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  "body": "\n## Statement\n\nLet $f \\colon (0, \\infty) \\to \\C$ be measurable and let $s = \\sigma + i\\tau$ with $\\sigma, \\tau \\in \\R$.\nDefine $h_\\sigma \\colon \\R \\to \\C$ by $h_\\sigma(u) = e^{-\\sigma u} f(e^{-u})$.\nThen $f$ is [[Mellin convergent|SGZZ4Y0D36]] at $s$ if and only if $h_\\sigma$ is [[integrable|X7GTF3D98F]] on $\\R$.\nIn that case\n$$\\mathcal{M}f(\\sigma + i\\tau) = \\widehat{h_\\sigma}\\left(\\frac{\\tau}{2\\pi}\\right),$$\nwhere $\\mathcal{M}f$ is the [[Mellin transform|N5PVGRJE1T]] and $\\widehat{h}(\\xi) = \\int_{\\R} h(u) e^{-2\\pi i u \\xi} \\, du$ is the [[Fourier transform|DPVM34ADZV]].\n\n## Remarks\n\nThe proof is the substitution $x = e^{-u}$, with $dx = -e^{-u} \\, du$, which maps $(0, \\infty)$ onto $\\R$ with reversed orientation.\nIt gives $\\int_0^\\infty x^{\\sigma - 1} \\abs{f(x)} \\, dx = \\int_{\\R} \\abs{h_\\sigma(u)} \\, du$, and\n$$\\int_0^\\infty x^{s-1} f(x) \\, dx = \\int_{\\R} e^{-su} f(e^{-u}) \\, du = \\int_{\\R} h_\\sigma(u) e^{-i\\tau u} \\, du.$$\nExample: $f(x) = e^{-x}$ and $\\sigma = 1$ give $h_1(u) = e^{-u} e^{-e^{-u}}$, and the statement says $\\widehat{h_1}(\\tau/2\\pi) = \\Gamma(1 + i\\tau)$.\nOther sign conventions use $x = e^{u}$, which replaces $\\tau / 2\\pi$ by $-\\tau / 2\\pi$.\nFor fixed $\\sigma$, the map $f \\mapsto h_\\sigma$ is a bijection, so every theorem on the Fourier transform on $\\R$ has a Mellin version on each vertical line.\n\n## Explanations\n\n### L1 human\n\nOn a logarithmic scale, multiplication becomes addition, and the powers $x^{s}$ become exponentials $e^{su}$.\nSo the Mellin transform, which integrates against powers, becomes the Fourier transform, which integrates against exponentials.\nThe real part of $s$ only adds a fixed weight $e^{-\\sigma u}$ to the function.\n\n### L2 human\n\nThis dictionary transports results from Fourier analysis.\nThe Riemann–Lebesgue lemma gives the decay of $\\mathcal{M}f(\\sigma + i\\tau)$ as $\\abs{\\tau} \\to \\infty$.\nThe Fourier inversion theorem and the Plancherel theorem give the Mellin inversion theorem and the Parseval formula for Mellin transforms.\nVarying $\\sigma$ as well, the same substitution identifies the Mellin transform with a two-sided Laplace transform of $u \\mapsto f(e^{-u})$.\nIt also identifies the strip of Mellin convergence with the strip of convergence of that Laplace transform.\n",
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