What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}x^n\ ?$

Question

Recently, the generating function of order 2 for the alternating harmonic series was calculated (What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n\ ?$).

I would like take the next step to the order 3. Is there a closed form function here as well?

Defining $\overline{H}_n = \sum_{k=1}^{n} (-1)^{k+1}/k$ and

$$g_q(z) = \sum_{n=0}^{\infty}\frac{z^n}{n^q} \overline{H}_n\tag{1}$$

I ask the

Question

Can you calculate the function defined by the sum $g_3(z)$, i.e. express it through known functions?

My effort so far

The alternating harmonic series has an integral representation

$$\overline{H}_n= \int_0^1 \frac{1-(-1)^n x^n}{x+1} \, dx\tag{2}$$

Hence we can easily calculate its generating function

$$g_0(z) = \sum_{n=1}^{\infty} \overline{H}_n z^n = \int_0^1 \left(\sum _{n=1}^{\infty } \frac{\left(1-(-1)^n x^n\right) z^n}{x+1}\right) \, dx\\=\int_0^1 \frac{z}{(1-z) (x z+1)} \, dx=\frac{\log (z+1)}{1-z}\tag{3}$$

The next orders can of $g(z)$ be generated successively by dividing by $z$ and integrating, i.e.

$$g_{q+1}(z)=\int_0^z \frac{ g_{q}(z)}{z}\,dz, q=0,1,2,...\tag{4}$$

Because $g_2$ is known we could just plug it into $(4)$ and integrate. The problem is, however, that $g_2$ already consists of about 20 summands, and hence in the first place there are about 20 integrals to calculate. In view of the exploding number of possible transformations (substitution, partial integration, utilizing relations between the polylog functions involved etc.) it is highly desirabable to keep the number of integrals to be "cracked" as small as possible.

For the case of order 3 I have now boiled it down to just one (!) integral having derived (by partial integration) this formula

$$g_3(z) = g_2(z) \log(z) -\frac{1}{2} g_1(z) \log(z)^2 +\frac{1}{2}i(z)\tag{5}$$

where the remaining integral is

$$i(z) = \int_0^z \frac{\log(t)^2 \log(1+t)} {t(1-t)} \,dt\tag{6}$$

The integral is convergent in the range $0<z<1$. In fact, the integrand has the expansion

$$\frac{\log(t)^2 \log(1+t)} {t(1-t)} \underset{t\to0}\simeq \log ^2(t)\left(1+\frac{t}{2}+\frac{5 t^2}{6}+ O(t^3)\right) $$

and close to $z=1$ as well

$$\frac{\log(t)^2 \log(1+t)} {t(1-t)} \underset{t\to1}\simeq (1 - t)\log(2) + (1 - t)^2 \left(-\frac{1}{2} + 2 \log(2)\right)+O((1 - t)^3)$$

For $z<0$ the integral becomes complex as we can see from the example case

$$\int_0^{-\frac{1}{2}} \log ^2(t) \, dt=\frac{1}{2} \left(\pi ^2-2-\log ^2(2)+2 i \pi (1+\log (2))-\log (4)\right)$$

This is surprising because the original power series for the generating function $(1)$ is convergent for $|z|<1$ and hence defines a real function. The problem is resolved considering that the complete expression containes other terms which (in some way) compensate the singularities.

Here I'm stuck and I was not able to solve the integral. But as I know that there are many experienced and skilled experts in this forum I'm confident they can solve $i(z)$.

Discussion

I have moved the text to https://math.stackexchange.com/a/3544006/198592

Answer 1

Incomplete solution

$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}x^n=x+\sum_{n=2}^\infty\frac{\overline{H}_n}{n^3}x^n$$

By using

$$\sum_{n=2}^\infty f(n)=\sum_{n=1}^\infty f(2n+1)+\sum_{n=1}^\infty f(2n)$$

we have

$$\Longrightarrow \sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}x^n=x+\sum_{n=1}^\infty\frac{\overline{H}_{2n+1}}{(2n+1)^3}x^{2n+1}+\sum_{n=1}^\infty\frac{\overline{H}_{2n}}{(2n)^3}x^{2n}$$

now use $$\overline{H}_{2n+1}=H_{2n+1}-H_n, \quad \overline{H}_{2n}=H_{2n}-H_n$$

We get that

$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}x^n=\color{blue}{x+\sum_{n=1}^\infty\frac{H_{2n+1}}{(2n+1)^3}x^{2n+1}+\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}x^{2n}}-\frac18\sum_{n=1}^\infty\frac{H_n}{n^3}x^{2n}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^3}x^{2n+1}$$

$$=\color{blue}{\sum_{n=1}^\infty\frac{H_n}{n^3}x^n}-\frac18\sum_{n=1}^\infty\frac{H_n}{n^3}x^{2n}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^3}x^{2n+1}$$

The first sum is already calculated here and the second sum is the same as the the first one but just replace $x$ with $x^2$. The last sum seems annoying but I will give it a try.

Answer 2

I have now moved partial results from the OP here.

Integral representation of the generating function

Using

$$\frac{1}{n^q} = \frac{1}{\Gamma (q)}\int_0^\infty t^{q-1} e^{-n t}\,dt\tag{7}$$

and $(1)$ and $(2)$ of the OP we find that (for $q\gt0$) we have an integral representation of the generating function of order $q$ for the alternating harmonic sum:

$$g_q(z) = g^{(i)}_q(z) = \frac{1}{\Gamma (q)} \int_0^{\infty }t^{q-1} \frac{\log \left(1+z e^{-t} \right)}{1-z e^{-t}}\, dt\tag{8}$$

Explicit solution

12.02.2020 Recently, in an incomplete solution (https://math.stackexchange.com/a/3542942/198592) for the case $q=3$ the problem was reduced to this (unsolved) remaining sum

$$s(x) = \sum_{n=1}^\infty\frac{H_n}{(2n+1)^3}x^{2n+1}\tag{9} $$

It is interesting that the integral form of this sum can be calculated explicitly by Mathematica (as a "monster") in terms of hypergeometric functions.

The integral form is obtained using $(7)$ and doing the sum under the integral wih the result

$$s_{i}(x) =-\frac{1}{2} \int_0^{\infty } \frac{t^2 \left(e^{-t} x \log \left(1-e^{-2 t} x^2\right)\right)}{1-e^{-2 t} x^2} \, dt\tag{10}$$

First we transform the integral letting $t\to -\log (z)$ into

$$-\frac{1}{2} \int_0^1 \frac{x z \log ^2(z) \log \left(1-x^2 z^2\right)}{2 \left(1-x^2 z^2\right)} \, dz\tag{11}$$

which can be simplified expanding $ \log(1-v^2) = \log(1+v)(1-v) = \log(1+v) + \log(1-v)$ and taking partial fractions to these remaining integrals

$$-\frac{1}{4} x \int_0^1 \frac{z \log ^2(z) \log (1\pm x z)}{x z\mp1} \, dz \tag{12}$$

which in turn are equivalent to the integral $(6)$ of my OP on which I got stuck.

So you are trying to crack the same nut using series.

Unfortuately, numerous attempts of partial integrations and substitutions always led me back to where I came from. Hence abandoning for a moments the idea that a solution in terms of polylogs can be found it is gratifying that in the form $(10)$ Mathematica returned the announced answer:

$$s(x) = -\frac{1}{64} x^3 \left(\sqrt{\pi } \left(\\\left( 2 \gamma ^2+\pi ^2+4 \gamma (\log (4)-2)+2 (\log (4)-4) \log (4)\right) \\ \, _3\tilde{F}_2^{(\{0,0,0\},\{0,1\},0)}\left(\left\{1,2,\frac{3}{2}\right\},\left\{\frac{5}{2},2\right\},x^2\right) \\ +2 \, _3\tilde{F}_2^{(\{0,0,0\},\{2,1\},0)}\left(\left\{1,2,\frac{3}{2}\right\},\left\{\frac{5}{2},2\right\},x^2\right) \\ -4 \gamma \, _3\tilde{F}_2^{(\{0,0,1\},\{0,1\},0)}\left(\left\{1,2,\frac{3}{2}\right\},\left\{\frac{5}{2},2\right\},x^2\right) \\ +8 \, _3\tilde{F}_2^{(\{0,0,1\},\{0,1\},0)}\left(\left\{1,2,\frac{3}{2}\right\},\left\{\frac{5}{2},2\right\},x^2\right) \\ +4 \, _3\tilde{F}_2^{(\{0,0,1\},\{1,1\},0)}\left(\left\{1,2,\frac{3}{2}\right\},\left\{\frac{5}{2},2\right\},x^2\right) \\ +2\, _3\tilde{F}_2^{(\{0,0,2\},\{0,1\},0)}\left(\left\{1,2,\frac{3}{2}\right\},\left\{\frac{5}{2},2\right\},x^2\right) \\ -4 (\gamma -2+\log (4)) \, _3\tilde{F}_2^{(\{0,0,0\},\{1,1\},0)}\left(\left\{1,2,\frac{3}{2}\right\},\left\{\frac{5}{2},2\right\},x^2\right) \\ -4 \log (4) \, _3\tilde{F}_2^{(\{0,0,1\},\{0,1\},0)}\left(\left\{1,2,\frac{3}{2}\right\},\left\{\frac{5}{2},2\right\},x^2\right)\right) \\ -8 \gamma \Phi \left(x^2,3,\frac{3}{2}\right)\right)$$

Here $\gamma$ is Euler's gamma, $\Phi$ is the Lerch function and $\, _3\tilde{F}_2$ is the regularized hypergeometric function defined as

$$\,_3\tilde{F}_2(a_1,a_2,a_3;b_1,b_2;z)=\frac{1}{\Gamma (b_1) \Gamma (b_2)}\sum _{n=0}^{\infty } \frac{(a_1)_n (a_2)_n (a_3)_n}{ (b_1)_n (b_2)_n}\frac{z^n}{n!}$$

with the Pochhammer symbol being defined as $(a)_n = \frac{\Gamma(a+n)}{\Gamma(a)}$.

The example

$$\, _3\tilde{F}_2^{(\{0,0,0\},\{0,1\},0)}\left(\left\{1,2,\frac{3}{2}\right\},\left\{\frac{5}{2},2\right\},x^2\right)$$

explains the adopted notation for the derivatives with respect to the parameters. In this case the first derivatives with respect to $b_2$ is taken at the value $b_2 = 2$ taken from the list of arguments.

As a simplified example of the action of such a derivative consider

$$\frac{\partial}{\partial b}(b)_n|_{b\to 1} = -\frac{\psi ^{(0)}(n+1)+\gamma }{(1)_n}=-\frac{H_n}{n!}$$

i.e. in this way harmonic sums appear.

Another example (without derivatives) which shows that the "monster" behaves quite well is

$$\,_3\tilde{F}_2\left(\left\{1,2,\frac{3}{2}\right\},\left\{\frac{5}{2},2\right\},z^2\right)=-\frac{4 \left(z-\operatorname{arctanh}(z)\right)}{\sqrt{\pi } z^3}$$