Rare sum with Trigamma function $\left( -1 \right)^{n-1}\left( \psi _{1}\left( n \right) \right)^{2}$

Question

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

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Here is my procedure, but I do not know how to calculate the last integral. Anyone have any ideas?

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

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

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

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

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

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

Answer 1

$$S=\sum_{n=1}^\infty (-1)^{n-1}(\psi^{(1)}(n))^2$$ Using the integral representation of trigamma function,

$$ \sum_{k=1}^\infty (-1)^{k-1} (\psi^{(1)}(k))^2 = \int_0^1 \int_0^1 \frac{\ln(y)\ln(x)}{(1-x)(1-y)}\sum_{k=1}^\infty (-1)^{k-1}(x)^{k-1}(y)^{k-1}\,dx\,dy $$

Using the below, $$\color{red}{\sum_{k=1}^\infty (-1)^{k-1}(x)^{k-1}(y)^{k-1}=\frac{1}{1+xy}}$$

$$ S= \int_0^1 \int_0^1 \frac{\ln(y)\ln(x)}{(1-x)(1-y)(1+xy)} dx\,dy$$

$$S = \int_0^1 \frac{\ln(x)}{1-x} \left( \int_0^1 \frac{\ln(y)}{(1-y)(1+xy)} \,dy \right) \,dx $$

Dealing with the inner integral,

$$ \implies\int_0^1 \frac{\ln(y)}{(1-y)(1+xy)} dy \underset{\text{partial fraction}}= \frac{x}{x+1} \int_0^1 \frac{\ln(y)}{1+xy} dy + \frac{1}{1+x}\, \int_0^1 \frac{\ln(y)}{1-y} dy $$

We know that $\frac{\text{Li}_2(-x)}{x}=\int_0^1 \frac{\ln(y)}{1+xy}\,dy$ (It's from standard definition)

We know that $-\zeta(2)=\int_0^1 \frac{\ln(y)}{1-y}\, dy$ (It's from here)

Therefore,

$$\implies \frac{\operatorname{Li}_2(-x) - \zeta(2)}{1+x}$$

Plug this result into our integral $S$,

$$ S = \int_0^1 \frac{\ln(x)\text{Li}_2(-x)}{1-x^2}\, dx - \zeta(2) \int_0^1 \frac{\ln(x)}{1-x^2}\, dx $$

We know that $\int_0^1 \frac{\ln(x)}{1-x^2}\, dx =-\frac{\pi^2}{8}$ (Took from here)

The only hard part is this integral,

$$\int_0^1 \frac{\ln(x)\operatorname{Li}_2(-x)}{1-x^2}\, dx$$

I applied partial fractions,

$$\implies \frac12\int_0^1 \frac{\ln(x)\operatorname{Li}_2(-x)}{1+x}\, dx+\frac12\int_0^1 \frac{\operatorname{Li}_2(-x)}{1-x}\, dx$$

Using the below formula from here,

$$\color{red}{\sum_{n=1}^\infty H_n^{(p)}x^n=\frac{\operatorname{Li}_p(x)}{1-x}\underset{\text{p=2, x=-x}}\implies \sum_{n=1}^\infty (-1)^nH_n^{(2)}x^n=\frac{\operatorname{Li}_2(-x)\ln(x)}{1+x}}$$

We have these,

$$\implies \frac12\sum_{n=1}^\infty (-1)^nH_n^{(2)}\int_0^1 x^n \ln(x)\, dx+\frac12\int_0^1 \frac{\operatorname{Li}_2(-x)\ln(x)}{1-x}\, dx$$

Now umm for this second integral, we can stick to using the definition

$$\implies \frac12\sum_{n=1}^\infty (-1)^nH_n^{(2)}\int_0^1 x^n \ln(x)\, dx+\frac12\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 \frac{x^n\ln(x)}{1-x}\, dx$$

We know that $\int_0^1 \frac{x^n \ln(x)}{1 - x} \, dx = H_n^{(2)} - \zeta(2)$

$$\implies \frac12\sum_{n=1}^\infty (-1)^nH_n^{(2)}\int_0^1 x^n \ln(x)\, dx+\frac12\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 \frac{x^n\ln(x)}{1-x}\, dx$$

$$\implies \frac{1}{2} \sum_{n=1}^{\infty} (-1)^n \left( \frac{1}{n^2} - H_n^{(2)} \right) \int_0^1 x^{n-1} \ln(x) \, dx + \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \left( H_n^{(2)} - \zeta (2) \right)$$

$$\implies -\frac{1}{2} \sum_{n=1}^{\infty} (-1)^n \left( \frac{1}{n^2} - H_n^{(2)} \right) \left( \frac{1}{n^2} \right) + \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \left( H_n^{(2)} - \zeta (2) \right)$$

Using the polylogarithm definition,

$$\implies \frac{1}{2} \sum_{n=1}^{\infty}(-1)^n \frac{H_n^{(2)}}{n^2} - \frac{1}{2} \operatorname{Li}_4 (-1) + \frac{1}{2} \sum_{n=1}^{\infty} (-1)^n\frac{H_n^{(2)}}{n^2} - \frac{1}{2} \zeta (2) \operatorname{Li}_2 (-1)$$

Using standard values I took from WA,

$$\implies \frac{17}{1440} \pi^4+\sum_{n=1}^{\infty} (-1)^n\frac{H_n^{(2)} }{n^2}$$

I'm a little lazy to come up with a nicer proof than the existing one, see here,

$$\sum_{n=1}^{\infty}(-1)^n\frac{H_n^{(2)}}{n^2}=-4\operatorname{Li}_4\left(\frac12\right)+\frac{51}{16}\zeta(4)-\frac72\ln(2)\zeta(3)+\ln^2(2)\zeta(2)-\frac16\ln^4(2)$$

So that completes our integral,

We can now add up all our results,

$$\sum_{n=1}^\infty (-1)^{n-1}(\psi^{(1)}(n))^2 = \frac{49}{720}\pi^4+ \zeta(2) \ln^2(2) - \frac{7}{2} \ln(2) \, \zeta(3) - \frac{1}{6} \ln^4(2) - 4\operatorname{Li}_4 \left(\frac{1}{2}\right)$$

Numerical checks are here and here