I uploaded this question [here][1] and [here][2] from my old account.

Let $\psi^{(1)}$ be the trigamma function defined by
\begin{equation}
\tag{1}
\psi_1(z) = -\int\limits_0^1 \frac{x^{z-1} \ln x}{1-x} \, dx.
\end{equation}

It follows that
\begin{equation}
\begin{split}
(\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.
\end{split} \tag{2}
\end{equation}

Following the [book by Cornel Ioan Vălean][3] we compute the order
\begin{equation}
\tag{3}
\mathcal{S} = \sum_{n=1}^\infty (-1)^{n-1} (\psi^{(1)}(n))^2.
\end{equation}


To this end, from (2) and (3) it follows that
\begin{align}
\mathcal{S} &= \sum_{n=1}^{+\infty} ( (\psi_1 (2n-1))^2-(\psi_1(2n))^2)\\
&=\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.
\end{align}


  [1]: https://math.stackexchange.com/q/2412715/1295548
  [2]: https://math.stackexchange.com/q/2411733/1295548
  [3]: https://link.springer.com/book/10.1007/978-3-030-02462-8