What is the closed form of $\sum_{n\geq 1}(-1)^{n-1}\psi'(n)^2$?

Question

This problem was proposed by Cornel Ioan Valean.

What is the closed form of $$ S=\sum_{n\geq 1}(-1)^{n+1}\psi'(n)^2 $$ ?

I recall that $\psi'(z)=\frac{d^2}{dz^2}\log\Gamma(z)=\sum_{m\geq 0}\frac{1}{(m+z)^2}$ for any $z>0$.
My attempt was to perform the following manipulation $$ \sum_{n\geq 1}(-1)^{n+1}\psi'(n)^2 = \sum_{\substack{m,n\geq 1 \\ \min(m,n)\text{ odd}}}\frac{1}{m^2 n^2} \tag{A}$$ in order to turn the original series into $$\begin{eqnarray*} 2\sum_{n\geq 0}\frac{\zeta(2)-H_{2n+1}^{(2)}}{(2n+1)^2}+\sum_{n\geq 0}\frac{1}{(2n+1)^4}&=&\frac{5\pi^4}{96}-2\sum_{n\geq 0}\frac{H_{2n+1}^{(2)}}{(2n+1)^2}\\&=&\frac{19\pi^4}{1440}+\frac{1}{2}\color{blue}{\sum_{n\geq 1}\frac{H_{2n}^{(2)}}{n^2}}\end{eqnarray*} \tag{B}$$ not so bad after all. My issue is that now I am not able to find a decent closed form for the last series. If we apply summation by parts we have: $$\begin{eqnarray*} \color{blue}{\sum_{n\geq 1}\frac{H_{2n}^{(2)}}{n^2}}&=&\frac{\pi^4}{36}-\sum_{m\geq 1}\frac{H_m^{(2)}}{(2m+2)^2}-\sum_{m\geq 1}\frac{H_m^{(2)}}{(2m+1)^2}\\&=&\frac{37 \pi^4}{1440}-\color{green}{\sum_{m\geq 1}\frac{H_{m}^{(2)}}{(2m+1)^2}}\end{eqnarray*}\tag{C} $$ but the green series does not seem to be really "better" than the blue one.
Maybe it is relevant to point out that $$ \color{green}{\sum_{m\geq 1}\frac{H_{m}^{(2)}}{(2m+1)^2}} = -\int_{0}^{1}\frac{\text{Li}_2(z^2)}{1-z^2}\log(z)\,dz.\tag{D}$$

Do you see a way to tackle the last series, or the original one through a different approach? Numerically, $S\approx 2.3949463369266426$.

Answer 1

I'll start from the last integral

By integration by parts

$$\int^1_0 \frac{\log(z)}{1-z^2}\mathrm{Li}_2(z^2)\, dz = -\frac{π^4}{144}+\int^1_0\frac{\log(1-z^2)(\log(z) \log(1 + z) + \mathrm{Li}_2(1 - z) + \mathrm{Li}_2(-z))}{z}dz$$

Separate to reduce to the evaluation fo three integrals

$$\int^1_0 \frac{\log(z)}{1-z^2}\mathrm{Li}_2(z^2)\, dz = -\frac{π^4}{144}+\color{Red}{I_1}+\color{blue}{I_2}+\color{green}{I_3}$$

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Let us start by $\color{Red}{I_1}$ \begin{align}\color{Red}{I_1}&=\int^1_0\frac{\log(1-z^2)\log(z) \log(1 + z) }{z}\,dz \\ &= \int^1_0\frac{\log(z)\log^2(1+z)}{z}\,dz+\int^1_0\frac{\log(1-z)\log(z) \log(1 + z)}{z}dz \\ &=\color{brown}{I_4}+\color{purple}{I_5}\end{align}

Note that $\color{brown}{I_4}$ has been already evaluated to

$$\color{brown}{I_4}=\int_0^1\frac{\log^2(1+z)\log(z)}z\mathrm dz =\frac{\pi^4}{24}-\frac16\ln^42+\frac{\pi^2}6\ln^22-\frac72\zeta(3)\ln2-4\operatorname{Li}_4\!\left(\tfrac12\right)$$

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Let us look for $\color{blue}{I_2}$ \begin{align}\color{blue}{I_2}=\int^1_0\frac{\log(1-z^2)\mathrm{Li}_2(1 - z)}{z}dz &= \int^1_0 \frac{\log(1-z^2)(\zeta(2)-\log(z)\log(1-z)-\mathrm{Li}_2(z))}{z}\,dz\\ &= -\frac{π^4}{72}-\int^1_0 \frac{\log(z)\log^2(1-z)}{z}\,dz-\color{purple}{I_5}\\&-\int^1_0 \frac{\log(1-z^2)\mathrm{Li}_2(z)}{z}\,dz \\&= -\int^1_0 \frac{\log(z)\log^2(1-z)}{z}\,dz-\int^1_0 \frac{\log(1+z)\mathrm{Li}_2(z)}{z}\,dz -\color{purple}{I_5}\\ &=3\zeta(4)-\zeta^2(2)-\int^1_0 \frac{\log(1+z)\mathrm{Li}_2(z)}{z}\,dz-\color{purple}{I_5}\\ &=-\frac{π^4}{120}+\int^1_0 \frac{\log(1-z)\mathrm{Li}_2(-z)}{z}\,dz-\color{purple}{I_5}\\ &=-\frac{π^4}{120}+\color{#5D8AA8}{I_6}-\color{purple}{I_5}\end{align}

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Now we need to evaluate $\color{green}{I_3}$

\begin{align}\color{green}{I_3}=\int^1_0\frac{\log(1-z^2)\mathrm{Li}_2(-z)}{z}dz &= \int^1_0 \frac{\log(1-z)\mathrm{Li}_2(-z)}{z}\,dz+\int^1_0 \frac{\log(1+z)\mathrm{Li}_2(-z)}{z}\,dz\\ &=\color{#5D8AA8}{I_6} -\frac{\pi^4}{288}\end{align}

The mysterious $\color{#5D8AA8}{I_6}$ is \begin{align}\color{#5D8AA8}{I_6}=\int^1_0 \frac{\log(1-z)\mathrm{Li}_2(-z)}{z} \,dz&=-\sum_{k=1}^\infty \frac{1}{k}\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\int^1_0 z^{k+n-1}\,dz\\ &=-\sum_{k=1}^\infty \frac{1}{k}\sum_{n=1}^\infty \frac{(-1)^n}{n^2(n+k)} \\&=-\sum_{n=1}^\infty \frac{(-1)^{n-1} H_n}{n^3}\\ &=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3) \end{align}

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Collecting the results together

\begin{align}\int^1_0 \frac{\log(z)}{1-z^2}\mathrm{Li}_2(z^2)\, dz &= -\frac{π^4}{144}+\color{Red}{I_1}+\color{blue}{I_2}+\color{green}{I_3} \\&=-\frac{π^4}{144}+\color{brown}{I_4}+\color{purple}{I_5}-\frac{π^4}{120}+\color{#5D8AA8}{I_6}-\color{purple}{I_5}+\color{#5D8AA8}{I_6} -\frac{\pi^4}{288} \\&=\frac{121 π^4}{1440} + \frac{1}{3} π^2 \log^2(2) - \frac{1}{3}\log^4(2) - 7 \log(2) ζ(3)- 8 \mathrm{Li}_4\left(\frac{1}{2}\right)\end{align}

Hence we get the final result

\begin{align}\sum_{n =1}^\infty (-1)^{n+1}\psi'(n)^2 &=\frac{49 π^4}{720} + \frac{1}{6} π^2 \log^2(2) - \frac{1}{6}\log^4(2) - \frac{7}{2} \log(2) ζ(3)- 4 \mathrm{Li}_4\left(\frac{1}{2}\right) \\&\approx 2.3949463369266426 \end{align}

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References

Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$

How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$

Answer 2

Assume that $$\sum_{n\geq1}\frac{H_{n}^{\left(2\right)}}{n^{2}}z^{n}=f\left(z\right),\,\left|z\right|\leq1.\tag{1}$$ Then if we take $z=\exp\left(\pi i\right)$ we get $$f\left(\exp\left(\pi i\right)\right)=\sum_{n\geq1}\frac{H_{n}^{\left(2\right)}}{n^{2}}\exp\left(n\pi i\right)=\sum_{n\geq1}\frac{H_{2n}^{\left(2\right)}}{\left(2n\right)^{2}}-\sum_{n\geq1}\frac{H_{2n-1}^{\left(2\right)}}{\left(2n-1\right)^{2}}$$ and taking $z=-\exp\left(\pi i\right)$ we have $$f\left(-\exp\left(\pi i\right)\right)=\sum_{n\geq1}\frac{H_{n}^{\left(2\right)}\left(-1\right)^{n}}{n^{2}}\exp\left(n\pi i\right)=\sum_{n\geq1}\frac{H_{2n}^{\left(2\right)}}{\left(2n\right)^{2}}+\sum_{n\geq1}\frac{H_{2n-1}^{\left(2\right)}}{\left(2n-1\right)^{2}}$$ hence $$\frac{1}{2}\sum_{n\geq1}\frac{H_{2n}^{\left(2\right)}}{n^{2}}=f\left(\exp\left(\pi i\right)\right)+f\left(-\exp\left(\pi i\right)\right).$$ So we have to find the closed form of $(1)$. From the generating function $$\sum_{n\geq1}H_{n}^{\left(2\right)}z^{n}=\frac{\textrm{Li}_{2}\left(z\right)}{1-z}$$ we have, twice integrating and dividing $z$, that $$\sum_{n\geq1}\frac{H_{n}^{\left(2\right)}}{n^{2}}z^{n}=3\textrm{Li}_{4}\left(z\right)+\frac{1}{2}\textrm{Li}_{3}^{2}\left(z\right)-2\sum_{n\geq1}\frac{H_{n}}{n^{3}}z^{n}\tag{2}$$ and the closed form of the series in $(2)$ can be found here, answers of Tunk-Fey and Robert Israel, so we have essentially done. I'm too lazy to do all the calculations.