Given the nth harmonic number $ H_n = \sum_{j=1}^{n} \frac{1}{j}$, we get from this post that apparently,

$$\sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n= S_{k-1,2}(z) + \rm{Li}_{\,k+1}(z)$$

for $-1\leq z\leq 1$, and with Nielsen generalized polylogarithm $S_{n,p}(z)$ and polylogarithm $\rm{Li}_n(z)$. Hence for small $k$,

$$\sum_{n=1}^{\infty}\frac{H_n}{n^2\, 2^n}= S_{1,2}\big(\tfrac12\big)+\rm{Li}_3\big(\tfrac12\big)$$

$$\sum_{n=1}^{\infty}\frac{H_n}{n^3\, 2^n}= S_{2,2}\big(\tfrac12\big)+\rm{Li}_4\big(\tfrac12\big)$$

$$\sum_{n=1}^{\infty}\frac{H_n}{n^4\, 2^n}= S_{3,2}\big(\tfrac12\big)+\rm{Li}_5\big(\tfrac12\big)$$

and so on. Explicitly, given $a=\ln 2$,

$$S_{1,2}\big(\tfrac12\big) +\tfrac1{6}a^3-\tfrac18 \zeta(3)=0 $$

$$S_{2,2}\big(\tfrac12\big) +\tfrac1{168}a^4+\tfrac17a^2\,\rm{Li}_2\big(\tfrac12\big)+\tfrac17a\,\rm{Li}_3\big(\tfrac12\big)-\tfrac18\zeta(4) = 0$$

which are discussed in this and this post. And by yours truly,

$$S_{3,2}\big(\tfrac12\big) -A+B = 0$$

$$A = \tfrac{41}{840}a^5+\tfrac5{21}a^3\,\rm{Li}_2\big(\tfrac12\big)+\tfrac47a^2\,\rm{Li}_3\big(\tfrac12\big)+a\,\rm{Li}_4\big(\tfrac12\big) + \rm{Li}_5\big(\tfrac12\big) $$


Q: What, however, is the explicit evaluation in ordinary polylogs of the next steps, namely $S_{4,2}\big(\tfrac12\big)$ and $S_{5,2}\big(\tfrac12\big)$?

P.S. Try as I might, they resist being evaluated and there are indications these higher order integrals may not be expressible by ordinary polylogs.


2 Answers 2


$\newcommand{\Li}{\mathrm{Li}}$You are probably correct that the sufficiently higher order sums do not have a closed form in terms of classical polylogarithms. In particular the $k=6$ sum should not have such an expression, although this claim rests on some big conjectures (the so-called Grothendeick Period Conjecture, or some sufficiently special-case). However the $k=5$ case does appear to have a closed form, as given below.

Essentially there is an algebraically defined version of (Nielsen) polylogs, the so-called motivic (Nielsen) polylogs. These objects have a much richer structure, and there is an invariant (the coproduct $\Delta$ and the cobracket $\delta$) which can distinguish Nielsen from classical polylogs. The cobracket $\delta\Li_n(x) = 0 $, whereas $ \delta S_{5,2}(x) = \Li_2(x) \wedge \Li_5(1) + \Li_4(x) \wedge \Li_3(1) \neq 0 $; the factors of the wedge are viewed modulo products. In particular, $ \delta S_{5,2}(\frac{1}{2}) = \Li_4(\frac{1}{2}) \wedge \zeta(3) \neq 0 $. (Can check that $L_4(\frac{1}{2}) / \zeta(4) \not\in \mathbb{Q} $, where $ L_4 $ is some single-valued version of $\Li_4$.)

However $ \delta S_{4,2}(x) = \Li_3(x) \wedge \Li_3(1) $, and since $ \Li_3(\frac{1}{2}) = \frac{7}{8} \zeta(3) - \frac{1}{2} \zeta(2) \log(2) + \frac{1}{6} \log(2)^3 $, one gets $$ \delta S_{4,2}\Big(\frac{1}{2}\Big) = \frac{7}{8} \zeta(3) \wedge \zeta(3) = 0 \,, $$ by the antisymmetry of the wedge. Another Big Conjecture from Goncharov claims that classical polylogs are exactly the kernel of this cobracket, so conjecturally we should be able to expression $ S_{4,2}(\frac{1}{2}) $ in terms of $ \Li_6 $.

In the answer I posted to another of your questions, I gave an explicit formula for $S_{4,2}(-1)$ and some explanation via the coproduct of how one obtains it (up to a rational numerically determined coefficient of $\zeta(6)$). One can also get a formula for $S_{4,2}(\frac{1}{2})$, and hence for sum for $k=5$, from the $S_{4,2}(-1)$ reduction. In particular \begin{align*} \sum_{n=1}^\infty\frac{(-1)^n H_n}{n^5} = {} & S_{4,2}(-1) - \frac{31}{32} \zeta(6) = \zeta(\overline{5}, 1) - \frac{31}{32} \zeta(6) \\ \overset{?}{=} {} & \frac{1}{13} \bigg(\frac{1}{3}\Li_6\Big(-\frac{1}{8}\Big)-162 \Li_6\Big(-\frac{1}{2}\Big)-126 \Li_6\Big(\frac{1}{2}\Big)\bigg) -\frac{4783 }{1248}\zeta (6) +\frac{3}{8} \zeta (3)^2 \\ & {}+\frac{31}{16} \zeta (5) \log(2) -\frac{15}{26} \zeta (4) \log ^2(2) +\frac{3}{104} \zeta (2) \log^4(2) -\frac{1}{208} \log ^6(2) \,, \\[2em] \sum_{n=1}^\infty\frac{H_n}{n^5 2^n} \overset{?}{=} {} & -\frac{1}{26} \left( \frac{1}{3}\Li_6\left(-\frac{1}{8}\right) -162\Li_6\left(-\frac{1}{2}\right) -204\Li_6\left(\frac{1}{2}\right)\right) -\frac{101 }{624} \zeta (6) -\frac{7}{16} \zeta (3)^2 \\ &+\Big(\Li_5\left(\frac{1}{2}\right) -\zeta (5)+\frac{1}{2}\zeta (2) \zeta (3) \Big) \log (2) +\frac{73}{208} \zeta (4) \log ^2(2)\\ &-\frac{1}{6} \zeta (3) \log^3(2) +\frac{17}{624} \zeta (2) \log ^4(2) -\frac{11}{6240}\log ^6(2) \,. \end{align*} Amusingly, since $ \Li_3(\phi^{-2}) = \frac{4}{5} \zeta(3) + \text{products} $, where $ \phi = \frac{1 + \sqrt{5}}{2} $ is the golden ratio, we can also conjecturally evaluate $S_{4,2}(\phi^{-2})$ (up to a rational but numerically determined coefficient of $\zeta(6)$). We have \begin{align*} \sum_{n=1}^\infty\frac{H_n}{n^5 \phi^{2n}} \overset{?}{=} {} & \frac{1}{396} \Big( 2 \Li_6\big( \phi ^{-6} \big) -128 \Li_6\big( \phi ^{-3}\big) +1197\Li_6\big(\phi ^{-2}\big) -576 \Li_6\big(\phi^{-1} \big) \Big) \\ &+\frac{35 }{99}\zeta (6) +\frac{2}{5} \zeta (3)^2 + \Li_5 \left(\phi ^{-2}\right) \log (\phi) -\zeta (5)\log (\phi ) \\ & +\frac{2}{11} \zeta (4) \log ^2(\phi ) -\zeta (3) \Li_3\left(\phi ^{-2}\right) +\frac{10}{33} \zeta (2) \log ^4(\phi ) -\frac{79}{990} \log ^6(\phi ) \,. \end{align*} The only uncertainty in these formulae should be the (necessarily rational) coefficient of $\zeta(6)$, all other terms are fixed by analysis of the coproduct.


Let $I$ denotes $\int_0^1\frac{\ln^4(1+x)\ln x}{x}\ dx$

I proved here

\begin{align} I&=-120\operatorname{Li}_6\left(\frac12\right)-72\ln2\operatorname{Li}_5\left(\frac12\right)-24\ln^22\operatorname{Li}_4\left(\frac12\right)+78\zeta(6)+\frac34\ln2\zeta(5)\\ &\quad-\frac32\ln^22\zeta(4)-3\ln^32\zeta(3)+2\ln^42\zeta(2)+12\zeta^2(3)-12\ln2\zeta(2)\zeta(3)\\ &\quad-\frac{17}{30}\ln^62+24\sum_{n=1}^\infty\frac{H_n}{n^52^n}\tag{1} \end{align}

On the other hand and by applying integration by parts we have

\begin{align} I=-2\int_0^1\frac{\ln^3(1+x)\ln^2x}{1+x}\ dx \end{align} You can find here the following identity

$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$

Replacing $x$ with $-x$ yields

$$-\frac{\ln^3(1+x)}{1+x}=\sum_{n=1}^\infty (-1)^nx^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$

Now we can write \begin{align} I&=2\sum_{n=1}^\infty (-1)^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)\int_0^1 x^n \ln^2x\ dx\\ &=-2\sum_{n=1}^\infty (-1)^n\left(H_{n-1}^3-3H_{n-1}H_{n-1}^{(2)}+2H_{n-1}^{(3)}\right)\int_0^1 x^{n-1} \ln^2x\ dx\\ &=-2\sum_{n=1}^\infty (-1)^n\left(H_{n}^3-3H_{n}H_{n}^{(2)}+2H_{n}^{(3)}+\frac{6H_n}{n^3}-\frac{3H_n^2}{n}+\frac{3H_n^{(2)}}{n}-\frac{6}{n^2}\right)\left(\frac{2}{n^3}\right)\\ \end{align}


\begin{align} I&=12\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^3}-4\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^3}-8\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^3}-24\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}\\ &\quad+12\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^4}-12\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^4}+24\sum_{n=1}^\infty\frac{(-1)^n}{n^6}\tag{2} \end{align}

We can see from $(1)$ and $(2)$ that our target sum is related to $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}$.

I am not sure if the alternating sums in $(2)$ have closed form or not but I am sure that $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}$ has no closed form because the power of the denominator is odd more than 3. So the target sum has no closed form unless $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}$ cancels out with other sums somehow which I doubt.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.