Let us now write down the recurrence relations for ${\bf H}^{(2)}_n(t)$. As usual we start from the integral representation. We have: \begin{eqnarray} &&{\bf H}^{(2)}_n(t) =\\ && \int\limits_0^t \frac{[\log(t/\xi)]^{n-1}}{(n-1)!} \cdot \underbrace{\frac{Li_2(\xi)}{1-\xi}}_{[Li_1(\xi) Li_2(\xi) - \int \frac{Li_1(\xi)^2}{\xi} d\xi]^{'}} d \xi +Li_{n+2}(t)=\\ && \frac{1_{n\ge 3}}{2 (n-3)!} \int\limits_0^t [\log(t/\xi)]^{n-3} \cdot \frac{Li_2(\xi)^2}{\xi} d\xi - \frac{1_{n\ge 2}}{(n-1)!} \int\limits_0^t [\log(t/\xi)]^{n-1} \cdot \frac{Li_1(\xi)^2}{\xi} d\xi + Li_{n+2}(t) +\\ && \left[Li_1(t) Li_2(t) - \int\limits_0^t \frac{Li_1(\xi)^2}{\xi} d \xi\right] 1_{n=1} + \frac{1}{2} Li_2(t)^2 1_{n=2} \end{eqnarray} In the bottom line we integrated by parts and expressed the result through integrals that contain squares of a poly-log only rather than products of two different poly-logs. Now we set $t=-1$. We can immediately say that from the two integrals on the right hand side the middle one has already been computed in my previous answer to this question. The first integral is harder but it has been calculated in here Generalized definite dilogarithm integral. . Then all we need to do is to write down the results , add then up and simplify which, despite appearances, is a tedious and mundane task. We have: \begin{eqnarray} &&\frac{1}{(n-1)!} \int\limits_0^1 [\log(1/\xi)]^{n-1} \cdot \frac{\log(1+\xi)^2}{\xi} d\xi = \\ &&\left(\frac{1}{2^n} - 1\right)(n+1) \zeta(n+2) + \sum\limits_{j=1}^{n-1} \left(-1-\frac{1}{2^n} + \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n+1-j)\\ && - 2 \sum\limits_{l=0}^{n-1} {\bf H}^{(l+1)}_{n+1-l} (-1) \end{eqnarray} Likewise: \begin{eqnarray} &&\frac{1}{2(n-3)!} \int\limits_0^1 [\log(1/\xi)]^{n-3} \cdot \frac{Li_2(-\xi)^2}{\xi} d\xi =\\ && \frac{1}{4} \left(1-\frac{1}{2^{n+1}}\right)(n-1)_{(2)} \zeta(n+2) + (n-2)\left(-\frac{1}{2} + \frac{1}{2^{n+1}}\right) \log(2) \zeta(n+1) + \\ &&\frac{1}{4} \left(1-\frac{1}{2^{n-1}}\right)\zeta(2) \zeta(n)+\frac{1}{2} \sum\limits_{l=3}^n (l-2) {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} Now since \begin{equation} Li_{n+2}(-1) = \left(-1+\frac{1}{2^{n+1}}\right) \zeta(n+2) \end{equation} for $n=0,1,\cdots$ this concludes the calculation. All we need to do is to collect all the different terms. Bringing everything together we get: \begin{eqnarray} &&{\bf H}^{(2)}_n(-1)=\\ &&\frac{1}{3\cdot 2^{n+3}} \left(\right.\\ &&3 \left(\left(2^{n+1}-1\right) n^2+\left(2^{n+1}-5\right) n+2^{n+2}-6\right)\zeta (n+2)\\ &&-4 \left(2^n-1\right) (n-2) \log (8) \zeta (n+1)\\ &&+\pi ^2 \left(2^n-2\right) \zeta (n)\\ &&\left.\right) +\\ &&\sum\limits_{j=1}^{n-1} (1+\frac{1}{2^n}-\frac{1}{2^{n-j-1}}) \zeta(1+j) \zeta(n+1-j) +\\ &&\sum\limits_{l=0}^n \left(2 \cdot 1_{0\le l \le 2} + (\frac{l}{2}+1)\cdot 1_{3 \le l \le n-1} + (\frac{l}{2}-1) 1_{l=n}\right)\cdot {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} for $n\ge 3$. In the case $n=1,2$ surface terms need to be taken into account. To be specific we have: \begin{eqnarray} {\bf H}^{(2)}_1(-1) &=& \frac{1}{12} \pi ^2 \log (2)-\zeta (3)\\ {\bf H}^{(2)}_2(-1) &=& -4 \text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{2} \zeta (3) \log (2)+\frac{17 \pi ^4}{480}-\frac{\log ^4(2)}{6}+\frac{1}{6} \pi ^2 \log ^2(2) \end{eqnarray}
Let us now write down the recurrence relations for ${\bf H}^{(2)}_n(t)$. As usual we start from the integral representation. We have: \begin{eqnarray} &&{\bf H}^{(2)}_n(t) =\\ && \int\limits_0^t \frac{[\log(t/\xi)]^{n-1}}{(n-1)!} \cdot \underbrace{\frac{Li_2(\xi)}{1-\xi}}_{[Li_1(\xi) Li_2(\xi) - \int \frac{Li_1(\xi)^2}{\xi} d\xi]^{'}} d \xi +Li_{n+2}(t)=\\ && \frac{1_{n\ge 3}}{2 (n-3)!} \int\limits_0^t [\log(t/\xi)]^{n-3} \cdot \frac{Li_2(\xi)^2}{\xi} d\xi - \frac{1_{n\ge 2}}{(n-1)!} \int\limits_0^t [\log(t/\xi)]^{n-1} \cdot \frac{Li_1(\xi)^2}{\xi} d\xi + Li_{n+2}(t) +\\ && \left[Li_1(t) Li_2(t) - \int\limits_0^t \frac{Li_1(\xi)^2}{\xi} d \xi\right] 1_{n=1} + \frac{1}{2} Li_2(t)^2 1_{n=2} \end{eqnarray} In the bottom line we integrated by parts and expressed the result through integrals that contain squares of a poly-log only rather than products of two different poly-logs. Now we set $t=-1$. We can immediately say that from the two integrals on the right hand side the middle one has already been computed in my previous answer to this question. The first integral is harder but it has been calculated in here Generalized definite dilogarithm integral. . Then all we need to do is to write down the results , add then up and simplify which, despite appearances, is a tedious and mundane task. We have: \begin{eqnarray} &&\frac{1}{(n-1)!} \int\limits_0^1 [\log(1/\xi)]^{n-1} \cdot \frac{\log(1+\xi)^2}{\xi} d\xi = \\ &&\left(\frac{1}{2^n} - 1\right)(n+1) \zeta(n+2) + \sum\limits_{j=1}^{n-1} \left(-1-\frac{1}{2^n} + \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n+1-j)\\ && - 2 \sum\limits_{l=0}^{n-1} {\bf H}^{(l+1)}_{n+1-l} (-1) \end{eqnarray} Likewise: \begin{eqnarray} &&\frac{1}{2(n-3)!} \int\limits_0^1 [\log(1/\xi)]^{n-3} \cdot \frac{Li_2(-\xi)^2}{\xi} d\xi =\\ && \frac{1}{4} \left(1-\frac{1}{2^{n+1}}\right)(n-1)_{(2)} \zeta(n+2) + (n-2)\left(-\frac{1}{2} + \frac{1}{2^{n+1}}\right) \log(2) \zeta(n+1) + \\ &&\frac{1}{4} \left(1-\frac{1}{2^{n-1}}\right)\zeta(2) \zeta(n)+\frac{1}{2} \sum\limits_{l=3}^n (l-2) {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} Now since \begin{equation} Li_{n+2}(-1) = \left(-1+\frac{1}{2^{n+1}}\right) \zeta(n+2) \end{equation} for $n=0,1,\cdots$ this concludes the calculation. All we need to do is to collect all the different terms. Bringing everything together we get: \begin{eqnarray} &&{\bf H}^{(2)}_n(-1)=\\ &&\frac{1}{3\cdot 2^{n+3}} \left(\right.\\ &&3 \left(\left(2^{n+1}-1\right) n^2+\left(2^{n+1}-5\right) n+2^{n+2}-6\right)\zeta (n+2)\\ &&-4 \left(2^n-1\right) (n-2) \log (8) \zeta (n+1)\\ &&+\pi ^2 \left(2^n-2\right) \zeta (n)\\ &&\left.\right) +\\ &&\sum\limits_{j=1}^{n-1} (1+\frac{1}{2^n}-\frac{1}{2^{n-j-1}}) \zeta(1+j) \zeta(n+1-j) +\\ &&\sum\limits_{l=0}^n \left(2 \cdot 1_{0\le l \le 2} + (\frac{l}{2}+1)\cdot 1_{3 \le l \le n-1} + (\frac{l}{2}-1) 1_{l=n}\right)\cdot {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} for $n\ge 3$. In the case $n=1,2$ surface terms need to be taken into account.
Let us now write down the recurrence relations for ${\bf H}^{(2)}_n(t)$. As usual we start from the integral representation. We have: \begin{eqnarray} &&{\bf H}^{(2)}_n(t) =\\ && \int\limits_0^t \frac{[\log(t/\xi)]^{n-1}}{(n-1)!} \cdot \underbrace{\frac{Li_2(\xi)}{1-\xi}}_{[Li_1(\xi) Li_2(\xi) - \int \frac{Li_1(\xi)^2}{\xi} d\xi]^{'}} d \xi +Li_{n+2}(t)=\\ && \frac{1_{n\ge 3}}{2 (n-3)!} \int\limits_0^t [\log(t/\xi)]^{n-3} \cdot \frac{Li_2(\xi)^2}{\xi} d\xi - \frac{1_{n\ge 2}}{(n-1)!} \int\limits_0^t [\log(t/\xi)]^{n-1} \cdot \frac{Li_1(\xi)^2}{\xi} d\xi + Li_{n+2}(t) +\\ && \left[Li_1(t) Li_2(t) - \int\limits_0^t \frac{Li_1(\xi)^2}{\xi} d \xi\right] 1_{n=1} + \frac{1}{2} Li_2(t)^2 1_{n=2} \end{eqnarray} In the bottom line we integrated by parts and expressed the result through integrals that contain squares of a poly-log only rather than products of two different poly-logs. Now we set $t=-1$. We can immediately say that from the two integrals on the right hand side the middle one has already been computed in my previous answer to this question. The first integral is harder but it has been calculated in here Generalized definite dilogarithm integral. . Then all we need to do is to write down the results , add then up and simplify which, despite appearances, is a tedious and mundane task. We have: \begin{eqnarray} &&\frac{1}{(n-1)!} \int\limits_0^1 [\log(1/\xi)]^{n-1} \cdot \frac{\log(1+\xi)^2}{\xi} d\xi = \\ &&\left(\frac{1}{2^n} - 1\right)(n+1) \zeta(n+2) + \sum\limits_{j=1}^{n-1} \left(-1-\frac{1}{2^n} + \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n+1-j)\\ && - 2 \sum\limits_{l=0}^{n-1} {\bf H}^{(l+1)}_{n+1-l} (-1) \end{eqnarray} Likewise: \begin{eqnarray} &&\frac{1}{2(n-3)!} \int\limits_0^1 [\log(1/\xi)]^{n-3} \cdot \frac{Li_2(-\xi)^2}{\xi} d\xi =\\ && \frac{1}{4} \left(1-\frac{1}{2^{n+1}}\right)(n-1)_{(2)} \zeta(n+2) + (n-2)\left(-\frac{1}{2} + \frac{1}{2^{n+1}}\right) \log(2) \zeta(n+1) + \\ &&\frac{1}{4} \left(1-\frac{1}{2^{n-1}}\right)\zeta(2) \zeta(n)+\frac{1}{2} \sum\limits_{l=3}^n (l-2) {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} Now since \begin{equation} Li_{n+2}(-1) = \left(-1+\frac{1}{2^{n+1}}\right) \zeta(n+2) \end{equation} for $n=0,1,\cdots$ this concludes the calculation. All we need to do is to collect all the different terms. Bringing everything together we get: \begin{eqnarray} &&{\bf H}^{(2)}_n(-1)=\\ &&\frac{1}{3\cdot 2^{n+3}} \left(\right.\\ &&3 \left(\left(2^{n+1}-1\right) n^2+\left(2^{n+1}-5\right) n+2^{n+2}-6\right)\zeta (n+2)\\ &&-4 \left(2^n-1\right) (n-2) \log (8) \zeta (n+1)\\ &&+\pi ^2 \left(2^n-2\right) \zeta (n)\\ &&\left.\right) +\\ &&\sum\limits_{j=1}^{n-1} (1+\frac{1}{2^n}-\frac{1}{2^{n-j-1}}) \zeta(1+j) \zeta(n+1-j) +\\ &&\sum\limits_{l=0}^n \left(2 \cdot 1_{0\le l \le 2} + (\frac{l}{2}+1)\cdot 1_{3 \le l \le n-1} + (\frac{l}{2}-1) 1_{l=n}\right)\cdot {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} for $n\ge 3$. In the case $n=1,2$ surface terms need to be taken into account. To be specific we have: \begin{eqnarray} {\bf H}^{(2)}_1(-1) &=& \frac{1}{12} \pi ^2 \log (2)-\zeta (3)\\ {\bf H}^{(2)}_2(-1) &=& -4 \text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{2} \zeta (3) \log (2)+\frac{17 \pi ^4}{480}-\frac{\log ^4(2)}{6}+\frac{1}{6} \pi ^2 \log ^2(2) \end{eqnarray}
Let us now write down the recurrence relations for ${\bf H}^{(2)}_n(t)$. As usual we start from the integral representation. We have: \begin{eqnarray} &&{\bf H}^{(2)}_n(t) =\\ && \int\limits_0^t \frac{[\log(t/\xi)]^{n-1}}{(n-1)!} \cdot \underbrace{\frac{Li_2(\xi)}{1-\xi}}_{[Li_1(\xi) Li_2(\xi) - \int \frac{Li_1(\xi)^2}{\xi} d\xi]^{'}} d \xi +Li_{n+2}(t)=\\ && \frac{1}{2 (n-3)!} \int\limits_0^t [\log(t/\xi)]^{n-3} \cdot \frac{Li_2(\xi)^2}{\xi} d\xi - \frac{1}{(n-1)!} \int\limits_0^t [\log(t/\xi)]^{n-1} \cdot \frac{Li_1(\xi)^2}{\xi} d\xi + Li_{n+2}(t) \end{eqnarray}\begin{eqnarray} &&{\bf H}^{(2)}_n(t) =\\ && \int\limits_0^t \frac{[\log(t/\xi)]^{n-1}}{(n-1)!} \cdot \underbrace{\frac{Li_2(\xi)}{1-\xi}}_{[Li_1(\xi) Li_2(\xi) - \int \frac{Li_1(\xi)^2}{\xi} d\xi]^{'}} d \xi +Li_{n+2}(t)=\\ && \frac{1_{n\ge 3}}{2 (n-3)!} \int\limits_0^t [\log(t/\xi)]^{n-3} \cdot \frac{Li_2(\xi)^2}{\xi} d\xi - \frac{1_{n\ge 2}}{(n-1)!} \int\limits_0^t [\log(t/\xi)]^{n-1} \cdot \frac{Li_1(\xi)^2}{\xi} d\xi + Li_{n+2}(t) +\\ && \left[Li_1(t) Li_2(t) - \int\limits_0^t \frac{Li_1(\xi)^2}{\xi} d \xi\right] 1_{n=1} + \frac{1}{2} Li_2(t)^2 1_{n=2} \end{eqnarray} In the bottom line we integrated by parts and expressed the result through integrals that contain squares of a poly-log only rather than products of two different poly-logs. Now we set $t=-1$. We can immediately say that from the two integrals on the right hand side the middle one has already been computed in my previous answer to this question. The first integral is harder but it has been calculated in here Generalized definite dilogarithm integral. . Then all we need to do is to write down the results , add then up and simplify which, despite appearances, is a tedious and mundane task. We have: \begin{eqnarray} &&\frac{1}{(n-1)!} \int\limits_0^1 [\log(1/\xi)]^{n-1} \cdot \frac{\log(1+\xi)^2}{\xi} d\xi = \\ &&\left(\frac{1}{2^n} - 1\right)(n+1) \zeta(n+2) + \sum\limits_{j=1}^{n-1} \left(-1-\frac{1}{2^n} + \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n+1-j)\\ && - 2 \sum\limits_{l=0}^{n-1} {\bf H}^{(l+1)}_{n+1-l} (-1) \end{eqnarray} Likewise: \begin{eqnarray} &&\frac{1}{2(n-3)!} \int\limits_0^1 [\log(1/\xi)]^{n-3} \cdot \frac{Li_2(-\xi)^2}{\xi} d\xi =\\ && \frac{1}{4} \left(1-\frac{1}{2^{n+1}}\right)(n-1)_{(2)} \zeta(n+2) + (n-2)\left(-\frac{1}{2} + \frac{1}{2^{n+1}}\right) \log(2) \zeta(n+1) + \\ &&\frac{1}{4} \left(1-\frac{1}{2^{n-1}}\right)\zeta(2) \zeta(n)+\frac{1}{2} \sum\limits_{l=3}^n (l-2) {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} Now since \begin{equation} Li_{n+2}(-1) = \left(-1+\frac{1}{2^{n+1}}\right) \zeta(n+2) \end{equation} for $n=0,1,\cdots$ this concludes the calculation. All we need to do is to collect all the different terms. Bringing everything together we get: \begin{eqnarray} &&{\bf H}^{(2)}_n(-1)=\\ &&\frac{1}{3\cdot 2^{n+3}} \left(\right.\\ &&3 \left(\left(2^{n+1}-1\right) n^2+\left(2^{n+1}-5\right) n+2^{n+2}-6\right)\zeta (n+2)\\ &&-4 \left(2^n-1\right) (n-2) \log (8) \zeta (n+1)\\ &&+\pi ^2 \left(2^n-2\right) \zeta (n)\\ &&\left.\right) +\\ &&\sum\limits_{j=1}^{n-1} (1+\frac{1}{2^n}-\frac{1}{2^{n-j-1}}) \zeta(1+j) \zeta(n+1-j) +\\ &&\sum\limits_{l=0}^n \left(2 \cdot 1_{0\le l \le 2} + (\frac{l}{2}+1)\cdot 1_{3 \le l \le n-1} + (\frac{l}{2}-1) 1_{l=n}\right)\cdot {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} for $n\ge 3$. In the case $n=1,2$ surface terms need to be taken into account.
Let us now write down the recurrence relations for ${\bf H}^{(2)}_n(t)$. As usual we start from the integral representation. We have: \begin{eqnarray} &&{\bf H}^{(2)}_n(t) =\\ && \int\limits_0^t \frac{[\log(t/\xi)]^{n-1}}{(n-1)!} \cdot \underbrace{\frac{Li_2(\xi)}{1-\xi}}_{[Li_1(\xi) Li_2(\xi) - \int \frac{Li_1(\xi)^2}{\xi} d\xi]^{'}} d \xi +Li_{n+2}(t)=\\ && \frac{1}{2 (n-3)!} \int\limits_0^t [\log(t/\xi)]^{n-3} \cdot \frac{Li_2(\xi)^2}{\xi} d\xi - \frac{1}{(n-1)!} \int\limits_0^t [\log(t/\xi)]^{n-1} \cdot \frac{Li_1(\xi)^2}{\xi} d\xi + Li_{n+2}(t) \end{eqnarray} In the bottom line we integrated by parts and expressed the result through integrals that contain squares of a poly-log only rather than products of two different poly-logs. Now we set $t=-1$. We can immediately say that from the two integrals on the right hand side the middle one has already been computed in my previous answer to this question. The first integral is harder but it has been calculated in here Generalized definite dilogarithm integral. . Then all we need to do is to write down the results , add then up and simplify which, despite appearances, is a tedious and mundane task. We have: \begin{eqnarray} &&\frac{1}{(n-1)!} \int\limits_0^1 [\log(1/\xi)]^{n-1} \cdot \frac{\log(1+\xi)^2}{\xi} d\xi = \\ &&\left(\frac{1}{2^n} - 1\right)(n+1) \zeta(n+2) + \sum\limits_{j=1}^{n-1} \left(-1-\frac{1}{2^n} + \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n+1-j)\\ && - 2 \sum\limits_{l=0}^{n-1} {\bf H}^{(l+1)}_{n+1-l} (-1) \end{eqnarray} Likewise: \begin{eqnarray} &&\frac{1}{2(n-3)!} \int\limits_0^1 [\log(1/\xi)]^{n-3} \cdot \frac{Li_2(-\xi)^2}{\xi} d\xi =\\ && \frac{1}{4} \left(1-\frac{1}{2^{n+1}}\right)(n-1)_{(2)} \zeta(n+2) + (n-2)\left(-\frac{1}{2} + \frac{1}{2^{n+1}}\right) \log(2) \zeta(n+1) + \\ &&\frac{1}{4} \left(1-\frac{1}{2^{n-1}}\right)\zeta(2) \zeta(n)+\frac{1}{2} \sum\limits_{l=3}^n (l-2) {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} Now since \begin{equation} Li_{n+2}(-1) = \left(-1+\frac{1}{2^{n+1}}\right) \zeta(n+2) \end{equation} for $n=0,1,\cdots$ this concludes the calculation. All we need to do is to collect all the different terms. Bringing everything together we get: \begin{eqnarray} &&{\bf H}^{(2)}_n(-1)=\\ &&\frac{1}{3\cdot 2^{n+3}} \left(\right.\\ &&3 \left(\left(2^{n+1}-1\right) n^2+\left(2^{n+1}-5\right) n+2^{n+2}-6\right)\zeta (n+2)\\ &&-4 \left(2^n-1\right) (n-2) \log (8) \zeta (n+1)\\ &&+\pi ^2 \left(2^n-2\right) \zeta (n)\\ &&\left.\right) +\\ &&\sum\limits_{j=1}^{n-1} (1+\frac{1}{2^n}-\frac{1}{2^{n-j-1}}) \zeta(1+j) \zeta(n+1-j) +\\ &&\sum\limits_{l=0}^n \left(2 \cdot 1_{0\le l \le 2} + (\frac{l}{2}+1)\cdot 1_{3 \le l \le n-1} + (\frac{l}{2}-1) 1_{l=n}\right)\cdot {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} for $n\ge 3$.
Let us now write down the recurrence relations for ${\bf H}^{(2)}_n(t)$. As usual we start from the integral representation. We have: \begin{eqnarray} &&{\bf H}^{(2)}_n(t) =\\ && \int\limits_0^t \frac{[\log(t/\xi)]^{n-1}}{(n-1)!} \cdot \underbrace{\frac{Li_2(\xi)}{1-\xi}}_{[Li_1(\xi) Li_2(\xi) - \int \frac{Li_1(\xi)^2}{\xi} d\xi]^{'}} d \xi +Li_{n+2}(t)=\\ && \frac{1_{n\ge 3}}{2 (n-3)!} \int\limits_0^t [\log(t/\xi)]^{n-3} \cdot \frac{Li_2(\xi)^2}{\xi} d\xi - \frac{1_{n\ge 2}}{(n-1)!} \int\limits_0^t [\log(t/\xi)]^{n-1} \cdot \frac{Li_1(\xi)^2}{\xi} d\xi + Li_{n+2}(t) +\\ && \left[Li_1(t) Li_2(t) - \int\limits_0^t \frac{Li_1(\xi)^2}{\xi} d \xi\right] 1_{n=1} + \frac{1}{2} Li_2(t)^2 1_{n=2} \end{eqnarray} In the bottom line we integrated by parts and expressed the result through integrals that contain squares of a poly-log only rather than products of two different poly-logs. Now we set $t=-1$. We can immediately say that from the two integrals on the right hand side the middle one has already been computed in my previous answer to this question. The first integral is harder but it has been calculated in here Generalized definite dilogarithm integral. . Then all we need to do is to write down the results , add then up and simplify which, despite appearances, is a tedious and mundane task. We have: \begin{eqnarray} &&\frac{1}{(n-1)!} \int\limits_0^1 [\log(1/\xi)]^{n-1} \cdot \frac{\log(1+\xi)^2}{\xi} d\xi = \\ &&\left(\frac{1}{2^n} - 1\right)(n+1) \zeta(n+2) + \sum\limits_{j=1}^{n-1} \left(-1-\frac{1}{2^n} + \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n+1-j)\\ && - 2 \sum\limits_{l=0}^{n-1} {\bf H}^{(l+1)}_{n+1-l} (-1) \end{eqnarray} Likewise: \begin{eqnarray} &&\frac{1}{2(n-3)!} \int\limits_0^1 [\log(1/\xi)]^{n-3} \cdot \frac{Li_2(-\xi)^2}{\xi} d\xi =\\ && \frac{1}{4} \left(1-\frac{1}{2^{n+1}}\right)(n-1)_{(2)} \zeta(n+2) + (n-2)\left(-\frac{1}{2} + \frac{1}{2^{n+1}}\right) \log(2) \zeta(n+1) + \\ &&\frac{1}{4} \left(1-\frac{1}{2^{n-1}}\right)\zeta(2) \zeta(n)+\frac{1}{2} \sum\limits_{l=3}^n (l-2) {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} Now since \begin{equation} Li_{n+2}(-1) = \left(-1+\frac{1}{2^{n+1}}\right) \zeta(n+2) \end{equation} for $n=0,1,\cdots$ this concludes the calculation. All we need to do is to collect all the different terms. Bringing everything together we get: \begin{eqnarray} &&{\bf H}^{(2)}_n(-1)=\\ &&\frac{1}{3\cdot 2^{n+3}} \left(\right.\\ &&3 \left(\left(2^{n+1}-1\right) n^2+\left(2^{n+1}-5\right) n+2^{n+2}-6\right)\zeta (n+2)\\ &&-4 \left(2^n-1\right) (n-2) \log (8) \zeta (n+1)\\ &&+\pi ^2 \left(2^n-2\right) \zeta (n)\\ &&\left.\right) +\\ &&\sum\limits_{j=1}^{n-1} (1+\frac{1}{2^n}-\frac{1}{2^{n-j-1}}) \zeta(1+j) \zeta(n+1-j) +\\ &&\sum\limits_{l=0}^n \left(2 \cdot 1_{0\le l \le 2} + (\frac{l}{2}+1)\cdot 1_{3 \le l \le n-1} + (\frac{l}{2}-1) 1_{l=n}\right)\cdot {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} for $n\ge 3$. In the case $n=1,2$ surface terms need to be taken into account.
Let us now write down the recurrence relations for ${\bf H}^{(2)}_n(t)$. As usual we start from the integral representation. We have: \begin{eqnarray} &&{\bf H}^{(2)}_n(t) =\\ && \int\limits_0^t \frac{[\log(t/\xi)]^{n-1}}{(n-1)!} \cdot \underbrace{\frac{Li_2(\xi)}{1-\xi}}_{[Li_1(\xi) Li_2(\xi) - \int \frac{Li_1(\xi)^2}{\xi} d\xi]^{'}} d \xi +Li_{n+2}(t)=\\ && \frac{1}{2 (n-3)!} \int\limits_0^t [\log(t/\xi)]^{n-3} \cdot \frac{Li_2(\xi)^2}{\xi} d\xi - \frac{1}{(n-1)!} \int\limits_0^t [\log(t/\xi)]^{n-1} \cdot \frac{Li_1(\xi)^2}{\xi} d\xi + Li_{n+2}(t) \end{eqnarray} In the bottom line we integrated by parts and expressed the result through integrals that contain squares of a poly-log only rather than products of two different poly-logs. Now we set $t=-1$. We can immediately say that from the two integrals on the right hand side the middle one has already been computed in my previous answer to this question. The first integral is harder but it has been calculated in here Generalized definite dilogarithm integral. . Then all we need to do is to write down the results , add then up and simplify which, despite appearances, is a tedious and mundane task. We have: \begin{eqnarray} &&\frac{1}{(n-1)!} \int\limits_0^1 [\log(1/\xi)]^{n-1} \cdot \frac{\log(1+\xi)^2}{\xi} d\xi = \\ &&\left(\frac{1}{2^n} - 1\right)(n+1) \zeta(n+2) + \sum\limits_{j=1}^{n-1} \left(-1-\frac{1}{2^n} + \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n+1-j)\\ && - 2 \sum\limits_{l=0}^{n-1} {\bf H}^{(l+1)}_{n+1-l} (-1) \end{eqnarray} Likewise: \begin{eqnarray} &&\frac{1}{2(n-3)!} \int\limits_0^1 [\log(1/\xi)]^{n-3} \cdot \frac{Li_2(-\xi)^2}{\xi} d\xi =\\ && \frac{1}{4} \left(1-\frac{1}{2^{n+1}}\right)(n-1)_{(2)} \zeta(n+2) + (n-2)\left(-\frac{1}{2} + \frac{1}{2^{n+1}}\right) \log(2) \zeta(n+1) + \\ &&\frac{1}{4} \left(1-\frac{1}{2^{n-1}}\right)\zeta(2) \zeta(n)+\frac{1}{2} \sum\limits_{l=3}^n (l-2) {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} Now since \begin{equation} Li_{n+2}(-1) = \left(-1+\frac{1}{2^{n+1}}\right) \zeta(n+2) \end{equation} for $n=0,1,\cdots$ this concludes the calculation. All we need to do is to collect all the different terms. Bringing everything together we get: \begin{eqnarray} &&{\bf H}^{(2)}_n(-1)=\\ &&\frac{1}{3\cdot 2^{n+3}} \left(3 \left(\left(2^{n+1}-1\right) n^2+\left(2^{n+1}-5\right) n+2^{n+2}-6\right) \zeta (n+2)+\pi ^2 \left(2^n-2\right) \zeta (n)-4 \left(2^n-1\right) (n-2) \log (8) \zeta (n+1)\right) +\\ &&\sum\limits_{j=1}^{n-1} (1+\frac{1}{2^n}-\frac{1}{2^{n-j-1}}) \zeta(1+j) \zeta(n+1-j) +\\ &&\sum\limits_{l=0}^n \left(2 \cdot 1_{0\le l \le 2} + (\frac{l}{2}+1)\cdot 1_{3 \le l \le n-1} + (\frac{l}{2}-1) 1_{l=n}\right)\cdot {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray}\begin{eqnarray} &&{\bf H}^{(2)}_n(-1)=\\ &&\frac{1}{3\cdot 2^{n+3}} \left(\right.\\ &&3 \left(\left(2^{n+1}-1\right) n^2+\left(2^{n+1}-5\right) n+2^{n+2}-6\right)\zeta (n+2)\\ &&-4 \left(2^n-1\right) (n-2) \log (8) \zeta (n+1)\\ &&+\pi ^2 \left(2^n-2\right) \zeta (n)\\ &&\left.\right) +\\ &&\sum\limits_{j=1}^{n-1} (1+\frac{1}{2^n}-\frac{1}{2^{n-j-1}}) \zeta(1+j) \zeta(n+1-j) +\\ &&\sum\limits_{l=0}^n \left(2 \cdot 1_{0\le l \le 2} + (\frac{l}{2}+1)\cdot 1_{3 \le l \le n-1} + (\frac{l}{2}-1) 1_{l=n}\right)\cdot {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} for $n\ge 3$.
Let us now write down the recurrence relations for ${\bf H}^{(2)}_n(t)$. As usual we start from the integral representation. We have: \begin{eqnarray} &&{\bf H}^{(2)}_n(t) =\\ && \int\limits_0^t \frac{[\log(t/\xi)]^{n-1}}{(n-1)!} \cdot \underbrace{\frac{Li_2(\xi)}{1-\xi}}_{[Li_1(\xi) Li_2(\xi) - \int \frac{Li_1(\xi)^2}{\xi} d\xi]^{'}} d \xi +Li_{n+2}(t)=\\ && \frac{1}{2 (n-3)!} \int\limits_0^t [\log(t/\xi)]^{n-3} \cdot \frac{Li_2(\xi)^2}{\xi} d\xi - \frac{1}{(n-1)!} \int\limits_0^t [\log(t/\xi)]^{n-1} \cdot \frac{Li_1(\xi)^2}{\xi} d\xi + Li_{n+2}(t) \end{eqnarray} In the bottom line we integrated by parts and expressed the result through integrals that contain squares of a poly-log only rather than products of two different poly-logs. Now we set $t=-1$. We can immediately say that from the two integrals on the right hand side the middle one has already been computed in my previous answer to this question. The first integral is harder but it has been calculated in here Generalized definite dilogarithm integral. . Then all we need to do is to write down the results , add then up and simplify which, despite appearances, is a tedious and mundane task. We have: \begin{eqnarray} &&\frac{1}{(n-1)!} \int\limits_0^1 [\log(1/\xi)]^{n-1} \cdot \frac{\log(1+\xi)^2}{\xi} d\xi = \\ &&\left(\frac{1}{2^n} - 1\right)(n+1) \zeta(n+2) + \sum\limits_{j=1}^{n-1} \left(-1-\frac{1}{2^n} + \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n+1-j)\\ && - 2 \sum\limits_{l=0}^{n-1} {\bf H}^{(l+1)}_{n+1-l} (-1) \end{eqnarray} Likewise: \begin{eqnarray} &&\frac{1}{2(n-3)!} \int\limits_0^1 [\log(1/\xi)]^{n-3} \cdot \frac{Li_2(-\xi)^2}{\xi} d\xi =\\ && \frac{1}{4} \left(1-\frac{1}{2^{n+1}}\right)(n-1)_{(2)} \zeta(n+2) + (n-2)\left(-\frac{1}{2} + \frac{1}{2^{n+1}}\right) \log(2) \zeta(n+1) + \\ &&\frac{1}{4} \left(1-\frac{1}{2^{n-1}}\right)\zeta(2) \zeta(n)+\frac{1}{2} \sum\limits_{l=3}^n (l-2) {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} Now since \begin{equation} Li_{n+2}(-1) = \left(-1+\frac{1}{2^{n+1}}\right) \zeta(n+2) \end{equation} for $n=0,1,\cdots$ this concludes the calculation. All we need to do is to collect all the different terms. Bringing everything together we get: \begin{eqnarray} &&{\bf H}^{(2)}_n(-1)=\\ &&\frac{1}{3\cdot 2^{n+3}} \left(3 \left(\left(2^{n+1}-1\right) n^2+\left(2^{n+1}-5\right) n+2^{n+2}-6\right) \zeta (n+2)+\pi ^2 \left(2^n-2\right) \zeta (n)-4 \left(2^n-1\right) (n-2) \log (8) \zeta (n+1)\right) +\\ &&\sum\limits_{j=1}^{n-1} (1+\frac{1}{2^n}-\frac{1}{2^{n-j-1}}) \zeta(1+j) \zeta(n+1-j) +\\ &&\sum\limits_{l=0}^n \left(2 \cdot 1_{0\le l \le 2} + (\frac{l}{2}+1)\cdot 1_{3 \le l \le n-1} + (\frac{l}{2}-1) 1_{l=n}\right)\cdot {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} for $n\ge 3$.
Let us now write down the recurrence relations for ${\bf H}^{(2)}_n(t)$. As usual we start from the integral representation. We have: \begin{eqnarray} &&{\bf H}^{(2)}_n(t) =\\ && \int\limits_0^t \frac{[\log(t/\xi)]^{n-1}}{(n-1)!} \cdot \underbrace{\frac{Li_2(\xi)}{1-\xi}}_{[Li_1(\xi) Li_2(\xi) - \int \frac{Li_1(\xi)^2}{\xi} d\xi]^{'}} d \xi +Li_{n+2}(t)=\\ && \frac{1}{2 (n-3)!} \int\limits_0^t [\log(t/\xi)]^{n-3} \cdot \frac{Li_2(\xi)^2}{\xi} d\xi - \frac{1}{(n-1)!} \int\limits_0^t [\log(t/\xi)]^{n-1} \cdot \frac{Li_1(\xi)^2}{\xi} d\xi + Li_{n+2}(t) \end{eqnarray} In the bottom line we integrated by parts and expressed the result through integrals that contain squares of a poly-log only rather than products of two different poly-logs. Now we set $t=-1$. We can immediately say that from the two integrals on the right hand side the middle one has already been computed in my previous answer to this question. The first integral is harder but it has been calculated in here Generalized definite dilogarithm integral. . Then all we need to do is to write down the results , add then up and simplify which, despite appearances, is a tedious and mundane task. We have: \begin{eqnarray} &&\frac{1}{(n-1)!} \int\limits_0^1 [\log(1/\xi)]^{n-1} \cdot \frac{\log(1+\xi)^2}{\xi} d\xi = \\ &&\left(\frac{1}{2^n} - 1\right)(n+1) \zeta(n+2) + \sum\limits_{j=1}^{n-1} \left(-1-\frac{1}{2^n} + \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n+1-j)\\ && - 2 \sum\limits_{l=0}^{n-1} {\bf H}^{(l+1)}_{n+1-l} (-1) \end{eqnarray} Likewise: \begin{eqnarray} &&\frac{1}{2(n-3)!} \int\limits_0^1 [\log(1/\xi)]^{n-3} \cdot \frac{Li_2(-\xi)^2}{\xi} d\xi =\\ && \frac{1}{4} \left(1-\frac{1}{2^{n+1}}\right)(n-1)_{(2)} \zeta(n+2) + (n-2)\left(-\frac{1}{2} + \frac{1}{2^{n+1}}\right) \log(2) \zeta(n+1) + \\ &&\frac{1}{4} \left(1-\frac{1}{2^{n-1}}\right)\zeta(2) \zeta(n)+\frac{1}{2} \sum\limits_{l=3}^n (l-2) {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} Now since \begin{equation} Li_{n+2}(-1) = \left(-1+\frac{1}{2^{n+1}}\right) \zeta(n+2) \end{equation} for $n=0,1,\cdots$ this concludes the calculation. All we need to do is to collect all the different terms. Bringing everything together we get: \begin{eqnarray} &&{\bf H}^{(2)}_n(-1)=\\ &&\frac{1}{3\cdot 2^{n+3}} \left(\right.\\ &&3 \left(\left(2^{n+1}-1\right) n^2+\left(2^{n+1}-5\right) n+2^{n+2}-6\right)\zeta (n+2)\\ &&-4 \left(2^n-1\right) (n-2) \log (8) \zeta (n+1)\\ &&+\pi ^2 \left(2^n-2\right) \zeta (n)\\ &&\left.\right) +\\ &&\sum\limits_{j=1}^{n-1} (1+\frac{1}{2^n}-\frac{1}{2^{n-j-1}}) \zeta(1+j) \zeta(n+1-j) +\\ &&\sum\limits_{l=0}^n \left(2 \cdot 1_{0\le l \le 2} + (\frac{l}{2}+1)\cdot 1_{3 \le l \le n-1} + (\frac{l}{2}-1) 1_{l=n}\right)\cdot {\bf H}^{(l+1)}_{n+1-l}(-1) \end{eqnarray} for $n\ge 3$.