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

$$\displaystyle{\psi _{1}\left( z \right)=-\int\limits_{0}^{1}{\frac{x^{z-1}\ln x}{1-x}dx}}$$

$$\displaystyle{\left( \psi _{1}\left( z \right) \right)^{2}=\left( -\int\limits_{0}^{1}{\frac{x^{z-1}\ln x}{1-x}dx} \right)\left( -\int\limits_{0}^{1}{\frac{y^{z-1}\ln y}{1-y}dy} \right)=\int\limits_{0}^{1}{\int\limits_{0}^{1}{\frac{\left( xy \right)^{z-1}\ln x\ln y}{\left( 1-x \right)\left( 1-y \right)}dxdy}}}$$

$$\displaystyle{\sum\limits_{n=1}^{+\infty }{\left( -1 \right)^{n-1}\left( \psi _{1}\left( n \right) \right)^{2}}=\sum\limits_{n=1}^{+\infty }{\left( \left( \psi _{1}\left( 2n-1 \right) \right)^{2}-\left( \psi _{1}\left( 2n \right) \right)^{2} \right)}}$$

$$\displaystyle{=\sum\limits_{n=1}^{+\infty }{\int\limits_{0}^{1}{\int\limits_{0}^{1}{\frac{\left( \left( xy \right)^{2n-2}-\left( xy \right)^{2n-1} \right)\ln x\ln y}{\left( 1-x \right)\left( 1-y \right)}dxdy}}}=\int\limits_{0}^{1}{\int\limits_{0}^{1}{\frac{\left( 1-xy \right)\ln x\ln y}{xy\left( 1-x \right)\left( 1-y \right)}\cdot \frac{\left( xy \right)^{2}}{1-\left( xy \right)^{2}}dxdy}}}$$

$$\displaystyle{=\int\limits_{0}^{1}{\int\limits_{0}^{1}{\frac{\ln x\ln y}{\left( 1-x \right)\left( 1-y \right)}\cdot \frac{xy}{1+xy}dxdy}}=-\frac{1}{6}\int\limits_{0}^{1}{\frac{6\text{Li}_{2}\left( -y \right)+\pi ^{2}y}{1-y^{2}}\ln ydy}}$$

$$\displaystyle{=-\frac{\pi ^{2}}{6}\int\limits_{0}^{1}{\frac{y}{1-y^{2}}\ln ydy}-\int\limits_{0}^{1}{\frac{\ln y\text{Li}_{2}\left( -y \right)}{1-y^{2}}dy}=\frac{\pi ^{2}}{6}\cdot \frac{\pi ^{2}}{24}-\int\limits_{0}^{1}{\frac{\ln y\text{Li}_{2}\left( -y \right)}{1-y^{2}}dy}}$$