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Numerical evidence suggests:

$$ \int_0^{\frac12}\int_0^{\frac12}\frac{1}{1-x^2-y^2} dy \, dx= \frac{G}{3}\qquad (1)$$

Couldn't find the indefinite integral, though maple simplifies (1) to

$$ \int _{0}^{1/2}\!-\arctan \left( 1/2\,{\frac {1}{\sqrt {-1+{x}^{2}}}} \right) {\frac {1}{\sqrt {-1+{x}^{2}}}}{dx}$$

Is (1) true?

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  • $\begingroup$ Rewriting the integrand as a geometric series might help. $\endgroup$ – Eckhard Sep 23 '14 at 14:16
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    $\begingroup$ Mathematica returns $2G$ for $\int_0^1 \int_0^1 (1-x^2-y^2)^{-1}dxdy$, as a side note. $\endgroup$ – Per Alexandersson Sep 23 '14 at 14:17
  • $\begingroup$ @PerAlexandersson for yours Maple returns complex number, might be wrong. $\endgroup$ – joro Sep 23 '14 at 14:21
  • $\begingroup$ @Joro: Yeah, Mathematica does something strange here I think. $\endgroup$ – Per Alexandersson Sep 23 '14 at 14:56
  • $\begingroup$ There is a simple formula for Catalan involving arctan: $$\int_0^1{\arctan x\over x}\,dx=G$$ It's given as Exercise 6.2.7 in Boros & Moll, Irresistible Integrals. They point to Adamchik, Integrals and series representations for Catalan's constant, and Bradley, Representations of Catalan's constant, for more formulas. The Adamchik reference might be cs.cmu.edu/~adamchik/articles/catalan/catalan.htm which has many formulas for $G$ but nothing quite like what you want. $\endgroup$ – Gerry Myerson Sep 24 '14 at 0:38
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Mathematica confirms the following:

Change the integral to polar coordinates to get

$$\frac{(1)}{2} = \int_0^{\pi/4} \int_0^{\sec(\theta)/2} \frac{1}{1-r^2}r\ dr\ d\theta = \frac{G}{6}.$$

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  • $\begingroup$ Thank you. Is there closed form for the indefinite integral? $\endgroup$ – joro Sep 23 '14 at 15:46
  • $\begingroup$ Yes, but it's a pretty large expression involving terms such as Log[1 + 1/2 (1 + I Sqrt[3]) E^(2 I t)] and PolyLog[2, -(1/2) (1 + I Sqrt[3]) E^(2 I t)] (where the outer integral runs from 0 to $t$). $\endgroup$ – Aeryk Sep 23 '14 at 17:19
  • $\begingroup$ Hm, it doesn't find closed form without integral? $\endgroup$ – joro Sep 23 '14 at 17:39
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There is also an interpretation of this integral in terms of hyperbolic geometry. In hyperbolic geometry, Catalan's constant $G$ is one quarter the (three dimensional) volume of a regular ideal octahedron (or the volume of an ideal tetrahedron with dihedral angles $\frac{\pi}{2}, \frac{\pi}{4}$, and $\frac{\pi}{4}$).

If $\Gamma$ is the group of orientation preserving isometries of a tessellation of $\mathbb{H}^3$ by regular ideal octahedra (aka $PGL(2,O_1)$), the quotient of $\mathbb{H}^3/\Gamma$ has volume: $$\frac{G}{6}=\int_0 ^\frac{1}{2}\int_0 ^\frac{1}{2}\int_\sqrt{1-x^2-y^2} ^\infty \frac{1}{z^3} dz dy dx = \frac{1}{2}\int_0 ^\frac{1}{2}\int_0 ^\frac{1}{2}\frac{1}{1-x^2-y^2} dy dx .$$

A decent reference for understanding this observation is Neumann and Reid's Notes on Adams' small volume orbifolds (page 312). Although not directly stated, their method relies on the observation that a certain manifold, the Whitehead link complement, is well known to be isometric to a regular ideal octahedron with faces identified in pairs, as noted above this means it has volume $4G$. The orbifold we are interested in is a 24 fold quotient of this manifold, and so it has volume $\frac{G}{6}$. The geometry of the orbifold is described in Colin Adams' paper Noncompact 3-Orbifolds of Small Volume (see figure 6(c) and Theorem 5.2).

However, an early reference is Borel's paper: Commensurability classes and volumes of hyperbolic 3-manifolds.

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  • $\begingroup$ Thank you. Is there closed form for the indefinite integral? $\endgroup$ – joro Sep 24 '14 at 9:27
  • $\begingroup$ It is not known whether $G$ is irrational. Does it matter for hyperbolic geometry if this volume is rational or not? $\endgroup$ – joro Sep 24 '14 at 9:49
  • $\begingroup$ I don't know of one. The closest thing I can think of in terms of resembling a closed form motivated by the hyperbolic geometry, which would involve Clausen function of order 2 (aka the Lobachevsky function). $\endgroup$ – Neil Hoffman Sep 24 '14 at 9:49
  • $\begingroup$ In some sense, it doesn't matter. Volume is often used a ``crude'' invariant to tell two hyperbolic manifolds apart. However, it would be intriguing to find a hyperbolic manifold with a rational volume. In fact, such a question is related to one of the last remaining open problems on W. Thurston's famous list: ams.org/journals/bull/1982-06-03/S0273-0979-1982-15003-0/… (See section 6 question 23). Looking at it in that light, it would be a very big deal if $G$ were rational. $\endgroup$ – Neil Hoffman Sep 24 '14 at 9:58
  • $\begingroup$ Thank you for the education. Is this known $\int_0^1 \int_0^1 (1-x^2-y^2)^{-1}dxdy=?$? According to comments Mathematica claims it $2G$. In Maple after the simplification get $2G -i \pi^2/4 $. Numerically in mpmath it appears unstable. $\endgroup$ – joro Sep 24 '14 at 12:43
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After some trick substitutions, I've put the integral in the form:

$$\int_0^{\pi/6}{\coth^{-1}{(2\,\cos{t})}dt}$$

Mathematica (version 8) then returns the exact value $\frac{G}{3}$. A nice definite integral for Catalan's constant, anyway.

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The proof of (1), sketched in http://www.maths.lancs.ac.uk/~jameson/catalan.pdf, goes as follows.

With the substitution $x=x^\prime/2,\,y=y^\prime/2$, the integral (1) takes the form (we have omitted the primes) $$I_2=\int_0^1\int_0^1\frac{dx\,dy}{4-x^2-y^2}.$$ Due to the $x\leftrightarrow y$ exchange symmetry of the integrand, $$I_2=2\iint\limits_\Delta \frac{dx\,dy}{4-x^2-y^2},$$ where the triangular integration domain $\Delta$ is the lower half of the unit square $0\le x\le 1,\,0\le y\le 1$. Let us introduce the polar coordinates, as suggested by Aeryk, $x=r\cos{\theta},\,y=r\sin{\theta}$. Then $$I_2=\int\limits_0^{\pi/4}d\theta\int\limits_0^{1/cos{\theta}}\frac{2r}{4-r^2}dr=\int\limits_0^{\pi/4}\ln{\frac{4\cos^2{\theta}}{4\cos^2{\theta}-1}}\,d\theta.$$ Now, by using $\sin{3\theta}=\sin{\theta}(3\cos^2{\theta}-\sin^2{\theta})=\sin{\theta}(4\cos^2{\theta}-1)$, we cen rewrite the above integral in the form $$I_2=\int\limits_0^{\pi/4}\ln{\frac{4\cos^2{\theta}\sin{\theta}}{\sin{3\theta}}}\,d\theta=\frac{\pi}{2}\ln{2}+2I_C+I_S-I_{3S},$$ where the integrals $$I_S=\int\limits_0^{\pi/4}\ln{\sin{\theta}}\,d\theta=-\frac{1}{2}\,G-\frac{\pi}{4}\,\ln{2},\;I_C=\int\limits_0^{\pi/4}\ln{\cos{\theta}}\,d\theta=\frac{1}{2}\,G-\frac{\pi}{4}\,\ln{2}$$ were calculated in http://www.maths.lancs.ac.uk/~jameson/catalan.pdf, while for $I_{3S}$ we have $$I_{3S}=\int\limits_0^{\pi/4}\ln{\sin{3\theta}}\,d\theta=\frac{1}{3} \int\limits_0^{3\pi/4}\ln{\sin{\theta}}\,d\theta=\frac{1}{3}\left[ \int\limits_0^{\pi/2}\ln{\sin{\theta}}\,d\theta+ \int\limits_{\pi/2}^{3\pi/4}\ln{\sin{\theta}}\,d\theta\right].$$ The substitution $\alpha=-(\pi/2-\theta)$ shows that the second integral equals to $I_C$, while for the first integral the following result can be found in http://www.maths.lancs.ac.uk/~jameson/catalan.pdf $$\int\limits_0^{\pi/2}\ln{\sin{\theta}}\,d\theta= \int\limits_0^{\pi/2}\ln{\cos{\theta}}\,d\theta=-\frac{\pi}{2}\,\ln{2}.$$ Therefore $$I_{3S}=\frac{1}{3}\left[-\frac{\pi}{2}\,\ln{2}+\frac{1}{2}\,G-\frac{\pi}{4}\,\ln{2}\right]=\frac{1}{6}\,G-\frac{\pi}{4}\,\ln{2},$$ and $$I_2=\frac{\pi}{2}\ln{2}+2\left(\frac{1}{2}\,G-\frac{\pi}{4}\,\ln{2}\right)+\left(-\frac{1}{2}\,G-\frac{\pi}{4}\,\ln{2}\right)-\left(\frac{1}{6}\,G-\frac{\pi}{4}\,\ln{2}\right)=\frac{G}{3}.$$ By the similar method, the following result was proved in http://www.maths.lancs.ac.uk/~jameson/catalan.pdf $$I_1=\int_0^1\int_0^1\frac{dx\,dy}{2-x^2-y^2}=G.$$

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  • $\begingroup$ Thanks. The paper references the question in 8.. $\endgroup$ – joro Feb 5 '17 at 7:51

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