# List of integrals or series for Gieseking's constant $\rm{Cl}_2\big(\tfrac{\pi}3\big)$?

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 $$\kappa$$?

• 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 ${_3 F _2}$ – Yuriy S Jun 12 '19 at 10:46
• @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. – Tito Piezas III Jun 12 '19 at 10:48
• For a whimsical introduction to Gieseking's constant $\kappa$, see "The Newest Inductee in the Number Hall of Fame" (1998) by Colin Adams where he discusses the connection of $\kappa$ to hyperbolic geometry. And that $$\kappa = \frac{9\sqrt3\,\zeta_m(2)}{2\pi^2} = 1.01494\dots$$ with $m=Q(\sqrt{-3})$ and $\zeta_m(s)$ is the Dedekind zeta function. – Tito Piezas III Jun 12 '19 at 12:40
• @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 $\kappa$. – Tito Piezas III Jun 12 '19 at 14:44
• 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. – Tito Piezas III Jun 13 '19 at 5:03

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 $$\kappa$$.

• \begin{align}\tan\left(\frac{\pi}{4}\right)=1\end{align} – FDP Jun 15 '19 at 9:29
• You have closed forms for $T_2(z)$ at $z\ne1$? I am very interested... what are they? – clathratus Jun 21 '19 at 2:47
• @clathratus: I've re-formatted the answer for clarity. Kindly see $(7)$ and $(8)$ and $(9)$. – Tito Piezas III Jun 21 '19 at 8:55

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.)

• Interesting, the appearance of $\ln 2$ and $\ln 3$. – Tito Piezas III Jun 12 '19 at 13:50
• @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. ;) – user90369 Jun 12 '19 at 13:54
• Is the denominator of $\ln 3$ really $2$? – Tito Piezas III Jun 12 '19 at 14:06
• @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. – user90369 Jun 12 '19 at 14:09
• @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. ;) – user90369 Jun 19 '19 at 19:24

$$\kappa=\frac{3\sqrt{3}}{2} \, _3F_2\left({\frac{1}{2},\frac{1}{2},\frac{1}{2}\atop \frac{3}{2},\frac{3}{2}};\frac{3}{4}\right)-\frac{\pi }{2} \log 3\tag{a}$$ 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.

• 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. – Tito Piezas III Jun 12 '19 at 12:06
• 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? – Tito Piezas III Jun 12 '19 at 14:14
• @TitoPiezasIII $$\int_0^{{\pi }/{6}} \frac{x \, dx}{\sin x \sqrt{1-3 \sin ^2x}}=\frac{K}{3}+\frac{\pi}{12} \log \left(2+\sqrt{3}\right)$$ – Nemo Jun 12 '19 at 19:47
• $$\int_0^{\frac{\pi }{2}} \log \left(\sin x+\sqrt{4-3 \sin ^2x}\right) \, dx=\frac43 K$$ – Nemo Jun 12 '19 at 21:00
• $$\int_0^{\frac{\pi }{6}} \log \left(\cos x+\sqrt{\cos^2x-1/2}\right) \, dx=\frac{K}{6}+\frac{\pi}{24}\ln 2$$ – Nemo Jun 12 '19 at 21:12

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: $$\mathrm{Cl}_2\left(\frac{\pi}{3}\right) = \kappa = \frac{f_1(6)}{24\sqrt{3}}$$

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

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

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

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

• Can we convert this to a BBP-type formula like here? – Tito Piezas III Jun 12 '19 at 15:28
• That's what I was thinking after seeing that, looking into it – Benedict W. J. Irwin Jun 12 '19 at 15:33
• 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 $N$. – user90369 Jun 12 '19 at 15:49
• 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. ;) – user90369 Jun 12 '19 at 17:15
• @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! – Raymond Manzoni Jun 12 '19 at 21:43

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.

• It turns out there may be an infinite family. See this related post – Tito Piezas III Jun 21 '19 at 8:16

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.

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.

• Also, $$2K= \,_3F_2\big(\tfrac12,1,1;\,\tfrac32,\tfrac32;\,1\big)$$ – Tito Piezas III Jun 26 '19 at 5:09

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.

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 $$E(x)=\,_3F_2\left({{1,x,x}\atop{1+x,1+x}};1\right)-\,_3F_2\left({{1,\frac12+x,\frac12+x}\atop{\frac32+x,\frac32+x}};1\right).$$

• 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 $$A = \sqrt{-1}\,\rm{Gi}_2^{2,1}\big(\tfrac1{\sqrt{-3}}\big) = \tfrac1{4\sqrt3} \Phi\big(\tfrac{-1}3,2,\tfrac12\big)$$ – Tito Piezas III Jun 26 '19 at 3:07
• Thanks to the relationship between $\rm{Gi}$ and hypergeometrics you pointed in this post, then we find another for $\kappa$ as, $$\kappa = \frac{2\sqrt3}5\,_3F_2\big(\tfrac12,\tfrac12,1; \tfrac32,\tfrac32; \tfrac{-1}3\big)+\pi\frac{\ln 3}{10}$$ – Tito Piezas III Jun 26 '19 at 3:47
• @TitoPiezasIII wonderful! – clathratus Jun 26 '19 at 5:49
• Potentially related: $$\int_0^1\int_0^1 \frac{dxdt}{x^2t^2+xt+1}=\frac{2}{\sqrt3}\mathrm{Cl}_2\left(\tfrac{2\pi}{3}\right)$$ – clathratus Oct 19 '19 at 19:44
• Okay it was related: $$\int_0^1\int_0^1 \frac{dxdt}{(xt)^2-xt+1}=\frac{2}{\sqrt3}\mathrm{Cl}_2\left(\tfrac{\pi}{3}\right)$$ – clathratus Oct 21 '19 at 15:40

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 $$\mathrm K=\frac9{10}\int_1^\infty\frac{t^4+1}{t^6+1}\ln t\ dt.$$

• 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, $$\int_1^\infty \frac{\ln(x)}{x^2+x+1}dx =\frac{4\sqrt3\,\color{red}\kappa}9$$ – Tito Piezas III Jul 1 '19 at 17:28