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  "body": "\n## Definition\n\nLet $f \\colon (0, \\infty) \\to \\R$ be [[infinitely differentiable|0V33BNKZ9S]], and let $f^{(k)}$ be its [[derivative of order $k$|04TKTF8KDK]], with $f^{(0)} = f$.\nThe function $f$ is *completely monotone* if\n$$(-1)^k f^{(k)}(x) \\ge 0$$\nfor every integer $k \\ge 0$ and every $x \u003e 0$.\n\n## Remarks\n\nExample: $f(x) = 1/x$ is completely monotone, since $f^{(k)}(x) = (-1)^k k! \\, x^{-k-1}$ and so $(-1)^k f^{(k)}(x) = k! \\, x^{-k-1} \u003e 0$.\nNon-example: $g(x) = e^{-x^2}$ satisfies the conditions for $k = 0$ and $k = 1$, but $g''(x) = (4 x^2 - 2) e^{-x^2} \u003c 0$ for $0 \u003c x \u003c 1/\\sqrt{2}$, so the condition fails for $k = 2$.\nSums, products and positive multiples of completely monotone functions are completely monotone.\nThe same definition is used on other open intervals; on $[0, \\infty)$ one also asks that $f$ be continuous at $0$.\nThe function $f$ is completely monotone on $(0, \\infty)$ if and only if $x \\mapsto f(-x)$ is [[absolutely monotonic|QQSPWH4NG9]] on $(-\\infty, 0)$.\n",
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