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  "body": "\n## Definition\n\nLet $f \\colon (0, \\infty) \\to \\C$ be measurable and let $s \\in \\C$.\nFor $x \u003e 0$, let $x^{s-1} = e^{(s-1)\\log x}$ be the [[complex power|0RNGDMHT7M]] with the real logarithm.\nAssume that $f$ is [[Mellin convergent|SGZZ4Y0D36]] at $s$, that is, $x \\mapsto x^{s-1} f(x)$ is integrable on $(0, \\infty)$.\nThe *Mellin transform* of $f$ at $s$ is the [[Lebesgue integral|DWJF7ABPR1]]\n$$\\mathcal{M}f(s) = \\int_0^\\infty x^{s-1} f(x) \\, dx.$$\nAs a function, $\\mathcal{M}f$ is defined on the set of $s$ at which $f$ is Mellin convergent.\n\n## Remarks\n\nSince $\\abs{x^{s-1}} = x^{\\operatorname{Re} s - 1}$, this set depends only on $\\operatorname{Re} s$ and is a union of vertical lines; it is a vertical strip, possibly empty or degenerate.\nExample: $f(x) = e^{-x}$.\nNear $0$, $\\abs{x^{s-1} e^{-x}} \\le x^{\\operatorname{Re} s - 1}$, which is integrable on $(0, 1)$ when $\\operatorname{Re} s \u003e 0$; near $\\infty$, the factor $e^{-x}$ beats every power.\nSo $\\mathcal{M}f(s) = \\Gamma(s)$ for $\\operatorname{Re} s \u003e 0$, the [[gamma function|0H5SYBSED9]].\nNon-example: $f(x) = 1$.\nIntegrability of $x^{\\operatorname{Re} s - 1}$ on $(0, 1)$ needs $\\operatorname{Re} s \u003e 0$, and on $(1, \\infty)$ it needs $\\operatorname{Re} s \u003c 0$, so $f$ is Mellin convergent at no $s$.\nOther notations are $\\tilde f(s)$, $F(s)$ and $\\set{\\mathcal{M} f}(s)$.\nSome authors call the open strip of convergence the fundamental strip and write it $\\langle \\alpha, \\beta \\rangle$.\n",
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