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  "body": "\n## Statement\n\nLet $f \\colon (0, \\infty) \\to \\C$ be integrable on every compact subinterval $[u, v] \\subset (0, \\infty)$.\nLet $a, b \\in \\R$ with $b \u003c a$, and assume that\n$$f(x) = O(x^{-a}) \\text{ as } x \\to \\infty, \\qquad f(x) = O(x^{-b}) \\text{ as } x \\to 0^+.$$\nThen $f$ is [[Mellin convergent|SGZZ4Y0D36]] at every $s \\in \\C$ with $b \u003c \\operatorname{Re} s \u003c a$.\nSo the [[Mellin transform|N5PVGRJE1T]] $\\mathcal{M}f(s)$ is defined on the open strip $b \u003c \\operatorname{Re} s \u003c a$.\n\nHere $O$ is [[big O notation|0TGV80E6A4]].\n\n## Remarks\n\nLet $\\sigma = \\operatorname{Re} s$.\nChoose $0 \u003c \\delta \u003c 1 \u003c R$ with $\\abs{f(x)} \\le C x^{-b}$ on $(0, \\delta]$ and $\\abs{f(x)} \\le C x^{-a}$ on $[R, \\infty)$.\nThen $x^{\\sigma - 1} \\abs{f(x)} \\le C x^{\\sigma - b - 1}$ on $(0, \\delta]$, which is integrable since $\\sigma - b \u003e 0$.\nAlso $x^{\\sigma - 1} \\abs{f(x)} \\le C x^{\\sigma - a - 1}$ on $[R, \\infty)$, which is integrable since $\\sigma - a \u003c 0$.\nOn $[\\delta, R]$, $f$ is integrable and $x^{\\sigma - 1}$ is bounded.\n\nThe largest open strip $\\alpha \u003c \\operatorname{Re} s \u003c \\beta$ on which $f$ is Mellin convergent is called the fundamental strip and is written $\\langle \\alpha, \\beta \\rangle$.\nExample: $f(x) = e^{-x}$ is $O(x^0)$ at $0$ and $O(x^{-a})$ at $\\infty$ for every $a$, so its fundamental strip is $\\langle 0, \\infty \\rangle$, where $\\mathcal{M}f = \\Gamma$.\nExample: $f(x) = 1/(1 + x)$ is $O(1)$ at $0$ and $O(x^{-1})$ at $\\infty$, giving the strip $\\langle 0, 1 \\rangle$.\nThe strict inequalities are needed: for this $f$ at $s = 1$, the bound $x^{\\sigma - a - 1} = x^{-1}$ is not integrable at $\\infty$, and indeed $\\int_0^\\infty dx/(1+x)$ diverges.\n\n## Explanations\n\n### L1 human\n\nThe Mellin integral has two dangerous ends, $0$ and $\\infty$.\nAssume that $f$ grows at most like $x^{-b}$ near $0$ and decays at least like $x^{-a}$ near $\\infty$.\nThen both ends are safe for every $s$ whose real part lies strictly between $b$ and $a$.\nSo the transform lives on a vertical strip whose edges are read off from the behaviour of $f$ at the two ends.\n\n### L2 human\n\nThis is the first step in every computation with Mellin transforms: one finds the strip before anything else.\nThe edges of the strip are tied to the asymptotics of $f$.\nConversely, under suitable conditions, poles of the continued transform beyond the strip give terms of the asymptotic expansion of $f$ at $0$ or at $\\infty$.\nOn the open strip the transform is also holomorphic, which is the companion result on differentiability of the Mellin transform.\nUnder the substitution $x = e^{-u}$, the statement becomes the familiar fact that a function with exponential decay at both ends of $\\R$ has a two-sided Laplace transform on a strip.\n",
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