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  "body": "\n## Statement\n\nLet $\\Gamma$ be the [[gamma function|0V3FZMVB4K]].\nThen\n$$\\Gamma\\!\\left(\\tfrac12\\right) = \\sqrt{\\pi}.$$\n\n## Remarks\n\nBy definition $\\Gamma(\\tfrac12) = \\int_0^\\infty t^{-1/2} e^{-t}\\, dt$.\nThe substitution $t = x^2$, $dt = 2x\\,dx$, turns this into $2\\int_0^\\infty e^{-x^2}\\,dx$, and the [[Gaussian integral|0DFT1GP1JP]] gives $2 \\cdot \\tfrac{\\sqrt{\\pi}}{2} = \\sqrt{\\pi}$.\nAnother proof uses the [[reflection formula|WF7KQAQTS2]] at $z = \\tfrac12$: $\\Gamma(\\tfrac12)^2 = \\pi/\\sin(\\tfrac{\\pi}{2}) = \\pi$, and $\\Gamma(\\tfrac12) \u003e 0$.\nWith the functional equation it gives all half-integer values, for example $\\Gamma(\\tfrac32) = \\tfrac{\\sqrt\\pi}{2}$ and $\\Gamma(-\\tfrac12) = -2\\sqrt{\\pi}$.\nIn Mathlib the statement is `Complex.Gamma_one_half_eq`.\n\n## Explanations\n\n### L1 human\n\nThe gamma function at one half equals the square root of $\\pi$, about $1.7725$.\nIn factorial language, this says that \"$(-\\tfrac12)!$\" is $\\sqrt{\\pi}$.\n\n### L2 human\n\nThe value is the gamma-function form of the Gaussian integral $\\int_{-\\infty}^\\infty e^{-x^2}\\,dx = \\sqrt{\\pi}$.\nIt is the reason $\\pi$ appears in the volume of balls in odd dimension and in the normalisation of the normal and chi-squared distributions.\n",
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