# Generating function of forth powers of harmonic numbers.

Let $x\in (-1,1)$ and let $n\ge 1$ be an integer. Now, let us define a following family of harmonic sums: \begin{eqnarray} S^{(n)}(x):= \sum\limits_{m=1}^\infty [H_m]^n \cdot x^m \end{eqnarray}

It is not hard to see that the following recursion relation holds true: \begin{eqnarray} S^{(n+1)}(x)=\int\limits_0^1 Li_1(t) \cdot \frac{d}{d t} \left( S^{(n)}(x t)\right) dt =-\int\limits_0^1 \frac{S^{(n)}(x t)-S^{(n)}(x)}{1-t} dt \end{eqnarray} Now by building up on the results in Generating function for cubes of Harmonic numbers where $S^{(3)}(x)$ was being derived in closed form we obtained the following: \begin{eqnarray} &&4(1-x) \cdot S^{(4)}(x)=\\ &&4 \text{Li}_2(x){}^2-12 \text{Li}_4(1-x)-28 \text{Li}_4(x)-32 \text{Li}_4\left(\frac{x}{x-1}\right)+16 \text{Li}_3(x) \log (1-x)+\\ &&16 \zeta (3) \log (1-x)+\frac{5}{2} \log ^4(1-x)-\frac{8}{3} \log (x) \log ^3(1-x)+\pi ^2 \log^2(1-x)+\frac{2 \pi ^4}{15}+\\ && 4\left\{ \begin{array}{rr} -\text{Li}_4(1-x)+\frac{1}{24} \log ^4(1-x)+\frac{1}{12} \pi ^2 \log ^2(1-x)+\frac{\pi ^4}{90} & \mbox{if $x\ge 0$}\\ \text{Li}_4\left(\frac{1}{1-x}\right)+\frac{1}{12} \log ^4(1-x)+\frac{1}{6} i \pi \log ^3(1-x)-\frac{1}{12} \pi ^2 \log ^2(1-x)-\frac{\pi ^4}{90} & \mbox{if $x<0$} \end{array} \right. \end{eqnarray} Running the code below:

x =.; {Normal[
Series[1/(
4 (1 - x)) ((2 \[Pi]^4)/15 + \[Pi]^2 Log[1 - x]^2 +
5/2 Log[1 - x]^4 - 8/3 Log[1 - x]^3 Log[x] +
4 PolyLog[2, x]^2 + 16 Log[1 - x] PolyLog[3, x] -
12 PolyLog[4, 1 - x] - 28 PolyLog[4, x] -
32 PolyLog[4, x/(-1 + x)] + 16 Log[1 - x] Zeta[3] +
4 (\[Pi]^4/   90 - PolyLog[4, 1 - x] +
1/12 \[Pi]^2 Log[1 - x]^2 + 1/24 Log[1 - x]^4 )), {x, 0, 5},
Assumptions -> 0 < x < 1]],
Normal[Series[
1/(4 (1 - x)) ((2 \[Pi]^4)/15 + \[Pi]^2 Log[1 - x]^2 +
5/2 Log[1 - x]^4 - 8/3 Log[1 - x]^3 Log[x] +
4 PolyLog[2, x]^2 + 16 Log[1 - x] PolyLog[3, x] -
12 PolyLog[4, 1 - x] - 28 PolyLog[4, x] -
32 PolyLog[4, x/(-1 + x)] + 16 Log[1 - x] Zeta[3] +
4 (-(\[Pi]^4/90) + PolyLog[4, 1/(1 - x)] -
1/12 \[Pi]^2 Log[1 - x]^2 + 1/12 Log[1 - x]^4 +
1/6 I Pi Log[1 - x]^3)), {x, 0, 5},
Assumptions -> -1 < x < 0]]}


produces the following result:

{x + (81 x^2)/16 + (14641 x^3)/1296 + (390625 x^4)/20736 + (
352275361 x^5)/12960000,
x + (81 x^2)/16 + (14641 x^3)/1296 + (390625 x^4)/20736 + (
352275361 x^5)/12960000}


as it should be.

Now, apart from the obvious question as to how the result looks like for generic values of $n$ I would like to learn more about the motivation for deriving those Euler sums. Clearly, I can see by myself, that it is very addictive to delve into those calculations and in most cases closed form results can really be obtained. However, what is the use of all that besides pure fun and exercise in integral calculus?

• Answering your last point: polylogarithmic integrals and Euler sums are involved in some interesting conjectures about the (hyper-)volume of hyperbolic manifolds, see en.wikipedia.org/wiki/Volume_conjecture – Jack D'Aurizio Jan 5 '18 at 17:23

One application is quantum field theory, where nested harmonic sums and polylogarithms are used in the calculation of massless quantities in QED and QCD.

See for instance

and the nice survey

all written by J. Blümlein.

• @FelixMarin: Thanks for the credit, Felix. :-) – Markus Scheuer Mar 8 '18 at 21:48