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  "body": "\n## Statement\n\nLet $\\mu$ be a measure on $\\R$ of the form $d\\mu(x) = f(x) \\, dx$, where $f \\ge 0$ is Lebesgue integrable.\nSuppose that every [[moment|01XAHPADBA]] $m_n = \\int x^n f(x) \\, dx$ is finite.\nIf\n$$\\int_{-\\infty}^{\\infty} \\frac{-\\ln f(x)}{1 + x^2} \\, dx \u003c \\infty,$$\nthen the [[Hamburger moment problem|TBNCFXTMMW]] for $\\mu$ is [[indeterminate|GCTN4BZ1ET]]: there is a positive measure $\\nu \\ne \\mu$ on $\\R$ with the same moments.\n\n## Remarks\n\nThe integral is well defined in $(-\\infty, \\infty]$, since $\\ln f \\le f$ where $f \\ge 1$ and $f$ is integrable.\nThe result is a corollary of a density theorem of Krein: the integral is infinite if and only if the exponential sums $\\sum_k a_k e^{i \\lambda_k x}$ with $\\lambda_k \\ge 0$ are dense in $L^2(\\mu)$.\nExample: $f(x) = c \\, e^{-\\abs{x}^{\\alpha}}$ with $0 \u003c \\alpha \u003c 1$ and $c \u003e 0$ has $-\\ln f(x) = \\abs{x}^{\\alpha} - \\ln c$, and $\\int \\abs{x}^{\\alpha}/(1 + x^2) \\, dx \u003c \\infty$ for $\\alpha \u003c 1$, so this density is not determined by its moments.\nFor $\\alpha \\ge 1$ the integral diverges, and in fact Carleman's condition shows that the moment problem is determinate.\nThe condition is sufficient for indeterminacy but not necessary.\n\n## Explanations\n\n### L1 human\n\nA density whose tails are very heavy, but not so heavy that the moments are infinite, can share all its moments with another distribution.\nKrein's condition measures heaviness of the tails by how slowly $-\\ln f$ grows.\nIf $-\\ln f(x)$ grows more slowly than $\\abs{x}$, in the averaged sense of the integral, the moments do not determine the distribution.\n\n### L2 human\n\nIt complements [[Carleman's condition|RVZ831CNYW]], which gives determinacy when the moments grow slowly.\nThe two meet at densities like $e^{-\\abs{x}^{\\alpha}}$: determinate for $\\alpha \\ge 1$ and indeterminate for $\\alpha \u003c 1$.\nVersions for the Stieltjes problem on $[0, \\infty)$ and converses under regularity hypotheses on $f$ are known.\n",
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