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  "body": "\n## Statement\n\nLet $f \\colon (0, \\infty) \\to \\C$ be measurable and let $g(x) = f(1/x)$ for $x \u003e 0$.\nLet $s \\in \\C$.\nThen $g$ is [[Mellin convergent|SGZZ4Y0D36]] at $s$ if and only if $f$ is Mellin convergent at $-s$.\nIn that case\n$$\\mathcal{M}g(s) = \\mathcal{M}f(-s),$$\nwhere $\\mathcal{M}$ is the [[Mellin transform|N5PVGRJE1T]].\n\n## Remarks\n\nThe proof is the substitution $u = 1/x$, with $dx = -u^{-2} \\, du$:\n$$\\int_0^\\infty x^{s-1} f(1/x) \\, dx = \\int_0^\\infty u^{1-s} f(u) \\, u^{-2} \\, du = \\int_0^\\infty u^{-s-1} f(u) \\, du.$$\nThe same computation with $\\abs{f}$ gives the equivalence of convergence.\nSo the strip of convergence is reflected in the imaginary axis: if $f$ converges for $\\alpha \u003c \\operatorname{Re} s \u003c \\beta$, then $g$ converges for $-\\beta \u003c \\operatorname{Re} s \u003c -\\alpha$.\nExample: $f(x) = e^{-x}$ gives $\\int_0^\\infty x^{s-1} e^{-1/x} \\, dx = \\Gamma(-s)$ for $\\operatorname{Re} s \u003c 0$.\n\n## Explanations\n\n### L1 human\n\nReplacing $x$ by $1/x$ swaps the roles of $0$ and $\\infty$.\nOn the Mellin side, this simply replaces $s$ by $-s$.\n\n### L2 human\n\nThis rule is used with functional equations of the form $\\theta(1/x) = \\sqrt{x}\\, \\theta(x)$, where the behaviour at $0$ is read from the behaviour at $\\infty$.\nIt also explains why the Mellin transform of a function and of its reflection have mirror strips of convergence.\nTogether with the rules for dilations and powers, it shows how the Mellin transform reacts to the maps $x \\mapsto a x^{b}$ with $a \u003e 0$ and $b \\ne 0$.\n",
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