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?

  • $\begingroup$ Rewriting the integrand as a geometric series might help. $\endgroup$
    – Eckhard
    Sep 23 '14 at 14:16
  • 1
    $\begingroup$ Mathematica returns $2G$ for $\int_0^1 \int_0^1 (1-x^2-y^2)^{-1}dxdy$, as a side note. $\endgroup$ 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. $\endgroup$ 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. $\endgroup$ Sep 24 '14 at 0:38

Mathematica confirms the following:

Change the integral to polar coordinates to get

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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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). $\endgroup$ 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. $\endgroup$ 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

After some trick substitutions, I've put the integral in the form:


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


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 $$I_1=\int_0^1\int_0^1\frac{dx\,dy}{2-x^2-y^2}=G.$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.