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  "body": "\n## Definition\n\nA function $k \\colon (0, \\infty) \\to \\R$ is *completely monotone* if it is infinitely differentiable and $(-1)^j k^{(j)}(t) \\ge 0$ for all integers $j \\ge 0$ and all $t \u003e 0$.\nLet $k \\colon (0, \\infty) \\to \\R$ and let $f$ be a given function on $[0, \\infty)$.\nConsider the [[Volterra equation of convolution type|SNFYTC1TTA]]\n$$x(t) + \\int_0^t k(t - s) \\, x(s) \\, ds = f(t), \\quad t \\ge 0,$$\nand its first-kind analogue $\\int_0^t k(t - s) \\, x(s) \\, ds = f(t)$.\nEach is a *Volterra equation with a completely monotone kernel* if:\n\n1. $k$ is completely monotone on $(0, \\infty)$;\n2. $k$ is integrable on $(0, 1)$, so that it is locally integrable on $[0, \\infty)$.\n\n## Remarks\n\nExample: $k(t) = t^{-1/2}$.\nIts derivatives are $k^{(j)}(t) = (-1)^j \\frac{1 \\cdot 3 \\cdots (2j - 1)}{2^j} t^{-1/2 - j}$, so $(-1)^j k^{(j)} \\ge 0$, and $\\int_0^1 t^{-1/2} \\, dt = 2$.\nSo [[Abel's integral equation|G4GWBYTW7B]] is a first-kind equation with a completely monotone kernel.\nNon-example: $k(t) = t$ is positive but increasing, so $(-1)^1 k'(t) = -1 \u003c 0$ and clause 1 fails.\nNon-example: $k(t) = 1/t$ is completely monotone but not integrable near $0$, so clause 2 fails.\nBy [[Bernstein's theorem|ZBS1JA62QP]], clause 1 holds exactly when $k(t) = \\int_{[0, \\infty)} e^{-st} \\, d\\mu(s)$ for a positive Borel measure $\\mu$ for which the integral converges for $t \u003e 0$.\nFor $k(t) = t^{-1/2}$ the measure is $d\\mu(s) = \\pi^{-1/2} s^{-1/2} \\, ds$.\n",
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