evaluation of $\sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n} H_{n+1}^{(2)}}{(n+1)^{2}}$ and other Euler sums

Question

I was trying to evaluate this famous integral $$\int_{0}^{1} \frac{\ln (x) \ln^{2}(1+x) \ln(1-x)}{x} \ dx $$

Here is my attempt so solve the integral

\begin{align} &\int_{0}^{1} \frac{\ln (x) \ln^{2}(1+x) \ln(1-x)}{x} \ dx = 2 \int_{0}^{1} \frac{\ln (x) \ln(1-x)}{x} \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{n+1} x^{n+1} \ dx \\ &= 2 \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{n+1} \int_{0}^{1} x^{n} \ln(x) \ln(1-x) \ dx \\ &= 2 \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{n+1} \frac{\partial }{\partial a \partial b} B(a,b) \Big|_{(a=n+1,b=1)} \\ &= 2 \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{n+1} B(n+1,1) \Big [ \Big( \psi(n+1)-\psi(n+2) \Big) \Big(\psi(1)-\psi(n+2) \Big)-\psi'(n+2)\Big] \\ &= 2 \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{(n+1)^{2}} \Big[ (H_{n} -H_{n+1} ) (-H_{n+1}) - \frac{\pi^{2}}{6} + H_{n+1}^{(2)} \Big] \\ &= 2 \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{(n+1)^{2}} \Big[\Big(-\frac{1}{n+1} \Big) (-H_{n+1}) - \frac{\pi^{2}}{6} + H_{n+1}^{(2)} \Big] \\ &= 2 \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n} H_{n+1}}{(n+1)^{3}} - \frac{\pi^{2}}{3} \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{(n+1)^{2}} + 2 \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n} H_{n+1}^{(2)}}{(n+1)^{2}} \\ &= 2 \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(H_{n})^{2}}{(n+1)^{3}} + 2 \sum_{n=1}^{\infty} (-1)^{k-1} \frac{H_{n}}{(n+1)^{4}} - \frac{\pi^{2}}{3} \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{(n+1)^{2}} + 2 \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n} H_{n+1}^{(2)}}{(n+1)^{2}} \end{align}

Evaluating the first Euler sum is probably quite difficult, and the fourth sum seems crazy. Maybe I made a mistake.

Note that: this is not a duplicate of the question because I want to know the evaluation of the last 4 Euler sums, not the main integral. Thank for reading. edit; wrong link

Answer 1

Sorry, no Euler sums here.

\begin{align}J&=\int_0^1\frac{\ln^2(1+x)\ln(1-x)\ln x}{x}dx\\ &=\frac{1}{6}\int_0^1\frac{\left(\ln^3\left(\frac{1-x}{1+x}\right)+\ln^3(1-x^2)-2\ln^3(1-x)\right)\ln x}{x}dx\\ &=\frac{1}{6}\underbrace{\int_0^1\frac{\ln^3\left(\frac{1-x}{1+x}\right)\ln x}{x}dx}_{=A}+\frac{1}{6}\underbrace{\int_0^1\frac{\ln^3(1-x^2)\ln x}{x}dx}_{u=x^2}-\frac{1}{3}\underbrace{\int_0^1\frac{\ln^3(1-x)\ln x}{x}dx}_{=B}\\ &=\frac{A}{6}+\frac{B}{24}-\frac{B}{3}=\frac{A}{6}-\frac{7B}{24}\\ A&\overset{u=\frac{1-x}{1+x}}=2\int_0^1\frac{\ln\left(\frac{1-u}{1+u}\right)\ln^3u}{1-u^2}du\\ &=-\underbrace{\int_0^1\frac{\ln(1+u)\ln^3u}{1-u}du}_{=J_1}-\underbrace{\int_0^1\frac{\ln(1+u)\ln^3u}{1+u}\!\!du}_{=J_2}+\underbrace{\int_0^1\frac{\ln(1-u)\ln^3u}{1+u}\!\!du}_{=J_3}+\underbrace{\int_0^1\frac{\ln(1-u)\ln^3u}{1-u}\!\!du}_{=J_4} \end{align} \begin{align}J_1&\overset{\text{IBP}}=\left[\left(\int_0^u\frac{\ln^3t}{1-t}\!\!dt\right)\ln(1+u)\right]_0^1-\int_0^1\int_0^1\frac{u\ln^3(tu)}{(1-tu)(1+u)}dtdu\\ &=-6\zeta(4)\ln 2+\int_0^1\int_0^1\left(\frac{\ln^3(tu)}{(1+t)(1+u)}-\frac{\ln^3(tu)}{(1+t)(1-tu)}\right)dtdu\\ &\overset{\text{Fubini}}=-6\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)-\frac{21}{2}\zeta(4)\ln 2-\int_0^1\frac{1}{t(1+t)}\left(\int_0^t\frac{\ln^3u}{1-u}du \right)dt\\ &\overset{\text{IBP}}=-\frac{9}{2}\zeta(2)\zeta(3)-\frac{33}{2}\zeta(4)\ln 2+\int_0^1\frac{\ln^4t}{1-t}dt+\ln 2\int_0^1\frac{\ln^3u}{1-u}du-J_1\\ &=\boxed{-\frac{9}{4}\zeta(2)\zeta(3)-\frac{45}{4}\zeta(4)\ln 2+12\zeta(5)}\\ \end{align} \begin{align}J_2&\overset{\text{IBP}}=\left[\left(\int_0^u\frac{\ln^3t}{1+t}dt\right)\ln(1+u)\right]_0^1-\int_0^1\int_0^1\frac{u\ln^3(tu)}{(1+tu)(1+u)}dtdu\\ &=-\frac{21}{4}\zeta(4)\ln 2+\int_0^1\int_0^1\left(\frac{\ln^3(tu)}{(1-t)(1+u)}-\frac{\ln^3(tu)}{(1-t)(1+tu)}\right)dtdu\\ &\overset{\text{Fubini}}=-\frac{21}{4}\zeta(4)\ln 2-\frac{15}{2}\zeta(2)\zeta(3)-6\zeta(4)\ln 2+\int_0^1\int_0^1\left(\frac{\ln^3u}{(1-t)(1+u)}-\frac{\ln^3(tu)}{(1-t)(1+tu)}\right)dtdu\\ &=-\frac{45}{4}\zeta(4)\ln 2-\frac{15}{2}\zeta(2)\zeta(3)+\int_0^1\left(\frac{1}{1-t}\left(\int_0^1\frac{\ln^3u}{1+u}du\right)-\frac{1}{t(1-t)}\left(\int_0^t\frac{\ln^3u}{1+u}du\right)\right)\\ &=-\frac{45}{4}\zeta(4)\ln 2-\frac{15}{2}\zeta(2)\zeta(3)+\int_0^1\left(\frac{1}{1-t}\left(\int_t^1\frac{\ln^3u}{1+u}du\right)-\frac{1}{t}\left(\int_0^t\frac{\ln^3u}{1+u}du\right)\right)\\ &\overset{\text{IBP}}=-\frac{45}{4}\zeta(4)\ln 2-\frac{15}{2}\zeta(2)\zeta(3)+J_3+\int_0^1\frac{\ln^4t}{1+t}dt=\boxed{-J3-\frac{45}{4}\zeta(4)\ln 2-\frac{15}{2}\zeta(2)\zeta(3)+\frac{45}{2}\zeta(5)}\\ \end{align} \begin{align}J_4&\overset{\text{IBP}}=\left[\left(\int_0^u\frac{\ln^3t}{1-t}dt-\int_0^1\frac{\ln^3t}{1-t}dt\right)\ln(1-u)\right]_0^1+\\&\int_0^1\int_0^1\left(\frac{u\ln^3(tu)}{(1-tu)(1-u)}-\frac{\ln^3(t)}{(1-t)(1-u)}\right)dtdu\\ &=\int_0^1\int_0^1\left(\frac{\ln^3(tu)}{(1-t)(1-u)}-\frac{\ln^3t}{(1-t)(1-tu)}-\frac{\ln^3t}{(1-t)(1-u)}\right)dtdu\\ &\overset{\text{Fubini}}=-12\zeta(2)\zeta(3)+\int_0^1\int_0^1\left(\frac{\ln^3 u}{(1-t)(1-u)}-\frac{\ln^3(tu)}{(1-t)(1-tu)}\right)dtdu\\ &=-12\zeta(2)\zeta(3)+\int_0^1\int_0^1\left(\frac{1}{1-t}\left(\int_t^1\frac{\ln^3u}{1-u}du\right)-\frac{1}{t}\left(\int_0^t\frac{\ln^3u}{1-u}du\right)\right)\\ &\overset{\text{IBP}}=-12\zeta(2)\zeta(3)-J_4+24\zeta(5)\\ &=\boxed{12\zeta(5)-6\zeta(2)\zeta(3)}\\ \end{align}

\begin{align}J_2&=\int_0^1\frac{\ln^4(1+u)}{4(1+u)}du-\underbrace{\int_0^1\frac{\ln^3(1+u)\ln u}{1+u}dx}_{\text{IBP}}+\underbrace{\int_0^1\frac{3\ln^2(1+u)\ln^2 u}{2(1+u)}dx}_{\text{IBP}}+\int_0^1\frac{\ln^4 u}{4(1+u)}dx-\\&\underbrace{\int_0^1\frac{\ln^4\left(\frac{u}{1+u}\right)}{1+u}dx}_{z=\frac{u}{1+u}}\\ &=\frac{\ln^52}{20}+\frac{1}{4}\underbrace{\int_0^1\frac{\ln^4(1+x)}{x}dx}_{z=\frac{1}{1+x}}-\underbrace{\int_0^1\frac{\ln^3(1+x)\ln x}{x}dx}_{=C}+\frac{45}{8}\zeta(5)-\frac{1}{4}\int_0^{\frac{1}{2}}\frac{\ln^4z}{1-z}dz\\ &=\frac{\ln^52}{20}+\frac{1}{4}\int_{\frac{1}{2}}^1\frac{\ln^4z}{z}dz+\frac{1}{4}\int_{\frac{1}{2}}^1\frac{\ln^4z}{1-z}dz-C+\frac{45}{8}\zeta(5)-\frac{1}{4}\int_0^{\frac{1}{2}}\frac{\ln^4z}{1-z}dz\\ &=\frac{\ln^52}{10}+\frac{93}{8}\zeta(5)-\frac{1}{2}\int_0^{\frac{1}{2}}\frac{\ln^4z}{1-z}dz-C\\ C&\overset{u=\frac{1}{1+x}}=\frac{\ln^52}{5}+\int_{\frac{1}{2}}^1\frac{\ln^4u}{1-u}du-\underbrace{\int_{\frac{1}{2}}^1\frac{\ln(1-u)\ln^3 u}{u}du}_{\text{IBP}}-\int_{\frac{1}{2}}^1\frac{\ln(1-u)\ln^3u}{1-u}du\\ &=-\frac{\ln^52}{20}+\frac{3}{4}\int_{\frac{1}{2}}^1\frac{\ln^4u}{1-u}du-\int_{\frac{1}{2}}^1\frac{\ln(1-u)\ln^3u}{1-u}du\\ &=18\zeta(5)-\frac{\ln^52}{20}-\frac{3}{4}\int_0^{\frac{1}{2}}\frac{\ln^4u}{1-u}du-J_4+\underbrace{\int_0^{\frac{1}{2}}\frac{\ln(1-u)\ln^3u}{1-u}du}_{=D}\\ D&=\int_0^{\frac{1}{2}}\frac{\ln^4u}{4(1-u)}du+\underbrace{\int_0^{\frac{1}{2}}\frac{3\ln^2(1-u)\ln^2u}{2(1-u)}du}_{\text{IBP}}-\underbrace{\int_0^{\frac{1}{2}}\frac{\ln^3(1-u)\ln u}{1-u}du}_{\text{IBP}}+\int_0^{\frac{1}{2}}\frac{\ln^4(1-u)}{4(1-u)}du-\\&\underbrace{\int_0^{\frac{1}{2}}\frac{\ln^4\left(\frac{u}{1-u}\right)}{4(1-u)}du}_{z=\frac{u}{1-u}}\\ &=\frac{1}{4}\int_0^{\frac{1}{2}}\frac{\ln^4u}{1-u}du+\frac{\ln^52}{2}+\underbrace{\int_0^{\frac{1}{2}}\frac{\ln^3(1-u)\ln u}{u}du}_{z=1-u}-\frac{\ln^52}{4}-\frac{1}{4}\underbrace{\int_0^{\frac{1}{2}}\frac{\ln^4(1-u)}{u}du}_{z=1-u}+\frac{\ln^52}{20}-\\&\frac{45}{8}\zeta(5)\\ &=\frac{1}{2}\int_0^{\frac{1}{2}}\frac{\ln^4u}{1-u}du+\frac{3\ln^52}{10}-\frac{93}{8}\zeta(5)+J_4-D\\ &=\frac{1}{4}\int_0^{\frac{1}{2}}\frac{\ln^4u}{1-u}du+\frac{3\ln^52}{20}-\frac{93}{16}\zeta(5)+\frac{J_4}{2}\\ C&=\frac{195}{16}\zeta(5)+\frac{\ln^52}{10}-\frac{1}{2}\int_0^{\frac{1}{2}}\frac{\ln^4u}{1-u}du-\frac{J_4}{2}\\ J_2&=-\frac{9}{16}\zeta(5)+\frac{J_4}{2}=\boxed{\frac{87}{16}\zeta(5)-3\zeta(2)\zeta(3)}\\ J_3&=\frac{369}{16}\zeta(5)-\frac{45}{4}\zeta(4)\ln 2-\frac{J_4}{2}-\frac{15}{2}\zeta(2)\zeta(3)=\boxed{\frac{273}{16}\zeta(5)-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)}\\ A&=\boxed{\frac{93}{8}\zeta(5)-\frac{21}{4}\zeta(2)\zeta(3)}\\ B&\overset{u=1-x}=J_4=\boxed{12\zeta(5-6\zeta(2)\zeta(3)}\\ J&=\frac{A}{6}-\frac{7B}{24}=\boxed{-\frac{25}{16}\zeta(5)+\frac{7}{8}\zeta(2)\zeta(3)} \end{align}