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.
Solution 1:
Here is how one can compute $\mathcal S_m$ for arbitrary $m$.
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 $k$.
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.
Solution 2:
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:
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.
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 $m$.
Solution 3:
And what about this one?
$$\mathcal{S}_6 \stackrel{?}{=} \ln^2(2) + \ln(2)\ln(3) -\frac{5\pi^2}{36} .$$
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: $$ 3\,\mathcal{S}_{4}+5\,\mathcal{S}_{1/2} = 3\,\mathcal{S}_{1/3}+5\,\mathcal{S}_{2}. $$
Solution 4:
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.