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  "body": "\n## Definition\n\nLet $f \\colon (0, \\infty) \\to \\C$ be a function whose [[Mellin transform|N5PVGRJE1T]] $f^*(s) = \\int_0^\\infty f(x)\\, x^{s-1}\\, dx$ converges absolutely in a [[fundamental strip|1JAA21FHHQ]] $\\alpha \u003c \\operatorname{Re} s \u003c \\beta$.\nLet $\\gamma \u003c \\alpha$, and assume that $f^*$ extends to a [[meromorphic function|0ZZ7ZYDR8W]] on the strip $\\gamma \u003c \\operatorname{Re} s \u003c \\beta$ with finitely many [[poles|066E7PJCYE]] $\\rho_1, \\dots, \\rho_m$ there.\nFor each pole $\\rho$, write the principal part of $f^*$ at $\\rho$ as\n$$\\sum_{k=0}^{K} \\frac{c_{\\rho, k}}{(s - \\rho)^{k+1}}.$$\nThe *pole expansion of $f$ at $0$* is\n$$\\sum_{j=1}^{m} \\sum_{k} c_{\\rho_j, k}\\, \\frac{(-1)^k}{k!}\\, x^{-\\rho_j} (\\log x)^k,$$\nwhere the inner sum runs over $0 \\le k \\le K$ for each pole.\nIt is the sum of the [[residues|0PHN07KVQT]] of $f^*(s)\\, x^{-s}$ at $\\rho_1, \\dots, \\rho_m$.\nThe expansion at $\\infty$ is defined in the same way from the poles to the right of the strip, with the opposite sign.\n\n## Remarks\n\nThe justification is the converse mapping theorem.\nAssume moreover that $f^*(s) = O(\\abs{s}^{-r})$ for some $r \u003e 1$ as $\\abs{\\operatorname{Im} s} \\to \\infty$, uniformly in the closed strip $\\gamma \\le \\operatorname{Re} s \\le c$ for some $c$ with $\\alpha \u003c c \u003c \\beta$.\nThen, as $x \\to 0^+$, $f(x)$ equals the sum above plus $O(x^{-\\gamma})$, so this sum is an [[asymptotic expansion|BJQTEPEZ9G]] of $f$ to that order.\nThe proof writes $f$ by the [[Mellin inversion theorem|XJWC13NY9E]] as an integral over the line $\\operatorname{Re} s = c$, moves the line to $\\operatorname{Re} s = \\gamma$, and collects residues.\nThe residue of $x^{-s} (s - \\rho)^{-k-1}$ at $\\rho$ is $x^{-\\rho} (-\\log x)^k / k!$, which gives the coefficients above.\n\nExample: $f(x) = e^{-x}$ has $f^*(s) = \\Gamma(s)$ on $\\operatorname{Re} s \u003e 0$.\nThe poles at $s = -n$ are simple with residue $(-1)^n / n!$, and $\\Gamma$ decays exponentially on vertical lines.\nSo for every $N$ one gets $e^{-x} = \\sum_{n=0}^{N} (-1)^n x^n / n! + O(x^{N+1/2})$ as $x \\to 0^+$, as Taylor's formula confirms.\n\nNon-example: $f(x) = 1$ for $0 \u003c x \\le 1$ and $f(x) = 0$ for $x \u003e 1$ has $f^*(s) = 1/s$ on $\\operatorname{Re} s \u003e 0$.\nThe pole at $0$ gives the term $1$, which is correct for $x \u003c 1$, but $1/s$ decays only like $\\abs{s}^{-1}$, so the decay hypothesis fails.\nThe jump of $f$ at $x = 1$ is what spoils the decay.\n",
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