Hypergeometric series for $\mathrm{Cl}_2(\pi/3)$

Question

I am trying to find a hypergeometric series for $\mathrm{Cl}_2(\pi/3)$, where $$\mathrm{Cl}_2(x)=-\int_0^x\log\left|2\sin\frac{t}2\right|dt=\sum_{k\geq1}\frac{\sin kx}{k^2}$$ Is the Clausen Function of order $2$.

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Context: I have been really interested in hypergeometric series lately, and took it on as a task to find a hypergeometric representation for the aforementioned constant.

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What I've done.

We define $$d_3(n)=(-1)^{\left\lfloor\frac{n}3\right\rfloor}$$ as well as $$\chi_3(n)=\text{sgn}\,\text{mod}(n,3)$$ With $$\text{sgn}\,n=\frac{n}{|n|}$$ And $\text{sgn}\,0:=0$. So we have that $$\mathrm{Cl}_2\left(\frac\pi3\right)=\frac{\sqrt3}2\sum_{n\geq1}d_3(n)\frac{\chi_3(n)}{n^2}$$ Because $$\sin\frac{n\pi}3=d_3(n)\chi_3(n)\frac{\sqrt3}2$$ But I have absolutely no idea how to convert this to a hypergeometric series. Could I have some help? Thanks.

Answer 1

This is Gieseking's constant and it does have a hypergeometric form,

$$\begin{aligned} \rm{Cl}_2\left(\frac{\pi}{3}\right) &= \frac{3\sqrt3}4\left(\sum_{n=0}^\infty\frac{1}{(3n+1)^2}-\sum_{n=0}^{\infty}\frac{1}{(3n+2)^2}\right)\\ &= \,_3 F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;\tfrac14\right)\\ &= 1.0149416\dots \end{aligned}$$

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Added edit: More generally,

$$\rm{Ls}_1\big(\tfrac\pi3\big) = \int_0^{\pi/3}\log^0\left(2\sin\frac{t}2\right)dt=0!\,(-1)^0\;_2 F_1\left(\tfrac12,\tfrac12;\tfrac32;\tfrac14\right)=\tfrac\pi3$$

$$\rm{Ls}_2\big(\tfrac\pi3\big) = \int_0^{\pi/3}\log^1\left(2\sin\frac{t}2\right)dt=1!\,(-1)^1\;_3 F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;\tfrac14\right)=-\rm{Cl}_2\left(\tfrac{\pi}{3}\right)$$

$$\rm{Ls}_3\big(\tfrac\pi3\big) = \int_0^{\pi/3}\log^2\left(2\sin\frac{t}2\right)dt=2!\,(-1)^2\;_4 F_3\left(\tfrac12,\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32,\tfrac32;\tfrac14\right) = \frac{7\pi^3}{216}$$

and closed-forms of $\rm{Ls}_n\big(\tfrac\pi3\big)$ for $n \leq 8$ (skipping $n=7$) are given here. Or,

$$\begin{aligned} \rm{Ls}_{n}\big(\tfrac\pi3\big) &= \int_0^{\pi/3}\ln^{n-1}\left(2\sin\frac{t}2\right)dt\\ &= -(n-1)!\,(-1)^n\sum_{k=0}^\infty\frac{1}{k!}\frac{\left(\big(\frac12\big)_k\right)^{n+1}}{\left(\big(\frac32\big)_k\right)^{n}}\frac1{4^k}\\ &= -(n-1)!\,(-1)^n\sum_{k=0}^\infty\frac{1}{k!}\frac{\big((2k-1)!!\big)^{n+1}}{\big((2k+1)!!\big)^{n}}\frac1{8^k}\\ &= -(n-1)!\,(-1)^n\sum_{k=0}^\infty\frac{\binom{2k}{k}}{(2k+1)^{n}} \frac1{16^k} \end{aligned}$$

where $(x)_k$ is the Pochhammer symbol, and the special case $\rm{Ls}_{2}\big(\tfrac\pi3\big)=-\rm{Cl}_{2}\big(\tfrac\pi3\big)$.

Answer 2

Well, you have already shown that $\text{Cl}_2\left(\frac{\pi}{3}\right)$ is proportional to $$ \sum_{n\geq 0}\frac{1}{(3n+1)^2}-\sum_{n\geq 0}\frac{1}{(3n+2)^2} =\int_{0}^{1}\frac{-\log(x)\,dx}{1+x+x^2}$$ i.e. to $$ \phantom{}_3 F_2\left(\tfrac{1}{3},\tfrac{1}{3},1;\tfrac{4}{3},\tfrac{4}{3};1\right)-\phantom{}_3 F_2\left(\tfrac{2}{3},\tfrac{2}{3},1;\tfrac{5}{3},\tfrac{5}{3};1\right) $$ and you may apply Thomae's theorem to derive many equivalent representations.