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  "body": "\n## Definition\n\nLet $f \\colon (0, \\infty) \\to \\C$ be measurable, let $s \\in \\C$ and let $m \\in \\C$.\nThe function $f$ *has Mellin transform $m$ at $s$* if both conditions hold:\n\n1. $f$ is [[Mellin convergent|SGZZ4Y0D36]] at $s$, that is, $x \\mapsto x^{s-1} f(x)$ is integrable on $(0, \\infty)$.\n2. The [[Mellin transform|N5PVGRJE1T]] of $f$ at $s$ equals $m$: $\\int_0^\\infty x^{s-1} f(x) \\, dx = m$.\n\n## Remarks\n\nExample: $f = \\mathbf{1}_{(0, 1]}$, the indicator function of $(0, 1]$, has Mellin transform $1/s$ at every $s$ with $\\operatorname{Re} s \u003e 0$.\nIndeed $\\int_0^1 x^{\\operatorname{Re} s - 1} \\, dx = 1/\\operatorname{Re} s \u003c \\infty$, and $\\int_0^1 x^{s-1} \\, dx = 1/s$.\nNon-example: the same $f$ at $s = 0$ fails clause 1, because $\\int_0^1 x^{-1} \\, dx$ diverges.\nSo $f$ does not have Mellin transform $m$ at $0$ for any $m$, although the function $1/s$ has a pole there and no value.\nIn Mathlib this predicate is `HasMellin f s m`.\n",
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