thumb|Catalan constant as area under the curve of arctanx /x
In mathematics, Catalan's constant is the alternating sum of the reciprocals of the odd square numbers:
: <math>G = \sum_{n=0}^{\infty} \frac{(-1)^{n{(2n+1)^2} = \frac{1}{1^2} - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} + \frac{1}{9^2} - \cdots.</math>
Its numerical value is approximately
: ,
and it is also equal to where is the Dirichlet beta function.
Catalan's constant was named after Eugène Charles Catalan, who found quickly-converging series for its calculation and published a memoir on it in 1865.
Uses
In low-dimensional topology, Catalan's constant is 1/4 of the volume of an ideal hyperbolic octahedron, and therefore 1/4 of the hyperbolic volume of the complement of the Whitehead link. It is 1/8 of the volume of the complement of the Borromean rings.
In combinatorics and statistical mechanics, it arises in connection with counting domino tilings, spanning trees, and Hamiltonian cycles of grid graphs.
In number theory, Catalan's constant appears in a conjectured formula for the asymptotic number of primes of the form <math>n^2+1</math> according to Hardy and Littlewood's Conjecture F. However, it is an unsolved problem (one of Landau's problems) whether there are even infinitely many primes of this form.
Catalan's constant also appears in the calculation of the mass distribution of spiral galaxies.
Properties
It is not known whether is irrational, let alone transcendental. has been called "arguably the most basic constant whose irrationality and transcendence (though strongly
suspected) remain unproven".
There exist however partial results. It is known that infinitely many of the numbers β(2n) are irrational, where β(s) is the Dirichlet beta function. In particular at least one of β(2), β(4), β(6), β(8), β(10) and β(12) must be irrational, where β(2) is Catalan's constant. These results by Wadim Zudilin and Tanguy Rivoal are related to similar ones given for the odd zeta constants ζ(2n+1).
Catalan's constant is known to be an algebraic period, which follows from some of the double integrals given below.
Series representations
Catalan's constant appears in the evaluation of several rational series including:<math display="block">\frac{\pi^2}{16}+\frac G2 = \sum_{n=0}^\infty \frac{1}{(4n+1)^2}.</math><math display="block">\frac{\pi^2}{16}-\frac G2 = \sum_{n=0}^\infty \frac{1}{(4n+3)^2}.</math>
The following two formulas involve quickly converging series, and are thus appropriate for numerical computation:
<math display="block">\begin{align}
G & = 3 \sum_{n=0}^\infty \frac{1}{2^{4n
\left(-\frac{1}{2(8n+2)^2}+\frac{1}{2^2(8n+3)^2}-\frac{1}{2^3(8n+5)^2}+\frac{1}{2^3(8n+6)^2}-\frac{1}{2^4(8n+7)^2}+\frac{1}{2(8n+1)^2}\right) \\
& \qquad -2 \sum_{n=0}^\infty \frac{1}{2^{12n
\left(\frac{1}{2^4(8n+2)^2}+\frac{1}{2^6(8n+3)^2}-\frac{1}{2^9(8n+5)^2}-\frac{1}{2^{10} (8n+6)^2}-\frac{1}{2^{12} (8n+7)^2}+\frac{1}{2^3(8n+1)^2}\right)
\end{align}</math>
and
<math display="block">G = \frac{\pi}{8}\log\left(2 + \sqrt{3}\right) + \frac{3}{8}\sum_{n=0}^\infty \frac{1}{(2n+1)^2 \binom{2n}{n.</math>
The theoretical foundations for such series are given by Broadhurst, for the first formula, and Ramanujan, for the second formula. The algorithms for fast evaluation of the Catalan constant were constructed by E. Karatsuba. Using these series, calculating Catalan's constant is now about as fast as calculating Apéry's constant, <math>\zeta(3)</math>.
:<math>G = \frac{1}{2}\sum_{k=0}^{\infty }\frac{(-8)^{k}(3k+2)}{(2k+1)^{3}{\binom{2k}{k^{3</math>
:<math>G = \frac{1}{64}\sum_{k=1}^{\infty }\frac{256^{k}(580k^2-184k+15)}{k^3(2k-1)\binom{6k}{3k}\binom{6k}{4k}\binom{4k}{2k</math>
:<math>G = -\frac{1}{1024}\sum_{k=1}^{\infty }\frac{(-4096)^k(45136k^4-57184k^3+21240k^2-3160k+165)}{k^3(2k-1)^3}\left( \frac{(2k)!^6(3k)!^3}{k!^3(6k)!^3} \right)</math>
All of these series have time complexity <math>O(n\log(n)^3)</math>. Some of these expressions include:
<math display="block">\begin{align}
G &= -\frac{1}{\pi i}\int_{0}^{\frac{\pi}{2 \ln\ln \tan x \ln \tan x \,dx \\[3pt]
G &= \iint_{[0,1]^2} \! \frac{1}{1+x^2 y^2} \,dx\, dy \\[3pt]
G &= \int_0^1\int_0^{1-x} \frac{1}{1 -x^2-y^2} \,dy\,dx \\[3pt]
G &= \int_1^\infty \frac{\ln t}{1 + t^2} \,dt \\[3pt]
G &= -\int_0^1 \frac{\ln t}{1 + t^2} \,dt \\[3pt]
G &= \frac{1}{2} \int_0^\frac{\pi}{2} \frac{t}{\sin t} \,dt \\[3pt]
G &= \int_0^\frac{\pi}{4} \ln \cot t \,dt \\[3pt]
G &= \frac{1}{2} \int_0^\frac{\pi}{2} \ln \left( \sec t +\tan t \right) \,dt \\[3pt]
G &= \int_0^1 \frac{\arccos t}{\sqrt{1+t^2 \,dt \\[3pt]
G &= \int_0^1 \frac{\operatorname{arcsinh} t}{\sqrt{1-t^2 \,dt \\[3pt]
G &= \frac{1}{2} \int_0^\infty \frac{\operatorname{arctan} t}{t\sqrt{1+t^2 \,dt \\[3pt]
G &= \frac{1}{2} \int_0^1 \frac{\operatorname{arctanh} t}{\sqrt{1-t^2 \,dt \\[3pt]
G &= \int_0^\infty \arccot e^{t} \,dt \\[3pt]
G &= \frac{1}{4} \int_0^ \csc \sqrt{t} \,dt \\[3pt]
G &= \frac{1}{16} \left(\pi^2 + 4\int_1^\infty \arccsc^2 t \,dt\right) \\[3pt]
G &= \frac{1}{2} \int_0^\infty \frac{t}{\cosh t} \,dt \\[3pt]
G &= \frac{\pi}{2} \int_1^\infty \frac{\left(t^4-6t^2+1\right)\ln\ln t}{\left(1+t^2\right)^3} \,dt \\[3pt]
G &= \frac{1}{2} \int_0^\infty \frac{\arcsin \left(\sin t\right)}{t} \,dt \\[3pt]
G &= 1 + \lim_{\alpha\to{1^-\!\left\{\int_0^{\alpha}\!\frac{\left(1+6t^2+t^4\right)\arctan{t{t\left(1-t^2\right)^2}\, dt
+ 2\operatorname{artanh}{\alpha} - \frac{\pi\alpha}{1-\alpha^2} \right\} \\[3pt]
G &= 1 - \frac18 \iint_{\R^2}\!\!\frac{x\sin\left(2xy/\pi\right)}{\,\left(x^2+\pi^2\right)\cosh x\sinh y\,} \,dx\,dy \\[3pt]
G &= \int_{0}^{\infty}\int_{0}^{\infty}\frac{\sqrt[4]{x} \left(\sqrt{x} \sqrt{y}-1\right)}{(x+1)^2 \sqrt[4]{y} (y+1)^2 \log (x y)}dxdy
\end{align}</math>
where the last three formulas are related to Malmsten's integrals.
If is the complete elliptic integral of the first kind, as a function of the elliptic modulus , then
<math display="block"> G = \tfrac{1}{2} \int_0^1 \mathrm{K}(k)\,dk </math>
If is the complete elliptic integral of the second kind, as a function of the elliptic modulus , then
<math display="block"> G = -\tfrac{1}{2}+\int_0^1 \mathrm{E}(k)\,dk </math>
With the gamma function , Catalan's constant is
<math display="block">\begin{align}
G &= \frac{\pi}{4} \int_0^1 \Gamma\left(1+\frac{x}{2}\right)\Gamma\left(1-\frac{x}{2}\right)\,dx \\
&= \frac{\pi}{2} \int_0^\frac12\Gamma(1+y)\Gamma(1-y)\,dy
\end{align}</math>
The integral
<math display="block"> G = \operatorname{Ti}_2(1)=\int_0^1 \frac{\arctan t}{t}\,dt </math>
is a known special function, called the inverse tangent integral, and was extensively studied by Srinivasa Ramanujan.
Relation to special functions
appears in values of the second polygamma function, also called the trigamma function, at fractional arguments:
:<math>G=\cfrac{1}{1+\cfrac{1^4}{8+\cfrac{3^4}{16+\cfrac{5^4}{24+\cfrac{7^4}{32+\cfrac{9^4}{40+\ddots</math>
The simple continued fraction is given by:
:<math>G=\cfrac{1}{1+\cfrac{1}{10+\cfrac{1}{1+\cfrac{1}{8+\cfrac{1}{1+\cfrac{1}{88+\ddots</math>
This continued fraction would have infinite terms if and only if <math>G</math> is irrational, which is still unresolved.
The following continued fraction representation gives (asymptotically) 2.08 new correct decimal places per cycle:
:<math>G=\frac{\frac{13}{2{Z_{0, Z_{k}=a(k)+\frac{b(k)}{Z_{k+1</math>
with
:<math>a(k)=3520k^6+5632k^5+2064k^4-384k^3-156k^2+16k+7</math>
:<math>b(k)=(2k+1)^4(2k+2)^4(20k^2-8k+1)(20k^2+72k+65)</math>
Known digits
The number of known digits of Catalan's constant has increased dramatically during the last decades. This is due both to the increase of performance of computers as well as to algorithmic improvements.
{| class="wikitable" style="margin: 1em auto 1em auto"
|+ Number of known decimal digits of Catalan's constant
! Date || Decimal digits || Computation performed by
|-
| 1832 ||align="right"| 16 || Thomas Clausen
|-
| 1858 ||align="right"| 19 || Carl Johan Danielsson Hill
|-
| 1864 ||align="right"| 14 || Eugène Charles Catalan
|-
| 1877 ||align="right"| 20 || James W. L. Glaisher
|-
| 1913 ||align="right"| 32 || James W. L. Glaisher
|-
| 1990 ||align="right"| || Greg J. Fee
|-
| 1996 ||align="right"| || Greg J. Fee
|-
| August 14, 1996 ||align="right"| || Greg J. Fee & Simon Plouffe
|-
| September 29, 1996 ||align="right"| || Thomas Papanikolaou
|-
| 1996 ||align="right"| || Thomas Papanikolaou
|-
| 1997 ||align="right"| || Patrick Demichel
|-
| January 4, 1998 ||align="right"| || Xavier Gourdon
|-
| 2001 ||align="right"| || Xavier Gourdon & Pascal Sebah
|-
| 2002 ||align="right"| || Xavier Gourdon & Pascal Sebah
|-
| October 2006 ||align="right"| || Shigeru Kondo & Steve Pagliarulo
|-
| August 2008 ||align="right"| || Shigeru Kondo & Steve Pagliarulo
|-
| April 16, 2009 ||align="right"| || Alexander J. Yee & Raymond Chan
|-
| April 12, 2016 ||align="right"| || Ron Watkins
|-
| September 6, 2020 ||align="right"| || Andrew Sun
|-
| March 9, 2022 ||align="right"| || Seungmin Kim