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  "body": "\n## Statement\n\nLet $H = \\set{s \\in \\C : \\operatorname{Re} s \u003e 0}$, and for $t \u003e 0$ let $t^{s-1} = e^{(s-1)\\ln t}$.\n\n1. For every $s \\in H$ the integral $\\int_0^\\infty t^{s-1} e^{-t} \\, dt$ converges [[absolutely|0F9VKJPWCA]].\n2. The function $G(s) = \\int_0^\\infty t^{s-1} e^{-t} \\, dt$ is [[holomorphic|0NAHXR1FRH]] on $H$.\n3. For real $s \u003e 0$, $G(s) = \\Gamma(s)$, where $\\Gamma$ is the [[gamma function|0V3FZMVB4K]].\n\nSo $G$ is the unique holomorphic extension of $\\Gamma$ from $(0, \\infty)$ to $H$, and $\\Gamma(s) = G(s)$ for all $s \\in H$.\n\n## Remarks\n\nAbsolute convergence follows from $\\abs{t^{s-1} e^{-t}} = t^{\\sigma - 1} e^{-t}$ with $\\sigma = \\operatorname{Re} s \u003e 0$, which is the real integrand.\nFor holomorphy, the truncated integrals $\\int_{1/n}^{n}$ are entire in $s$ and converge to $G$ uniformly on each strip $a \\le \\operatorname{Re} s \\le b$ with $0 \u003c a \u003c b$, so the limit is holomorphic.\nUniqueness comes from the [[identity theorem|0YBVKMYX0W]], since $(0, \\infty)$ has accumulation points in the connected set $H$.\nThe integral diverges at $t = 0$ when $\\operatorname{Re} s \\le 0$; the extension of $\\Gamma$ beyond $H$ uses the functional equation $\\Gamma(s + 1) = s\\,\\Gamma(s)$ instead.\nIn Mathlib the statement is `Complex.Gamma_eq_integral`.\n\n## Explanations\n\n### L1 human\n\nEuler's integral is the most common formula for the gamma function.\nIt works not only for positive real numbers but for every complex number with positive real part.\nFor example, at $s = 1$ it gives $\\int_0^\\infty e^{-t}\\,dt = 1 = \\Gamma(1)$.\n\n### L2 human\n\nThe integral is the gamma function's Mellin transform description: $\\Gamma(s)$ is the Mellin transform of $e^{-t}$.\nMany identities, such as the formula linking the [[beta function|QHNWVSC6AQ]] with $\\Gamma$, start from this integral and a change of variables.\nThe integral alone only defines $\\Gamma$ on the right half-plane; the [[meromorphic|0NSD1TPHR0]] extension to $\\C$ comes from the functional equation.\n",
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