Let $p$ and $q$ be integers such that $p\ge 1$ and $q\ge 2$ and let $t\in(-1,1)$. We consider a following integral. \begin{equation} {\mathcal A}^{(p,2)}_q(t) := \int\limits_0^1 [\log(\xi)]^p \frac{[Li_q(t \xi)]^2}{\xi} d\xi \end{equation} where $Li_{.}(.)$ is the poly-logarithm. By generalizing the approach from Definite Dilogarithm integral $\int^1_0 \frac{\operatorname{Li}_2^2(x)}{x}\, dx $ we computed the following: \begin{eqnarray} &&{\mathcal A}^{(p,2)}_2(1) = (-1)^{p-1} p! \cdot \\ &&\left[ \sum\limits_{l=1}^{p+1} (p+2-l) {\bf H}^{(l)}_{p+5-l}(1) - \sum_{l=2}^{p+1} (p+2-l) \zeta(l) \zeta(p+5-l) - \zeta(2) \zeta(p+3)\right] \end{eqnarray} here ${\bf H}^{(l)}_n(t) := \sum\limits_{m=1}^\infty H_m^{(l)}/m^n \cdot t^m$. Now, the obvious question is what is the result for generic $t$ and generic $q\ge 2$.
2 Answers
We state the result for $q=2$ and for arbitrary $t$. We have: \begin{eqnarray} &&{\mathcal A}^{(p,2)}_2(t) = (-1)^{p-1} p!\cdot \\ &&\left[ \sum\limits_{l=1}^{p+1}(p+2-l)\sum\limits_{k=1}^\infty \frac{t^k}{k^l} \zeta(5-l+p,k)- \sum\limits_{l=2}^{p+1} (p+2-l) Li_l(t) Li_{p+5-l}(1) - Li_2(t) Li_{p+3}(t) - (p+1)Li_1(t) [Li_{p+4}(1) - Li_{p+4}(t)] \right] = \\ &&(-1)^p p! \cdot \\ &&\left[ -\binom{p+2}{2} Li_{p+5}(t) - (p+1) Li_1(t) Li_{p+4}(t) + Li_2(t) Li_{p+3} (t) + \sum\limits_{l=1}^{p+1} (p+2-l) {\bf H}^{(p+5-l)}_l(t) \right] \end{eqnarray} Here $\zeta()$ is the Hurwitz zeta function. The generalization for arbitrary $q \ge 1$ is pretty straightforward so we only state the final result. We have: \begin{eqnarray} &&{\mathcal A}^{(p,2)}_q(t) = (-1)^p p! \cdot \\ &&\left( \sum\limits_{l_1=0}^q \binom{q+p-l_1}{p} \left\{Li_{l_1}(t) \cdot 1_{l_1 \ge 1} - 1 \cdot 1_{l_1=0}\right\}\cdot Li_{2q+p+1-l_1}(t) (-1)^{q-l_1} + \sum\limits_{l_1=1}^{p+1} \binom{q+p-l_1}{q-1} (-1)^q {\bf H}^{(2q+p+1-l_1)}_{l_1}(t) \right) \end{eqnarray} Here both $q\ge 1$ and $p\ge 1$ and $t\in(-1,1)$ and ${\bf H}^{(q)}_p(t) := \sum\limits_{m=1}^\infty H_m^{(q)}/m^p \cdot t^m$.
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This answer is only valid for $t=-1$ and for $p=0$.We have: \begin{eqnarray} &&{\mathcal A}^{(0,2)}_q(-1) := \int\limits_0^1 \frac{[Li_q(-\xi)]^2}{\xi} d\xi\\ &&=\frac{1}{2^{2 q-2}} \int\limits_0^1 \frac{[Li_q(\xi^2)]^2}{\xi} d\xi - \frac{2}{2^{q-1}} \int\limits_0^1 \frac{Li_q(\xi^2)Li_q(\xi)}{\xi}d\xi + \int\limits_0^1 \frac{[Li_q(\xi)]^2}{\xi}d\xi \\ &&=\left(\frac{1}{2^{2q-1}}-1\right)\cdot \int\limits_0^1 \frac{[Li_q(\xi)]^2}{\xi}d\xi - 2 \int\limits_0^1 \frac{Li_q(-\xi) Li_q(\xi)}{\xi} d\xi \end{eqnarray} In the second line from the top we used the identity $Li_q(-\xi) = (1/2^{q-1})\cdot Li_q(\xi^2) - Li_q(\xi)$ valid for $q=1,2,\cdots$ and then in the subsequent line we changed variables accordingly and simplified the result. The first integral on the rhs has already been computed above and it reads: \begin{eqnarray} &&\int\limits_0^1 \frac{[Li_q(\xi)]^2}{\xi}d\xi \\ &&=\sum\limits_{l=2}^q \zeta(l) \zeta(2q+1-l)(-1)^{q-l} + (-1)^{q-1} {\bf H}^{(1)}_{2 q}(+1)\\ &&= \sum\limits_{l=2}^q \zeta(l) \zeta(2q+1-l)(-1)^{q-l} +(q+1) (-1)^{q-1} \zeta(1+2 q) + \frac{(-1)^{q}}{2} \sum\limits_{l=2}^{2q-1} \zeta(l) \zeta(2q+1-l) \end{eqnarray} The second integral is computed by integrating by parts $(q+1)$ times and expanding the remaining integral in a series and integrating term by term. We have: \begin{eqnarray} &&\int\limits_0^1 \frac{Li_q(-\xi)Li_q(\xi)}{\xi}d\xi =\\ && \sum\limits_{l=0}^{q-1} Li_{q-l}(-1) Li_{q+1+l}(1) (-1)^l + (-1)^q \sum\limits_{l=1}^{2q-1} Li_{l}(-1) \zeta(2q+1-l)(-1)^{l-1} + (-1)^{q+1} {\bf H}^{(1)}_{2 q}(-1)=\\ && \sum\limits_{l=0}^{q-1} (-1+\frac{1}{2^{q-l-1}}) \zeta(q-l) \zeta(q+1+l) (-1)^l + (-1)^q \sum\limits_{l=1}^{2q-1} (-1+\frac{1}{2^{l-1}})\zeta(l) \zeta(2q+1-l)(-1)^{l-1} + (-1)^{q+1} {\bf H}^{(1)}_{2 q}(-1) \end{eqnarray} This finishes the computation. The generalization to arbitrary non-negative integer $p$ is pretty straightforward and we will provide it later on. The result reads: \begin{eqnarray} &&{\mathcal A}^{(p,2)}_q(-1)=(-1)^p p! \cdot\left\{\right.\\ &&\sum\limits_{l=2}^q \binom{q+p-l}{p} \left[1+\frac{1}{2^{2q-1+p}} - \frac{1}{2^{2q-1+p-l}}\right]\cdot \zeta(2q+p+1-l)\zeta(l)(-1)^{q-l}+\\ &&\sum\limits_{l=2}^{p+1} \binom{q+p-l}{q-1} \left[1+\frac{1}{2^{2q-1+p}} - \frac{1}{2^{2q-1+p-l}}\right]\cdot \zeta(2q+p+1-l)\zeta(l)(-1)^{q}+\\ &&\sum\limits_{l=1}^{p+1} \binom{q+p-l}{q-1} (-1)^{q-1} \left[(\frac{1}{2^{2q-1+p}}-1)\cdot {\bf H}^{(l)}_{2q+p+1-l}(+1) - 2 {\bf H}^{(l)}_{2q+p+1-l}(-1)\right]\\ &&\left.\right\} \end{eqnarray} for $p\ge 0$ and for $q\ge 1$.
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