In mathematics, the lemniscate constant ϖ is a transcendental mathematical constant that is the ratio of the perimeter of Bernoulli's lemniscate to its diameter, analogous to the definition of π for the circle.^[1] Equivalently, the perimeter of the lemniscate ${\displaystyle (x^{2}+y^{2})^{2}=x^{2}-y^{2}}$ is 2ϖ. The lemniscate constant is closely related to the lemniscate elliptic functions and approximately equal to 2.62205755.^[2] It also appears in evaluation of the gamma and beta function at certain rational values. The symbol ϖ is a cursive variant of π; see Pi § Variant pi.

Sometimes the quantities 2ϖ or ϖ/2 are referred to as the lemniscate constant.^[3][4]

As of 2024 over 1.2 trillion digits of this constant have been calculated.^[5]

Gauss's constant, denoted by G, is equal to ϖ /π ≈ 0.8346268^[6] and named after Carl Friedrich Gauss, who calculated it via the arithmetic–geometric mean as ${\displaystyle 1/M{\bigl (}1,{\sqrt {2}}{\bigr )}}$.^[7] By 1799, Gauss had two proofs of the theorem that ${\displaystyle M{\bigl (}1,{\sqrt {2}}{\bigr )}=\pi /\varpi }$ where ${\displaystyle \varpi }$ is the lemniscate constant.^[8]

John Todd named two more lemniscate constants, the first lemniscate constant A = ϖ/2 ≈ 1.3110287771 and the second lemniscate constant B = π/(2ϖ) ≈ 0.5990701173.^[9][10][11]

The lemniscate constant ${\displaystyle \varpi }$ and Todd's first lemniscate constant ${\displaystyle A}$ were proven transcendental by Carl Ludwig Siegel in 1932 and later by Theodor Schneider in 1937 and Todd's second lemniscate constant ${\displaystyle B}$ and Gauss's constant ${\displaystyle G}$ were proven transcendental by Theodor Schneider in 1941.^[9][12][13] In 1975, Gregory Chudnovsky proved that the set ${\displaystyle \{\pi ,\varpi \}}$ is algebraically independent over ${\displaystyle \mathbb {Q} }$, which implies that ${\displaystyle A}$ and ${\displaystyle B}$ are algebraically independent as well.^[14][15] But the set ${\displaystyle {\bigl \{}\pi ,M{\bigl (}1,1/{\sqrt {2}}{\bigr )},M'{\bigl (}1,1/{\sqrt {2}}{\bigr )}{\bigr \}}}$ (where the prime denotes the derivative with respect to the second variable) is not algebraically independent over ${\displaystyle \mathbb {Q} }$.^[16] In 1996, Yuri Nesterenko proved that the set ${\displaystyle \{\pi ,\varpi ,e^{\pi }\}}$ is algebraically independent over ${\displaystyle \mathbb {Q} }$.^[17]

Usually, ${\displaystyle \varpi }$ is defined by the first equality below, but it has many equivalent forms:^[18]

$${\displaystyle {\begin{aligned}\varpi &=2\int _{0}^{1}{\frac {\mathrm {d} t}{\sqrt {1-t^{4}}}}={\sqrt {2}}\int _{0}^{\infty }{\frac {\mathrm {d} t}{\sqrt {1+t^{4}}}}=\int _{0}^{1}{\frac {\mathrm {d} t}{\sqrt {t-t^{3}}}}=\int _{1}^{\infty }{\frac {\mathrm {d} t}{\sqrt {t^{3}-t}}}\\[6mu]&=4\int _{0}^{\infty }{\Bigl (}{\sqrt[{4}]{1+t^{4}}}-t{\Bigr )}\,\mathrm {d} t=2{\sqrt {2}}\int _{0}^{1}{\sqrt[{4}]{1-t^{4}}}\mathop {\mathrm {d} t} =3\int _{0}^{1}{\sqrt {1-t^{4}}}\,\mathrm {d} t\\[2mu]&=2K(i)={\tfrac {1}{2}}\mathrm {B} {\bigl (}{\tfrac {1}{4}},{\tfrac {1}{2}}{\bigr )}={\tfrac {1}{2{\sqrt {2}}}}\mathrm {B} {\bigl (}{\tfrac {1}{4}},{\tfrac {1}{4}}{\bigr )}={\frac {\Gamma (1/4)^{2}}{2{\sqrt {2\pi }}}}={\frac {2-{\sqrt {2}}}{4}}{\frac {\zeta (3/4)^{2}}{\zeta (1/4)^{2}}}\\[5mu]&=2.62205\;75542\;92119\;81046\;48395\;89891\;11941\ldots ,\end{aligned}}}$$

where K is the complete elliptic integral of the first kind with modulus k, Β is the beta function, Γ is the gamma function and ζ is the Riemann zeta function.

The lemniscate constant can also be computed by the arithmetic–geometric mean ${\displaystyle M}$,

$${\displaystyle \varpi ={\frac {\pi }{M{\bigl (}1,{\sqrt {2}}{\bigr )}}}.}$$

Gauss's constant is typically defined as the reciprocal of the arithmetic–geometric mean of 1 and the square root of 2, after his calculation of ${\displaystyle M{\bigl (}1,{\sqrt {2}}{\bigr )}}$ published in 1800:^[19]$${\displaystyle G={\frac {1}{M{\bigl (}1,{\sqrt {2}}{\bigr )}}}}$$John Todd's lemniscate constants may be given in terms of the beta function B: $${\displaystyle {\begin{aligned}A&={\frac {\varpi }{2}}={\tfrac {1}{4}}\mathrm {B} {\bigl (}{\tfrac {1}{4}},{\tfrac {1}{2}}{\bigr )},\\[3mu]B&={\frac {\pi }{2\varpi }}={\tfrac {1}{4}}\mathrm {B} {\bigl (}{\tfrac {1}{2}},{\tfrac {3}{4}}{\bigr )}.\end{aligned}}}$$

$${\displaystyle \beta '(0)=\log {\frac {\varpi }{\sqrt {\pi }}}}$$

which is analogous to

$${\displaystyle \zeta '(0)=\log {\frac {1}{\sqrt {2\pi }}}}$$

where ${\displaystyle \beta }$ is the Dirichlet beta function and ${\displaystyle \zeta }$ is the Riemann zeta function.^[20]

Analogously to the Leibniz formula for π, $${\displaystyle \beta (1)=\sum _{n=1}^{\infty }{\frac {\chi (n)}{n}}={\frac {\pi }{4}},}$$ we have^{[21][22][23][24][25]} $${\displaystyle L(E,1)=\sum _{n=1}^{\infty }{\frac {\nu (n)}{n}}={\frac {\varpi }{4}}}$$ where ${\displaystyle L}$ is the L-function of the elliptic curve ${\displaystyle E:\,y^{2}=x^{3}-x}$ over ${\displaystyle \mathbb {Q} }$; this means that ${\displaystyle \nu }$ is the multiplicative function given by $${\displaystyle \nu (p^{n})={\begin{cases}p-{\mathcal {N}}_{p},&p\in \mathbb {P} ,\,n=1\\[5mu]0,&p=2,\,n\geq 2\\[5mu]\nu (p)\nu (p^{n-1})-p\nu (p^{n-2}),&p\in \mathbb {P} \setminus \{2\},\,n\geq 2\end{cases}}}$$ where ${\displaystyle {\mathcal {N}}_{p}}$ is the number of solutions of the congruence $${\displaystyle a^{3}-a\equiv b^{2}\,(\operatorname {mod} p),\quad p\in \mathbb {P} }$$ in variables ${\displaystyle a,b}$ that are non-negative integers (${\displaystyle \mathbb {P} }$ is the set of all primes). Equivalently, ${\displaystyle \nu }$ is given by $${\displaystyle F(\tau )=\eta (4\tau )^{2}\eta (8\tau )^{2}=\sum _{n=1}^{\infty }\nu (n)q^{n},\quad q=e^{2\pi i\tau }}$$ where ${\displaystyle \tau \in \mathbb {C} }$ such that ${\displaystyle \operatorname {\Im } \tau >0}$ and ${\displaystyle \eta }$ is the eta function.^[26][27][28] The above result can be equivalently written as $${\displaystyle \sum _{n=1}^{\infty }{\frac {\nu (n)}{n}}e^{-2\pi n/{\sqrt {32}}}={\frac {\varpi }{8}}}$$ (the number ${\displaystyle 32}$ is the conductor of ${\displaystyle E}$) and also tells us that the BSD conjecture is true for the above ${\displaystyle E}$.^[29] The first few values of ${\displaystyle \nu }$ are given by the following table; if ${\displaystyle 1\leq n\leq 113}$ such that ${\displaystyle n}$ doesn't appear in the table, then ${\displaystyle \nu (n)=0}$: $${\displaystyle {\begin{array}{|c|c|c|c|}\hline n&\nu (n)&n&\nu (n)\\\hline 1&1&53&14\\\hline 5&-2&61&-10\\\hline 9&-3&65&-12\\\hline 13&6&73&-6\\\hline 17&2&81&9\\\hline 25&-1&85&-4\\\hline 29&-10&89&10\\\hline 37&-2&97&18\\\hline 41&10&101&-2\\\hline 45&6&109&6\\\hline 49&-7&113&-14\\\hline \end{array}}}$$

Let ${\displaystyle \Delta }$ be the minimal weight level ${\displaystyle 1}$ new form. Then^[30] $${\displaystyle \Delta (i)={\frac {1}{64}}\left({\frac {\varpi }{\pi }}\right)^{12}.}$$ The ${\displaystyle q}$-coefficient of ${\displaystyle \Delta }$ is the Ramanujan tau function.

Viète's formula for π can be written:

$${\displaystyle {\frac {2}{\pi }}={\sqrt {\frac {1}{2}}}\cdot {\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {\frac {1}{2}}}}}\cdot {\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {\frac {1}{2}}}}}}}\cdots }$$

An analogous formula for ϖ is:^[31]

$${\displaystyle {\frac {2}{\varpi }}={\sqrt {\frac {1}{2}}}\cdot {\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\bigg /}\!{\sqrt {\frac {1}{2}}}}}\cdot {\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\Bigg /}\!{\sqrt {{\frac {1}{2}}+{\frac {1}{2}}{\bigg /}\!{\sqrt {\frac {1}{2}}}}}}}\cdots }$$

The Wallis product for π is:

$${\displaystyle {\frac {\pi }{2}}=\prod _{n=1}^{\infty }\left(1+{\frac {1}{n}}\right)^{(-1)^{n+1}}=\prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)={\biggl (}{\frac {2}{1}}\cdot {\frac {2}{3}}{\biggr )}{\biggl (}{\frac {4}{3}}\cdot {\frac {4}{5}}{\biggr )}{\biggl (}{\frac {6}{5}}\cdot {\frac {6}{7}}{\biggr )}\cdots }$$

An analogous formula for ϖ is:^[32]

$${\displaystyle {\frac {\varpi }{2}}=\prod _{n=1}^{\infty }\left(1+{\frac {1}{2n}}\right)^{(-1)^{n+1}}=\prod _{n=1}^{\infty }\left({\frac {4n-1}{4n-2}}\cdot {\frac {4n}{4n+1}}\right)={\biggl (}{\frac {3}{2}}\cdot {\frac {4}{5}}{\biggr )}{\biggl (}{\frac {7}{6}}\cdot {\frac {8}{9}}{\biggr )}{\biggl (}{\frac {11}{10}}\cdot {\frac {12}{13}}{\biggr )}\cdots }$$

A related result for Gauss's constant (${\displaystyle G=\varpi /\pi }$) is:^[33]

$${\displaystyle {\frac {\varpi }{\pi }}=\prod _{n=1}^{\infty }\left({\frac {4n-1}{4n}}\cdot {\frac {4n+2}{4n+1}}\right)={\biggl (}{\frac {3}{4}}\cdot {\frac {6}{5}}{\biggr )}{\biggl (}{\frac {7}{8}}\cdot {\frac {10}{9}}{\biggr )}{\biggl (}{\frac {11}{12}}\cdot {\frac {14}{13}}{\biggr )}\cdots }$$

An infinite series discovered by Gauss is:^[34]

$${\displaystyle {\frac {\varpi }{\pi }}=\sum _{n=0}^{\infty }(-1)^{n}\prod _{k=1}^{n}{\frac {(2k-1)^{2}}{(2k)^{2}}}=1-{\frac {1^{2}}{2^{2}}}+{\frac {1^{2}\cdot 3^{2}}{2^{2}\cdot 4^{2}}}-{\frac {1^{2}\cdot 3^{2}\cdot 5^{2}}{2^{2}\cdot 4^{2}\cdot 6^{2}}}+\cdots }$$

The Machin formula for π is ${\textstyle {\tfrac {1}{4}}\pi =4\arctan {\tfrac {1}{5}}-\arctan {\tfrac {1}{239}},}$ and several similar formulas for π can be developed using trigonometric angle sum identities, e.g. Euler's formula ${\textstyle {\tfrac {1}{4}}\pi =\arctan {\tfrac {1}{2}}+\arctan {\tfrac {1}{3}}}$. Analogous formulas can be developed for ϖ, including the following found by Gauss: ${\displaystyle {\tfrac {1}{2}}\varpi =2\operatorname {arcsl} {\tfrac {1}{2}}+\operatorname {arcsl} {\tfrac {7}{23}}}$, where ${\displaystyle \operatorname {arcsl} }$ is the lemniscate arcsine.^[35]

The lemniscate constant can be rapidly computed by the series^[36][37]

${\displaystyle \varpi =2^{-1/2}\pi {\biggl (}\sum _{n\in \mathbb {Z} }e^{-\pi n^{2}}{\biggr )}^{2}=2^{1/4}\pi e^{-\pi /12}{\biggl (}\sum _{n\in \mathbb {Z} }(-1)^{n}e^{-\pi p_{n}}{\biggr )}^{2}}$

where ${\displaystyle p_{n}={\tfrac {1}{2}}(3n^{2}-n)}$ (these are the generalized pentagonal numbers). Also^[38]

${\displaystyle \sum _{m,n\in \mathbb {Z} }e^{-2\pi (m^{2}+mn+n^{2})}={\sqrt {1+{\sqrt {3}}}}{\dfrac {\varpi }{12^{1/8}\pi }}.}$

In a spirit similar to that of the Basel problem,

${\displaystyle \sum _{z\in \mathbb {Z} [i]\setminus \{0\}}{\frac {1}{z^{4}}}=G_{4}(i)={\frac {\varpi ^{4}}{15}}}$

where ${\displaystyle \mathbb {Z} [i]}$ are the Gaussian integers and ${\displaystyle G_{4}}$ is the Eisenstein series of weight ⁠${\displaystyle 4}$⁠ (see Lemniscate elliptic functions § Hurwitz numbers for a more general result).^[39]

A related result is

${\displaystyle \sum _{n=1}^{\infty }\sigma _{3}(n)e^{-2\pi n}={\frac {\varpi ^{4}}{80\pi ^{4}}}-{\frac {1}{240}}}$

where ${\displaystyle \sigma _{3}}$ is the sum of positive divisors function.^[40]

In 1842, Malmsten found

${\displaystyle \beta '(1)=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {\log(2n+1)}{2n+1}}={\frac {\pi }{4}}\left(\gamma +2\log {\frac {\pi }{\varpi {\sqrt {2}}}}\right)}$

where ${\displaystyle \gamma }$ is Euler's constant and ${\displaystyle \beta (s)}$ is the Dirichlet-Beta function.

The lemniscate constant is given by the rapidly converging series

$${\displaystyle \varpi =\pi {\sqrt[{4}]{32}}e^{-{\frac {\pi }{3}}}{\biggl (}\sum _{n=-\infty }^{\infty }(-1)^{n}e^{-2n\pi (3n+1)}{\biggr )}^{2}.}$$

The constant is also given by the infinite product

${\displaystyle \varpi =\pi \prod _{m=1}^{\infty }\tanh ^{2}\left({\frac {\pi m}{2}}\right).}$

Also^[41]

${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{6635520^{n}}}{\frac {(4n)!}{n!^{4}}}={\frac {24}{5^{7/4}}}{\frac {\varpi ^{2}}{\pi ^{2}}}.}$

A (generalized) continued fraction for π is $${\displaystyle {\frac {\pi }{2}}=1+{\cfrac {1}{1+{\cfrac {1\cdot 2}{1+{\cfrac {2\cdot 3}{1+{\cfrac {3\cdot 4}{1+\ddots }}}}}}}}}$$ An analogous formula for ϖ is^[10] $${\displaystyle {\frac {\varpi }{2}}=1+{\cfrac {1}{2+{\cfrac {2\cdot 3}{2+{\cfrac {4\cdot 5}{2+{\cfrac {6\cdot 7}{2+\ddots }}}}}}}}}$$

Define Brouncker's continued fraction by^[42] $${\displaystyle b(s)=s+{\cfrac {1^{2}}{2s+{\cfrac {3^{2}}{2s+{\cfrac {5^{2}}{2s+\ddots }}}}}},\quad s>0.}$$ Let ${\displaystyle n\geq 0}$ except for the first equality where ${\displaystyle n\geq 1}$. Then^[43][44] $${\displaystyle {\begin{aligned}b(4n)&=(4n+1)\prod _{k=1}^{n}{\frac {(4k-1)^{2}}{(4k-3)(4k+1)}}{\frac {\pi }{\varpi ^{2}}}\\b(4n+1)&=(2n+1)\prod _{k=1}^{n}{\frac {(2k)^{2}}{(2k-1)(2k+1)}}{\frac {4}{\pi }}\\b(4n+2)&=(4n+1)\prod _{k=1}^{n}{\frac {(4k-3)(4k+1)}{(4k-1)^{2}}}{\frac {\varpi ^{2}}{\pi }}\\b(4n+3)&=(2n+1)\prod _{k=1}^{n}{\frac {(2k-1)(2k+1)}{(2k)^{2}}}\,\pi .\end{aligned}}}$$ For example, $${\displaystyle {\begin{aligned}b(1)&={\frac {4}{\pi }},&b(2)&={\frac {\varpi ^{2}}{\pi }},&b(3)&=\pi ,&b(4)&={\frac {9\pi }{\varpi ^{2}}}.\end{aligned}}}$$

In fact, the values of ${\displaystyle b(1)}$ and ${\displaystyle b(2)}$, coupled with the functional equation $${\displaystyle b(s+2)={\frac {(s+1)^{2}}{b(s)}},}$$ determine the values of ${\displaystyle b(n)}$ for all ${\displaystyle n}$.

Simple continued fractions for the lemniscate constant and related constants include^[45][46] $${\displaystyle {\begin{aligned}\varpi &=[2,1,1,1,1,1,4,1,2,\ldots ],\\[8mu]2\varpi &=[5,4,10,2,1,2,3,29,\ldots ],\\[5mu]{\frac {\varpi }{2}}&=[1,3,4,1,1,1,5,2,\ldots ],\\[2mu]{\frac {\varpi }{\pi }}&=[0,1,5,21,3,4,14,\ldots ].\end{aligned}}}$$

The lemniscate constant ϖ is related to the area under the curve ${\displaystyle x^{4}+y^{4}=1}$. Defining ${\displaystyle \pi _{n}\mathrel {:=} \mathrm {B} {\bigl (}{\tfrac {1}{n}},{\tfrac {1}{n}}{\bigr )}}$, twice the area in the positive quadrant under the curve ${\displaystyle x^{n}+y^{n}=1}$ is ${\textstyle 2\int _{0}^{1}{\sqrt[{n}]{1-x^{n}}}\mathop {\mathrm {d} x} ={\tfrac {1}{n}}\pi _{n}.}$ In the quartic case, ${\displaystyle {\tfrac {1}{4}}\pi _{4}={\tfrac {1}{\sqrt {2}}}\varpi .}$

In 1842, Malmsten discovered that^[47]

$${\displaystyle \int _{0}^{1}{\frac {\log(-\log x)}{1+x^{2}}}\,dx={\frac {\pi }{2}}\log {\frac {\pi }{\varpi {\sqrt {2}}}}.}$$

Furthermore, $${\displaystyle \int _{0}^{\infty }{\frac {\tanh x}{x}}e^{-x}\,dx=\log {\frac {\varpi ^{2}}{\pi }}}$$

and^[48]

$${\displaystyle \int _{0}^{\infty }e^{-x^{4}}\,dx={\frac {\sqrt {2\varpi {\sqrt {2\pi }}}}{4}},\quad {\text{analogous to}}\,\int _{0}^{\infty }e^{-x^{2}}\,dx={\frac {\sqrt {\pi }}{2}},}$$ a form of Gaussian integral.

The lemniscate constant appears in the evaluation of the integrals

$${\displaystyle {\frac {\pi }{\varpi }}=\int _{0}^{\frac {\pi }{2}}{\sqrt {\sin(x)}}\,dx=\int _{0}^{\frac {\pi }{2}}{\sqrt {\cos(x)}}\,dx}$$

$${\displaystyle {\frac {\varpi }{\pi }}=\int _{0}^{\infty }{\frac {dx}{\sqrt {\cosh(\pi x)}}}}$$

John Todd's lemniscate constants are defined by integrals:^[9]

$${\displaystyle A=\int _{0}^{1}{\frac {dx}{\sqrt {1-x^{4}}}}}$$

$${\displaystyle B=\int _{0}^{1}{\frac {x^{2}\,dx}{\sqrt {1-x^{4}}}}}$$

The lemniscate constant satisfies the equation^[49]

$${\displaystyle {\frac {\pi }{\varpi }}=2\int _{0}^{1}{\frac {x^{2}\,dx}{\sqrt {1-x^{4}}}}}$$

Euler discovered in 1738 that for the rectangular elastica (first and second lemniscate constants)^[50][49]

$${\displaystyle {\textrm {arc}}\ {\textrm {length}}\cdot {\textrm {height}}=A\cdot B=\int _{0}^{1}{\frac {\mathrm {d} x}{\sqrt {1-x^{4}}}}\cdot \int _{0}^{1}{\frac {x^{2}\mathop {\mathrm {d} x} }{\sqrt {1-x^{4}}}}={\frac {\varpi }{2}}\cdot {\frac {\pi }{2\varpi }}={\frac {\pi }{4}}}$$

Now considering the circumference ${\displaystyle C}$ of the ellipse with axes ${\displaystyle {\sqrt {2}}}$ and ${\displaystyle 1}$, satisfying ${\displaystyle 2x^{2}+4y^{2}=1}$, Stirling noted that^[51]

$${\displaystyle {\frac {C}{2}}=\int _{0}^{1}{\frac {dx}{\sqrt {1-x^{4}}}}+\int _{0}^{1}{\frac {x^{2}\,dx}{\sqrt {1-x^{4}}}}}$$

Hence the full circumference is

$${\displaystyle C={\frac {\pi }{\varpi }}+\varpi =3.820197789\ldots }$$

This is also the arc length of the sine curve on half a period:^[52]

$${\displaystyle C=\int _{0}^{\pi }{\sqrt {1+\cos ^{2}(x)}}\,dx}$$

Analogously to $${\displaystyle 2\pi =\lim _{n\to \infty }\left|{\frac {(2n)!}{\mathrm {B} _{2n}}}\right|^{\frac {1}{2n}}}$$ where ${\displaystyle \mathrm {B} _{n}}$ are Bernoulli numbers, we have $${\displaystyle 2\varpi =\lim _{n\to \infty }\left({\frac {(4n)!}{\mathrm {H} _{4n}}}\right)^{\frac {1}{4n}}}$$ where ${\displaystyle \mathrm {H} _{n}}$ are Hurwitz numbers.

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• Sequences A014549, A053002, and A062539 in OEIS
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• "Gauss's constant and where it occurs". www.johndcook.com. 2021-10-17.