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

Solution 1:

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 $\kappa$.

Solution 2:

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

Solution 3:

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.)

Solution 4:

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

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

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

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

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