Quadratic Euler sums $\sum _{n=1}^{\infty } \frac{(-1)^{n-1} \widetilde H(n)^3}{2 n+1}$

Answering part of the question of @Ali Shather in a comment I have derived an integral representation for the second sum.

Result

We write

$$s_2 = \sum _{n=1}^{\infty } \frac{(-1)^{n-1}\widetilde H(n) H(n)^2}{2 n+1} = \int_0^1 i_2(u) \,du\tag{1}$$

with the integrand

$$i_2(u) = \sum _{n=1}^{\infty }(-1)^{n-1}u^{2 n}\widetilde H(n) H(n)^2\tag{2}$$

Notice that the denominator $2n+1$ is generated by the integration over $u$.

I have calculated the sum in $(2)$ with a the following result which is composed of $\text{Li}_3$, $\text{Li}_2$, $\zeta(3)$ and $\log$s,

$$i_{2}(u)= -\frac{1}{24 \left(u^2+1\right)}\left(-24 \text{Li}_3\left(1-u^2\right)+6 \text{Li}_3\left(\left(\frac{1-u^2}{u^2+1}\right)^2\right)-24 \text{Li}_3\left(\frac{1}{u^2+1}\right)+24 \text{Li}_3\left(\frac{u^2}{u^2+1}\right)-24 \text{Li}_3\left(\frac{2 u^2}{u^2+1}\right)+24 \text{Li}_3\left(\frac{1}{2} \left(u^2+1\right)\right)+24 \log (2) \text{Li}_2\left(1-u^2\right)+24 \text{Li}_2\left(\frac{1}{u^2+1}\right) \log \left(\frac{2}{u^2}+2\right)-24 \text{Li}_2\left(\frac{u^2}{u^2+1}\right) \log \left(\frac{u^2}{u^2+1}\right)+24 \text{Li}_2\left(\frac{2 u^2}{u^2+1}\right) \log \left(\frac{2 u^2}{u^2+1}\right)-24 \text{Li}_2\left(\frac{1-u^2}{u^2+1}\right) \log \left(\frac{1}{u^2}-u^2\right)-24 \text{Li}_2\left(-\frac{1-u^2}{u^2+1}\right) \log \left(\frac{1-u^2}{u^2+1}\right)+24 \text{Li}_2\left(\frac{1}{2} \left(u^2+1\right)\right) \log \left(\frac{2}{u^2+1}\right)+8 \log ^3\left(u^2+1\right)+60 \log (2) \log ^2\left(u^2+1\right)+8 \log ^2(2) \left(\log (2)-3 \log \left(u^2+1\right)\right)+8 \log \left(1-u^2\right) \left(-6 \left(\log \left(1-u^2\right)+2 \log \left(\frac{1}{2} \left(u^2+1\right)\right)\right) \log (u)+\log \left(1-u^2\right) \left(\log \left(\frac{1}{8} \left(1-u^2\right)\right)+3 \log \left(u^2+1\right)\right)+12 \log ^2(u)\right)-96 \log (2) \log (u) \log \left(u^2+1\right)-4 \pi ^2 \log \left(\frac{4}{1-u^2}\right)+21 \zeta (3)\right)\tag{3}$$

The corresponding Mathematica expression is provided in the appendix.

For a first check, the numeric evaluation of the integral gives with WorkingPrecision 30:

$$N(s_2) \simeq 0.9497576952344695293296230|77234\tag{4}$$

@metamorphory claimed to have calculated many more digits

$$0.9497576952344695293296230|8973659924...\tag{5}$$

Before the $"|"$ both values conincide. I take this as a strong ndication that my expression for $s_2$ is correct.

Derivation

In order to calculate the sum in $(2)$ we have several choices for the harmonic numbers involved. I found it useful to leave one factor $H_n$ as is and replace the other two be their integral representations

$$H_n \to \int_0^1 \frac{1-x^n}{1-x}\tag{6a}$$

$$\widetilde H(n) \to \int_0^1 \frac{1-(-y)^n}{1+y}\tag{6b}$$

The sum in $(5)$ can then be done resulting in the following double integral for the integrand

$$i(u) = \int_{[0,1]^2} \left(\frac{\log \left(u^2+1\right)}{\left(u^2+1\right) (x+1) (1-y)}+\frac{\log \left(1-u^2 x\right)}{(x+1) (1-y) \left(u^2 x-1\right)}-\frac{\log \left(u^2 y+1\right)}{(x+1) (1-y) \left(u^2 y+1\right)}-\frac{\log \left(1-u^2 x y\right)}{(x+1) (1-y) \left(u^2 x y-1\right)}\right)\,dx\,dy\tag{7}$$

It turned out that it is convernient to take the $y$-integral first, for which Mathematica gives luckily directly (without the intermedaiate step of finding an antidrivative) this expression

$$i_y(u,x)=\frac{1}{6 \left(u^2+1\right) (x+1) \left(u^2 x-1\right)}\left(\text{Li}_2\left(\frac{1}{u^2+1}\right) \left(6-6 u^2 x\right)+6 \left(u^2+1\right) \text{Li}_2\left(1-u^2 x\right)+\pi ^2 \left(u^2 (x-1)-2\right)+\left(9-9 u^2 x\right) \log ^2\left(u^2+1\right)+12 \left(u^2 x-1\right) \log (u) \log \left(u^2+1\right)+6 \left(u^2+1\right) \log \left(\frac{u^2 x}{1-u^2 x}\right) \log \left(1-u^2 x\right)\right)$$

For the final $x$-integral Mathematica (again directly) gives a long expression containing several complex terms, some explicit (factor i in front), others implicit (polylogs and logs with arguments outside the range leading to real expressions).

A careful analysis guided by the aim to make all terms real quatities using transformation formulas of polylogs leads to a lot of cancellations of complex terms and in the end leaves for the complex part of $i(u)$ this expression with a factor $i$ in front

$$z=-12 \text{Li}_2\left(\frac{1}{u^2+1}\right)+12 \text{Li}_2\left(\frac{1-u^2}{u^2+1}\right)-12 \text{Li}_2\left(\frac{u^2-1}{u^2+1}\right)+12 \text{Li}_2\left(1-\frac{1}{u^2}\right)-6 \log ^2\left(u^2+1\right)+24 \log (u) \log \left(\frac{u^2}{u^2+1}\right)+24 \log (u) \log \left(u^2+1\right)-24 \log ^2(u)+\pi ^2-12 \log ^2(2)+\log (4096) \log (2)$$

This, however, is numerically zero, and it is also simplified to exactly zero.

The rest of $i$ is the expression $i_2$ given in $(2)$ which contains only real terms.

This ends the derivation.

Discussion

Trying the final $u$-integration we encounter several non-trivial integrals.

One of the simpler ones is

$$\int_0^1 \frac{\log (u) \log (1-u)}{u^2+1} \, du=-\Im\left(\text{Li}_3\left(\frac{1}{2}-\frac{i}{2}\right)\right)-\frac{\pi ^3}{128}-\frac{1}{32} \pi \log ^2(2)$$

To do the $u$-integral completely and hereby confirm the closed form provided in the OP seems to require a huge amount of calculations.

Maybe the integrand can be significantly simplified which might facilitate the integration.

Facing the difficulties it is gratifying that the closed form was found by @Edit profile and settings using experimental mathematics and based on the high precision numerics provided by @metamorphy.

Appendix

The Mathematica expression of the integrand is

i2 = -(1/(
   24 (1 + u^2))) (-4 \[Pi]^2 Log[-(4/(-1 + u^2))] + 
    8 Log[2]^2 (Log[2] - 3 Log[1 + u^2]) - 
    96 Log[2] Log[u] Log[1 + u^2] + 60 Log[2] Log[1 + u^2]^2 + 
    8 Log[1 + u^2]^3 + 
    8 Log[1 - u^2] (12 Log[u]^2 - 
       6 Log[u] (Log[1 - u^2] + 2 Log[1/2 (1 + u^2)]) + 
       Log[1 - u^2] (Log[-(1/8) (-1 + u) (1 + u)] + 
          3 Log[1 + u^2])) + 24 Log[2] PolyLog[2, 1 - u^2] + 
    24 Log[2 + 2/u^2] PolyLog[2, 1/(1 + u^2)] - 
    24 Log[u^2/(1 + u^2)] PolyLog[2, u^2/(1 + u^2)] + 
    24 Log[(2 u^2)/(1 + u^2)] PolyLog[2, (2 u^2)/(1 + u^2)] + 
    24 Log[2/(1 + u^2)] PolyLog[2, 1/2 (1 + u^2)] - 
    24 Log[-1 + 2/(1 + u^2)] PolyLog[2, 1 - 2/(1 + u^2)] - 
    24 Log[1/u^2 - u^2] PolyLog[2, -1 + 2/(1 + u^2)] - 
    24 PolyLog[3, 1 - u^2] - 24 PolyLog[3, 1/(1 + u^2)] + 
    24 PolyLog[3, u^2/(1 + u^2)] - 24 PolyLog[3, (2 u^2)/(1 + u^2)] + 
    24 PolyLog[3, 1/2 (1 + u^2)] + 
    6 PolyLog[3, (1 - 2/(1 + u^2))^2] + 21 Zeta[3])

Tags:

Sequences And Series

Harmonic Numbers

Closed Form

Euler Sums

Related

limit related to the Lambert function Integrating $\sin(x)x^2$ by parts, why do we only add $C$ at the end? Question about product of generating functions in a proof that for positive integer $n$, $\sum_{k=0}^n(-1)^k\binom nk\binom{2n-k}n=1$. Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$ Question about Spivak's proof of how to use u-substitution when the derivative of the inner function does not appear in the integral Solve the following equation in integers $x,y:$ $x^2+6xy+8y^2+3x+6y=2.$ A sequence of quadratic equations Limit $\lim_{n\to \infty} \frac{1}{n^{n+1}}\sum_{k=1}^n k^p$ Is there a smooth, preferably analytic function that grows faster than any function in the sequence $e^x, e^{e^x}, e^{e^{e^x}}...$ Countability of a sequence of natural numbers Characteristic classes, Möbius strip, and the cylinder Why is it the (group) morphisms that matter?

Recent Posts