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  "body": "\n## Example\n\nLet $f(x) = e^{-x}$ for $x \u003e 0$.\nIts [[Mellin transform|N5PVGRJE1T]] is the [[gamma function|0H5SYBSED9]]:\n$$\\mathcal{M}f(s) = \\int_0^\\infty x^{s-1} e^{-x} \\, dx = \\Gamma(s), \\qquad \\operatorname{Re} s \u003e 0.$$\nThe right side is [[Euler's integral|Z8D1CZFTWP]], which is the usual definition of $\\Gamma$ on the half-plane $\\operatorname{Re} s \u003e 0$.\n\nCheck of [[Mellin convergence|SGZZ4Y0D36]]: let $\\sigma = \\operatorname{Re} s$.\nOn $(0, 1]$, $x^{\\sigma - 1} e^{-x} \\le x^{\\sigma - 1}$, which is integrable exactly when $\\sigma \u003e 0$.\nOn $[1, \\infty)$, $x^{\\sigma - 1} e^{-x} \\le C_\\sigma e^{-x/2}$ for a constant $C_\\sigma$, which is integrable for every $\\sigma$.\nFor $\\sigma \\le 0$, $x^{\\sigma - 1} e^{-x} \\ge e^{-1} x^{\\sigma - 1}$ on $(0, 1]$, which is not integrable.\nSo $f$ is Mellin convergent exactly on the half-plane $\\operatorname{Re} s \u003e 0$.\n\nIntegration by parts gives $\\Gamma(s + 1) = s \\Gamma(s)$, and $\\Gamma(1) = \\int_0^\\infty e^{-x} \\, dx = 1$, so $\\Gamma(n + 1) = n!$.\nBy the dilation rule, $\\int_0^\\infty x^{s-1} e^{-ax} \\, dx = a^{-s} \\Gamma(s)$ for $a \u003e 0$.\nBy the [[Mellin inversion theorem|XJWC13NY9E]], for every $c \u003e 0$ and $x \u003e 0$,\n$$e^{-x} = \\frac{1}{2\\pi i} \\int_{c - i\\infty}^{c + i\\infty} \\Gamma(s) x^{-s} \\, ds.$$\n\n## Explanations\n\n### L1 human\n\nThe gamma function, which extends the factorial, is the simplest Mellin transform.\nOne weighs the decaying exponential $e^{-x}$ against the powers $x^{s-1}$ and integrates.\nAt $s = 1, 2, 3, \\dots$ the answers are $0!, 1!, 2!, \\dots$.\nThe integral only makes sense when $\\operatorname{Re} s \u003e 0$, because otherwise the power blows up too fast at $0$.\n",
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