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.

Answer 3

A solution using Abel's summation as suggested by Cornel.

Let $\ \displaystyle S\ $ denote $\ \displaystyle \sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^2}\ $ and by using Abel's summation:

$\displaystyle\sum_{k=1}^n a_k b_k=A_nb_{n+1}+\sum_{k=1}^{n}A_k\left(b_k-b_{k+1}\right)\ $ where $\ \displaystyle A_n=\sum_{i=1}^n a_i\ $

and letting let $\ \displaystyle a_k=\frac{1}{(2k+1)^2}\ $ , $\ \displaystyle b_k=H_k^{(2)}$, we get

\begin{align} \sum_{k=1}^n\frac{H_k^{(2)}}{(2k+1)^2}&=\sum_{i=1}^n\frac{H_{n+1}^{(2)}}{(2i+1)^2}-\sum_{k=1}^n\frac{1}{(k+1)^2}\left(\sum_{i=1}^k\frac{1}{(2i+1)^2}\right)\\ &=\sum_{i=1}^n\frac{H_{n+1}^{(2)}}{(2i+1)^2}-\sum_{k=1}^n\frac{1}{(k+1)^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}+\frac{1}{(2k+1)^2}-1\right) \end{align} Letting $n$ approach $\infty$, we get \begin{align} S&=\zeta(2)\sum_{i=1}^\infty\frac{1}{(2i+1)^2}-\sum_{k=1}^\infty\frac{1}{(k+1)^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}\right)\\ &\quad-\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}+\sum_{k=1}^\infty\frac1{(k+1)^2}\\ &=\zeta(2)\left(\frac34\zeta(2)-1\right)-\sum_{k=1}^\infty\frac{1}{k^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}-\frac{1}{(2k-1)^2}\right)\\ &\quad-\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}+\zeta(2)-1\\ &=\frac{15}8\zeta(4)-1-\sum_{k=1}^\infty\frac{1}{k^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}\right)+\sum_{k=1}^\infty\frac{1}{k^2(2k-1)^2}-\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}\\ &=\frac{15}8\zeta(4)-1-\sum_{k=1}^\infty\frac{1}{k^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}\right)+1\\ &\quad+\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}-\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}\\ &=\frac{15}8\zeta(4)-\sum_{k=1}^\infty\frac{1}{k^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}\right)\\ &=\frac{15}8\zeta(4)-4\sum_{k=1}^\infty\frac{H_{2k}^{(2)}}{(2k)^2}+\frac14\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2}\\ &=\frac{15}8\zeta(4)-4\left(\frac12\sum_{k=1}^\infty\frac{H_{k}^{(2)}}{k^2}+\frac12\sum_{k=1}^\infty\frac{(-1)^kH_k^{(2)}}{k^2}\right)+\frac14\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2}\\ &=\frac{15}8\zeta(4)-\frac74\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2}-2\sum_{k=1}^\infty\frac{(-1)^kH_k^{(2)}}{k^2} \end{align} By plugging $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^nH_n^{(2)}}{n^2}=-4\operatorname{Li}_4\left(\frac12\right)+\frac{51}{16}\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac16\ln^42\ $

( proved here ) and $\ \displaystyle\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2}=\frac74\zeta(4)\ $, we get the closed form of $\ S$

Answer 4

Solution: As we know, the integral representation of the trigamma function is

$$ \psi^{(1)}(k) = \int_0^1 \frac{x^{k-1} \ln(x)}{1-x} \, dx $$

Using this, we get:

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

Now, summing over $k$, we have:

\begin{align} \sum_{k=1}^{\infty} (-1)^{k-1} \left( \psi^{(1)}(k) \right)^2 & = \int_0^1 \int_0^1 \frac{\ln(y) \ln(x)}{(1-x)(1-y)} \, dx \, dy \sum_{k=1}^{\infty} (-xy)^{k-1} \\ & = \int_0^1 \int_0^1 \frac{\ln(y) \ln(x)}{(1-x)(1-y)(1+xy)} \, dx \, dy \\ & = \int_0^1 \frac{\ln(x)}{1-x} \left( \int_0^1 \frac{\ln(y)}{(1-y)(1+xy)} \, dy \right) dx \end{align}

Let

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

We now compute the individual sums:

\begin{align} A &= \frac{x}{x+1} \sum_{n=1}^{\infty} (-x)^{n-1} \int_0^1 y^{n-1} \ln(y) \, dy + \frac{1}{x+1} \sum_{n=1}^{\infty} \int_0^1 y^{n-1} \ln(y) \, dy \\ &= \frac{x}{x+1} \sum_{n=1}^{\infty} (-x)^{n-1} \left( \frac{-1}{n^2} \right) + \frac{1}{x+1} \sum_{n=1}^{\infty} \left( \frac{-1}{n^2} \right) \\ &= \frac{1}{x+1} \sum_{n=1}^{\infty} \frac{(-x)^n}{n^2} - \frac{\zeta(2)}{x+1} = \frac{Li_2(-x)}{x+1} - \frac{\zeta(2)}{x+1} \end{align}

Using this result in the previous expression, we get:

\begin{align} S &= \int_0^1 \frac{\ln(x)}{1-x} \left( \frac{1}{1+x} \left( Li_2(-x) - \zeta(2) \right) \right) dx \\ &= \int_0^1 \frac{\ln(x) Li_2(-x)}{1-x^2} \, dx - \zeta(2) \int_0^1 \frac{\ln(x)}{1-x^2} \, dx \end{align}

Now, let us compute:

$$ B = \int_0^1 \frac{\ln(x)}{1-x^2} \, dx = \frac{1}{2} \int_0^1 \frac{\ln(x)}{1-x} \, dx + \frac{1}{2} \int_0^1 \frac{\ln(x)}{1+x} \, dx $$

We can simplify the sums:

\begin{align} B &= \frac{1}{2} \sum_{n=1}^{\infty} \int_0^1 x^{n-1} \ln(x) \, dx + \frac{1}{2} \sum_{n=1}^{\infty} (-1)^{n-1} \int_0^1 x^{n-1} \ln(x) \, dx \\ &= \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{-1}{n^2} \right) + \frac{1}{2} \sum_{n=1}^{\infty} (-1)^{n-1} \left( \frac{-1}{n^2} \right) \end{align}

\begin{aligned} & = \frac{-\zeta(2)}{2} + \frac{-\zeta(2)}{4} = \frac{-3 \zeta(2)}{4} = -\frac{\pi^2}{8} \ldots \ldots \ldots \ldots . . \, (5) \\ & \text{Let } C = \int_0^1 \frac{\ln(x) \, Li_2(-x)}{1 - x^2} \, dx \\ & = \frac{1}{2} \int_0^1 \frac{\ln(x) \, Li_2(-x)}{1+x} \, dx + \frac{1}{2} \int_0^1 \frac{\ln(x) \, Li_2(-x)}{1-x} \, dx \\ & = \frac{1}{2} \sum_{n=1}^{\infty} (-1)^n H_n^{(2)} \int_0^1 x^n \ln(x) \, dx + \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \int_0^1 \frac{x^n \ln(x)}{1-x} \, dx \\ & = \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) \\ & = \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) \\ & = \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^n H_n^{(2)}}{n^2} - \frac{1}{2} Li_4(-1) + \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^n H_n^{(2)}}{n^2} - \frac{1}{2} \zeta(2) Li_2(-1) \\ & = \sum_{n=1}^{\infty} \frac{(-1)^n H_n^{(2)}}{n^2} + \frac{17 \pi^4}{1440} \end{aligned} $$

We have:

$$ \sum_{n=1}^{\infty} \frac{(-1)^n H_n^{(2)}}{n^2} = \frac{51 \pi^4}{1440} - \frac{7}{2} \ln(2) \zeta(3) + \frac{\pi^2}{6} \ln^2(2) - \frac{1}{6} \ln^4(2) - 4 Li_4\left( \frac{1}{2} \right) $$

Then,

$$ C = \frac{17 \pi^4}{360} - \frac{7}{2} \ln(2) \zeta(3) + \frac{\pi^2}{6} \ln^2(2) - \frac{1}{6} \ln^4(2) - 4 Li_4\left( \frac{1}{2} \right) $$

Plugging (5) and (6) into (4), we get:

$$ S = \frac{49 \pi^4}{720} - \frac{7}{2} \ln(2) \zeta(3) + \frac{\pi^2}{6} \ln^2(2) - \frac{1}{6} \ln^4(2) - 4 Li_4\left( \frac{1}{2} \right) $$