Completely monotone function

Remarks

Example: $f(x) = 1/x$ is completely monotone, since $f^{(k)}(x) = (-1)^k k! \, x^{-k-1}$ and so $(-1)^k f^{(k)}(x) = k! \, x^{-k-1} > 0$. Non-example: $g(x) = e^{-x^2}$ satisfies the conditions for $k = 0$ and $k = 1$, but $g''(x) = (4 x^2 - 2) e^{-x^2} < 0$ for $0 < x < 1/\sqrt{2}$, so the condition fails for $k = 2$. Sums, products and positive multiples of completely monotone functions are completely monotone. The same definition is used on other open intervals; on $[0, \infty)$ one also asks that $f$ be continuous at $0$. The function $f$ is completely monotone on $(0, \infty)$ if and only if $x \mapsto f(-x)$ is absolutely monotonic on $(-\infty, 0)$.