeta(3)\ln{x}+\frac{1}{2}{\rm Li}_2^2(x)+\color{blue}{\int\frac{-\ln^2(1-x)\ln{x}}{x}{\rm d}x}+\color{\orange}{2\int\frac{{\rm Li}_3(1-x)-{\rm Li}_2(1-x)\ln(1-x)}{x}{\rm d}x}} \end{align} The blue integral is \begin{align} &\ \ \ \ \ \color{blue}{\int\frac{-\ln^2(1-x)\ln{x}}{x}{\rm d}x}\\ &=-\frac{1}{2}\ln^2{x}\ln^2(1-x)-\int\frac{\ln^2{x}\ln(1-x)}{1-x}{\rm d}x\\ &=-\frac{1}{2}\ln^2{x}\ln^2(1-x)+\sum^\infty_{n=1}H_n\int x^n\ln^2{x} \ {\rm d}x\\ &=-\frac{1}{2}\ln^2{x}\ln^2(1-x)+\sum^\infty_{n=1}H_n\partial^2_n\frac{x^{n+1}}{n+1}\\ &=\color\grey{-\frac{1}{2}\ln^2{x}\ln^2(1-x)+\ln^2{x}\sum^\infty_{n=1}\frac{H_nx^{n+1}}{n+1}}-2\ln{x}\sum^\infty_{n=1}\frac{H_nx^{n+1}}{(n+1)^2}+2\sum^\infty_{n=1}\frac{H_{n}x^{n+1}}{(n+1)^3}\\ &=\color{blue}{2\ln{x}{\rm Li}_3(x)-2{\rm Li}_4(x)-2\ln{x}\sum^\infty_{n=1}\frac{H_n}{n^2}x^n+2\sum^\infty_{n=1}\frac{H_n}{n^3}x^n} \end{align} The orange integral is \begin{align} &\ \ \ \ \ \ \color{orange}{2\int\frac{{\rm Li}_3(1-x)-{\rm Li}_2(1-x)\ln(1-x)}{x}{\rm d}x}\\ &=2{\rm Li}_3(1-x)\ln{x}-2{\rm Li}_2(1-x)\ln{x}\ln(1-x)+2\int\frac{\ln(1-x)\ln^2{x}}{1-x}{\rm d}x\\ &=\color{orange}{2{\rm Li}_3(1-x)\ln{x}-2{\rm Li}_2(1-x)\ln{x}\ln(1-x)-\ln^2{x}\ln^2(1-x)-4\ln{x}{\rm Li}_3(x)+4{\rm Li}_4(x)+4\ln{x}\sum^\infty_{n=1}\frac{H_n}{n^2}x^n-4\sum^\infty_{n=1}\frac{H_n}{n^3}x^n} \end{align} So \begin{align} & \ \ \ \ \ \sum^\infty_{n=1}\frac{H_n^{(2)}}{n^2}x^n\\ &=3{\rm Li}_4(x)+2{\rm Li}_3(1-x)\ln{x}-2{\rm Li}_3(x)\ln{x}-2\zeta(3)\ln{x}+\frac{1}{2}{\rm Li}_2^2(x)-2{\rm Li}_2(1-x)\ln{x}\ln(1-x)-\ln^2{x}\ln^2(1-x)+2\ln{x}\sum^\infty_{n=1}\frac{H_n}{n^2}x^n-2\sum^\infty_{n=1}\frac{H_n}{n^3}x^n+C \end{align} Therefore \begin{align} & \ \ \ \ \ \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n^2}\\ &=3{\rm Li}_4(-1)+\color\grey{2{\rm Li}_3(2)\ln(-1)-2{\rm Li}_3(-1)\ln(-1)-2\zeta(3)\ln(-1)}\\ &+\frac{1}{2}{\rm Li}_2^2(-1)\color\grey{-2{\rm Li}_2(2)\ln(-1)\ln(2)-\ln^2(-1)\ln^2{2}+2\ln(-1)\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^2}}-2\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^3}\\ &=\frac{17\pi^4}{480}-4{\rm Li}_4\left(\frac{1}{2}\right)-\frac{1}{6}\ln^4{2}-\frac{7}{2}\zeta(3)\ln{2}+\frac{\pi^2}{6}\ln^2{2} \end{align} The grey terms miraculously cancel.

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Step 3a: Evaluating $\displaystyle 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}}{j(j+1)^3}$ \begin{align} & \ \ \ \ \ 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}}{j(j+1)^3}\\ &=2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}}{j}-\frac{(-1)^jH_{j+1}}{(j+1)^3}-\frac{(-1)^jH_{j+1}}{(j+1)^2}-\frac{(-1)^jH_{j+1}}{j+1}\\ &=\small{2\sum^\infty_{j=1}\frac{(-1)^jH_{j}}{j}+2\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)}+2\sum^\infty_{j=1}\frac{(-1)^jH_j}{j^3}+2+2\sum^\infty_{j=1}\frac{(-1)^jH_j}{j^2}+2+2\sum^\infty_{j=1}\frac{(-1)^jH_j}{j^3}+2}\\ &=4{\rm Li}_4\left(\frac{1}{2}\right)-\frac{11\pi^4}{180}+\frac{1}{6}\ln^4{2}+\frac{7}{2}\zeta(3)\ln{2}-\frac{5}{4}\zeta(3)-\frac{\pi^2}{6}\ln^2{2}-\frac{\pi^2}{3}+2\ln^2{2}-4\ln{2}+8 \end{align}

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Step 3b: Evaluating $\displaystyle -\frac{\pi^2}{3}\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^2}$ \begin{align} -\frac{\pi^2}{3}\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^2} &=-\frac{\pi^2}{3}\sum^\infty_{j=1}\frac{(-1)^j}{j}-\frac{(-1)^j}{(j+1)^2}-\frac{(-1)^j}{j+1}\\ &=\frac{\pi^2}{3}\ln{2}+\frac{\pi^4}{36}-\frac{\pi^2}{3}+\frac{\pi^2}{3}\ln{2}-\frac{\pi^2}{3}\\ &=\frac{\pi^4}{36}+\frac{2\pi^2}{3}\ln{2}-\frac{2\pi^2}{3} \end{align}

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Step 3c: Evaluating $\displaystyle 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(2)}}{j(j+1)^2}$ \begin{align} & \ \ \ \ \ 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(2)}}{j(j+1)^2}\\ &=2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(2)}}{j}-\frac{(-1)^jH_{j+1}^{(2)}}{(j+1)^2}-\frac{(-1)^jH_{j+1}^{(2)}}{j+1}\\ &=4\sum^\infty_{j=1}\frac{(-1)^jH_{j}^{(2)}}{j}+2\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^2}+2\sum^\infty_{j=1}\frac{(-1)^jH_j^{(2)}}{j^2}+2+2\\ &=-8{\rm Li}_4\left(\frac{1}{2}\right)+\frac{17\pi^4}{240}-\frac{1}{3}\ln^4{2}-7\zeta(3)\ln{2}-4\zeta(3)+\frac{\pi^2}{3}\ln^2{2}+\frac{\pi^2}{3}\ln{2}-\frac{\pi^2}{6}-4\ln{2}+8\\ \end{align}

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Step 3d: Evaluating $\displaystyle -2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)}$ \begin{align} -2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)} &=-2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j}+2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j+1}\\ &=4\zeta(3)\ln{2}-2\zeta(3)\\ \end{align}

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Step 3e: Evaluating $\displaystyle 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j(j+1)}$ \begin{align} 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j(j+1)} &=2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j}-2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j+1}\\ &=4\sum^\infty_{j=1}\frac{(-1)^jH_{j}^{(3)}}{j}+2\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^3}+2\\ &=-\frac{19\pi^4}{360}+3\zeta(3)\ln{2}-\frac{3}{2}\zeta(3)-\frac{\pi^2}{6}-4\ln{2}+8 \end{align}

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Step 4: Obtaining the final result

Summing the results from steps 3a, 3b, 3c, 3d and 3e gives $$\int^1_0\ln(1+x)\ln(1-x)\ln^2{x} \ {\rm d}x=24-\frac{4\pi^2}3-\frac{11\pi^4}{720}-12\ln2\\+2\ln^22-\frac16\ln^42+\pi ^2\ln2+\frac{\pi^2}6\ln^22-4\operatorname{Li}_4\!\left(\tfrac12\right)-\frac{35}4\zeta(3)+\frac72\zeta(3)\ln2.$$ hence completing the proof.

Answer 2

using an identity proved by Cornel Ioan Valean and it can be found in his book " Almost impossible integrals, sums, and series": $$\ln(1-x)\ln(1+x)=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)x^{2n} $$

multiply both sides by $\ln^2x$ then integrate \begin{align} I&=\int_0^1\ln(1-x)\ln(1+x)\ln^2\ dx=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\int_0^1x^{2n}\ln^2x\ dx\\ &=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\left(\frac2{(2n+1)^3}\right)\\ &=-2\sum_{n=1}^\infty\frac{H_{2n}}{n(2n+1)^3}+2\sum_{n=1}^\infty\frac{H_n}{n(2n+1)^3}-\sum_{n=1}^\infty\frac{1}{n^2(2n+1)^3}\\ &=-2\sum_{n=1}^\infty\frac{H_n}{n(n+1)^3}-2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n(n+1)^3}+2\sum_{n=1}^\infty\frac{H_n}{n(2n+1)^3}-\sum_{n=1}^\infty\frac{1}{n^2(2n+1)^3}\\ &=-2S_1-2S_2+2S_3-S_4 \end{align} The first sum: \begin{align} S_1=\sum_{n=1}^\infty\frac{H_n}{n(n+1)^3}&=\sum_{n=1}^\infty\frac{H_n}{n(n+1)}-\sum_{n=1}^\infty\frac{H_n}{(n+1)^2}-\sum_{n=1}^\infty\frac{H_n}{(n+1)^3}\\ &=\zeta(2)-\zeta(3)-\frac14\zeta(4) \end{align} The second sum: \begin{align} S_2&=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n(n+1)^3}=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n(n+1)}-\sum_{n=1}^\infty\frac{(-1)^nH_n}{(n+1)^2}-\sum_{n=1}^\infty\frac{(-1)^nH_n}{(n+1)^3}\\ &=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n(n+1)}-\operatorname{Li}_3(-1)+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}-\operatorname{Li}_4(-1)+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}\\ &=\left(\ln^22-\frac12\zeta(2)\right)+\frac34\zeta(3)+\frac78\zeta(4)+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3} \end{align} using the generating function$$\sum_{n=1}^\infty\frac{x^nH_n}{n^2}=\frac12\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\zeta(3)$$ we get$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}=-\frac58\zeta(3)$$ I was able here to prove$$\begin{align} \sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42 \end{align}$$ therefore$$S_2=2\operatorname{Li_4}\left(\frac12\right)-\frac{15}8\zeta(4)+\frac18\zeta(3)-\frac12\zeta(2)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42+\ln^22$$

The third sum: using the following identity $$\sum_{n=1}^\infty\frac{H_n}{n(n+k)}=\frac{H_k^2}{2k}+\frac{H_k^{(2)}}{2k}-\frac{H_k}{k^2}+\frac{\zeta(2)}{k}$$ differentiate both sides with respect to $k$ twice then let $k=1/2$ \begin{align} S_3=\sum_{n=1}^\infty\frac{H_n}{n(2n+1)^3}&=\frac18\sum_{n=1}^\infty\frac{H_n}{n(n+1/2)^3}\\ &=\frac72\ln2\zeta(3)+3\ln2\zeta(2)-\frac{45}{16}\zeta(4)-\frac72\zeta(3)+2\ln^22 \end{align}

The fouth sum: \begin{align} S_4=\sum_{n=1}^\infty\frac1{n^2(2n+1)^3}&=\sum_{n=1}^\infty\left(\frac1{n^2}+\frac{8}{(2n+1)^2}+\frac{4}{(2n+1)^3}-6\left(\frac1n-\frac{1}{n+1/2}\right)\right)\\ &=\zeta(2)+(6\zeta(2)-8)+\left(\frac72\zeta(3)-4\right)-6H_{1/2}\\ &=\frac72\zeta(3)+7\zeta(2)+12\ln2-24 \end{align} combining $S_1$, $S_2$, $S_3$ and $S_4$ , we have $$I=24-8\zeta(2)-\frac{35}{4}\zeta(3)-\frac{11}{8}\zeta(4)-12\ln2+2\ln^22-\frac16\ln^42+6\ln2\zeta(2)+\ln^22\zeta(2)+\frac72\ln2\zeta(3)-4\operatorname{Li}_4\left(\frac12\right)$$

Answer 3

This problem gives a lot of insight into how to tackle such questions and gives opportunity to learn a lot so we will post an answer. We start from the well known expansion: \begin{equation} \log(1-x) = -\sum\limits_{k=1}^\infty \frac{x^k}{k} \end{equation} By replacing $x$ by $-x$ and multiplying the results we get: \begin{eqnarray} \log(1-x) \log(1+x) &=& \sum\limits_{k=1}^\infty \frac{x^k}{k} \cdot \left(1+(-1)^k \right) \cdot \tilde{H}_{k-1} \\ &=& \sum\limits_{k=1}^\infty \frac{x^{2 k}}{2 k} \cdot 2 \cdot \tilde{H}_{2k-1} \end{eqnarray} where $\tilde{H}_n := \sum\limits_{k=1}^n (-1)^k/k$ are "alternating sign" harmonic numbers.

Now, we bear in mind that multiplying a power of $x$ with a power of $\log(x)$ and integrating the result from zero to one gives -- by reduction to the Gamma function -- a neat closed form result. In fact we have: \begin{equation} I:=\int\limits_0^1 \log(1-x) \log(1+x) \log(x)^2 dx = \sum\limits_{k=1}^\infty \frac{2!}{k (2 k+1)^3} \tilde{H}_{2k-1} \end{equation} In order to proceed we need to find some expression for our alternating sum harmonic numbers. This is easy: \begin{equation} \tilde{H}_{2k-1} = \sum\limits_{k_1=1}^\infty (-1)^{k_1} \left(\frac{1}{k_1} + \frac{1}{k_1+2k-1} \right) = -\log(2) + \sum\limits_{k_1=1}^\infty \frac{(-1)^{k_1}}{k_1+2k-1} \end{equation} Inserting this to the equation one line above we have: \begin{eqnarray} I = -\log(2) \cdot \left( 12 - \frac{\pi^2}{2} -2 \log(4) - \frac{7}{2} \zeta(3)\right) + \sum\limits_{k=1}^\infty \frac{2!}{k(2 k+1)^3} \sum\limits_{k_1=1}^\infty \frac{(-1)^{k_1}}{k_1+2 k-1} \end{eqnarray} Here we refrained from giving the details of the calculation because the sum in question has been already dealt with in this website. The result follows from reducing the summand into simple fractions and using the series expansion of polylogarithms, for example. In the remaining double sum we do the usual. Firstly we substitute for the denominator in the second sum and then we swap the order of the sums. We have: \begin{eqnarray} &&I = -\log(2) \cdot \left( 12 - \frac{\pi^2}{2} -2 \log(4) - \frac{7}{2} \zeta(3)\right) + \\ &&\sum\limits_{k_1=1}^\infty \frac{(-1)^{k_1+1}}{k_1} \sum\limits_{k=1}^{\lfloor \frac{k_1}{2} \rfloor} \left( -\frac{4}{2 k+1}-\frac{4}{(2 k+1)^2}-\frac{4}{(2 k+1)^3}+\frac{2}{k}\right) \end{eqnarray} Now, there are only two double sums that remain. We will evaluate here the harder one leaving the easier one for later since the calculations are pretty much the same. We have: \begin{eqnarray} && {\mathfrak S}_p:=\sum\limits_{k_1=1}^\infty \frac{(-1)^{k_1+1}}{k_1}\sum\limits_{k=1}^{\lfloor \frac{k_1}{2} \rfloor } \frac{1}{(1+2 k)^p} \\ &&= \frac{(-1)^{p-1}}{(p-1)!} \int\limits_0^1 \sum\limits_{k_1=1}^\infty \frac{(-1)^{k_1+1}}{k_1}\left(\xi^2+\xi^4+\cdots+ \xi^{2 \lfloor \frac{k_1}{2} \rfloor} \right) \cdot \log(\xi)^{p-1} d\xi \\ &&= \frac{(-1)^{p-1}}{(p-1)!} \int\limits_0^1 \sum\limits_{k_1=1}^\infty \frac{(-1)^{k_1+1}}{k_1} \left( \frac{\xi^2 - (\xi^2)^{\lfloor \frac{k_1}{2} \rfloor+1}}{1-\xi^2} \right) \cdot \log(\xi)^{p-1} d\xi \\ &&= \frac{(-1)^{p-1}}{(p-1)!} \int\limits_0^1 \left( \frac{\xi^2 \log(2) - (\xi\, \mbox{arctanh}(\xi)+\xi^2/2 \log(1-\xi^2))}{1-\xi^2} \right) \cdot \log(\xi)^{p-1} d\xi \\ &&= \frac{(-1)^{p-1}}{(p-1)!} \int\limits_0^1 \left( \frac{\xi^2 \log(2) - (\xi/2 (1+\xi) \log(1+\xi) - \xi/2 (1-\xi) \log(1-\xi))}{1-\xi^2} \right) \cdot \log(\xi)^{p-1} d\xi \\ &&\frac{(-1)^{p-1}}{(p-1)!} \int\limits_0^1 \left( \frac{\xi^2}{1-\xi^2} \log(2) - \frac{1}{2} \frac{\xi}{1-\xi} \log(1+\xi)+\frac{1}{2} \frac{\xi}{1+\xi} \log(1-\xi) \right) \cdot \log(\xi)^{p-1} d\xi \end{eqnarray} The nice thing is that we can actually evaluate all the integrals above in terms of zeta function values (or multivariate zeta function values that mostly reduce to single to the former). We have: \begin{eqnarray} \int\limits_0^1 \frac{\xi^2}{1-\xi^2} [\log(\xi)]^{p-1} d\xi&=& \left.\frac{d^{p-1}}{d \theta^{p-1}}\int\limits_0^1 \left( -\frac{1-\xi^{2+\theta}}{1-\xi^2} + \frac{1}{1-\xi^2} \right) d \xi \right|_{\theta=0} \\ &=& -\left.\frac{d^{p-1}}{d \theta^{p-1}} \frac{1}{2} \left(H_{\frac{\theta+1}{2}} + 2 \log(2) \right) \right|_{\theta=0} \\ &=& (-1)^{p-1} (p-1)! \frac{1}{2^p} \left( \zeta(p) - H_{\frac{1}{2}}^{(p)} \right) \end{eqnarray} The remaining integrals are harder but still feasible. We have: \begin{eqnarray} &&\int\limits_0^1 \frac{\xi}{1-\xi} \log(1+\xi) \log(\xi)^{p-1} d\xi\\ &&= \left.\frac{d^{p-1}}{d \theta^{p-1}} \int\limits_0^1 \frac{\xi^{1+\theta}}{1-\xi} \log(1+\xi) d\xi \right|_{\theta=0}\\ &&= \left.\frac{d^{p-1}}{d \theta^{p-1}} \int\limits_0^1 \frac{1-\xi^{1+\theta}}{1-\xi} (-\log(1+\xi)) d\xi\right|_{\theta=0}\\ &&= \left.\frac{d^{p-1}}{d \theta^{p-1}} \sum\limits_{j=1}^\infty \frac{(-1)^j}{j} \left(H_{\theta+1+j} - H_j\right) \right|_{\theta=0}\\ &&= (-1)^{p-1}(p-1)! \sum\limits_{j=1}^\infty \frac{(-1)^j}{j} \left(\zeta(p)-H_{1+j}^{(p)}\right) \\ &&= (-1)^{p-1}(p-1)! \left( \sum\limits_{j=1}^\infty \frac{(-1)^j}{j(j+1)^p} + \sum\limits_{\infty > j_1 > j \ge 1} \frac{1}{j_1^p} \frac{(-1)^j}{j} \right) \\ &&= (-1)^{p-1}(p-1)! \left( -\log(2)- \sum\limits_{q=1}^p \left(-1+(1-2^{1-q}) \zeta(q) \right) + \zeta\left( \begin{array}{rr} p,&& 1 \\ 1, && -1 \end{array} \right) \right) \\ \end{eqnarray} The remaining integrals reads: \begin{eqnarray} &&\int\limits_0^1 \frac{\xi}{1+\xi} \log(1-\xi) [\log(\xi)]^{p-1} d\xi\\ &&=\sum\limits_{j=1}^\infty (-1)^j \tilde{H}_{j-1} \cdot \int\limits_0^1 \xi^j \cdot [\log(\xi)]^{p-1} d \xi \\ &&=\sum\limits_{j=1}^\infty (-1)^j \left(\tilde{H}_{j}-\frac{1}{j}\right) \cdot (-1)^{p-1} (p-1)! \frac{1}{(j+1)^p} \\ &&=(-1)^{p-1}(p-1)! \left( -\sum\limits_{j=1}^\infty \frac{(-1)^j}{j (j+1)^p} + \sum\limits_{\infty > j_1 > j \ge 1} \frac{1}{j_1^p} \cdot \frac{(-1)^{j}}{j} \right)\\ &&= (-1)^{p-1}(p-1)! \left( +\log(2)+ \sum\limits_{q=1}^p \left(-1+(1-2^{1-q}) \zeta(q) \right) + \zeta\left( \begin{array}{rr} p,&& 1 \\ 1, && -1 \end{array} \right) \right) \\ \end{eqnarray} Now note that in the last expression on the right hand side for the sum ${\mathfrak S}_p$ the two integrals that we have just evaluated above enter with opposite signs and therefore the nasty multivariate zeta function term cancels out and the sum in question reduces to single zeta function values only! Now the only thing that remains is to calculate the second sum : \begin{eqnarray} {\mathcal S} := \sum\limits_{k_1=1}^\infty \frac{(-1)^{k_1}}{k_1} \sum\limits_{k=1}^{\lfloor \frac{k_1}{2}\rfloor} \frac{1}{k} \end{eqnarray} I guess it is clear that that sum will be computed using similar techniques as above. We will finish this project as soon as possible.

Answer 4

I dabbled with this a little and managed to get it in terms of some series involving digamma and its derivatives.

Yet again, series involving digamma. How in the world can we evaluate these?. They do indeed have interesting closed forms in terms of polylogs, zeta, log, etc.

There has got to be a way, rather it be Euler sums or whatever.

One of them is

$$\sum_{k=1}^{\infty}\frac{1}{2k(k+1)^{2}}\left(\psi_{1}\left(k/2+3/2\right)-\psi_{1}\left(k/2+1\right)\right)$$.

there are some already known results that could be used. i.e. $$\sum_{k=1}^{\infty}\psi_{1}\left(k/2+1\right)x^{k}=\frac{\pi^{2}x(x+3)}{6(1-x^{2})}-\frac{4Li_{2}(x)}{1-x^{2}}$$.

I think I can derive the other ones, but it would be tedious, and right now I don't have the time nor the gumption to do so. If anyone would like to delve, please go ahead. I find these polygamma sums interesting and have found little on them. This is why I would like to do some research and write something up. It may have to wait a while though.

I think I managed to derive $$\sum_{n=1}^{\infty}\psi_{1}\left(\frac{n}{2}+\frac{3}{2}\right)x^{n}=\frac{x\pi^{2}}{12(x+1)}-\frac{Li_{2}(x)}{x(x^{2}-1)}-\frac{\pi^{2}x}{12(1-x)}$$

Another, and perhaps better, approach would be to write it as:

$\displaystyle 2ab=(a+b)^{2}-a^{2}-b^{2}$

$$2I=\int_{0}^{1}\log^{2}(x)\log^{2}(1-x^{2})dx-\int_{0}^{1}\log^{2}(x)\log^{2}(1+x)dx-\int_{0}^{1}\log^{2}(x)\log^{2}(1-x)dx$$, then perhaps use known euler sum generating functions.

i.e. $$\log^{2}(1-x)=2\sum_{k=1}^{\infty}\frac{H_{n}}{n+1}x^{n+1}$$

and $$\log^{2}(1+x)=2\sum_{n=1}^{\infty}\frac{(-1)^{n}H_{n}}{n}x^{n}+2Li_{2}(1+x)$$

and $$\log^{2}(1-x^{2})=2\sum_{n=1}^{\infty}\frac{H_{n}}{n+1}x^{2n+2}$$

which, for the far right integral, results in $$4\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)(n+2)^{3}}=-8\zeta(3)+24-\frac{4\pi^{2}}{3}-\zeta(4)$$.

do the others in a similar fashion, then sum them up.

Like you super, I had trouble with one sum. Other than that it all works out nicely. I know what you mean by an error. It sure would be easy in all this mess. One little negative or anything and it's wrong. Then, you have to root it out :)

$\displaystyle 2ab=(a+b)^{2}-a^{2}-b^{2}$

$$2I=\int_{0}^{1}\log^{2}(x)\log^{2}(1-x^{2})dx-\int_{0}^{1}\log^{2}(x)\log^{2}(1+x)dx-\int_{0}^{1}\log^{2}(x)\log^{2}(1-x)dx$$

Using the designated gen funcs on the appropriate log integral above:

$$\int_{0}^{1}\log^{2}(x)\log^{2}(1-x^{2})dx\tag{1}$$

$$=4\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)(2n+3)^{3}}=14\zeta(3)\log(2)-28\zeta(3)+8\log^{2}(2)+2\pi^{2}\log(2)-48\log(2)-\frac{\pi^{4}}{8}-\frac{14\pi^{2}}{3}+96$$

$$\int_{0}^{1}\log^{2}(x)\log^{2}(1-x)dx=4\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)(n+2)^{3}}=-8\zeta(3)+24-\frac{4}{3}\pi^{2}-\frac{\pi^{4}}{90}\tag{2}$$

Here is the ickiest one:

$$\int_{0}^{1}\log^{2}(x)\log^{2}(1+x)dx=4\sum_{n=1}^{\infty}\frac{(-1)^{n}H_{n}}{n(n+1)^{3}}-2\int_{0}^{1}\log^{2}(x)Li_{2}(x+1)dx\tag{3}$$

$$=7\log(2)\zeta(3)-\frac{5}{2}\zeta(3)+8Li_{4}(1/2)+1/3\log^{4}(2)-\frac{\pi^{2}}{3}\log^{2}(2)+4\log^{2}(2)-24\log(2)-\frac{\pi^{4}}{12}-\frac{2\pi^{2}}{3}+24$$

combining them all and dividing by 2, returns the result.

$\displaystyle 2I=(1)-(3)-(2)$

The above Euler sums can be done using residues. Though, (3) would probably prove the hardest to get along with.

The sum could probably be done using $\pi csc(\pi z)$ in its kernel. The dilog integral I am not sure about. I did not attempt it. I ran this one through Mathematica.

Take for example: $$4\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)(n+2)^{3}}$$

Use the kernel $\displaystyle (\gamma+\psi(-z))^{2}$ with $\displaystyle r(n)=\frac{1}{(n+1)(n+2)^{3}}$

$\displaystyle Res(0)=-5/16$

$\displaystyle Res(-1)=0$

$\displaystyle Res(-2)=-6+\frac{2\pi^{2}}{3}+2\zeta(3)-\frac{\pi^{4}}{36}$

$\displaystyle Res(n)=\frac{1}{(z-n)^{2}}+\frac{2H_{n}}{z-n}+\cdot\cdot\cdot $

$\displaystyle =-\sum_{n=1}^{\infty}\frac{4n+5}{(n+1)^{2}(n+2)^{4}}+2\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)(n+2)^{3}}+\cdot\cdot\cdot $

$\displaystyle =\frac{-91}{16}+2\zeta(3)+\frac{\pi^{4}}{30}+2H$

sum up the residues:

$$4\left(-5/16-6+\frac{2\pi^{2}}{3}+2\zeta(3)-\frac{\pi^{4}}{36}-91/16+2\zeta(3)+\frac{\pi^{4}}{30}+2H\right)=0$$

$$H=-8\zeta(3)+24-\frac{4\pi^{2}}{3}-\frac{\pi^{4}}{90}$$