While investigating the sum $$S=\sum_{n=1}^\infty \frac{\psi_0(n)} {n^2}$$ I used the integral representation of the Digamma function to get$$S=\int_0^\infty \frac{\pi^2 e^{-x}}{6x}-\frac{ \operatorname{Li}_2(e^{-x})}{1-e^{-x}}dx$$ I have no clue on how to evaluate this integral as it involves the Dilogarithm function and tricky limits at the endpoints of the integral. If anyone can help me solve this integral I would greatly appreciate it.
EDIT: I believe that $S$ can also be expressed as $$-\frac{\gamma \pi^2}{6}+\int_0^1\left(\frac{\pi^2}{6}-\operatorname{Li}_2(x)\right)\frac{dx}{1-x}$$
This sum can be evaluated by rewriting the digamma function in terms of the harmonic numbers, then applying a famous formula due to Euler (eqn 23):
$$\begin{align*} \sum_{n=1}^\infty\frac{\psi(n)}{n^2} &=\sum_{n=1}^\infty\frac{H_{n-1}-\gamma}{n^2} \\ &= -\frac{\pi^2\gamma}{6}+\sum_{n=1}^\infty\frac{H_{n-1}}{n^2}\\ &= -\frac{\pi^2\gamma}{6}+\sum_{n=1}^\infty\frac{H_{n}-\frac{1}{n}}{n^2}\\ &= -\frac{\pi^2\gamma}{6}-\zeta(3)+\sum_{n=1}^\infty\frac{H_{n}}{n^2}\\ &= \zeta(3)-\frac{\pi^2\gamma}{6} \end{align*}$$