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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 $\kappa$?

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  • $\begingroup$ I can derive the ${_4 F _3}$ representation from the series, but not the ${_3 F _2}$ representation. As for the $\arcsin$ integral, I derived it from the integral representation of ${_3 F _2}$ $\endgroup$
    – Yuriy S
    Jun 12, 2019 at 10:46
  • $\begingroup$ @YuriyS: Feel free to write an answer. I'm hoping for multiple answers as this will give us alternative perspectives to look at this constant. $\endgroup$ Jun 12, 2019 at 10:48
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    $\begingroup$ @BenedictW.J.Irwin: A place to search might be number theoretic properties of the Catalan $K$. It might have a faint analogous version in the Gieseking $\kappa$. $\endgroup$ Jun 12, 2019 at 14:44
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    $\begingroup$ To highlight the Gieseking's constant $\kappa$ "new inductee" status (per Colin Adam's article), we find Mathworld's article on the figure eight knot which discusses $V=2\kappa$ a lot doesn't even link to Gieseking's constant. $\endgroup$ Jun 13, 2019 at 5:03
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    $\begingroup$ @TitoPiezasIII Kindly see here for a proof of $$\sum_{n\ge1}\frac{\zeta(2n)}{(2q)^{2n}n(2n+1)}=1+\ln\frac{q}{\pi}-\frac{q}{\pi}\mathrm{Cl}_2(\tfrac{\pi}{q}).$$ $\endgroup$
    – clathratus
    Dec 9, 2019 at 20:11

10 Answers 10

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I. From this list of integrals and elsewhere for Catalan's constant $K=\rm{Cl}_2\big(\frac\pi2\big)$, I've now found ELEVEN (so far) that have a Gieseking $\kappa=\rm{Cl}_2\big(\frac\pi3\big)$ cubic analogue:


$$K= -\int_0^{\pi/2} \ln\left(2\sin \frac{x}2\right)\,dx\\ \kappa= -\int_0^{\pi/3} \ln\left(2\sin \frac{x}2\right)\,dx\tag1$$


$$K= -\frac2\pi\int_0^{\pi/2} x\ln\left(2\sin \frac{x}2\right)\,dx\,+\frac{35}{16}\frac{\zeta(3)}{\pi}\\ \kappa= -\frac3\pi\int_0^{\pi/3}x\ln\left(2\sin\frac{x}2\right)\,dx\,+2\frac{\zeta(3)}\pi\tag2$$


$$K= \frac12\int_0^{\pi/2} x\csc x\,dx\qquad \\ \kappa= \frac35\int_0^{\pi/3} x\csc x\,dx\;+\frac{\pi\ln3}{10}\tag3$$


$$K= \int_0^{\pi/4} \ln\left(\cot x\right)\,dx\\ \kappa= \frac65\int_0^{\pi/6} \ln\left(\cot x\right)\,dx\tag4$$


$$K= 2\int_0^{\sin(\pi/4)}\frac{\arcsin(x)}x dx\;-\frac{\pi\ln2}4\\ \kappa= 2\int_0^{\sin(\pi/6)}\frac{\arcsin(x)}x dx\qquad \tag5$$


$$K= -2\int_\color{red}1^{\cos(\pi/4)}\frac{\arccos(x)}x dx\;+\frac{\pi\ln2}4\\ \kappa= -3\int_\color{red}1^{\cos(\pi/6)}\frac{\arccos(x)}x dx\;+\frac{\pi\ln3}{4}\tag6$$


$$K= \int_0^{\tan(\pi/4)}\frac{\arctan(x)}x dx\qquad \\ \kappa= \frac65\int_0^{\tan(\pi/6)}\frac{\arctan(x)}x dx\;+\frac{\pi\ln3}{10}\tag7$$


$$K= \int_0^{1/\tan(\pi/4)}\frac{\arctan(x)}x dx\qquad \\ \kappa= \frac65\int_0^{1/\tan(\pi/6)}\frac{\arctan(x)}x dx\;-\frac{\pi\ln3}{5}\tag8$$


$$K= \frac{2}{\pi}\int_0^{\tan(\pi/4)}\frac{\arctan^2(x)}x dx+\frac{7\zeta(3)}{4\pi}\qquad \\ \kappa= \frac{18}{5\pi}\int_0^{\tan(\pi/6)}\frac{\arctan^2(x)}x dx+\frac{7\zeta(3)}{4\pi}+\frac{\pi\ln3}{20}\tag9$$


$$K= -\int_0^{\tan(\pi/4)}\frac{\ln x}{1+x^2} dx\\ \kappa= -\frac65\int_0^{\tan(\pi/6)}\frac{\ln x}{1+x^2} dx\tag{10}$$


$$K= -2\int_0^{2\sin(\pi/4)}\frac{\ln x}{\sqrt{4-x^2}} dx\\ \kappa= -2\int_0^{2\sin(\pi/6)}\frac{\ln x }{\sqrt{4-x^2}} dx\tag{11}$$


P.S. Note that $(7)$ and $(8)$ is the inverse tangent integral,

$$T_2(z)= \int_0^{z}\frac{\arctan(x)}x dx $$

hence $T_2(1)= K$, while both $T_2(1/\sqrt3)$ and $T_2(\sqrt3)$ involve $\kappa$.

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    $\begingroup$ \begin{align}\tan\left(\frac{\pi}{4}\right)=1\end{align} $\endgroup$
    – FDP
    Jun 15, 2019 at 9:29
  • $\begingroup$ You have closed forms for $T_2(z)$ at $z\ne1$? I am very interested... what are they? $\endgroup$
    – clathratus
    Jun 21, 2019 at 2:47
  • $\begingroup$ @clathratus: I've re-formatted the answer for clarity. Kindly see $(7)$ and $(8)$ and $(9)$. $\endgroup$ Jun 21, 2019 at 8:55
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$$ \kappa=\frac{3\sqrt{3}}{2} \, _3F_2\left({\frac{1}{2},\frac{1}{2},\frac{1}{2}\atop \frac{3}{2},\frac{3}{2}};\frac{3}{4}\right)-\frac{\pi }{2} \log 3\tag{a} $$ Ramanujan's Notebooks I, chapter 9, Entry 16. (a) is a companion to (2) from Tito's list.

$$ \kappa=\frac35\int_0^{\pi/2}\log \left(\sqrt{3} \sin x+\sqrt{4-\sin ^2x}\right)dx\tag{b} $$ $$ \kappa=\frac{3\sqrt3}{5}\int_0^{\pi/2}\frac{x~dx}{\sin x \sqrt{4-\cos ^2x}}\tag{c} $$ $$ \kappa=3\sqrt3 \int_0^{{\pi }/{2}} \frac{\sin x\cdot\log \left(\cot \frac{x}{2}\right)}{4-\sin ^2x}\, dx\tag{d} $$ (b), (c) and (d) are due to Lobachevskii, see Gradsteyn and Ryzhik, eq. 4.228.1.

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    $\begingroup$ Regarding your post, you'll find Colin Adams' "The Newest Inductee in the Number Hall of Fame" (1998) quite interesting as it is from a hyperbolic geometry perspective as your post is also. $\endgroup$ Jun 12, 2019 at 12:06
  • $\begingroup$ Note that Catalan $K$ and Gieseking $\kappa$ have $$K =-\int_0^{\pi/2}\ln\left(2\sin\frac{x}2\right)dx$$ $$\kappa =-\int_0^{\pi/3}\ln\left(2\sin\frac{x}2\right)dx$$ Do any of Lobachevskii's integrals have a Catalan version? $\endgroup$ Jun 12, 2019 at 14:14
  • $\begingroup$ @TitoPiezasIII $$ \int_0^{{\pi }/{6}} \frac{x \, dx}{\sin x \sqrt{1-3 \sin ^2x}}=\frac{K}{3}+\frac{\pi}{12} \log \left(2+\sqrt{3}\right)$$ $\endgroup$ Jun 12, 2019 at 19:47
  • $\begingroup$ $$ \int_0^{\frac{\pi }{2}} \log \left(\sin x+\sqrt{4-3 \sin ^2x}\right) \, dx=\frac43 K$$ $\endgroup$ Jun 12, 2019 at 21:00
  • $\begingroup$ $$ \int_0^{\frac{\pi }{6}} \log \left(\cos x+\sqrt{\cos^2x-1/2}\right) \, dx=\frac{K}{6}+\frac{\pi}{24}\ln 2$$ $\endgroup$ Jun 12, 2019 at 21:12
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Instead of series, hypergeometric, and integral representations we can also use $products$.

Then Catalan’s constant and Gieseking’s constant have the same base.

Let $~\displaystyle Q_1(x):=\lim_{n\to\infty}\frac{e^{xn} n^{-\frac{x^2}{2}}}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)^k}~$ .

Catalan constant : $\hspace{1cm}\displaystyle \sum\limits_{k=1}^\infty\frac{(-1)^{k-1}}{(2k-1)^2}= \frac{\pi}{2}\left(1-\frac{\ln 2}{2} + 4 \ln\frac{Q_1\left(\frac{1}{4}\right)}{ Q_1\left(-\frac{1}{4}\right)}\right)$

Gieseking constant : $\enspace\displaystyle \int\limits_0^{\frac{2\pi}{3}}\ln\left(2\cos\frac{x}{2}\right)\,dx = \pi\left(1-\frac{\ln 3}{2} + 3 \ln\frac{Q_1\left(\frac{1}{3}\right)}{ Q_1\left(-\frac{1}{3}\right)}\right)$

You can see that here, page 26.

$\,$

(Note to the link: The right side of $(5)(a)$ has to be multiplicated by $3$. But it's not relevant here.)

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  • $\begingroup$ Interesting, the appearance of $\ln 2$ and $\ln 3$. $\endgroup$ Jun 12, 2019 at 13:50
  • $\begingroup$ @TitoPiezasIII : That's a consequence of the definition of $Q_1$ . E.g. using Barnes G-function the formula is of course a bit different. ;) $\endgroup$
    – user90369
    Jun 12, 2019 at 13:54
  • $\begingroup$ Is the denominator of $\ln 3$ really $2$? $\endgroup$ Jun 12, 2019 at 14:06
  • $\begingroup$ @TitoPiezasIII : Yes. (Sorry, I've deleted the last comment, I mixed something.) The proof is on the next side. So, it should be right. Or let's test it by a program. $\endgroup$
    – user90369
    Jun 12, 2019 at 14:09
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    $\begingroup$ @Jam : No, but it's equivalent to Barnes G-function. And $Q_n$ is a type of a generalisation of the gamma function, it exist others too. ;) $\endgroup$
    – user90369
    Jun 19, 2019 at 19:24
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This makes a nice comparison $$ K = \frac{1}{160}\left[ \psi_1\left(\frac{1}{12}\right) + \psi_1\left(\frac{5}{12}\right) - \psi_1\left(\frac{7}{12}\right) -\psi_1\left(\frac{11}{12}\right) \right] $$ $$ \kappa = \frac{\sqrt{3}}{72}\left[ \psi_1\left(\frac{1}{6}\right) + \psi_1\left(\frac{2}{6}\right) - \psi_1\left(\frac{4}{6}\right) - \psi_1\left(\frac{5}{6}\right) \right] $$

Can imagine a class of constants of the form $$ C = Af(N) = A\left[ \psi_1\left(\frac{1}{N}\right) + \psi_1\left(\frac{N/2-1}{N}\right) - \psi_1\left(\frac{N/2+1}{N}\right) - \psi_1\left(\frac{N-1}{N}\right) \right] $$ for simple/interesting $A$.

Edit:: We can write with ($N=4$) $$ K = \frac{f(4)}{16\sqrt{4}} $$ and with $N=3$ $$ \kappa = \frac{f(3)}{24 \sqrt{3}} $$ which reiterates the $Catalan,4$, $Gieseking,3$ link form the $Q_1$ answer above.

Edit:: 26/06/2019 I have found on Wikipedia - Trigamma Function that: $$ \psi_1\left(\frac{p}{q}\right)=\frac{\pi^2}{2\sin^2(\pi p/q)}+2q\sum_{m=1}^{(q-1)/2}\sin\left(\frac{2\pi mp}{q}\right)\textrm{Cl}_2\left(\frac{2\pi m}{q}\right) $$ and also $$ \operatorname{Cl}_{2m}\left( \frac{q\pi}{p}\right)= \frac{1}{(2p)^{2m}(2m-1)!} \, \sum_{j=1}^{p} \sin\left(\tfrac{qj\pi}{p}\right)\, \left[\psi_{2m-1}\left(\tfrac{j}{2p}\right)+(-1)^q\psi_{2m-1}\left(\tfrac{j+p}{2p}\right)\right] $$

If we refine the definition to $$ f_k(N) = \left[ \psi_1\left(\frac{k}{N}\right) + \psi_1\left(\frac{N/2-k}{N}\right) - \psi_1\left(\frac{N/2+k}{N}\right) - \psi_1\left(\frac{N-k}{N}\right) \right] $$

Wolfram|Alpha gives us that: $$ \mathrm{Cl}_2\left(\frac{\pi}{3}\right) = \kappa = \frac{f_1(6)}{24\sqrt{3}} $$

$$ \mathrm{Cl}_2\left(\frac{\pi}{4}\right) = \frac{1}{2\cdot 4^2} \left( \frac{f_2(8)}{4} + \frac{f_1(8)}{\sqrt{8}} \right) = \frac{1}{2\cdot 4^2} \left( 8K + \frac{f_1(8)}{\sqrt{8}} \right) $$

$$ \mathrm{Cl}_2\left(\frac{\pi}{5}\right) = \frac{1}{2^3 \cdot 5^2} \left ( \sqrt{\frac{1}{2}(5-\sqrt{5})}f_1(10) + \sqrt{\frac{1}{2}(5+\sqrt{5})}f_2(10) \right) $$

$$ \mathrm{Cl}_2\left(\frac{\pi}{6}\right) = \frac{1}{2^5 \cdot 3} \left(64 K + \frac{f_1(6)}{\sqrt{3}} \right) = \frac{2}{3}K + \frac{1}{4}\kappa $$

$$ \mathrm{Cl}_2\left(\frac{\pi}{7}\right) = \frac{1}{2^2 \cdot 7^2}\left( \sin\left(\frac{\pi}{7}\right)f_1(14) + \cos\left(\frac{3\pi}{14}\right)f_2(14) + \cos\left(\frac{\pi}{14}\right)f_3(14) \right) $$

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    $\begingroup$ Can we convert this to a BBP-type formula like here? $\endgroup$ Jun 12, 2019 at 15:28
  • $\begingroup$ That's what I was thinking after seeing that, looking into it $\endgroup$ Jun 12, 2019 at 15:33
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    $\begingroup$ I suspect that exactly here $A$ makes a simple clear comparison of the constants mostly impossible. One should not forget that this is about cyclic series, and they have their own laws, with an $A$ in front of the bracket you will not get very far. $A$ is probably a complicate function of $N$, not a simple expression when it comes to comparisons of $C$ with respect to $N$. $\endgroup$
    – user90369
    Jun 12, 2019 at 15:49
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    $\begingroup$ If you want to match the series, then it might be better to choose $~\displaystyle C = B(N)\frac{\sqrt{N}^3}{4}\sum\limits_{n=0}^\infty\left(\frac{1}{(Nn+1)^2}-\frac{1}{(N(n+1)-1)^2}\right)$ with $\displaystyle B(4)=\frac{1}{2}$ and $B(3)=1$. But a little warning: It's almost always wrong to conclude a generalization from two special cases. ;) $\endgroup$
    – user90369
    Jun 12, 2019 at 17:15
  • $\begingroup$ @TitoPiezasIII: concerning BBP formulae see the paper by Kunle Adegoke, Lafont and Layeni "A Class of Binary BBP-type Formulas in General Degrees" (referenced too in Bayley's Compendium at the end). See too $(30),\;(31)$ at MathWorld. Nice list btw! $\endgroup$ Jun 12, 2019 at 21:43
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Formulas for Gieseking's constant $\kappa$ which uses only ONE hypergeometric function are,


$$\kappa= \,_3F_2\big(\tfrac12,\tfrac12,\tfrac12;\,\tfrac32,\tfrac32;\,\tfrac14\big)\tag{1a}$$

$$\kappa=-\tfrac1{36}\,_3F_2\big(\tfrac32,\tfrac32,\tfrac32;\,\tfrac52,\tfrac52;\,\tfrac14\big)+\tfrac13\pi\tag{1b}$$


$$\kappa= \tfrac{3\sqrt3}{2}\,_3F_2\big(\tfrac12,\tfrac12,\tfrac12;\,\tfrac32,\tfrac32;\,\tfrac34\big)-\tfrac12\pi\ln 3\tag{2a}$$

$$\kappa= -\tfrac{\sqrt3}{8}\,_3F_2\big(\tfrac32,\tfrac32,\tfrac32;\,\tfrac52,\tfrac52;\,\tfrac34\big) -\tfrac12\pi\ln 3+\pi\tag{2b}$$


$$\kappa= \tfrac{2\sqrt3}{5}\,_3F_2\big(1,\tfrac12,\tfrac12;\,\tfrac32,\tfrac32;\,\tfrac{-1}3\big) +\tfrac1{10}\pi\ln 3\tag{3a}$$

$$\kappa= \tfrac{4}{45\sqrt3}\,_3F_2\big(2,\tfrac32,\tfrac32;\,\tfrac52,\tfrac52;\,\tfrac{-1}3\big)+\tfrac1{10}\pi\ln 3+\tfrac15\pi\tag{3b}$$


$$\kappa= \tfrac{3\sqrt3}{10}\,_3F_2\big(1,1,\tfrac12;\,\tfrac32,\tfrac32;\,\tfrac34\big) +\tfrac1{10}\pi\ln 3\tag{4a}$$

$$\kappa= -\tfrac{\sqrt3}{10}\,_3F_2\big(2,2,\tfrac32;\,\tfrac52,\tfrac52;\,\tfrac34\big)+\tfrac1{10}\pi\ln 3+\tfrac25\pi\tag{4b}$$


$$\kappa= \tfrac{2\sqrt3}5\,_4F_3\big(1,1,2,\tfrac12;\,\tfrac54,\tfrac64,\tfrac74;\,\tfrac34\big)\tag{5}$$


Note that the (#b) can be derived from the (#a), respectively, as described in this post. However, there might be more non-derived examples. (See also the answer below using binomials for more hypergeometrics.)

P.S. I'm trying to check $\,_3F_2\big(1,1,1;\,\tfrac32,\tfrac32;z\big)$ but no luck so far.

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  • $\begingroup$ Also, $$2K= \,_3F_2\big(\tfrac12,1,1;\,\tfrac32,\tfrac32;\,1\big)$$ $\endgroup$ Jun 26, 2019 at 5:09
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BBP-type series

We look for a BBP-type formula for $\kappa$ with base $b^k$ such that $b\neq \pm1$. Turns out $b=\pm\frac1{3^m}$ will do. Courtesy of Manzoni's comment, we find such a formula in this paper.

$$\kappa = \frac1{3^{3/2}} \small\sum_{k=0}^\infty \left(-\frac1{3^3}\right)^k \left(\frac{3^2}{(6k+1)^2}-\frac{3^2}{(6k+2)^2}-\frac{3\times4}{(6k+3)^2}-\frac3{(6k+4)^2}+\frac1{(6k+5)^2}\right)$$

which is also found in the Mathworld's Figure Eight knot. In the same article (which discusses $V=2\kappa$ but doesn't mention Gieseking's constant at all), Mathworld further gives,

$$\kappa\; =\frac1{3^{9/2}} \small\sum_{k=0}^\infty \left(\frac1{3^6}\right)^k \left(\frac{3^5}{(12k+1)^2}-\frac{3^5}{(12k+2)^2}-\frac{3^4\times4}{(12k+3)^2}-\dots-\frac1{(12k+11)^2}\right)$$

$$\kappa\; =\; \frac1{3^{21/2}} \small\sum_{k=0}^\infty \left(\frac1{3^{12}}\right)^k \left(\frac{3^{11}}{(24k+1)^2}-\frac{3^{11}}{(24k+2)^2}-\frac{3^{10}\times4}{(24k+3)^2}-\dots-\frac1{(24k+23)^2}\right)$$

and it is tempting to speculate this pattern continues.

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  • $\begingroup$ It turns out there may be an infinite family. See this related post $\endgroup$ Jun 21, 2019 at 8:16
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Using binomials, this paper (on pp. 10-11) gives,

$$\kappa=\frac{3\sqrt3}{4}\sum_{n=1}^\infty \frac{15n-4}{n^3\binom{2n}{n}^2\binom{3n}{n}}\,(-27)^{n-1}$$

$$\kappa=\frac{3\sqrt3}{4}\sum_{n=1}^\infty \frac{5535n^3 - 4689n^2 + 1110n - 80}{n^3(3n-1)(3n-2)\binom{6n}{3n}^2\binom{3n}{n}}\,(-27)^{n-1}$$


Also, based on insights from this post, we use the general identity,

$$\frac4z\sum_{\color{red}{n=0}}^\infty\frac{\binom{2n}n}{(2n+1)^{m+a}}\frac1{z^n}-\sum_{n=1}^\infty\frac{\binom{2n}n}{(2n-1)^{m+a}}\frac1{z^n}=\sum_{n=1}^\infty\frac{\binom{2n}n}{(2n-1)^{\color{blue}{m+a+1}}}\frac1{z^n}\tag1$$

where $z=2^{m+2}$ to generate more formulas using known ones. For example, let $m=2$ hence $z=16$.

Let $a=0$ and from #2 in the main list, we know,

$$\frac4{16}\sum_{\color{red}{n=0}}^\infty\frac{\binom{2n}n}{(2n+1)^2}\frac1{16^n} = \frac{\kappa}4$$ But it can also be shown that, $$\sum_{n=1}^\infty\frac{\binom{2n}n}{(2n-1)^{2}}\frac1{16^n}=\frac{\sqrt3}2+\frac{\pi}{12}-1$$ thus the RHS must then be, $$\sum_{n=1}^\infty\frac{\binom{2n}n}{(2n-1)^{\color{blue}3}}\frac1{16^n}=\frac{\kappa}4-\frac{\sqrt3}2-\frac{\pi}{12}+1$$

Similarly, let $a=1$. We then find that,

$$\qquad\sum_{n=1}^\infty\frac{\binom{2n}n}{(2n-1)^{\color{blue}4}}\frac1{16^n}=-\frac{\kappa}4+\frac{\sqrt3}2+\frac{\pi}{12}-1+\frac{7\pi^3}{864}$$

though it gets problematic to evaluate the LHS of $(1)$ the higher we go.

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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.

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If we consider the function $$\mathrm{Gi}_s^{p,q}(z)=\sum_{k\ge0}\frac{z^{pk+q}}{(pk+q)^s}=\int_0^z \frac{\mathrm{Gi}_{s-1}^{p,q}(x)}{x}dx$$ then $$\kappa=\frac{\sqrt3}{2}\left(\mathrm{Gi}_2^{6,1}(1)+\mathrm{Gi}_2^{6,2}(1)-\mathrm{Gi}_2^{6,4}(1)-\mathrm{Gi}_2^{6,5}(1)\right),$$ or equivalently $$\kappa=\frac{1}{4\sqrt3}\sum_{j=0}^{m-1}\sum_{u=1}^{5}(u-3)(u^2-6u+2)\,_3F_2\left[{{1,\frac{u+6j}{6m},\frac{u+6j}{6m}}\atop{\frac{u+6j+6m}{6m},\frac{u+6j+6m}{6m}}}; 1\right]$$ for any $m\in\Bbb N$.

On the other hand, $$\mathrm K=\sum_{j=0}^{m-1}\sum_{u=1}^{3}(2-u)\ _3F_2\left[{{1,\frac{u+4j}{4m},\frac{u+4j}{4m}}\atop{\frac{u+4m+4j}{4m},\frac{u+4m+4j}{4m}}};1\right]$$ for any $m\in\Bbb N$.

See here for more details on the $\mathrm{Gi}$ function.

Note that the above formulae come mostly from the fact that $$\sum_{k\ge0}f(k)=\sum_{j=0}^{m-1}\sum_{k\ge0}f(mk+j)$$ for $m\in\Bbb N$.

Edit:

Also, for all $n\in\Bbb N$, $$\kappa=2^n\sum_{r=1}^{2^n\cdot3-1}\sin\left(\tfrac{r\pi}{2^n\cdot3}\right)E\left(\tfrac{r}{2^{n+1}\cdot3}\right)+\sum_{k=1}^{n}2^k\sum_{j=1}^{2^k\cdot3-1}(-1)^j\sin\left(\tfrac{\pi j}{2^k\cdot 3}\right)E\left(\tfrac{j}{2^{k+1}\cdot3}\right)$$ and $$\mathrm{K}=2^n\sum_{r=1}^{2^{n+1}-1}\sin\left(\tfrac{r\pi}{2^{n+1}}\right)E\left(\tfrac{r}{2^{n+2}}\right)+\sum_{k=1}^{n}2^k\sum_{j=1}^{2^{k+1}-1}(-1)^j \sin\left(\tfrac{\pi j}{2^{k+1}}\right)E\left(\tfrac{j}{2^{k+2}}\right)$$ where $$E(x)=\,_3F_2\left({{1,x,x}\atop{1+x,1+x}};1\right)-\,_3F_2\left({{1,\frac12+x,\frac12+x}\atop{\frac32+x,\frac32+x}};1\right).$$

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    $\begingroup$ More concisely, since there is a relationship between this function $\rm{Gi}$, the Lerch transcendent $\Phi$, and the inverse tangent integral $\rm{Ti}$, then using $(7)$ in my answer, we get, $$\kappa = \frac65A +\frac{\pi \ln 3}{10}$$ where $$A = \sqrt{-1}\,\rm{Gi}_2^{2,1}\big(\tfrac1{\sqrt{-3}}\big) = \tfrac1{4\sqrt3} \Phi\big(\tfrac{-1}3,2,\tfrac12\big)$$ $\endgroup$ Jun 26, 2019 at 3:07
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    $\begingroup$ Thanks to the relationship between $\rm{Gi}$ and hypergeometrics you pointed in this post, then we find another for $\kappa$ as, $$\kappa = \frac{2\sqrt3}5\,_3F_2\big(\tfrac12,\tfrac12,1; \tfrac32,\tfrac32; \tfrac{-1}3\big)+\pi\frac{\ln 3}{10}$$ $\endgroup$ Jun 26, 2019 at 3:47
  • $\begingroup$ @TitoPiezasIII wonderful! $\endgroup$
    – clathratus
    Jun 26, 2019 at 5:49
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    $\begingroup$ Potentially related: $$\int_0^1\int_0^1 \frac{dxdt}{x^2t^2+xt+1}=\frac{2}{\sqrt3}\mathrm{Cl}_2\left(\tfrac{2\pi}{3}\right)$$ $\endgroup$
    – clathratus
    Oct 19, 2019 at 19:44
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    $\begingroup$ Okay it was related: $$\int_0^1\int_0^1 \frac{dxdt}{(xt)^2-xt+1}=\frac{2}{\sqrt3}\mathrm{Cl}_2\left(\tfrac{\pi}{3}\right)$$ $\endgroup$
    – clathratus
    Oct 21, 2019 at 15:40
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If I'm not mistaken, $$\kappa=\frac{\sqrt3}{2}\int_1^\infty \frac{(t^3-1)(t+1)}{t^6-1}\ln t\ dt$$ and similarly $$\mathrm K=\frac9{10}\int_1^\infty \frac{(t^6-1)(t^4+1)}{t^{12}-1}\ln t\ dt$$ (I am using $\mathrm K$ to denote Catalan's Constant). These integrals simplify to $$\kappa=\frac{\sqrt3}{2}\int_1^\infty\frac{\ln t\ dt}{t^2-t+1}$$ and $$\mathrm K=\frac9{10}\int_1^\infty\frac{t^4+1}{t^6+1}\ln t\ dt.$$

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    $\begingroup$ Note also that $$\int_0^1 \frac{\ln^2(x^2+x+1)}{x}dx =\frac{4\pi\,\color{red}\kappa}9-\frac{2\zeta(3)}3$$ which is discussed in this post. Also, a variant to the one you gave, $$\int_1^\infty \frac{\ln(x)}{x^2+x+1}dx =\frac{4\sqrt3\,\color{red}\kappa}9$$ $\endgroup$ Jul 1, 2019 at 17:28

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