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  "body": "\n## Description\n\nThe *lemniscate constant* is $\\varpi = 2.62205755429\\ldots$, defined by\n$$\\varpi = 2\\int_0^1 \\frac{dt}{\\sqrt{1 - t^4}}.$$\nIt plays for the lemniscate of Bernoulli $(x^2 + y^2)^2 = x^2 - y^2$ the role that $\\pi$ plays for the unit circle.\nThe lemniscate has total arc length $2\\varpi$, as the circle has length $2\\pi$.\n\n1. With the [[gamma function|0H5SYBSED9]], $\\varpi = \\dfrac{\\Gamma(1/4)^2}{2\\sqrt{2\\pi}}$.\n2. With the [[complete elliptic integral of the first kind|RM54CPPE12]], $\\varpi = \\sqrt2\\,K(1/\\sqrt2)$.\n3. With the arithmetic–geometric mean $M$, $\\varpi = \\pi / M(1, \\sqrt2)$; the number $\\varpi/\\pi = 1/M(1, \\sqrt 2) = 0.8346\\ldots$ is Gauss's constant.\n4. $\\varpi$ is a [[transcendental number|0VTQXTCR2G]].\n\n## Remarks\n\nCheck of point 1: the substitution $s = t^4$ gives $\\int_0^1 (1 - t^4)^{-1/2}\\,dt = \\frac14 B(1/4, 1/2) = \\frac14\\,\\Gamma(1/4)\\Gamma(1/2)/\\Gamma(3/4)$.\nThe reflection formula $\\Gamma(1/4)\\Gamma(3/4) = \\pi/\\sin(\\pi/4) = \\pi\\sqrt2$ and $\\Gamma(1/2) = \\sqrt\\pi$ turn this into $\\Gamma(1/4)^2/(4\\sqrt{2\\pi})$, which is $\\varpi/2$.\nCheck of point 2: with $t = \\cos\\theta$ and $1 - \\cos^4\\theta = \\sin^2\\theta\\,(1 + \\cos^2\\theta)$, one gets $\\int_0^1 dt/\\sqrt{1 - t^4} = \\int_0^{\\pi/2} d\\theta/\\sqrt{1 + \\cos^2\\theta} = \\int_0^{\\pi/2} d\\theta/\\sqrt{2 - \\sin^2\\theta} = K(1/\\sqrt2)/\\sqrt2$.\nThe [[lemniscate sine|TQDZQ964AZ]] first reaches $1$ at $\\varpi/2$, as the sine first reaches $1$ at $\\pi/2$.\n\n## Explanations\n\n### L1 human\n\nEveryone knows $\\pi$, the half-length of the unit circle.\nThe lemniscate constant is the analogous number for the figure-eight lemniscate: half its length is $\\varpi \\approx 2.622$.\nIt turns up in elliptic integrals, in the arithmetic–geometric mean of $1$ and $\\sqrt 2$, and in values of the gamma function at $1/4$.\n",
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