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  "body": "\n## Statement\n\nLet $\\Gamma$ be the [[gamma function|0V3FZMVB4K]], let $\\operatorname{Log}$ be the principal [[complex logarithm|02QGGD5G87]], and for $z \\in \\C \\setminus (-\\infty, 0]$ let $z^{z - 1/2} = e^{(z - 1/2)\\operatorname{Log} z}$.\nFix $\\delta$ with $0 \u003c \\delta \u003c \\pi$.\nThen, as $\\abs{z} \\to \\infty$ with $\\abs{\\arg z} \\le \\pi - \\delta$,\n$$\\Gamma(z) = \\sqrt{2\\pi}\\; z^{z - 1/2} \\, e^{-z} \\left(1 + O\\!\\left(\\frac{1}{\\abs{z}}\\right)\\right),$$\nwhere the constant in $O$ depends only on $\\delta$.\nIn particular $\\Gamma(z)$ is [[asymptotically equivalent|0QR92XBV6H]] to $\\sqrt{2\\pi}\\, z^{z - 1/2} e^{-z}$ in this sector.\n\n## Remarks\n\nThe proof starts from [[Binet's formula|AGBWPB994F]], which writes $\\operatorname{Log}\\Gamma(z) - (z - \\tfrac12)\\operatorname{Log} z + z - \\tfrac12\\ln(2\\pi)$ as an integral that is $O(1/\\abs{z})$ for $\\operatorname{Re} z \u003e 0$; rotating the line of integration extends this bound to the whole sector.\nAnother route is Laplace's method applied to Euler's integral $\\Gamma(x + 1) = \\int_0^\\infty e^{x \\ln t - t}\\,dt$, whose integrand peaks at $t = x$.\nFor real $x \\to \\infty$ it gives $\\Gamma(x + 1) \\sim \\sqrt{2\\pi x}\\,(x/e)^x$; at integers this is [[Stirling's formula|0KC0F51T6A]] for $n!$.\nThe sector condition is needed: on the negative real axis $\\Gamma$ has poles, so no such estimate holds near it.\nThe full asymptotic series for $\\operatorname{Log}\\Gamma(z)$ is [[Stirling's series|K29QYWB6A5]].\n\n## Explanations\n\n### L1 human\n\nStirling's formula tells how big $\\Gamma(z)$ is when $z$ is large.\nIt extends the classical estimate $n! \\approx \\sqrt{2\\pi n}\\,(n/e)^n$ from whole numbers to real and complex arguments.\n\n### L2 human\n\nThe complex form gives the decay of $\\Gamma$ on vertical lines: for fixed real $x$, $\\abs{\\Gamma(x + iy)} \\sim \\sqrt{2\\pi}\\,\\abs{y}^{x - 1/2} e^{-\\pi\\abs{y}/2}$ as $\\abs{y} \\to \\infty$.\nThis decay is what makes Mellin–Barnes integrals and the inverse Mellin transform converge, and it is used to estimate the Riemann zeta function through its functional equation.\nDividing two instances gives $\\Gamma(x + a)/\\Gamma(x) \\sim x^a$ as $x \\to \\infty$.\n",
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