# A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$

Let $H_q$ denote harmonic numbers (generalized to a non-integer index $q$): $$H_q=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+q}\right)=\int_0^1\frac{1-x^q}{1-x}dx=\gamma+\psi(q+1),\tag1$$ where $\psi(z)=\Gamma'(z)/\Gamma(z)$ is the digamma function.

My goal is to evaluate the following series: $$\mathcal S_m=\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/m}}n.\tag2$$ Using the integral representation from $(1)$ we can get equivalent integral forms: $$\mathcal S_m=\int_0^1\frac{\ln(1+\sqrt[m]x)-\ln2}{1-x}\,dx=m\int_0^1\frac{\ln(1+z)-\ln2}{1-z^m}\,z^{m-1}dz.\tag3$$ Here are some simple cases: \begin{align}&\mathcal S_1=\frac{\ln^22}2-\frac{\pi^2}{12}\hspace{7.7em}\color{maroon}{\mathcal S_2=\ln^22-\frac{\pi^2}{12}}\\\\&\color{blue}{\mathcal S_3=\frac{3\ln^22}2-\frac{\pi^2}9+\frac12\,\operatorname{Li}_2\!\left(\tfrac14\right)}\hspace{2em}\color{green}{\mathcal S_4=\frac{7\ln^22}4-\frac{5\pi^2}{48}}\end{align}\tag4 For $\mathcal S_5$ the integral can be found using Mathematica (there is even a closed-form antiderivative, so it should be possible in principle to prove it by differentiation), but the result takes tens of thousands characters to write down (you can see it here), and Mathematica cannot do much simplification on it (here is a simplified result).

But I was able to conjecture a much simpler closed form that fits numerically with a high precision:

$$\mathcal S_5\stackrel{\color{gray}?}=\frac{\ln^22}2-\frac{\ln^25}4+\ln2\cdot\ln5-\frac12\,\operatorname{Li}_2\!\left(\tfrac15\right)-\operatorname{Li}_2\!\left(\frac{\sqrt5-1}2\right)\tag{\diamond}$$

I hope there is a way to prove this result manually without going through huge intermediate expressions, but so far I have not found it.

• Perhaps unhelpful comment: if $F(t) = \sum_{n=1}^\infty (-1)^n \log \Gamma(n t)$, then $S_m + \gamma\log 2 = F'(\frac1m)$. – Greg Martin Aug 26 '15 at 3:05
• Probably writing $1-z^{5}=\left(1-z\right)\left(2z^{2}+\left(1-\sqrt{5}\right)z+2\right)\left(2 z^{2}+\left(1+\sqrt{5}\right)z+2\right)$ and making a boring partial fractions decomposition it is possible to get more tractable integrals. – Marco Cantarini Aug 26 '15 at 7:01
• Even the $m=3$ part starts to get messy if one works only by hand. Btw. are you also interested in solutions for $m<5$? – tired Aug 26 '15 at 8:25
• A lot of interesting results and conjectures here (and at least one boring comment). – marty cohen Aug 26 '15 at 18:07

Here is how one can compute $\mathcal S_m$ for arbitrary $m$.

1. Your formula (3) can be rewritten as $$\mathcal{S}_m=-m\int_0^1\frac{\ln\frac{1+z}{2}}{z^m-1}z^{m-1}dz=-m\sum_{k=0}^{m-1}\alpha_{km}\int_0^1\frac{\ln\frac{1+z}{2}}{z-e^{2\pi i k/m}}dz,$$ where $$\alpha_{km}=\lim_{\quad z\to\; \exp{\frac{2\pi i k}m}}\frac{z^{m-1}\left(z-e^{2\pi i k/m}\right)}{z^m-1}=\frac{e^{2\pi i k(m-1)/m}}{\prod_{n\ne k}\left(e^{2\pi i k/m}-e^{2\pi i n/m}\right)}=\frac1m.$$ Note in particular that the last expression is independent of $k$.

2. The remaining integrals can be computed in terms of polylogarithms: $$I\left(\zeta\right)=\int_0^1\frac{\ln\frac{1+z}{2}}{z-\zeta}dz= \operatorname{Li}_2\left(\frac{2}{1+\zeta}\right)- \operatorname{Li}_2\left(\frac{1}{1+\zeta}\right)+ \ln2\ln\frac{\zeta}{1+\zeta}.\tag{1}$$ We have in particular $$I\left(1\right)=\frac{\pi^2}{12}-\frac{\ln^22}{2},\qquad I\left(-1\right)=-\frac{\ln^22}{2}.$$

This implies that $$\mathcal{S}_m=-\sum_{k=0}^{m-1}I\left(e^{2\pi i k/m}\right), \tag{2}$$ with $I\left(\zeta\right)$ defined by (1). It is clear that under the sum the elementary pieces of $I(\zeta)$ simplify. It might happen that something nice happens also with the dilogarithmic ones.

Update 1 (how to simplify one of the two sums of dilogarithms to an elementary expression):

• This formula for $\operatorname{Li}_2\left(e^{2\pi i \mathbb{Q}}\right)$ implies that $$\sum_{k=0}^{m-1}\operatorname{Li}_2\left(e^{2\pi i k/m}\right)=\frac{\pi^{2}}{6m},\qquad \sum_{k=0}^{m-1}\operatorname{Li}_2\left(-e^{2\pi i k/m}\right)=\begin{cases}\quad \frac{\pi^{2}}{6m},\quad & m\text{ even}, \\ -\frac{\pi^{2}}{12m},\quad & m \text{ odd}. \end{cases}$$
• We also have the identity $\operatorname{Li}_2\left(z\right)=-\operatorname{Li}_2\left(\frac{z}{z-1}\right)-\frac12\ln^2\left(1-z\right)$, which can be rewritten as $$\operatorname{Li}_2\left(\frac{1}{\zeta+1}\right)=-\operatorname{Li}_2\left(-\zeta^{-1}\right)-\frac12\ln^2\frac{\zeta}{\zeta+1}.$$

• Combining both results, we obtain for odd $m$ $$\sum_{k=0}^{m-1}\operatorname{Li}_2\left(\frac{1}{1+e^{2\pi i k/m}}\right)= \frac{\pi^2}{12m}-\frac12\sum_{k=0}^{m-1}\ln^2\left(1+e^{-2\pi i k /m}\right),$$ and for even $m$ $$\sum_{k=0 | k\neq \frac{m}{2}}^{m-1}\operatorname{Li}_2\left(\frac{1}{1+e^{2\pi i k/m}}\right)= \frac{\pi^2(m-1)}{6m}-\frac12\sum_{k=0| k\neq \frac{m}{2}}^{m-1}\ln^2\left(1+e^{-2\pi i k /m}\right).$$

Update 2 (simplification of the remaining sum for even $m$):

We can use again the identity $\operatorname{Li}_2\left(z\right)=-\operatorname{Li}_2\left(\frac{z}{z-1}\right)-\frac12\ln^2\left(1-z\right)$ to show that $$\operatorname{Li}_2\left(\frac{2}{1+\zeta}\right)+\operatorname{Li}_2\left(\frac{2}{1-\zeta}\right)=-\frac12\ln^2\frac{\zeta-1}{\zeta+1}.$$ For even $m$, if $\zeta$ is an $m$th root of unity, then so is $-\zeta$, which simplifies the second sum to an elementary expression: $$\sum_{k=0 | k\neq \frac{m}{2}}^{m-1}\operatorname{Li}_2\left(\frac{2}{1+e^{2\pi i k/m}}\right)=\frac{\pi^2}{12}-\frac{\ln^2 2}{2}-\frac12\sum_{k=1}^{\frac{m}{2}-1}\ln^2\frac{e^{2\pi i k/m}-1}{e^{2\pi i k/m}+1}.$$ Altogether, this leads to evaluation \begin{align*} \mathcal{S}_{2n}&=\frac{\ln 2\ln 8n^2}{2}-\frac{\pi^2}{12n}+\frac12 \sum_{k=1}^{n-1}\ln^2\frac{e^{\pi i k/n}-1}{e^{\pi i k/n}+1}-\frac12\sum_{k=0| k\neq n}^{2n-1}\ln^2\left(1+e^{-\pi i k /n}\right)=\\ &=\ln 2\ln 2n-\frac{\pi^2\left(n^2+1\right)}{24n}-\sum_{k=1}^{n-1}\ln\left(2\sin\frac{\pi k}{2n}\right)\ln\left(2\cos\frac{\pi k}{2n}\right). \tag{$\spadesuit$} \end{align*}

Update 3 (partial simplification for odd $m$)

Let us denote $m=2n+1$. We will use the identity $$\operatorname{Li}_2\left(\frac{2}{1+\zeta}\right)=\operatorname{Li}_2\left(\frac{\zeta+1}{\zeta-1}\right)+\frac12 \ln^2 \frac{\zeta-1}{\zeta+1} -\ln\left(-\frac{2}{1+\zeta}\right)\ln\frac{\zeta-1}{\zeta+1}-\frac{\pi^2}{6}. \tag{3}$$ Now the key three facts are that

• If $\zeta$ is $\zeta^m=1$, then so is $\zeta^{-1}$.

• Under replacement $\zeta\leftrightarrow \zeta^{-1}$, the dilogarithm argument on the right of (3) changes its sign.

• There is an identity $\operatorname{Li}_2\left( z\right)+ \operatorname{Li}_2\left( - z\right)=\frac{1}{2}\operatorname{Li}_2\left( z^2\right)$.

Put altogether, this leads to \begin{align*}\sum_{k=0}^{m-1}\operatorname{Li}_2\left(\frac{2}{1+e^{\frac{2\pi i k}{m}}}\right)=&\frac12\sum_{k=1}^n\operatorname{Li}_2\left(-\cot^2\frac{\pi k}{2n+1}\right)+\frac{4n^2+3n+2}{2n+1}\cdot\frac{\pi^2}{12}+\\ &+\sum_{k=1}^n\ln\left(\tan\frac{\pi k}{2n+1}\right) \ln\left(\frac12\sin\frac{2\pi k}{2n+1}\right). \end{align*} Combining this with the previous results of Update 1, we finally arrive at \begin{align} \nonumber\mathcal{S}_{2n+1}=& -\frac12\sum_{k=1}^n\operatorname{Li}_2\left(-\cot^2\frac{\pi k}{2n+1}\right)-\sum_{k=1}^n\ln^2\sin\frac{\pi k}{2n+1} +\\ &+\left(n+\frac12\right)\ln^2 2-\frac{\left(2n^2+n+1\right)\pi^2}{12\left(2n+1\right)}.\tag{$\clubsuit$} \end{align} Remark. For $n=2$ (i.e. $m=5$) this sum contains two dilogarithms $\operatorname{Li}_2\left(-1\pm\frac{2}{\sqrt5}\right)$. One should be able to reduce them to just one $\operatorname{Li}_2\left(\frac15\right)$ using a suitable identity.

• Great answer! Have you been able to simplify the sum of dilogarithms of complex arguments for $m=5$? – Vladimir Reshetnikov Aug 26 '15 at 21:17
• @VladimirReshetnikov Only partially. I see how to simplify the sum of $\operatorname{Li}_2 \left( \frac{1}{1+\zeta_k}\right)$ to an elementary expression for arbitrary $m$. But the sum of first terms looks more difficult. – Start wearing purple Aug 26 '15 at 21:23
• (+1) it would be interesting to know if and how the dilogs boil down to elementary form whenever $m$ is even – tired Aug 26 '15 at 23:14
• @VladimirReshetnikov For odd $m=2N+1$, there seems to be a way to reduce the remaining sum of $2N+1$ dilogarithms to $N$ dilogarithms. That means $2$ dilogarithms for $m=5$. However I don't know if their reduction to single $\operatorname{Li}_2 \left(\frac15\right)$ is an accident or this can be generalized for higher values of $m$. – Start wearing purple Aug 27 '15 at 15:36
• I was able to reach the same conclusion, although a slightly different sum of the form $$\sum_{k=1}^{(m-1)/2} \operatorname{Li}_2(-\tan^2(\pi k/m)).$$ And I was also able to check Vladimir's result for $m = 5$ using pentagon identity. In our case, it reads $$\mathrm{Li}_2\left(-1-\tfrac{2}{\sqrt{5}}\right) + \mathrm{Li}_2\left(-1+\tfrac{2}{\sqrt{5}}\right) = \mathrm{Li}_2\left(\tfrac{1}{5}\right) + \mathrm{Li}_2\left(-\tfrac{3+\sqrt{5}}{2}\right) + \mathrm{Li}_2\left(-\tfrac{2}{3+\sqrt{5}}\right) + \text{[some log term]}.$$ – Sangchul Lee Aug 28 '15 at 10:39

This is not a novel result, but rather an alternate derivation of @Start wearing purple's result. I tried to keep everything simple and explained, which resulted in a slightly verbose solution. So if you are not interested in details, you may follow only the tagged equations until Step 3.

Step 1 (Reduction of Integral). We begin with the following formula:

$$\mathcal{S}_m = \int_{0}^{1} \frac{\log(1-x^m)}{1+x} \, dx = \sum_{\omega \ : \ \omega^m = 1} \int_{0}^{1} \frac{\log(1 - \omega x)}{1+x} \, dx. \tag{1}$$

In order to work with the RHS, we consider it as a function of $\omega \in \Bbb{C} \setminus (1, \infty)$. Using the differentiation under the integral sign technique, we find that

\begin{align*} \int_{0}^{1} \frac{\log(1 - \omega x)}{1+x} \, dx &= \int_{0}^{\omega} \left( \frac{d}{dz} \int_{0}^{1} \frac{\log(1 - zx)}{1+x} \, dx \right) \, dz\\ &= \int_{0}^{\omega} \left( \int_{0}^{1} \left( \frac{1}{1+x} - \frac{1}{1-zx}\right) dx \right) \frac{dz}{1+z} \\ &= \int_{0}^{\omega} \left( \frac{\log(1-z)}{z} - \frac{\log\left(\frac{1-z}{2}\right)}{1+z} \right) \, dz \\ &= -\operatorname{Li}_2(\omega) + I(\omega), \end{align*}

where the integral w.r.t. $z$ is taken inside the region $\Bbb{C}\setminus(1, \infty)$ and the function $I(\omega)$ is defined on $\Bbb{C}\setminus(1,\infty)$ as

$$I(\omega) := -\int_{0}^{\omega} \frac{\log\big(\frac{1-z}{2}\big)}{1+z} \, dz. \tag{*}$$

Step 2 (Properties of $I(\omega)$). Why we consider this integral is that it satisfies the following two properties:

1. Using the identity $\text{(1)}$ and the relation we have developed, $\mathcal{S}_m$ is written as $$\mathcal{S}_m =\sum_{\omega \ : \ \omega^m = 1} (I(\omega) - \operatorname{Li}_2(\omega)) = -\frac{\zeta(2)}{m} + \sum_{\omega \ : \ \omega^m = 1} I(\omega). \tag{2}$$ This follows from the multiplication theorem for polylogarithm.

2. More importantly, for $\omega \in \Bbb{C}$ outside $(-\infty, 1) \cup (1, \infty)$, we have $$I(-\omega) + I(\omega) = \log^2 2 - \log\big(\tfrac{1+\omega}{2}\big)\log\big(\tfrac{1-\omega}{2}\big). \tag{3}$$ Here, we take a convention that the log part is $0$ when $\omega = \pm 1$. This is consistent both with the limit as $\omega \to 1$ or $\omega \to -1$ and with the actual known values of $I(1)+I(-1)$. Notice also that $\text{(3)}$ easily follows from integration by parts: $$I(-\omega) = \left[ -\log\big(\tfrac{1+z}{2}\big)\log\big(\tfrac{1-z}{2}\big) \right]_{0}^{-\omega} - \int_{0}^{-\omega} \frac{\log\big(\frac{1+z}{2}\big)}{1-z} \, dz.$$

Step 3 (Formula for $\mathcal{S}_{2p}$). When $m = 2p$, as @Start wearing purple observed, $\omega^m = 1$ implies $(-\omega)^m = 1$. Thus combining $\text{(2)}$ and $\text{(3)}$, we have

\begin{align*} \mathcal{S}_{2p} &= -\frac{\pi^2}{12p} + \frac{1}{2} \sum_{\omega \ : \ \omega^{2p} = 1} ( I(\omega) + I(-\omega)) \\ &= -\frac{\pi^2}{12p} + \frac{1}{2} \sum_{\omega \ : \ \omega^{2p} = 1} \left( \log^2 2 - \log\left(\frac{1+\omega}{2}\right)\log\left(\frac{1-\omega}{2}\right) \right). \end{align*}

(Here, as pointed out in Step 2, we consider $\log(\frac{1+\omega}{2})\log(\frac{1-\omega}{2}) = 0$ when $\omega = \pm 1$.) Finally, we can simplify the last summation by considering only $\omega$ with $\Im(\omega) > 0$ as follows:

\begin{align*} \mathcal{S}_{2p} &= p\log^2 2 - \frac{\pi^2}{12p} - \Re \sum_{\substack{\omega \ : \ \omega^{2p} = 1 \\ \Im(\omega) > 0}} \log\left(\frac{1+\omega}{2}\right)\log\left(\frac{1-\omega}{2}\right) \\ &= p\log^2 2 - \frac{\pi^2}{12p} - \Re \sum_{k=1}^{p-1} \log\left(\frac{1+e^{ik\pi/p}}{2}\right)\log\left(\frac{1+e^{i(k-p)\pi/p}}{2}\right) \\ &= p\log^2 2 - \frac{\pi^2}{12p} - \sum_{k=1}^{p-1} \left( \log\left(\cos\frac{\pi k}{2p}\right)\log\left(\sin\frac{\pi k}{2p}\right) + \frac{\pi^2}{4p^2}k(p-k) \right) \\ &= p\log^2 2 - \frac{(p^2+1)\pi^2}{24p} - \sum_{k=1}^{p-1} \log\left(\cos\frac{\pi k}{2p}\right)\log\left(\sin\frac{\pi k}{2p}\right). \end{align*}

This solution relies on the symmetry between $I(\omega)$ and $I(-\omega)$, so I doubt that this will work for odd $m$.

• Very nice solution. In the odd case I can simplify the dilogs using that if $\omega^{m}=1$, then also $\left(\omega^{-1}\right)^m=0$. What remains is $\frac{m-1}{2}$ dilogs with real negative arguments. – Start wearing purple Aug 28 '15 at 8:24

$$\mathcal{S}_6 \stackrel{?}{=} \ln^2(2) + \ln(2)\ln(3) -\frac{5\pi^2}{36} .$$

Furthermore, you probably know that

$$\operatorname{Li}_2\left(\frac{\sqrt5 - 1}{2}\right) = \frac{\pi^2}{10} - \ln^2\left(\varphi\right) = \frac{\pi^2}{10} - \ln^2(2) -\ln^2\left(1+\sqrt{5}\right) + 2\ln(2)\ln\left(1+\sqrt{5}\right),$$

where $\varphi = \tfrac{1}{2}\left(1+\sqrt5\right)$ is the golden ratio.

At last, it seems to me there are always some $$\operatorname{Li}_2\left({\tfrac12}\right) = \frac{\pi^2}{12} - \frac{1}{2}\ln^2(2)$$ and $$\operatorname{Li}^2_1\left({\tfrac12}\right) = \ln^2(2)$$ behind the scenes.

I've found some relations between particular $\mathcal{S}_m$ values. For example: \begin{align} \mathcal{S}_{1} &=\tfrac{9}{4}\mathcal{S}_{2}-\mathcal{S}_4\\ \mathcal{S}_{1} &=\tfrac{9}{5}\mathcal{S}_{1/3}-\tfrac{8}{5}\mathcal{S}_{1/2}\\ \mathcal{S}_{1} &= \tfrac{1}{2}\mathcal{S}_{1/2}+\tfrac{3}{8}\mathcal{S}_2\\ \mathcal{S}_{1} &=\tfrac{3}{7}\mathcal{S}_{1/3}+\tfrac{2}{7}\mathcal{S}_2\\ \mathcal{S}_{1} &=\tfrac{3}{5}\mathcal{S}_{1/2}+\tfrac{1}{5}\mathcal{S}_4\\ \mathcal{S}_{1} &=\tfrac{27}{55}\mathcal{S}_{1/3}+\tfrac{8}{55}\mathcal{S}_4\\ \mathcal{S}_{1} &=\mathcal{S}_{1/2}+\mathcal{S}_{4} -\tfrac{3}{2}\mathcal{S}_{2}\\ \mathcal{S}_{1} &=\tfrac{1}{10}\mathcal{S}_{4}+\tfrac{9}{10}\mathcal{S}_{1/3} - \tfrac{1}{2}\mathcal{S}_{1/2}\\ \end{align} or my favorite one: $$3\,\mathcal{S}_{4}+5\,\mathcal{S}_{1/2} = 3\,\mathcal{S}_{1/3}+5\,\mathcal{S}_{2}.$$

• I remembered that there was a closed form for $\operatorname{Li}_2(\phi^{-2})$, but for some reason I forgot that $\operatorname{Li}_2(\phi^{-1})$ had it as well, and did not even check. Thanks for the reminder! – Vladimir Reshetnikov Aug 26 '15 at 17:34
• Also, $\mathcal S_8\stackrel?=\frac{21}8\ln^2 2+\frac12\ln^2(1+\sqrt2)-\frac{17}{96}\pi^2$ and $\mathcal S_{12}\stackrel?=\frac74\ln^2 2+\ln2\cdot\ln3+\frac12\ln^2(2+\sqrt3)-\frac{37}{144}\pi^2$. – Vladimir Reshetnikov Aug 26 '15 at 17:51
• And $\mathcal S_{10}\stackrel?=3\ln^2 2+2\ln^2(1+\sqrt5)+\ln2\cdot\ln5-4\ln2\cdot\ln(1+\sqrt5)-\frac{13}{60}\pi^2$. – Vladimir Reshetnikov Aug 26 '15 at 18:05
• $\mathcal S_{16}\stackrel?=\frac12 \ln^2 2+\ln^2 \cos\frac\pi{16} + \ln^2 \cos\frac{3\pi}{16} + \ln^2 \sin\frac\pi{16} + \ln^2\sin\frac{3\pi}{16} - \frac{65}{192}\pi^2$. – Vladimir Reshetnikov Aug 26 '15 at 23:02
• $\mathcal S_{14}\stackrel?=3\left(\ln^2\sin\frac\pi7+\ln^2\sin\frac{2\pi}7+\ln^2\sin\frac{3\pi}7\right) - 8\ln^22 - \frac14\ln^27 + 4\ln2\cdot\ln7 - \frac{25}{84}\pi^2$. It seems that some pattern starts to emerge... – Vladimir Reshetnikov Aug 26 '15 at 23:33

This is not the full solution, it's too long for a comment, and gives some intuition of @user153012's observation about the appearance of $\operatorname{Li_2(\frac12)}$ and $\ln^2 2$.
We can see that $$S_m=\sum_{n,k=1}^{\infty}\frac{(-1)^n}{k(mk+n)}=\sum_{n,k=1}^{\infty}\frac{(-1)^n}{k}\int_0^1 x^{mk+n-1}dx=\int_0^1 \frac{\ln(1-x^m)}{1+x}$$ Now we can divide the integral into $m$ integrals of the form $\displaystyle \int_0^1 \frac{\ln(\alpha-x)}{1+x}dx$ where $\alpha$ is a $m$th root of unity.

But $$\int_0^1 \frac{\ln(a-x)}{1+x}dx=\int_1^2 \frac{\ln(a+1-x)}{x}dx=\ln2\ln(a+1)+\int_1^2 \frac{\ln(1-\frac{x}{a+1})}{x}dx\\=\ln2\ln(a+1)+\operatorname{Li_2}\left(\frac{1}{a+1}\right)-\operatorname{Li_2}\left(\frac{2}{a+1}\right).$$ Now , $1$ is a root of unity, so the closed form possibly conatains $\ln2\ln(1+1 )$ and $\operatorname{Li_2}\left(\frac{1}{1+1}\right).$

Here is the general case for odd numbers. Let $m$ be odd, and let $\alpha$ be its associated primitive $m$th root of unity. We'll evaluate $\displaystyle S_m=\int_0^1 \frac{\ln(1-x^m)}{1+x}=\ \sum_{k=0}^{m-1} \ln2\ln(\alpha^k+1)+\operatorname{Li_2}\left(\frac{1}{\alpha^k+1}\right)-\operatorname{Li_2}\left(\frac{2}{\alpha^k+1}\right).$

The first sum is $\,\,\ln2\ln(\prod_{k=0}^{m-1}(1+\alpha^k))=\ln^2 2$, which follows from letting $x=-1$ in $\displaystyle \prod_{k=0}^{m-1}(\alpha^k-x)=1-x^m$ $\,\,\,\,\,\,\,\,$($m$ being odd promises that $-1$ isn't a root of unity.)

The second sum is $$\sum_{k=0}^{m-1}\operatorname{Li_2}\left(\frac{1}{\alpha^k+1}\right)=\operatorname{Li_2}(\frac12)+\frac{\pi^2}{6}\frac{m-1}{2}-\left(\ln\left(\frac{1}{\alpha+1}\right)\ln\left(\frac{1}{\alpha^{m-2}}\right)+\cdots+\ln\left(\frac{1}{\alpha^{(m-1)/2}+1}\right)\ln\left(\frac{1}{\alpha^{(m+1)/2}}\right)\right)\\=\operatorname{Li_2}(\frac12)+\frac{\pi^2}{6}\frac{m-1}{2}-\sum_{k=0}^{(m-1)/2} \left(\ln^2\left(\frac{\sec(\frac{k\pi}{m})}{2}\right)+\frac{k^2\pi^2}{m^2}\right)$$.

Proof: apart from $\alpha^0=1$ which gives $\operatorname{Li_2}\left(\frac12\right)$, we can pair up $(m-1)/2$ pairs of conjugate roots. These conjugates satisfy $\displaystyle \frac{1}{1+\alpha^k}=1-\frac{1}{1+\alpha^{m-k}}$, and using the identity $\operatorname{Li_2}(x)+\operatorname{Li_2}(1-x)=\frac{\pi^2}{6}-\ln(x)\ln(1-x)$ the first equality follows.

Now, $\displaystyle \frac1{1+\alpha^k}=\frac1{1+e^{2\pi i k/m}}=\frac12-\frac12 i \tan\left(\frac{\pi k}{m}\right)$, and $\displaystyle \frac1{1+\alpha^{m-k}}=\frac12+\frac12 i \tan\left(\frac{\pi k}{m}\right)$

and so $$\ln\left(\frac1{1+\alpha^k}\right)\ln\left(\frac1{1+\alpha^{m-k}}\right)=\ln^2\left(\frac{\sec(\frac{k\pi}{m})}{2}\right)+\frac{k^2\pi^2}{m^2}$$

The last sum is hard. Maybe we can again pair up excluding $1$ and using dilogarithm identities to simplify things, but i can't see it.

General solution

I have found a systematic path to a general solution.

It turns out the the generalized problem is easier to solve and that, as a by-product, it reveals clearly the structure of the problem.

So for natural $$m$$ consider the sum

$$s_{m} = \sum_{n=1}^\infty\frac{(-1)^n\,H_{n/m}}n\tag{1}$$

Doing the $$k$$-sum after replacing $$H_{z}=\int_{0}^{1} \frac{1-x^z}{1-x}\,dx, z\to\frac{k}{m}$$ we can write

$$\sum _{k=1}^{\infty } \frac{(-1)^k \left(1-x^{k/m}\right)}{k (1-x)}=\int_0^1 \frac{-\log (2)+\log \left(1+x^{1/m}\right)}{1-x}\,dx \\\overset{x\to t^m}=-\int_0^1 m t^{m-1}\left(\frac{\log(2)-\log(1+t)}{1-t^m} \right)\,dt \\ =\log(2)\log(1-t^m) + m \int t^{m-1}\left(\frac{\log(1+t)}{1-t^m}\right)\,dt\tag{2}$$

Here in the last line we have switches to indeinite integrals to avoid spurious divergencies.

Integration by parts of the last integral with

$$u = \int m \frac{t^{m-1}}{1-t^m}\,dt = - \log(1-t^m), v = \log(1+t)\tag{3}$$

gives

$$m \int t^{m-1}\left(\frac{\log(1+t)}{1-t^m}\right)\,dt\\ = - \log(1-t^m)\log(1+t) + \int \frac{\log(1-t^m)}{1+t}\,dt\tag{4}$$

Collecting what we have up to now in

$$f_{m}(t) = - \log(1-t^m)\log(\frac{1+t}{2}) + \int \frac{\log(1-t^m)}{1+t}\,dt \tag{5}$$

so that

$$s_m = f_{m}(t\to 1)-f_{m}(t\to0)= \int_0^1 \frac{\log(1-t^m)}{1+t}\,dt\tag{6}$$

because the product of the logs vanishes at both ends. For what follows it will be convenient technically to proceed with the indefinite version $$f_m(t)$$.

The integral can be calculated. Writing

$$1-t^m = \prod_{k=1}^{m} (\alpha_{m,k} - t)$$

where the roots are

$$\alpha_{m,k} = \exp(2 i \pi \frac{k}{m}), k=1..m\tag{7}$$

In what follows we will drop the index $$m$$ in $$\alpha$$ for simplicity.

Now the log can be written as

$$\log(1-t^m) = \log(\prod_{k=1}^{m}(\alpha_{m,k}-t))= \sum_{k=1}^{m}\log(\alpha_{m,k}-t)\tag{8}$$

and the integral boils down to terms of the form

$$\int \frac{\log (\alpha -t)}{t+1} \, dt =\operatorname{Li}_2\left(\frac{\alpha -t}{\alpha +1}\right)+\log \left(\frac{t+1}{\alpha +1}\right) \log (\alpha -t)\tag{9}$$

Now observing that we can write

$$- \log(1-t^m)\log\left(\frac{1+t}{2}\right)= -\log\left(\frac{1+t}{2}\right)\sum_{k=1}^{m} \log(\alpha_k - t)\tag{10}$$

the indefinite integral $$(5)$$ can be written as

$$f_{m}(t) = \sum_{k=1}^m \left(\operatorname{Li}_2\left(\frac{\alpha_{k} -t}{\alpha_{k} +1}\right) \\ +\log \left(\frac{t+1}{\alpha_{k} +1}\right) \log (\alpha_{k} -t)-\log\left(\frac{1+t}{2}\right) \log(\alpha_k - t)\right)\tag{11}$$

To make this expression definite (and finite) we follow $$(6)$$ and take the value at $$t=1$$ and subtract its value at $$t=0$$

$$\phi_{m}(t) = f_{m}(t)- f_{m}(0)$$

This results in a sum over

$$A_{k} = \operatorname{Li}_2\left(\frac{\alpha_{k}-1}{\alpha_{k}+1}\right)-\operatorname{Li}_2\left(\frac{\alpha_{k}}{\alpha_{k}+1}\right)-\log\left(1-\frac{1}{\alpha_{k}}\right)\log\left(\frac{1+\alpha_{k}}{2}\right)\tag{12}$$

There's a still little technical difficulty to circumvent: it turns out that inserting the value of the index $$k$$ as an integer leads to a divergent behaviour of some $$A_{k}$$.

Hence we understand $$A_{k}$$ as the limit from below $$A_{k}=\lim_{\epsilon\to +0} \,A_{k-\epsilon}$$

Now we are ready to write down the closed expression for the sum $$(1)$$

$$s_{m} = \sum_{k=1}^{m} A_{k}\tag{13}$$

The general structure of the solution is clear: the sum is composed of quadratic $$\log$$s and linear $$\operatorname{Li}_2$$s. No higher polylogs appear.

Remark 1: The main remaining task would be a possible simplification of the polylog function $$\operatorname{Li}_2(z)$$ of a composite and complex arguments.

Remark 2: It seems possible to do a similar analysis for any rational $$m=\frac{p}{q}$$.

Remark 3: as we have encountered only antiderivatives before the final limits, we can apply the same reasoning to find the generating function

$$g_{m}(z) = \sum_{k=1}^{\infty}\frac{z^k}{k} H_{\frac{k}{m}}$$

Evaluation of specific cases

Now we turn to the harvest. But I ask for some patience.

Original post

With the help of Mathematica I find a long expression which is numerically correct but needs plenty of simplification in order to become "culturally acceptable".

Doing the $$k$$-sum after replacing $$H_{z}=\int_{0}^{1} \frac{1-x^z}{1-x}\,dx, z\to\frac{k}{5}$$ gives the integral

$$i= \int_0^1 \frac{\log (2)-\log \left(\sqrt{x}+1\right)}{x-1} \, dx$$

Letting $$x\to t^5$$ the integrand of $$dt$$ becomes

$$\frac{5 t^4 \log \left(\frac{2}{t+1}\right)}{t^5-1}=\frac{5 t^4 \log (2)}{t^5-1}-\frac{5 t^4 \log (t+1)}{t^5-1}$$

To avoid divergences we continue with the indefinite integral.

The first integral is

$$i_1= \int \frac{5 t^4 \log (2)}{t^5-1} = \log (2) \log \left(t^5-1\right)$$

and the second one is done by Mathematica returning a lengthy expression.

The we form the sum $$i=i_1 + i_2$$ and calculate the limits $$t\to0$$ and $$t\to1$$ to find this horrendeous expression

$$i = -\text{Li}_2\left(\frac{\sqrt{-1}}{-1+\sqrt{-1}}\right)+\text{Li}_2\left(\frac{1+\sqrt{-1}}{-1+\sqrt{-1}}\right)-\text{Li}_2\left(\frac{(-1)^{2/5}}{1+(-1)^{2/5}}\right)+\text{Li}_2\left(\frac{-1+(-1)^{2/5}}{1+(-1)^{2/5}}\right)-\text{Li}_2\left(\frac{(-1)^{3/5}}{-1+(-1)^{3/5}}\right)+\text{Li}_2\left(\frac{1+(-1)^{3/5}}{-1+(-1)^{3/5}}\right)-\text{Li}_2\left(\frac{(-1)^{4/5}}{1+(-1)^{4/5}}\right)+\text{Li}_2\left(\frac{-1+(-1)^{4/5}}{1+(-1)^{4/5}}\right)+\frac{43 \pi ^2}{60}+\frac{\log ^2(2)}{2}+\log (5) \log (2)-\log \left(\sqrt{-1}-1\right) \log \left(1+\sqrt{-1}\right)-\log \left((-1)^{2/5}-1\right) \log \left(1+(-1)^{2/5}\right)-\log \left((-1)^{3/5}-1\right) \log \left(1+(-1)^{3/5}\right)-\log \left((-1)^{4/5}-1\right) \log \left(1+(-1)^{4/5}\right)+\frac{1}{5} i \pi \left(\log \left(\sqrt{-1}-1\right)+5 \log \left(1+\sqrt{-1}\right)+2 \log \left(1+(-1)^{2/5}\right)+3 \log \left((-1)^{3/5}-1\right)+5 \log \left(1+(-1)^{3/5}\right)+4 \log \left(1+(-1)^{4/5}\right)\right)$$

The numeric values of which is $$N(i) = -0.152666...$$

• This approach merely restates the problem, rather than truly answering it... which is not to say that it is useless, but rather that it is incomplete. In order for completion to occur, one has to supply it with various lists of polylogarithmic formulas, identities, and interrelations. – Lucian Feb 20 '20 at 15:45
• @ Lucian We're getting philosophical: isn't every result, once it is understood, merely a restatement of the original problem? But you are right the result is incomplete as is does not prove (or disprove) the conjecture. But even then we can't be sure that we have found the "final" result which can't be simplified further. I am full of admiration for the people excelling in simplifying the polylogs. – Dr. Wolfgang Hintze Feb 20 '20 at 20:55
• A restatement, perhaps, but not a mere restatement. – Lucian Feb 21 '20 at 0:03

Introduction

This is not a new solution but a statement (too long for a comment)about the performance of Mathematica:

The author states:

"For S=5 the integral can be found using Mathematica (there is even a closed-form antiderivative, so it should be possible in principle to prove it by differentiation),

(a) but the result takes tens of thousands characters to write down (you can see it here), and

(b) Mathematica cannot do much simplification on it (here is a simplified result)."

I cannot confirm the statements (a) and (b), and a made a short systematic study of the performance for appropriate commands.

Calculations

I defined $S_m$ by the second integral form in (3) of the OP.

The cases m=1..4 are immediately solved in the form given in the OP.

Starting at m=5 I let Mathematica calculate the antiderivative and took the limits according to the fundamental theorem of calculus.

Up to m=10 the execution time remained well under a minute, and the results can be put on no more than two pages.

Performance data and the lengths of the expressions as given by LeafCount were sampled for m=1..10. Notice that the number of lines is approximately given by LeafCount/70.

Here's the command executed:

TableForm[
Table[{m,
x = AbsoluteTiming[
Integrate[m (Log[1 + z] - Log)/(1 - z^m) z^(m - 1), z]];
x[], LeafCount[x[]],
s = AbsoluteTiming[(Limit[x[], z -> 1] //
Quiet - Limit[x[], z -> 0] // Quiet)]; s[],
LeafCount[s[]], ss = AbsoluteTiming[Simplify[s[]]]; ss[],
LeafCount[ss[]], tt = x[] + s[] + ss[] }, {m, 1, 10}],
"t_SF", "LC(SF)", "t_TOT"}}]


Here

$t_{AD}$ - time to compute the antiderivative
$LC(AD)$ - length of antidrivative
$t_{Lim}$ - time to perform the Limits
$LC(Lim)$ - length of result of Limits
$t_{SF}$ - time to simplify
$LC(SF)$ - length of simplified Expression
$t_{TOT}$ - total execution time

The results (obtained with a fresh kernel) are summarized here

$$\begin{array}{cccccccc} m & t_{AD} & LC(AD) & t_{Lim} & LC(Lim) & t_{SF} & LC(FS) & t_{TOT}\\ 1 & 0.0583683 & 39 & 0.390468 & 28 & 0.00373215 & 28 & 0.452568 \\ 2 & 0.0430864 & 51 & 0.383368 & 30 & 0.00122568 & 30 & 0.42768 \\ 3 & 0.191909 & 167 & 4.35969 & 84 & 0.0553141 & 78 & 4.60691 \\ 4 & 0.0604302 & 121 & 1.30459 & 82 & 0.0130596 & 76 & 1.37808 \\ 5 & 1.05841 & 352 & 7.80234 & 479 & 0.266691 & 216 & 9.12744 \\ 6 & 0.276604 & 285 & 8.0831 & 169 & 0.0814213 & 163 & 8.44113 \\ 7 & 2.04412 & 585 & 17.424 & 2661 & 8.48519 & 2215 & 27.9533 \\ 8 & 0.549521 & 287 & 48.6261 & 266 & 0.151403 & 260 & 49.327 \\ 9 & 3.01488 & 401 & 22.6227 & 444 & 0.448261 & 374 & 26.0859 \\ 10 & 2.93397 & 1490 & 30.7236 & 2051 & 5.38636 & 1636 & 39.0439 \\ \end{array}$$