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  "body": "\n## Definition\n\nLet $c \\in \\R$ and let $\\varphi$ be a complex function defined on the line $\\operatorname{Re} s = c$.\nAssume that $y \\mapsto \\varphi(c + iy)$ is [[integrable|X7GTF3D98F]] on $\\R$.\nFor $x \u003e 0$ and $s \\in \\C$, let $x^{-s} = e^{-s \\log x}$ be the [[complex power|0RNGDMHT7M]].\nThe *inverse Mellin transform* of $\\varphi$ along the line $\\operatorname{Re} s = c$ is the function on $(0, \\infty)$ given by\n$$\\mathcal{M}^{-1}_c \\varphi(x) = \\frac{1}{2\\pi i} \\int_{c - i\\infty}^{c + i\\infty} x^{-s} \\varphi(s) \\, ds = \\frac{1}{2\\pi} \\int_{-\\infty}^{\\infty} x^{-c - iy} \\varphi(c + iy) \\, dy.$$\n\n## Remarks\n\nThe [[Lebesgue integral|DWJF7ABPR1]] converges absolutely for every $x \u003e 0$, because $\\abs{x^{-c - iy}} = x^{-c}$.\nThe result usually depends on $c$.\nWhen $\\varphi$ is holomorphic on a vertical strip and decays fast enough, Cauchy's theorem shows that it does not change as $c$ moves inside the strip.\nExample: $\\varphi = \\Gamma$ and $c \u003e 0$.\nStirling's formula gives $\\abs{\\Gamma(c + iy)} = O(\\abs{y}^{c - 1/2} e^{-\\pi \\abs{y}/2})$, so $\\varphi$ is integrable on the line, and $\\mathcal{M}^{-1}_c \\Gamma(x) = e^{-x}$.\nNon-example: $\\varphi(s) = 1/s$ and $c \u003e 0$.\nThen $\\abs{\\varphi(c + iy)} = (c^2 + y^2)^{-1/2}$ is not integrable on $\\R$, so the definition does not apply; the integral exists only as a limit of integrals over $[c - iT, c + iT]$.\nIt is common to drop the index $c$ and write $\\mathcal{M}^{-1} \\varphi$.\n",
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    "name": "Inverse Mellin transform",
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        "url": "https://en.wikipedia.org/wiki/Mellin_inversion_theorem"
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  "msc": [
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  "name": "Inverse Mellin transform",
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