Let's state with $\psi^{(1)}$ the trigamma. Calculate the order: $$\mathcal{S} = \sum_{n=1}^\infty (-1)^{n-1} (\psi^{(1)}(n))^2$$ (Cornel Ioan Valean)

I uploaded this question here https://math.stackexchange.com/q/2412715/1295548 and https://math.stackexchange.com/q/2411733/1295548 from my old account

$$ \psi_1(z) = -\int\limits_0^1 \frac{x^{z-1} \ln x}{1-x} \, dx $$

\begin{align} (\psi_1(z))^2 & =\left( -\int\limits_0^1 \frac{x^{z-1}\ln x}{1-x} \, dx \right) \left( -\int\limits_0^1 \frac{y^{z-1}\ln y}{1-y} \, dy \right) = \int\limits_0^1 \int\limits_0^1 \frac{(xy)^{z-1} \ln x\ln y}{(1-x)(1-y)} \, dx \, dy \\[8pt] & \sum_{n=1}^{+\infty} (-1)^{n-1} (\psi_1(n))^2 = \sum_{n=1}^{+\infty} ( (\psi_1 (2n-1))^2-(\psi_1(2n))^2) \end{align}

$$=\sum_{n=1}^{+\infty} \int\limits_0^1 \int\limits_0^1 \frac{( (xy)^{2n-2}-(xy)^{2n-1} ) \ln x\ln y}{(1-x) (1-y)} \, dx \, dy = \int\limits_0^1 \int\limits_0^1 \frac{(1-xy)\ln x\ln y}{xy(1-x)(1-y)}\cdot \frac{(xy)^2}{1-(xy)^2} \, dx \, dy $$

$$=\int\limits_0^1 \int\limits_0^1 \frac{\ln x\ln y}{(1-x) (1-y)} \cdot \frac{xy}{1+xy} \, dx \, dy = -\frac{1}{6} \int\limits_0^1 \frac{6 \operatorname{Li}_2 (-y)+\pi^2 y}{1-y^2} \ln y \, dy $$

$$=-\frac{\pi ^{2}}{6}\int\limits_0^1 \frac{y}{1-y^2}\ln y \, dy - \int\limits_0^1 \frac{\ln y\operatorname{Li}_2 (-y)}{1-y^2} \, dy = \frac{\pi^2}{6} \cdot \frac{\pi^2}{24}-\int\limits_0^1 \frac{\ln y \operatorname{Li}_2 (-y)}{1-y^2} \, dy $$