What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n\ ?$

Question

Is there closed form for

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

where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number.

My approach,

In this paper page $95$ Eq $(5)$ we have

$$\sum_{n=1}^\infty \overline{H}_n\frac{x^n}{n}=\operatorname{Li}_2\left(\frac{1-x}{2}\right)-\operatorname{Li}_2(-x)-\ln2\ln(1-x)-\operatorname{Li}_2\left(\frac12\right)$$

Divide both sides by $x$ then integrate we get

$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n=\int\frac{\operatorname{Li}_2\left(\frac{1-x}{2}\right)}{x}\ dx-\operatorname{Li}_3(-x)+\ln2\operatorname{Li}_2(x)-\operatorname{Li}_2\left(\frac12\right)\ln x$$

and my question is how to find the remaining integral? Thanks

Maybe you wonder why I have it as an indefinite integral, I meant so as I am planning to plug $x=0$ to find the constant after we find the closed form of the integral if possible.

I tried Mathematica, it gave

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Edit

With help of $Mathematica$ I was able to find

\begin{align} \sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n&=-\frac13\ln^3(2)+\frac12\ln^2(2)\ln(1-x)-\frac12\zeta(2)\ln(x)+\frac32\ln^2(2)\ln(x)\\ &\quad-\ln(2)\ln(x)\ln(1-x)-\frac12\ln(2)\ln^2(x)-\frac12\ln^2(2)\ln(1-x)\\ &\quad-\ln^2(2)\left(\frac{x}{1+x}\right)+\ln(2)\ln\left(\frac{x}{1+x}\right)[\ln(1-x)+\ln(x)]\\ &\quad+\ln(x)\ln(1-x)\ln(1+x)+\ln(x)\operatorname{Li}_2\left(\frac{1-x}{2}\right)+\ln\left(\frac{x}{1+x}\right)\operatorname{Li}_2(x)\\ &\quad+\ln(1+x)\operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{x}{1+x}\right)\ln\left(\frac{2x}{1+x}\right)-\operatorname{Li}_2\left(\frac{2x}{1+x}\right)\ln\left(\frac{2x}{1+x}\right)\\ &\quad+\operatorname{Li}_2\left(\frac{1+x}{2}\right)\ln\left(\frac{x}{2}\right)-\ln\left(\frac{x}{1+x}\right)\operatorname{Li}_2\left(\frac{1+x}{2}\right)-\operatorname{Li}_3(x)-\operatorname{Li}_3\left(\frac{x}{1+x}\right)\\ &\quad+\operatorname{Li}_3\left(\frac{2x}{1+x}\right)-\operatorname{Li}_3\left(\frac{1+x}{2}\right)-\operatorname{Li}_3(-x)+\ln(2)\operatorname{Li}_2(x)+\frac{7}{8}\zeta(3) \end{align}

Answer 1

Here is a better way

From the question body we have

$$\sum_{n=1}^\infty \overline{H}_n\frac{x^n}{n}=\operatorname{Li}_2\left(\frac{1-x}{2}\right)-\operatorname{Li}_2\left(\frac12\right)-\operatorname{Li}_2(-x)-\ln2\ln(1-x)$$

Divide both sides by $x$ then integrate

$$\sum_{n=1}^\infty \overline{H}_n\frac{x^n}{n^2}=\underbrace{\int\frac{\operatorname{Li}_2\left(\frac{1-x}{2}\right)-\operatorname{Li}_2\left(\frac12\right)}{x}\ dx}_{IBP}-\operatorname{Li}_3(-x)+\ln2\operatorname{Li}_2(x)$$

$$=\ln x\left[\operatorname{Li}_2\left(\frac{1-x}{2}\right)-\operatorname{Li}_2\left(\frac12\right)\right]-\int\ln x\left[\frac{\ln(1+x)-\ln2}{1-x}\right]\ dx-\operatorname{Li}_3(-x)+\ln2\operatorname{Li}_2(x)$$

$$\small{=\ln x\left[\operatorname{Li}_2\left(\frac{1-x}{2}\right)-\operatorname{Li}_2\left(\frac12\right)\right]-\underbrace{\int\frac{\ln x\ln(1+x)}{1-x}\ dx}_{\large f(x)}+\ln2\operatorname{Li}_2(1-x)-\operatorname{Li}_3(-x)+\ln2\operatorname{Li}_2(x)}\tag1$$

For $f(x)$, use $2ab=a^2+b^2-(a-b)^2$ where $a=\ln x$ and $b=\ln(1+x)$

$$2f(x)=\int\frac{\ln^2 x}{1-x}\ dx+\int\frac{\ln^2(1+x)}{1-x}\ dx-\int\frac{\ln^2\left(\frac{x}{1+x}\right)}{1-x}\ dx\tag2$$

Apply integration by parts twice for the first integral

$$\int\frac{\ln^2 x}{1-x}\ dx=\boxed{2\operatorname{Li}_3(x)-2\ln x\operatorname{Li}_2(x)-\ln(1-x)\ln^2x}$$

Set $\frac{1}{1+x}=y$ for the second integral

$$\int\frac{\ln^2(1+x)}{1-x}\ dx=\int\frac{\ln^2y}{y(1-2y)}\ dy$$

$$=\int\frac{\ln^2y}{y}\ dy+2\int\frac{\ln^2y}{1-2y}\ dy$$

$$=\frac13\ln^3y+2\left[\operatorname{Li}_3(2y)-\ln y\operatorname{Li}_2(2y)-\frac12\ln^2y\ln(1-2y)\right]$$

$$=\boxed{2\operatorname{Li}_3\left(\frac{2}{1+x}\right)+2\ln(1+x)\operatorname{Li}_2\left(\frac{2}{1+x}\right)-\ln^2(1+x)\ln\left(\frac{x-1}{x+1}\right)-\frac13\ln^3(1+x)}$$

For the third integral, set $\frac{x}{1+x}=t$

$$\int\frac{\ln^2\left(\frac{x}{1+x}\right)}{1-x}\ dx=\int\frac{\ln^2t}{(1-t)(1-2y)}$$

$$=2\int\frac{\ln^2t}{1-2t} dt-\int\frac{\ln^2t}{1-t} dt$$

$$=2\operatorname{Li}_3(2t)-2\operatorname{Li}_3(t)+2\ln t\left[\operatorname{Li}_2(t)-\operatorname{Li}_2(2t)\right]+\ln^2t\left[\ln(1-t)-\ln(1-2t)\right]$$

$$=\boxed{\small{2\operatorname{Li}_3\left(\frac{2x}{1+x}\right)-2\operatorname{Li}_3\left(\frac{x}{1+x}\right)+2\ln\left(\frac{x}{1+x}\right)\left[\operatorname{Li}_2\left(\frac{x}{1+x}\right)-\operatorname{Li}_2\left(\frac{2x}{1+x}\right)\right]-\ln^2\left(\frac{x}{1+x}\right)\ln(1-x)}\ }$$

Plug the boxed results in $(2)$ then divide by $2$

$$f(x)=\operatorname{Li}_3(x)+\operatorname{Li}_3\left(\frac{2}{1+x}\right)-\operatorname{Li}_3\left(\frac{2x}{1+x}\right)+\operatorname{Li}_3\left(\frac{x}{1+x}\right)-\ln x\operatorname{Li}_2(x)\\ +\ln(1+x)\operatorname{Li}_2\left(\frac{2}{1+x}\right)-\ln\left(\frac{x}{1+x}\right)\left[\operatorname{Li}_2\left(\frac{x}{1+x}\right)-\operatorname{Li}_2\left(\frac{2x}{1+x}\right)\right]\\ -\frac12\ln^2(1+x)\ln\left(\frac{x-1}{1+x}\right)+\frac12\ln(1-x)\ln^2\left(\frac{x}{1+x}\right)-\frac16\ln^3(1+x)-\frac12\ln^2x\ln(1-x)$$

Now plug $f(x)$ in $(1)$

$$\small{\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n=\ln x\left[\operatorname{Li}_2\left(\frac{1-x}{2}\right)-\operatorname{Li}_2\left(\frac12\right)\right]-f(x)+2\ln2\operatorname{Li}_2(1-x)-\operatorname{Li}_3(-x)+\ln2\operatorname{Li}_2(x)+C}$$

set $x=0$ and notice that $f(0)=\operatorname{Li}_3(2)$

$$0=-\operatorname{Li}_3(2)+\ln2\zeta(2)+C\Longrightarrow C=\operatorname{Li}_3(2)-\ln2\zeta(2)$$

Therefore

$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n$$ $$=-\operatorname{Li}_3(x)-\operatorname{Li}_3(-x)-\operatorname{Li}_3\left(\frac{2}{1+x}\right)+\operatorname{Li}_3\left(\frac{2x}{1+x}\right)-\operatorname{Li}_3\left(\frac{x}{1+x}\right)+\ln x\operatorname{Li}_2(x)\\ -\ln(1+x)\operatorname{Li}_2\left(\frac{2}{1+x}\right)+\ln\left(\frac{x}{1+x}\right)\left[\operatorname{Li}_2\left(\frac{x}{1+x}\right)-\operatorname{Li}_2\left(\frac{2x}{1+x}\right)\right]\\ +\ln x\left[\operatorname{Li}_2\left(\frac{1-x}{2}\right)-\operatorname{Li}_2\left(\frac12\right)\right]+\ln2\operatorname{Li}_2(1-x)+\ln2\operatorname{Li}_2(x)\\ +\frac12\ln^2(1+x)\ln\left(\frac{x-1}{1+x}\right)-\frac12\ln(1-x)\ln^2\left(\frac{x}{1+x}\right)\\+\frac16\ln^3(1+x)+\frac12\ln^2x\ln(1-x)+\operatorname{Li}_3(2)-\ln2\zeta(2)$$

Answer 2

This is a long comment to https://math.stackexchange.com/a/3523732/198592 which just provides my result for comparison.

Let $\overline{H}_n=\sum_{k=1}^{n}(-1)^{k+1}\frac{1}{k}$ be the alternating harmonic sum and define the generating function of order $q=0,1,2,...$ as

$$g_{q}(x) = \sum_{n=1}^\infty\frac{\overline{H}_n}{n^q}x^n\tag{1}$$

For $q=2$ I have obtained

$$\begin{align} {g}_2(x)& =-\operatorname{Li}_3\left(\frac{x+1}{2}\right)-\operatorname{Li}_3(-x)-\operatorname{Li}_3(x)-\operatorname{Li}_3\left(\frac{x}{x+1}\right)+\operatorname{Li}_3\left(\frac{2 x}{x+1}\right)\\ & +\log (2) \operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{x+1}{2}\right) \left(\log (x)-\log \left(\frac{2 x}{x+1}\right)\right)\\ & + \operatorname{Li}_2\left(\frac{1}{2}-\frac{x}{2}\right) \log (x)+\operatorname{Li}_2(x) \left(\log \left(\frac{x}{x+1}\right)+\log (x+1)\right)\\ & +\left(\operatorname{Li}_2\left(\frac{x}{x+1}\right) -\operatorname{Li}_2\left(\frac{2 x}{x+1}\right)\right) \log \left(\frac{2 x}{x+1}\right)+\frac{1}{2} \log ^2(2) \log (x)\\ & +\frac{1}{2} \log ^2\left(\frac{2 x}{x+1}\right) \left(\log \left(\frac{1-x}{2}\right)+\log \left(\frac{1}{x+1}\right)-\log \left(-\frac{x-1}{x+1}\right)\right)\\ & +\log (2) \log (x) \log \left(\frac{2 x}{x+1}\right)-\frac{1}{2} \log (2) \log (x) (\log (x)-2 \log (x+1)+\log (4))\\ & -\frac{1}{12} \pi ^2 \log (x)+\log \left(\frac{1-x}{2}\right) \log \left(\frac{x+1}{2}\right) \log (x)+\frac{7 \zeta (3)}{8}+\frac{\log ^3(2)}{6} \end {align}\tag{2}$$

Notice the appearance of $\zeta(3)$ which I don't see in your expression.

Here is the graph of the g.f.

I find the following boundary values

$$g_2(+1) = \frac{1}{4} \pi ^2 \log (2)-\frac{\zeta (3)}{4}\simeq 1.40976$$ $$g_2(-1) = -\frac{1}{4} \pi ^2 \log (2)+\frac{5 \zeta (3)}{8}\simeq -0.958987$$

The values at $\pm \frac{1}{2}$ are somewhat too long to be provided here at the moment.

To facilitate comparison here is the Mathematica statement

g2[x_]:=Log[2]^3/6 - 1/12 \[Pi]^2 Log[x] + 1/2 Log[2]^2 Log[x] + 
 Log[2] Log[x] Log[(2 x)/(1 + x)] + 
 1/2 (Log[(1 - x)/2] + Log[1/(1 + x)] - 
    Log[-((-1 + x)/(1 + x))]) Log[(2 x)/(1 + x)]^2 + 
 Log[(1 - x)/2] Log[x] Log[(1 + x)/2] - 
 1/2 Log[2] Log[x] (Log[4] + Log[x] - 2 Log[1 + x]) + 
 Log[x] PolyLog[2, 1/2 - x/2] + 
 Log[2] PolyLog[2, x] + (Log[x/(1 + x)] + Log[1 + x]) PolyLog[2, x] + 
 Log[(2 x)/(
   1 + x)] (PolyLog[2, x/(1 + x)] - 
    PolyLog[2, (2 x)/(1 + x)]) + (Log[x] - 
    Log[(2 x)/(1 + x)]) PolyLog[2, (1 + x)/2] - PolyLog[3, -x] - 
 PolyLog[3, x] - PolyLog[3, x/(1 + x)] + PolyLog[3, (2 x)/(1 + x)] - 
 PolyLog[3, (1 + x)/2] + (7 Zeta[3])/8

Answer 3

Starring with Landens identity

$$\operatorname{Li}_2(1-t)+\operatorname{Li}_2\left(\frac{t-1}{t}\right)=-\frac12\ln^2t$$

set $1-t=\frac{1-x}{2}$ we get

$$\operatorname{Li}_2\left(\frac{1-x}{2}\right)=-\operatorname{Li}_2\left(-\frac{1-x}{1+x}\right)-\frac12\ln^2\left(\frac{1+x}{2}\right)$$

$$\Longrightarrow \int\frac{\operatorname{Li}_2\left(\frac{1-x}{2}\right)}{x}\ dx=-\int\frac{\operatorname{Li}_2\left(-\frac{1-x}{1+x}\right)}{x}\ dx-\frac12\int\frac{\ln^2\left(\frac{1+x}{2}\right)}{x}\ dx=-\mathcal{J}-\frac12\mathcal{K}$$

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For $\mathcal{J}$, set $\frac{1-x}{1+x}=y$

$$\mathcal{J}=-2\int\frac{\operatorname{Li}_2(-y)}{1-y^2}\ dy\overset{IBP}{=}\ln\left(\frac{1-y}{1+y}\right)\operatorname{Li}_2(-y)+\int\frac{\ln\left(\frac{1-y}{1+y}\right)\ln(1+y)}{y}\ dy$$

$$=\ln\left(\frac{1-y}{1+y}\right)\operatorname{Li}_2(-y)+\underbrace{\int\frac{\ln(1-y)\ln(1+y)}{y}\ dy}_{\mathcal{\large J}_1}-\underbrace{\int\frac{\ln^2(1+y)}{y}\ dy}_{\mathcal{\large J}_2}$$

for $\mathcal{J}_1$, use $\ln(1-y)\ln(1+y)=\frac14\ln^2(1-y^2)-\frac14\ln^2\left(\frac{1-y}{1+y}\right)$

$$ \mathcal{J}_1=\frac14\underbrace{\int\frac{\ln^2(1-y^2)}{y}\ dy}_{y^2\to t}-\frac14\underbrace{\int\frac{\ln^2\left(\frac{1-y}{1+y}\right)}{y}\ dy}_{\frac{1-y}{1+y}=u}$$

$$=\frac18\int\frac{\ln^2(1-t)}{t}\ dt+\frac12\int\frac{\ln^2u}{1-u^2}\ du$$

I managed here to prove

$$\int\frac{\ln^2(1-t)}{t}dt=\ln(1-t)\left[\operatorname{Li}_2(1-t)-\operatorname{Li}_2(t)+\zeta(2))\right]-2\operatorname{Li}_3(1-t)\tag{*}$$

substitute $t=y^2$ back

$$\frac14\int\frac{\ln^2(1-y^2)}{y}dt=\frac18\int\frac{\ln^2(1-t)}{t}\ dy$$

$$=\frac18\ln(1-y^2)\left[\operatorname{Li}_2(1-y^2)-\operatorname{Li}_2(y^2)+\zeta(2)\right]-\frac14\operatorname{Li}_3(1-y^2)\tag1$$

As for the second integral,

$$\int\frac{\ln^2u}{1-u^2}\ du\overset{IBP}{=}\frac12\ln\left(\frac{1+u}{1-u}\right)\ln u-\int\frac{\ln\left(\frac{1+u}{1-u}\right)\ln u}{u}\ du$$

$$=\frac12\ln\left(\frac{1+u}{1-u}\right)\ln u-\int\frac{\ln(1+u)\ln u}{u}\ du+\int\frac{\ln(1-u)\ln u}{u}\ du$$

$$=\frac12\ln\left(\frac{1+u}{1-u}\right)\ln u-[-\operatorname{Li}_3(-u)\ln u+\operatorname{Li}_3(-u)]+[-\operatorname{Li}_3(u)\ln u+\operatorname{Li}_3(u)]$$

$$=\frac12\ln\left(\frac{1+u}{1-u}\right)\ln u+\operatorname{Li}_3(-u)\ln u-\operatorname{Li}_3(-u)-\operatorname{Li}_3(u)\ln u+\operatorname{Li}_3(u)$$

substitute $u=\frac{1-y}{1+y}$ back

$$-\frac14\int\frac{\ln^2\left(\frac{1-y}{1+y}\right)}{y}\ dy=\frac12\int\frac{\ln^2u}{1-u^2}\ du$$

$$=-\frac14\ln y\ln\left(\frac{1-y}{1+y}\right)+\frac12\operatorname{Li}_3\left(-\frac{1-y}{1+y}\right)\ln\left(\frac{1-y}{1+y}\right)$$

$$-\frac12\operatorname{Li}_3\left(-\frac{1-y}{1+y}\right)-\frac12\operatorname{Li}_3\left(\frac{1-y}{1+y}\right)\ln\left(\frac{1-y}{1+y}\right)-\frac12\operatorname{Li}_3\left(\frac{1-y}{1+y}\right)\tag2$$

Combine $(1)$ and $(2)$ to get $\mathcal{J}_1$

$$\mathcal{J}_1=\frac18\ln(1-y^2)\left[\operatorname{Li}_2(1-y^2)-\operatorname{Li}_2(y^2)+\zeta(2)\right]-\frac14\operatorname{Li}_3(1-y^2)-\frac14\ln y\ln\left(\frac{1-y}{1+y}\right)$$ $$+\frac12\operatorname{Li}_3\left(-\frac{1-y}{1+y}\right)\ln\left(\frac{1-y}{1+y}\right)-\frac12\operatorname{Li}_3\left(-\frac{1-y}{1+y}\right)-\frac12\operatorname{Li}_3\left(\frac{1-y}{1+y}\right)\ln\left(\frac{1-y}{1+y}\right)-\frac12\operatorname{Li}_3\left(\frac{1-y}{1+y}\right)$$

For $\mathcal{J}_2$, use $(*)$

$$\small{\mathcal{J}_2=\int\frac{\ln^2(1+y)}{y}\ dy\overset{y=-t}{=}-\int\frac{\ln^2(1-t)}{t}\ dt=-\ln(1+y)\left[\operatorname{Li}_2(1+y)-\operatorname{Li}_2(-y)+\zeta(2)\right]+2\operatorname{Li}_3(1+y)}$$

Similarly

$$\mathcal{K}=\int\frac{\ln^2\left(\frac{1+x}{2}\right)}{x}\ dx$$

$$=\int\frac{\ln^2(1+x)}{x}\ dx-2\ln2\int\frac{\ln(1+x)}{x}\ dx+\ln^22\int\frac{1}{x}\ dx$$

$$=-\ln(1+x)\left[\operatorname{Li}_2(1+x)-\operatorname{Li}_2(-x)+\zeta(2))\right]+2\operatorname{Li}_3(1+x)+2\ln2\operatorname{Li}_2(-x)+\ln^22\ln x$$

and what is left is only combining results and simplification but that's too tedious so I am just going to leave it as it and I am not sure if my calculations is right or not.

Answer 4

From this paper page $101$ we have

$$\sum_{n=1}^\infty\overline{H}_n\frac{x^{n+1}}{(n+1)^2}=\operatorname{Li}_3\left(\frac{2x}{1+x}\right)-\operatorname{Li}_3\left(\frac{x}{1+x}\right)-\operatorname{Li}_3\left(\frac{1+x}{2}\right)-\operatorname{Li}_3(x)$$ $$+\ln(1+x)\left[\operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{1}{2}\right)+\frac12\ln 2\ln(1+x)\right]+\operatorname{Li}_3\left(\frac{1}{2}\right)$$

but

$$\sum_{n=1}^\infty\overline{H}_n\frac{x^{n+1}}{(n+1)^2}=\sum_{n=0}^\infty\overline{H}_n\frac{x^{n+1}}{(n+1)^2}=\sum_{n=1}^\infty\overline{H}_{n-1}\frac{x^n}{n^2},\quad \overline{H}_{n-1}=\overline{H}_n+\frac{(-1)^n}{n}$$

$$=\sum_{n=1}^\infty\overline{H}_{n}\frac{x^n}{n^2}+\operatorname{Li}_3(-x)$$

Thus

$$\sum_{n=1}^\infty\overline{H}_{n}\frac{x^n}{n^2}=\operatorname{Li}_3\left(\frac{2x}{1+x}\right)-\operatorname{Li}_3\left(\frac{x}{1+x}\right)-\operatorname{Li}_3\left(\frac{1+x}{2}\right)-\operatorname{Li}_3(-x)-\operatorname{Li}_3(x)$$ $$+\ln(1+x)\left[\operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{1}{2}\right)+\frac12\ln 2\ln(1+x)\right]+\operatorname{Li}_3\left(\frac{1}{2}\right)$$