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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]].\nThe function $f$ is *Mellin convergent at $s$* if the function $x \\mapsto x^{s-1} f(x)$ is [[integrable|X7GTF3D98F]] on $(0, \\infty)$ with respect to Lebesgue measure, that is,\n$$\\int_0^\\infty x^{\\operatorname{Re} s - 1} \\abs{f(x)} \\, dx \u003c \\infty.$$\n\n## Remarks\n\nThe two forms agree because $\\abs{x^{s-1}} = x^{\\operatorname{Re} s - 1}$.\nSo Mellin convergence at $s$ depends only on $\\operatorname{Re} s$.\nIf $f$ is Mellin convergent at $\\sigma_1$ and $\\sigma_2$ with $\\sigma_1 \u003c \\sigma_2$, then it is Mellin convergent at every $s$ with $\\sigma_1 \\le \\operatorname{Re} s \\le \\sigma_2$, because $x^{\\sigma - 1} \\le x^{\\sigma_1 - 1} + x^{\\sigma_2 - 1}$ for $\\sigma_1 \\le \\sigma \\le \\sigma_2$.\nExample: $f(x) = 1/(1 + x)$ at $s = 1/2$.\nHere $\\int_0^\\infty x^{-1/2}/(1 + x) \\, dx = \\pi$, by the substitution $x = t^2$, so $f$ is Mellin convergent at $1/2$; the same estimates show it is Mellin convergent exactly when $0 \u003c \\operatorname{Re} s \u003c 1$.\nNon-example: the same $f$ at $s = 1$.\nThe integral $\\int_0^\\infty dx/(1 + x)$ diverges at $\\infty$, so $f$ is not Mellin convergent at $1$.\n",
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  "msc": [
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  "name": "Mellin convergence",
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