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  "body": "\n## Summary\n\nA function $f$ on $(0, \\infty)$ has Mellin transform $m$ at a point $s$ when the integral $\\int_0^\\infty x^{s-1} f(x) \\, dx$ converges absolutely and its value is $m$.\nThe notion bundles the [[Mellin convergence|SGZZ4Y0D36]] of $f$ at $s$ with the value of the [[Mellin transform|N5PVGRJE1T]].\nIt is convenient in proofs, because a formula such as $\\mathcal{M}f(s) = \\Gamma(s)$ then carries its convergence claim with it.\n\n## Explanations\n\n### L1 human\n\nWriting $\\mathcal{M}f(s) = m$ quietly assumes that the integral makes sense.\nSaying that $f$ has Mellin transform $m$ at $s$ makes this explicit: the integral converges absolutely, and it equals $m$.\nFor example, $e^{-x}$ has Mellin transform $\\Gamma(2) = 1$ at $s = 2$, since $\\int_0^\\infty x e^{-x} \\, dx = 1$.\nThe constant function $1$ has no Mellin transform at any $s$, so it has Mellin transform $m$ for no pair $(s, m)$.\n",
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