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Improved logical exposition and math formatting. Converted links to hyperlinks and added hyperlink to Valean's book.
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MathMax
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MathMax
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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/2412715/1295548here and https://math.stackexchange.com/q/2411733/1295548here 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} 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}

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

$$=\int\limits_0^1 \int\limits_0^1 \frac{\ln x\ln y}{(1-x) (1-y)} \cdot \frac{xy}{1+xy} \, dx \, dy = -\frac{1}{6} \int\limits_0^1 \frac{6 \operatorname{Li}_2 (-y)+\pi^2 y}{1-y^2} \ln y \, dy $$ Following the book by Cornel Ioan Vălean we compute the order \begin{equation} \tag{3} \mathcal{S} = \sum_{n=1}^\infty (-1)^{n-1} (\psi^{(1)}(n))^2. \end{equation}

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

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}

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

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

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

I uploaded this question here and here 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 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}

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The MathJax usage here was unbelievable abominable. It still needs further work after this.
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Michael Hardy
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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

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

$$(\psi_1(z))^2 =\left( -\int\limits_0^1 \frac{x^{z-1}\ln x}{1-x} \, dx \right) \left( -\int\limits_0^1 \frac{y^{z-1}\ln y}{1-y} \, dy \right) = \int\limits_0^1 \int\limits_0^1 \frac{(xy)^{z-1} \ln x\ln y}{(1-x)(1-y)} \, dx \, dy $$$$ \psi_1(z) = -\int\limits_0^1 \frac{x^{z-1} \ln x}{1-x} \, dx $$

$$\sum_{n=1}^{+\infty} (-1)^{n-1} (\psi_1(n))^2 = \sum_{n=1}^{+\infty} ( (\psi_1 (2n-1))^2-(\psi_1(2n))^2) $$\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}

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

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

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

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

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

$$(\psi_1(z))^2 =\left( -\int\limits_0^1 \frac{x^{z-1}\ln x}{1-x} \, dx \right) \left( -\int\limits_0^1 \frac{y^{z-1}\ln y}{1-y} \, dy \right) = \int\limits_0^1 \int\limits_0^1 \frac{(xy)^{z-1} \ln x\ln y}{(1-x)(1-y)} \, dx \, dy $$

$$\sum_{n=1}^{+\infty} (-1)^{n-1} (\psi_1(n))^2 = \sum_{n=1}^{+\infty} ( (\psi_1 (2n-1))^2-(\psi_1(2n))^2) $$

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

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

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

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}

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

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

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

The MathJax usage here was unbelievable abominable. It still needs further work after this.
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Michael Hardy
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Conjectured closed form of $\int\limits_{0}^{1}{\frac$\int\limits_0^1 \frac{\ln y\texty \operatorname{Li}_{2}\left(_2 (-y \right)}{1-y^{2}}dyy^2}$ \, dy$

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

$$\displaystyle{\psi _{1}\left( z \right)=-\int\limits_{0}^{1}{\frac{x^{z-1}\ln x}{1-x}dx}}$$$$\psi_1(z) = -\int\limits_0^1 \frac{x^{z-1} \ln x}{1-x} \, dx $$

$$\displaystyle{\left( \psi _{1}\left( z \right) \right)^{2}=\left( -\int\limits_{0}^{1}{\frac{x^{z-1}\ln x}{1-x}dx} \right)\left( -\int\limits_{0}^{1}{\frac{y^{z-1}\ln y}{1-y}dy} \right)=\int\limits_{0}^{1}{\int\limits_{0}^{1}{\frac{\left( xy \right)^{z-1}\ln x\ln y}{\left( 1-x \right)\left( 1-y \right)}dxdy}}}$$$$(\psi_1(z))^2 =\left( -\int\limits_0^1 \frac{x^{z-1}\ln x}{1-x} \, dx \right) \left( -\int\limits_0^1 \frac{y^{z-1}\ln y}{1-y} \, dy \right) = \int\limits_0^1 \int\limits_0^1 \frac{(xy)^{z-1} \ln x\ln y}{(1-x)(1-y)} \, dx \, dy $$

$$\displaystyle{\sum\limits_{n=1}^{+\infty }{\left( -1 \right)^{n-1}\left( \psi _{1}\left( n \right) \right)^{2}}=\sum\limits_{n=1}^{+\infty }{\left( \left( \psi _{1}\left( 2n-1 \right) \right)^{2}-\left( \psi _{1}\left( 2n \right) \right)^{2} \right)}}$$$$\sum_{n=1}^{+\infty} (-1)^{n-1} (\psi_1(n))^2 = \sum_{n=1}^{+\infty} ( (\psi_1 (2n-1))^2-(\psi_1(2n))^2) $$

$$\displaystyle{=\sum\limits_{n=1}^{+\infty }{\int\limits_{0}^{1}{\int\limits_{0}^{1}{\frac{\left( \left( xy \right)^{2n-2}-\left( xy \right)^{2n-1} \right)\ln x\ln y}{\left( 1-x \right)\left( 1-y \right)}dxdy}}}=\int\limits_{0}^{1}{\int\limits_{0}^{1}{\frac{\left( 1-xy \right)\ln x\ln y}{xy\left( 1-x \right)\left( 1-y \right)}\cdot \frac{\left( xy \right)^{2}}{1-\left( xy \right)^{2}}dxdy}}}$$$$=\sum_{n=1}^{+\infty} \int\limits_0^1 \int\limits_0^1 \frac{( (xy)^{2n-2}-(xy)^{2n-1} ) \ln x\ln y}{(1-x) (1-y)} \, dx \, dy = \int\limits_0^1 \int\limits_0^1 \frac{(1-xy)\ln x\ln y}{xy(1-x)(1-y)}\cdot \frac{(xy)^2}{1-(xy)^2} \, dx \, dy $$

$$\displaystyle{=\int\limits_{0}^{1}{\int\limits_{0}^{1}{\frac{\ln x\ln y}{\left( 1-x \right)\left( 1-y \right)}\cdot \frac{xy}{1+xy}dxdy}}=-\frac{1}{6}\int\limits_{0}^{1}{\frac{6\text{Li}_{2}\left( -y \right)+\pi ^{2}y}{1-y^{2}}\ln ydy}}$$$$=\int\limits_0^1 \int\limits_0^1 \frac{\ln x\ln y}{(1-x) (1-y)} \cdot \frac{xy}{1+xy} \, dx \, dy = -\frac{1}{6} \int\limits_0^1 \frac{6 \operatorname{Li}_2 (-y)+\pi^2 y}{1-y^2} \ln y \, dy $$

$$\displaystyle{=-\frac{\pi ^{2}}{6}\int\limits_{0}^{1}{\frac{y}{1-y^{2}}\ln ydy}-\int\limits_{0}^{1}{\frac{\ln y\text{Li}_{2}\left( -y \right)}{1-y^{2}}dy}=\frac{\pi ^{2}}{6}\cdot \frac{\pi ^{2}}{24}-\int\limits_{0}^{1}{\frac{\ln y\text{Li}_{2}\left( -y \right)}{1-y^{2}}dy}}$$$$=-\frac{\pi ^{2}}{6}\int\limits_0^1 \frac{y}{1-y^2}\ln y \, dy - \int\limits_0^1 \frac{\ln y\operatorname{Li}_2 (-y)}{1-y^2} \, dy = \frac{\pi^2}{6} \cdot \frac{\pi^2}{24}-\int\limits_0^1 \frac{\ln y \operatorname{Li}_2 (-y)}{1-y^2} \, dy $$

Conjectured closed form of $\int\limits_{0}^{1}{\frac{\ln y\text{Li}_{2}\left( -y \right)}{1-y^{2}}dy}$

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

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

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

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

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

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

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

Conjectured closed form of $\int\limits_0^1 \frac{\ln y \operatorname{Li}_2 (-y)}{1-y^2} \, dy$

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

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

$$(\psi_1(z))^2 =\left( -\int\limits_0^1 \frac{x^{z-1}\ln x}{1-x} \, dx \right) \left( -\int\limits_0^1 \frac{y^{z-1}\ln y}{1-y} \, dy \right) = \int\limits_0^1 \int\limits_0^1 \frac{(xy)^{z-1} \ln x\ln y}{(1-x)(1-y)} \, dx \, dy $$

$$\sum_{n=1}^{+\infty} (-1)^{n-1} (\psi_1(n))^2 = \sum_{n=1}^{+\infty} ( (\psi_1 (2n-1))^2-(\psi_1(2n))^2) $$

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

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

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

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Martin.s
Martin.s
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