Closed form of the sum $s_4 = \sum_{n=1}^{\infty}(-1)^n \frac{H_{n}}{(2n+1)^4}$

Question

I am interested to know if the following sum has a closed form

$$s_4 = \sum_{n=1}^{\infty}(-1)^n \frac{H_{n}}{(2n+1)^4}\tag{1}$$

I stumbled on this question while studying a very useful book about harmonic series and logarithmic integrals whichh has appeared recently [1]. Checking it for possible missing entries I was led to consider this family alternating Euler sums

$$s(a) = \sum_{n=1}^{\infty}(-1)^n \frac{H_{n}}{(2n+1)^a}$$

where $H_{n} =\sum_{k=1}^{n}\frac{1}{k}$ is the harmonic number, as well as the corresponding integrals

$$i(a) = \int_0^1 \frac{\log ^{a-1}\left(\frac{1}{x}\right) \log \left(x^2+1\right)}{x^2+1} \, dx$$

These are related by

$$i(a) = - \Gamma(a) s(a)$$

As listed in detail in the appendix, closed forms exist for all odd $a=2q+1$. For even $a=2q$ only the case $a=2, q=1$ is known.

Hence, the natural question is to aks for a closed form for the smallest case open up to now, $a = 4$.

What did I do so far

The integral $i(a)$ can be found by differentiation with respect to a parameter $u$ from the generating integral

$$g_i(u) = \int_0^1 \frac{t^u \log \left(t^2+1\right)}{t^2+1} \, dt$$

which is evaluated by Mathematica in terms of hypergeometric functions as follows

$$g_i(u) = \frac{1}{4} \left(-\frac{2^{\frac{u+5}{2}} \, _3F_2\left(\frac{1}{2}-\frac{u}{2},\frac{1}{2}-\frac{u}{2},\frac{1}{2}-\frac{u}{2};\frac{3}{2}-\frac{u}{2},\frac{3}{2}-\frac{u}{2};\frac{1}{2}\right)}{(u-1)^2}\\-2 \pi H_{-\frac{u}{2}-\frac{1}{2}} \sec \left(\frac{\pi u}{2}\right)-\log (4) B_{\frac{1}{2}}\left(\frac{1-u}{2},\frac{u+1}{2}\right)\right)$$

As the parameter $u$ appears in 7 places each derivative generates a factor 7 in the length of the result.

Unless someone comes up with a very clever idea to simplify the hypergeometric expressions this path seems to be hopeless.

Appendix: known closed forms

For positive integer values of $a$ the following results have been obtained:

a) for odd $a=2q+1$

the closed form was calculated in [1], 4.1.15 (4.91) as:

$$s(2q+1) = (2q+1)\beta(2q+2) + \frac{\pi}{(2q)! 4^{q+1}}\lim_{m\to \frac12 }\frac{\mathrm{d}^{2q}}{\mathrm{d} m^{2q}} \frac{\psi(1-m) + \gamma}{\sin(m\pi)}$$

Here

$$\beta(z)=\sum_{k=1}^{\infty}\frac{(-1)^k}{2k+1)^z}$$

is Dirichelt's beta function.

As for a simplification of r.h.s. see https://math.stackexchange.com/a/4139359/198592

b) for even $a=2q$

there is a closed form just for $a=2, i.e. q=1$, found in [1] 4.5.5 (4.187)

$$s(2) = 2 \;\Im \text{Li}_3(1-i)+ \frac{3\pi^3}{32}+\frac{\pi}{8}\log^2(2) -\log(2) G$$

where $G = \beta(2)$ is Catalan's constant.

References

[1] Ali Shadhar Olaikhan, "An introduction to harmonic series and logarithmic integrals", April 2021, ISBN 978-1-7367360-0-5

Answer 1

This constant mentioned by OP probably cannot be expressed in terms of ordinary polylogarithm, at least not the familiar $\operatorname{Li}_n(1/2), \operatorname{Li}_n((1\pm i)/2), \operatorname{Li}_n(1 \pm i)$. However, we do have a hypergeometric representation.

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The series given by OP is $-6$ times the integral $$\int_0^1 \frac{\log ^{3}\left(x\right) \log \left(x^2+1\right)}{x^2+1} \, dx = -6 \Im(\text{QMZ}(4,\{4,1\},\{1,0\}))-6 \Im(\text{QMZ}(4,\{4,1\},\{1,2\}))$$ here $$\text{QMZ}(4,\{4,1\},\{1,0\}) = \sum_{n>m\geq 1}\frac{i^n}{n^4m}$$ $$\text{QMZ}(4,\{4,1\},\{1,2\}) = \sum_{n>m\geq 1}\frac{i^n (-1)^m}{n^4m}$$

In the paper, it is proved that ($C$ is Catalan constant, $\zeta(\cdot,\cdot)$ is Hurwitz's zeta)

so we can solve for the underlined terms, and conclude that the integral, hence original series, can be written as two ${_6F_5}$.

Answer 2

The interest of pisco and FDP encouraged me try out simple transformations on the integral.

1) Integration by parts

IBP with

$$U=\int \frac{\log \left(t^2+1\right)}{t \left(t^2+1\right)} \, dt = -\frac{\text{Li}_2\left(-t^2\right)}{2}-\frac{1}{4} \log ^2\left(t^2+1\right)$$

$$V=t \log ^3(t)$$

leads to

$$\begin{align}i(3) =& -3 (4 C+2 i \text{Li}_3(-i)-2 i \text{Li}_3(i)+2 i \text{Li}_4(-i)-2 i \text{Li}_4(i)\\&+\pi -16+\log (4))+\frac{3}{4} A+\frac{1}{4}B\end{align}\tag{s1.1}$$

Where C = Catalan's constant and

$$A = \int_0^1 \log ^2(t) \log ^2\left(t^2+1\right) \, dt\tag{s1.2}$$

$$B = \int_0^1 \log ^3(t) \log ^2\left(t^2+1\right) \, dt\tag{s1.3}$$

2) Substitution of integration variable

Letting $x\to \tan(\phi)$ we obtain

$$i(3) = \int_0^{\frac{\pi }{4}} \log ^3(\tan (\phi )) \log \left(\sec ^2(\phi )\right) \, d\phi\tag{s2.1}$$

Observing $\log \left(\sec ^2(\phi )\right) = - 2 \log \left(\cos(\phi)\right) $ and expanding $\log ^3(\tan (\phi ))=\left(\log(\sin(\phi)) -\log(\cos(\phi))\right)^3 $

we end up with four nice integrals of the type

$$i(p,q) = \int_0^{\frac{\pi }{4}} \log ^p(\cos (\phi )) \log ^q(\sin (\phi )) \, d\phi\tag{s2.2}$$

I was able (via the antiderivative, using Mathematica) to solve only this one:

$$\begin{align}i(4,0)=\int_0^{\frac{\pi }{4}} \log ^4(\cos (\phi )) \, d\phi =-\frac{1}{480} \pi \left(15 \left(48 \zeta (3) \log (4)+\log ^4(4)\right)+19 \pi ^4+30 \pi ^2 \log ^2(4)\right)+\frac{1}{192} \left(48 \left(48 \sqrt{2} \, _6F_5\left(\{\frac{1}{2}\}^6;\{\frac{3}{2}\}^5;\frac{1}{2}\right)\\ +\log (2) \left(24 \sqrt{2} \, _5F_4\left(\{\frac{1}{2}\}^5;\{\frac{3}{2}\}^4,\frac{3}{2};\frac{1}{2}\right)\\ +\sqrt{2} \log (64) \, _4F_3\left(\{\frac{1}{2}\}^4;\{\frac{3}{2}\}^3;\frac{1}{2}\right)\\ -2 i \text{Li}_2\left(-\frac{1+i}{\sqrt{2}}\right) \log ^2(2)+2 i \text{Li}_2\left(1-\frac{1+i}{\sqrt{2}}\right) \log ^2(2)\right)\right)\\ -11 i \pi ^2 \log ^3(2)+3 \pi \left(\log (2)+8 \log \left(1+\frac{1+i}{\sqrt{2}}\right)\right) \log ^3(2)\right)\end{align}\tag{s2.3}$$

Numerically we have

$$i(4,0) \simeq 0.00115068$$.

Notice that the components in $(s2.3)$ are similar to those of @pisco's answer: hypergeometric functions of the order up to (6/5), and (poly)logs. No derivatives of the hypergeometric functions appear.

The other three integrals $i(3,1)$, $i(2,2)$, and $i(3,1)$ have resisted up to now and are open to attacks.

3) Collecting knowledge and tentative bottom line

Being grateful for the hints I have consulted this article here and the references therein, and the answer of pisco.

I have not found any expression for e.g. $i(4,0)$ which does not contain at least one hypergeometric function.

Hence may well be that the list of admissible components for "closed forms" must be enlarged to include hypergeometric functions.

After all, also roots of polynomials can't always be expressed in terms of closed forms, called radicals, if the order surpasses a critical value ...