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  "body": "\n## Example\n\nFor $x \u003e 0$ let $\\psi(x) = \\sum_{n=1}^{\\infty} e^{-\\pi n^2 x}$.\nWith the [[theta function|2F5CYQJVAF]] $\\theta(x) = \\sum_{n \\in \\Z} e^{-\\pi n^2 x}$, one has $\\psi(x) = (\\theta(x) - 1)/2$.\nFor $\\operatorname{Re} s \u003e 1$, the [[Mellin transform|N5PVGRJE1T]] of $\\psi$ at $s/2$ is\n$$\\int_0^\\infty x^{s/2 - 1} \\psi(x) \\, dx = \\pi^{-s/2} \\, \\Gamma\\left(\\frac{s}{2}\\right) \\zeta(s),$$\nwhere $\\Gamma$ is the [[gamma function|0H5SYBSED9]] and $\\zeta$ is the [[Riemann zeta function|1XWBY4N0VT]].\n\nCheck: for each $n \\ge 1$ the dilation rule gives\n$$\\int_0^\\infty x^{s/2 - 1} e^{-\\pi n^2 x} \\, dx = (\\pi n^2)^{-s/2} \\, \\Gamma\\left(\\frac{s}{2}\\right) = \\pi^{-s/2} \\, \\Gamma\\left(\\frac{s}{2}\\right) n^{-s}.$$\nLet $\\sigma = \\operatorname{Re} s \u003e 1$.\nThe sum over $n$ of the integrals of the absolute values is $\\pi^{-\\sigma/2} \\Gamma(\\sigma/2) \\sum_n n^{-\\sigma} \u003c \\infty$.\nSo, by Fubini's theorem, the sum and the integral may be exchanged, and the sum of the right sides is $\\pi^{-s/2} \\Gamma(s/2) \\zeta(s)$.\n\nThe series $\\psi$ is Mellin convergent at $s/2$ exactly for $\\operatorname{Re} s \u003e 1$.\nNear $\\infty$, $\\psi(x) \\le C e^{-\\pi x}$; near $0$, $\\psi(x) \\sim \\frac{1}{2} x^{-1/2}$, which follows from the identity $\\theta(1/x) = \\sqrt{x}\\, \\theta(x)$.\n\n## Explanations\n\n### L1 human\n\nThe theta series adds up Gaussian bumps $e^{-\\pi n^2 x}$, one for each whole number $n$.\nTaking its Mellin transform turns each bump into a multiple of $n^{-s}$, so the whole series turns into the zeta function, times a gamma factor.\nThis is how Riemann connected the symmetry $\\theta(1/x) = \\sqrt{x}\\, \\theta(x)$ of the theta function with the [[functional equation of the zeta function|NFXKNDJX9C]].\nSplitting the integral at $x = 1$ and using that symmetry gives the analytic continuation of $\\zeta$ and the symmetry under $s \\mapsto 1 - s$.\n",
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