TITLE: List of integrals or series for Gieseking's constant $\rm{Cl}_2\big(\tfrac{\pi}3\big)$?
QUESTION [31 upvotes]: Catalan's constant $K$ can be defined as,
$$K = \text{Cl}_2\big(\tfrac{\pi}2\big) = \Im\, \rm{Li}_2\big(e^{\pi i/2}\big)= \sum_{n=0}^\infty\left(\frac1{(4n+1)^2}-\frac1{(4n+3)^2}\right)=0.91596\dots$$
It seems to have a natural cubic analogue called Gieseking's constant $\kappa$ (or kappa, by analogy), but apparently is (not as well-known) known under different names,
$$\kappa = \rm{Cl}_2\big(\tfrac{\pi}3\big)=\tfrac32\rm{Cl}_2\big(\tfrac{2\pi}3\big) = \Im\, \rm{Li}_2\big(e^{\pi i/3}\big)= \tfrac32\Im\, \rm{Li}_2\big(e^{2\pi i/3}\big)= 1.01494\dots$$
and the Gieseking manifold has volume $\kappa = 1.01494\dots$ while the hyperbolic volume of the knot complement of the figure eight knot is $V=2\kappa = 2.029788\dots$. Below are some series and hypergeometric representations of $\kappa$ by various people including yours truly,
$$\kappa=\frac{3\sqrt3}4\sum_{n=0}^\infty\left(\frac1{(3n+1)^2}-\frac1{(3n+2)^2}\right)\tag1$$
$$\kappa=\sum_{n=0}^\infty \frac{\binom {2n}n}{(2n+1)^2} \left(\frac1{16}\right)^n = \,_3F_2\big(\tfrac12,\tfrac12,\tfrac12;\,\tfrac32,\tfrac32;\,\tfrac14\big)\tag{2a}$$
$$\frac{2\,\kappa}{3\sqrt3}+\frac{\pi\ln3}{3\sqrt3}=\sum_{n=0}^\infty \frac{\binom {2n}n}{(2n+1)^2} \left(\frac3{16}\right)^n = \,_3F_2\big(\tfrac12,\tfrac12,\tfrac12;\,\tfrac32,\tfrac32;\,\tfrac34\big)\tag{2b}$$
$$\pi\,\kappa=\frac32\sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} +2\zeta(3)\tag3$$
$$\kappa=\frac{\sqrt3}{10}\sum_{n=1}^\infty \frac{48^n}{n(2n-1)\binom{2n}{n}\binom{4n}{2n}} = \frac{2\sqrt3}5\,_4F_3\big(\tfrac12,1,1,2;\,\tfrac54,\tfrac32,\tfrac74;\,\tfrac34\big)\tag4$$
$$\kappa=\frac{-1}{12\sqrt3}\sum_{n=1}^\infty \frac{(15n-4)(-27)^n}{n^3\binom{2n}{n}^2\binom{3n}{n}}\tag5$$
$$\kappa=\frac{-1}{10\sqrt3}\sum_{n=1}^\infty \frac{(5n-1)(-144)^n}{n^3\binom{2n}{n}^2\binom{4n}{2n}}\tag6$$
and integrals,
$$\kappa =-\int_0^{\pi/3}\ln\left(2\sin\frac{x}2\right)dx\tag7$$
$$\kappa =\int_0^{2\pi/3}\ln\left(2\cos\frac{x}2\right)dx\tag8$$
$$\kappa = \sqrt3\int_0^\infty x K_0^3(x) dx\tag9$$
$$\kappa =2\int_0^{1/2}\frac{\arcsin(x)}x dx\tag{10}$$
$$\kappa = \frac35\int_0^{{\pi }/{3}} \frac{x \left({\sqrt{3}-{\sin x}}\right) dx}{\sin x \cdot \sqrt{3-2 \sqrt{3} \sin x}}\tag{11a}$$
$$\kappa = \frac{3\sqrt3}5\int_0^{{\pi }/{3}} \frac{(2-\sqrt3\sin x)(x-\sin x\cos x)\, dx}{\sin^3 x \cdot \sqrt{3-2 \sqrt{3} \sin x}}\tag{11b}$$
and involving harmonic numbers $H_n$,
$$8\,\kappa = 9\sqrt3\sum_{n=1}^\infty \frac{H_n}{\binom{2n}{n}} -4\pi+2\pi\ln3\tag{12}$$
$$\quad 8\,\kappa = 6\sqrt3\sum_{n=1}^\infty \frac{H_n}{\binom{2n}{n}n} -\frac{\pi^2}{\sqrt3}+2\pi\ln3\tag{13}$$
$$\pi\,\kappa = \frac3{10}\sum_{n=1}^\infty \frac{17H_n+H_{2n}}{\binom{2n}{n}n^2}\quad\quad\tag{14}$$
and their equivalent forms after some transformations. Note that $K_n(x)$ is the modified Bessel function of the second kind. Some of these have not been proven rigorously.
Relevant links are: (1), (2), (3), (4),(5), (6), (7),(8), (9), (10),(11a), (11b), (12), (14).

Q: What other series, hypergeometric, and integral representations are there for Gieseking's constant $\kappa$?

REPLY [2 votes]: We have 2 complementary pairs,
$$\begin{aligned}
&\sum_{n=0}^\infty \frac{\binom {2n}n}{(2n+1)^2} \left(\frac1{16}\right)^n = \,_3F_2\big(\tfrac12,\tfrac12,\tfrac12;\,\tfrac32,\tfrac32;\,\tfrac14\big)=\kappa\\
&\sum_{n=0}^\infty \frac{\binom {2n}n}{(2n+1)^2} \left(\frac3{16}\right)^n = \,_3F_2\big(\tfrac12,\tfrac12,\tfrac12;\,\tfrac32,\tfrac32;\,\tfrac34\big)=\frac{\kappa}{3\sqrt3}+\frac{\pi\ln3}{3\sqrt3}
\end{aligned}$$
and,
$$\begin{aligned}
&\sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n}=\frac12 \,_4F_3\big(1,1,1,1;\,\tfrac32,2,2;\,\tfrac14\big) = \frac{2\pi\,\kappa}3-\frac{4\zeta(3)}3\\
&\sum_{n=1}^\infty \frac{3^n}{n^3\,\binom {2n}n}=\frac32 \,_4F_3\big(1,1,1,1;\,\tfrac32,2,2;\,\tfrac34\big) = \frac{8\pi\,\kappa}9-\frac{26\zeta(3)}9+\frac{2\pi^2\ln3}9\\
\end{aligned}$$
The first three were mentioned in the original post, but the fourth is new and its general form is discussed in this post. However, another in the post,
$$\tfrac{2\sqrt3}5\,_4F_3\big(\tfrac12,1,1,2;\,\tfrac54,\tfrac32,\tfrac74;\,\tfrac34\big)=\kappa \qquad$$
doesn't seem to have a known complement.