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  "body": "\n## Statement\n\nLet $\\operatorname{Log}\\Gamma$ be the [[log-gamma function|JT95B44G4G]] and $\\operatorname{Log}$ the principal [[complex logarithm|02QGGD5G87]].\nFor $\\operatorname{Re} z \u003e 0$ put\n$$\\mu(z) = \\operatorname{Log}\\Gamma(z) - \\left(z - \\tfrac12\\right)\\operatorname{Log} z + z - \\tfrac12 \\ln(2\\pi).$$\nThen for every $z$ with $\\operatorname{Re} z \u003e 0$:\n\n1. (Binet's first formula) $$\\mu(z) = \\int_0^\\infty \\left(\\frac12 - \\frac1t + \\frac{1}{e^t - 1}\\right)\\frac{e^{-zt}}{t}\\, dt.$$\n2. (Binet's second formula) $$\\mu(z) = 2\\int_0^\\infty \\frac{\\arctan(t/z)}{e^{2\\pi t} - 1}\\, dt,$$ where $\\arctan(t/z) = \\frac{1}{2i}\\operatorname{Log}\\dfrac{1 + it/z}{1 - it/z}$ is the principal branch.\n\n## Remarks\n\nThe integrand of the first formula is bounded near $t = 0$, since $\\frac12 - \\frac1t + \\frac{1}{e^t - 1} = \\frac{t}{12} + O(t^3)$.\nThe first formula comes from integrating the integral representation of the [[digamma function|HWRRXB0YXS]].\nThe second comes from the Abel–Plana summation formula, or from the first by a partial fraction expansion of $\\frac{1}{e^t - 1}$.\nThe function $\\mu$ is called Binet's function; it is positive and decreasing on $(0, \\infty)$ and tends to $0$ as $x \\to \\infty$.\nThis limit is [[Stirling's formula for the gamma function|53750WSJ6K]] on the real axis, and expanding $\\frac12 - \\frac1t + \\frac{1}{e^t - 1}$ in powers of $t$ gives [[Stirling's series|K29QYWB6A5]] term by term.\nFor example, the first term $\\frac{t}{12}$ contributes $\\int_0^\\infty \\frac{1}{12} e^{-zt}\\,dt = \\frac{1}{12z}$, the first correction in Stirling's series.\n\n## Explanations\n\n### L1 human\n\nStirling's formula approximates $\\ln\\Gamma(z)$ by a simple expression.\nBinet's formulas give the exact error of this approximation as an integral.\nFrom the integrals one can read off how small the error is and in which direction it goes.\n\n### L2 human\n\nBinet's first formula is a Laplace transform, so Watson's lemma turns the Taylor series of its integrand directly into the asymptotic expansion of $\\mu(z)$.\nThe second formula is convenient for error bounds in the right half-plane, because the kernel $\\frac{1}{e^{2\\pi t} - 1}$ is explicit and positive.\nBoth show that $\\mu(x) \\to 0$ monotonically, which gives sharp bounds such as $0 \u003c \\mu(x) \u003c \\frac{1}{12x}$ for $x \u003e 0$.\n",
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