The closed form for $\sum_{n=1}^\infty \frac{H_{n/2}}{n^2}x^n$

Question

Is there a closed form for
$$\sum_{n=1}^\infty \frac{H_{n/2}}{n^2}x^n\ ?$$

Where $H_{n/2}=\int_0^1\frac{1-x^{n/2}}{1-x}\ dx$ is the harmonic number.

I managed to find the closed form but had hard time finding the constant.

My trial

I was able to prove

$$\sum_{n=1}^\infty \frac{H_{n/2}}{n}x^n=\operatorname{Li}_2\left(\frac{1}{1-x}\right)+\operatorname{Li}_2\left(\frac{1}{1+x}\right)-\operatorname{Li}_2\left(\frac{1-x}{1+x}\right)$$ $$+\ln(1-x)\ln(1+x)+\ln^2(1-x)-2\ln(x)\ln(1-x)-i\pi\ln(1-x)-\zeta(2)=f(x)$$

If we divide both sides by $x$ then integrate we get

$$\sum_{n=1}^\infty \frac{H_{n/2}}{n^2}x^n=\int\frac{f(x)}{x}\ dx$$

Wolfram gave

and after tedious manual simplifications I found

$$\int\frac{f(x)}{x}\ dx=\operatorname{Li}_3\left(\frac{1+x}{1-x}\right)-\operatorname{Li}_3\left(\frac{1+x}{x-1}\right)+\operatorname{Li}_3\left(\frac{1+x}{2x}\right)-\operatorname{Li}_3\left(\frac{1+x}{x}\right)-\operatorname{Li}_3\left(\frac{1+x}{2}\right)$$ $$-\operatorname{Li}_3(1+x)-2\operatorname{Li}_3(1-x)+\operatorname{Li}_3(x)$$

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

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

$$+\ln\left(\frac{1+x}{2}\right)\operatorname{Li}_2\left(\frac{1+x}{2}\right)+\ln(1+x)\operatorname{Li}_2(1+x)+\ln(2x)\operatorname{Li}_2(x)-2\ln(x)\operatorname{Li}_2(-x)$$ $$-\ln(x-1)\operatorname{Li}_2(1-x)+3\ln(1-x)\operatorname{Li}_2(1-x)+\ln2[\operatorname{Li}_2(1-x)+\operatorname{Li}_2(-x)]$$

$$+\ln(x)\ln^2(1-x)-\ln^2(x)\ln(1+x)-2\ln^2(x)\ln(1-x)+\ln^2(x)\ln(1+x)$$ $$+2\ln(x)\ln(1-x)\ln(1+x)+\frac12\ln2\ln^2(x)+\ln^22\ln(x)$$

$$+\frac{i\pi}{2}\left[\ln^2(1+x)+\ln^2\left(\frac{1+x}{1-x}\right)-4\ln(1-x)\ln(1+x)+2\operatorname{Li}_2(x)\right]-\zeta(2)\ln(x)+\color{red}{C}$$

I hope the closed form has no mistake or typo. I set $x=0,1$ to find the constant but failed, any idea? . Thank you

Answer 1

We present here the details of calculating the closed form of the generating function.

$$s(z) = \sum_{n=1}^{\infty}\frac{z^n}{n^2} H_{n/2}\tag{1}$$

I have given partial results already in a comment.

In contrast to that of the OP in which a constant C appears the present calculation is complete.

We proceed step by step with the generating functions up to the quantity in question $g_{2}(z)$.

We shall do this with Mathematica taking care that theses two conditions are met

a) $g(z=0) = 0$

This is a necassary condition for the integration in the next step to be convergent at $0$.

b) $g(z)$ is real for $-1<z<1$

This almost always produces "nicer" expressions, i.e. they are better integrable in the next step than the "rough" expressions.

$$g_0(z) = \sum_{n=1}^{\infty}z^n H_{n/2}=\frac{z \log (4)+2 \log (1-z)}{z^2-1}\tag{2}$$

$$g_{1}(z) =\sum_{n=1}^{\infty}\frac{z^n}{n} H_{n/2} =\int_0^z \frac{g_0(t)}{t}\,dt\tag{3}$$

$$g_{2}(z) =\sum_{n=1}^{\infty}\frac{z^n}{n^2} H_{n/2} =\int_0^z \frac{g_1(t)}{t}\,dt\tag{4}$$

The indefinite integral using Integrate[] related to $g_1(z)$ is

$$g_{1,i}(z) = \int \frac{g_0(z)}{z}\,dz=\operatorname{Li}_2\left(\frac{1-z}{2}\right)+2 \text{Li}_2(z)+\frac{1}{2} \log ^2(1-z)+\log (z+1) \log (1-z)-\log (2) \log (z+1)$$

Subtracting the value at $z=0$ which is $\frac{1}{12} \left(\pi ^2-6 \log ^2(2)\right)$ gives for the definite integral $(3)$ the following expression

$$g_1(z) = \operatorname{Li}_2\left(\frac{1-z}{2}\right)+2 \operatorname{Li}_2(z)+\frac{1}{2} \log ^2(1-z)+\\ \log (z+1) \log (1-z)-\log (2) \log (z+1)+\frac{1}{12} \left(6 \log ^2(2)-\pi ^2\right)\tag{3a}$$

This expression meets the "nicety"- conditions requested.

Now the next step. The indefinite integral becomes

$$g_{2,i}(z) = \int \frac{g_1(z)}{z}\,dz=\text{expression with length 28}$$

Subtracting the value at $z=0$ which is $g_{2,i}(z=0) = -\frac{17 \zeta (3)}{8}-\frac{1}{6} \log ^3(2)$ gives an expression as the sum of 30 terms (in order to save typing labour (and errors), I have provided also the Mathematica expression in the apendix)

$$g_2(z) = \text{sum of 30 terms, see appendix}\tag{4a}$$

Here is the graph of $g_2$

Special values which have already been given in a comment are

$$g_2(z=+1) = \lim_{z\to 1^-} \, g_{2}(z)\\ = \frac{1}{4} \left(-4 \text{Li}_3(2)+9 \zeta (3)-2 i \pi \log ^2(2)+\pi ^2 \log (2)\right)= \frac{11}{8} \zeta (3)\tag{5}$$

$$\\g_2(z=-1) = \lim_{z\to -1^+} \, g_{2}(z)\\= \frac{1}{8} \left(-16 \text{Li}_3(2)+11 \zeta (3)-4 i \pi \log (2) \log (4)+\pi ^2 \log (16)\right)\\=-\frac{3}{8} \zeta (3)\tag{6}$$

Going from the immediate result of the limit to the final result we have used the transformation formulas for the polylog functions (see e.g. https://en.wikipedia.org/wiki/Polylogarithm).

Discussion

Splitting the sum into even and odd summands we have

$$g_2(z) =g_{2,e}(z)+g_{2,o}(z) $$

Since we have $g_2(z)$, and $g_{2,e}(z)$ is easily calculated with the result

$$g_{2,e}(z)=\frac{1}{4} \left(\operatorname{Li}_3\left(x^2\right)-\operatorname{Li}_3\left(1-x^2\right)+\operatorname{Li}_2\left(1-x^2\right) \log \left(1-x^2\right)\\ +\log (x) \log ^2\left(1-x^2\right)+\zeta (3)\right)\tag{7}$$

we have also obtained the more complicated sum

$$g_{2,o}(z) =\sum_{m=1}^{\infty} \frac{z^{2m-1}}{2m-1} H_{m-\frac{1}{2}} \\ =g_{2}(z)-g_{2,e}(z)\tag{8} $$

Appendix

Mathematica expression of $g_{2}(z)$

Notice that the transformation to a "nice", i.e. to an all-real summands-expression, still has to be done (my task):

g2[z]=Log[2]^3/6 - 1/12 \[Pi]^2 Log[z] + 1/2 Log[2]^2 Log[z] + 
 1/2 Log[1 - z]^2 Log[z] + Log[2] Log[z] Log[(2 z)/(1 + z)] + 
 1/2 (Log[(1 - z)/2] + Log[1/(1 + z)] - 
    Log[-((-1 + z)/(1 + z))]) Log[(2 z)/(1 + z)]^2 + 
 Log[(1 - z)/2] Log[z] Log[(1 + z)/2] - 
 1/2 Log[2] Log[z] (Log[4] + Log[z] - 2 Log[1 + z]) + 
 Log[1 - z] Log[z] Log[1 + z] + 
 1/2 (-Log[-z] + Log[z]) Log[
   1 + z] (-2 Log[1 - z] + Log[1 + z]) + (Log[-z] - Log[z]) Log[
   1 + z] Log[(1 + z)/(1 - z)] + 
 1/2 (Log[2/(1 - z)] + Log[z] - Log[-((2 z)/(1 - z))]) Log[(1 + z)/(
   1 - z)]^2 + 
 Log[1 - z] PolyLog[2, 
   1 - z] + (Log[1 + z] - Log[(1 + z)/(1 - z)]) PolyLog[2, 1 - z] + 
 Log[z] PolyLog[2, 1/2 - z/2] + 
 Log[2] PolyLog[2, -z] + (Log[z/(1 + z)] + Log[1 + z]) PolyLog[2, z] +
  Log[(2 z)/(
   1 + z)] (PolyLog[2, z/(1 + z)] - 
    PolyLog[2, (2 z)/(1 + z)]) + (Log[z] - 
    Log[(2 z)/(1 + z)]) PolyLog[2, (1 + z)/
   2] + (Log[1 - z] + Log[(1 + z)/(1 - z)]) PolyLog[2, 1 + z] + 
 Log[(1 + z)/(
   1 - z)] (PolyLog[2, -((1 + z)/(1 - z))] - 
    PolyLog[2, (1 + z)/(1 - z)]) - 2 PolyLog[3, 1 - z] + 
 PolyLog[3, z] - PolyLog[3, z/(1 + z)] + PolyLog[3, (2 z)/(1 + z)] - 
 PolyLog[3, (1 + z)/2] - PolyLog[3, 1 + z] - 
 PolyLog[3, -((1 + z)/(1 - z))] + PolyLog[3, (1 + z)/(1 - z)] + (
 17 Zeta[3])/8

Answer 2

Using $g_1(z)$ from @Dr. Wolfgang Hintze solution above

$$\small{\sum_{n=1}^\infty\frac{H_{n/2}}{n}z^n=\operatorname{Li}_2\left(\frac{1-z}{2}\right)-\operatorname{Li}_2\left(\frac{1}{2}\right)+2 \text{Li}_2(z)+\frac{1}{2} \ln ^2(1-z)+\ln (z+1) \ln (1-z)-\log (2) \ln (z+1)}$$

By algebraic identities, we have

$$\frac{1}{2} \log ^2(1-z)+\log (z+1) \log (1-z)=\frac{1}{2} \log ^2(1-z^2)-\frac{1}{2} \log ^2(1+z)$$

so

$$\small{\sum_{n=1}^\infty\frac{H_{n/2}}{n}z^n=\operatorname{Li}_2\left(\frac{1-z}{2}\right)-\operatorname{Li}_2\left(\frac{1}{2}\right)+2 \text{Li}_2(z)+\frac{1}{2} \ln ^2(1-z^2)-\frac{1}{2} \ln ^2(1+z)}-\ln(2)\ln(1+z)$$

And in this paper page $95$ Eq $(5)$ we have

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

Subtracting the two generalizations we have

$$\small{\sum_{n=1}^\infty\frac{H_{n/2}}{n}z^n-\sum_{n=1}^\infty \frac{\overline{H}_n}{n}z^n=2 \text{Li}_2(z)+\text{Li}_2(-z)+\frac{1}{2} \ln ^2(1-z^2)-\frac{1}{2} \ln ^2(1+z)+\ln(2)\ln\left(\frac{1-z}{1+z}\right)}$$

Now divide both sides by $z$ then $\int_0^x$ we get

$$\sum_{n=1}^\infty\frac{H_{n/2}}{n^2}x^n-\sum_{n=1}^\infty \frac{\overline{H}_n}{n^2}x^n$$ $$=2\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)+\frac12\underbrace{\int_0^x\frac{\ln^2(1-z^2)}{z}\ dz}_{\large I_1}-\frac12\underbrace{\int_0^x\frac{\ln^2(1+z)}{z}\ dz}_{\large I_2}$$ $$+\ln(2)[\operatorname{Li}_2(-x)-\operatorname{Li}_2(x)]$$

$I_1$ and $I_2$ can be found in the book Almost Impossible Integrals, Sums and Series page 3.

$$I_1=\int_0^x\frac{\ln^2(1-z^2)}{z}\ dz=\frac12\int_0^{x^2}\frac{\ln^2(1-t)}{t}\ dt$$

$$=\ln(x)\ln^2(1-x^2)+\ln(1-x^2)\operatorname{Li}_2(1-x^2)-\operatorname{Li}_3(1-x^2)+\zeta(3)$$

$$I_2=\ln(x)\ln^2(1+x)-\frac23\ln^3(1+x)-2\ln(1+x)\operatorname{Li}_2\left(\frac{1}{1+x}\right)-2\operatorname{Li}_3\left(\frac{1}{1+x}\right)+2\zeta(3)$$

$\sum_{n=1}^\infty \frac{\overline{H}_n}{n^2}x^n$ is already calculated here

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

Combine all results we get

$$\sum_{n=1}^\infty\frac{H_{n/2}}{n^2}x^n=\operatorname{Li}_3\left(\frac{2x}{1+x}\right)-\operatorname{Li}_3\left(\frac{x}{1+x}\right)-\operatorname{Li}_3\left(\frac{1+x}{2}\right)+\operatorname{Li}_3\left(\frac{1}{1+x}\right)$$

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

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

$$-\frac12\ln(x)\ln^2(1+x)+\frac13\ln^3(1+x)+\frac12\ln(x)\ln^2(1-x^2)-\frac12\zeta(3)+\operatorname{Li}_3\left(\frac{1}{2}\right)$$

Answer 3

Similar approach but more independent

Again using $g_1(z)$ proved by @Dr. Wolfgang Hintze

$$\small{\sum_{n=1}^\infty\frac{H_{n/2}}{n}z^n=\operatorname{Li}_2\left(\frac{1-z}{2}\right)-\operatorname{Li}_2\left(\frac{1}{2}\right)+2 \text{Li}_2(z)+\frac{1}{2} \ln ^2(1-z^2)-\frac{1}{2} \ln ^2(1+z)}-\ln(2)\ln(1+z)$$

Divide both sides by $z$ then integrate

$$\small{\sum_{n=1}^\infty\frac{H_{n/2}}{n^2}z^n=\underbrace{\int\frac{\operatorname{Li}_2\left(\frac{1-z}{2}\right)-\operatorname{Li}_2\left(\frac{1}{2}\right)}{z}}_{\large I_1}+2 \text{Li}_3(z)+\frac{1}{2} \underbrace{\int\frac{\ln ^2(1-z^2)}{z}}_{\large I_2}-\frac{1}{2}\underbrace{\int\frac{\ln ^2(1+z)}{z}}_{\large I_3}+\ln(2)\operatorname{Li}_2(-z)}$$

$I_2$ and $I_3$ can be found using the indefinite integral

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

$$\Longrightarrow I_2=\int\frac{\ln ^2(1-z^2)}{z}\ dz\overset{z^2=t}{=}\frac12\int\frac{\ln ^2(1-t)}{t}\ dt$$

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

$$\Longrightarrow I_3=\int\frac{\ln ^2(1+z)}{z}\ dz\overset{-z=t}{=}\int\frac{\ln ^2(1-t)}{t}\ dt$$

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

For $I_1$, integrate by parts

$$I_1=\ln z\left[\operatorname{Li}_2\left(\frac{1-z}{2}\right)-\operatorname{Li}_2\left(\frac12\right)\right]-\int\ln z\left[\frac{\ln(1+z)-\ln2}{1-z}\right]\ dz$$

$$=\ln z\left[\operatorname{Li}_2\left(\frac{1-z}{2}\right)-\operatorname{Li}_2\left(\frac12\right)\right]-\underbrace{\int\frac{\ln z\ln(1+z)}{1-z}\ dz}_{\large f(z)}+\ln2\operatorname{Li}_2(1-z)$$

where $f(z)$ is already calculated here

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

combine the results of the three integrals we get

$$\sum_{n=1}^\infty\frac{H_{n/2}}{n^2}z^n=2\operatorname{Li}_3(z)+\ln z\left[\operatorname{Li}_2\left(\frac{1-z}{2}\right)-\operatorname{Li}_2\left(\frac12\right)\right]+\ln2[\operatorname{Li}_2(1-z)+\operatorname{Li}_2(-z)]$$

$$-f(z)+\frac14\ln(1-z^2)\left[\operatorname{Li}_2(1-z^2)-\operatorname{Li}_2(z^2)+\zeta(2)\right]-\frac12\operatorname{Li}_3(1-z^2)$$

$$-\frac12\ln(1+z)\left[\operatorname{Li}_2(1+z)-\operatorname{Li}_2(-z)+\zeta(2)\right]+\operatorname{Li}_3(1+z)+C$$

To find the constant $C$, set $z=0$ and note that $f(0)=\operatorname{Li}_3(2)$

$$0=\operatorname{Li}_3(2)+2\ln2\operatorname{Li}_2(2)+\frac12\zeta(3)+C\Longrightarrow C=-\operatorname{Li}_3(2)-2\ln2\operatorname{Li}_2(2)-\frac12\zeta(3)$$

Therefore

$$\sum_{n=1}^\infty\frac{H_{n/2}}{n^2}z^n=2\operatorname{Li}_3(z)+\ln z\left[\operatorname{Li}_2\left(\frac{1-z}{2}\right)-\operatorname{Li}_2\left(\frac12\right)\right]+\ln2[\operatorname{Li}_2(1-z)+\operatorname{Li}_2(-z)]$$

$$-f(z)+\frac14\ln(1-z^2)\left[\operatorname{Li}_2(1-z^2)-\operatorname{Li}_2(z^2)+\zeta(2)\right]-\frac12\operatorname{Li}_3(1-z^2)$$

$$-\frac12\ln(1+z)\left[\operatorname{Li}_2(1+z)-\operatorname{Li}_2(-z)+\zeta(2)\right]+\operatorname{Li}_3(1+z)-\operatorname{Li}_3(2)-2\ln2\operatorname{Li}_2(2)-\frac12\zeta(3)$$