Integral involving a dilogarithm versus an Euler sum.

By using the reflection formula for the dilogarithmic function we simplify the right hand side as follows: \begin{eqnarray} rhs &=& \int\limits_0^1 \frac{Li_2(x)}{x} \cdot \frac{[\log(x)]^{q-1}}{(1-x)} dx &+& \int\limits_0^1 \frac{[\log(x)]^{q-1} [\log(1-x)]^2}{x(1-x)} dx \\ &=& \int\limits_0^1 Li_2(x) \cdot [\log(x)]^{q-1}\cdot \left( \frac{1}{x} + \frac{1}{1-x}\right) dx &+& \left({\mathcal I}^{(2,q-1)} + {\mathcal I}^{(q-1,2)}\right) \\ &=& (-1)^{q-1}(q-1)! \left(Li_{q+2}(1)-Li_{q+2}(1)+\sum\limits_{n=1}^\infty \frac{H_n^{(2)}}{n^q}\right) &+&\left({\mathcal I}^{(2,q-1)} + {\mathcal I}^{(q-1,2)}\right) \\ \end{eqnarray} In the second line we introduced: \begin{equation} {\mathcal I}^{(q,p)} := \int\limits_{0}^1 \frac{[\log(1-x)]^q [\log(x)]^p}{x} dx \end{equation} and in the last line we expanded the terms $Li_2(x)/x$ and $Li_2(x)/(1-x)$ in a series in $x$ and integrated term by term. Therefore we get: \begin{equation} \sum\limits_{n=1}^\infty \frac{H_n^2}{n^q} = \sum\limits_{n=1}^\infty \frac{H_n^{(2)}}{n^q} + \frac{(-1)^{q-1}}{(q-1)!} \cdot \left({\mathcal I}^{(2,q-1)} + {\mathcal I}^{(q-1,2)}\right) \end{equation} The quantities ${\mathcal I}^{(q,p)}$ have been calculated in Compute an integral containing a product of powers of logarithms. . We have: \begin{eqnarray} &&\sum\limits_{m=1}^\infty \frac{[H_m]^2 - H_m^{(2)}}{m^q} = \\ &&\frac{(-1)^{q-1}}{(q-1)!} \left(\right.\\ && -(\frac{1}{q}+\frac{1}{3}) \Psi^{(q+1)}(1) + \sum\limits_{j=1}^{q-1} (\frac{1}{q} \binom{q}{j} + \frac{1_{q\ge 3}}{2}(\binom{q}{j}-1_{j\notin[2,q-2]})) \Psi^{(j)}(1) \Psi^{(q-j)}(1)-\\ &&\frac{1}{3} \sum\limits_{1\le j < j_1 \le q-2} \frac{(q-1)!}{j!(j_1-j)!(q-1-j_1)!} \Psi^{(j)}(1) \Psi^{(j_1-j)}(1) \Psi^{(q-1-j_1)}(1)\\ && \left.\right) \end{eqnarray} for $q\ge 2$. Particular cases of the sum above are given in http://algo.inria.fr/flajolet/Publications/FlSa98.pdf .