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  "body": "\n## Statement\n\nLet $f \\colon (0, \\infty) \\to \\C$ be locally integrable, and let $x \u003e 0$ satisfy\n$$\\int_0^\\infty \\frac{\\abs{f(t)}}{x + t}\\, dt \u003c \\infty.$$\nLet $(\\mathcal{L} f)(s) = \\int_0^\\infty e^{-st} f(t)\\, dt$ denote the [[Laplace transform|0V49JEZW06]].\nThen $(\\mathcal{L} f)(s)$ converges absolutely for every $s \u003e 0$, the function $\\mathcal{L} f$ is integrable against $e^{-xs}\\, ds$ on $(0, \\infty)$, and\n$$\\big(\\mathcal{L}(\\mathcal{L} f)\\big)(x) = \\int_0^\\infty \\frac{f(t)}{x + t}\\, dt,$$\nthe [[Stieltjes transform|RG1HFAYD98]] of $f$ at $x$.\n\n## Remarks\n\nThe proof is one application of Fubini's theorem.\nFor $x, t \u003e 0$ one has $\\int_0^\\infty e^{-xs} e^{-st}\\, ds = \\frac{1}{x + t}$.\nSo $\\int_0^\\infty \\int_0^\\infty e^{-xs} e^{-st} \\abs{f(t)}\\, dt\\, ds = \\int_0^\\infty \\frac{\\abs{f(t)}}{x + t}\\, dt \u003c \\infty$ by Tonelli, and Fubini allows the exchange of the two integrals.\nThe inner integral $\\int_0^\\infty e^{-st} \\abs{f(t)}\\, dt$ is then finite for almost every $s$, and since it decreases in $s$ it is finite for every $s \u003e 0$.\nThe hypothesis cannot be dropped: for $f = 1$ the right side diverges for every $x$, while $\\mathcal{L} f(s) = 1/s$ is not integrable against $e^{-xs}$ near $s = 0$.\n\nExample: $f(t) = t^{-1/2}$.\nThen $\\mathcal{L} f(s) = \\sqrt{\\pi / s}$ and $\\mathcal{L}(\\mathcal{L} f)(x) = \\sqrt{\\pi}\\, \\Gamma(1/2)\\, x^{-1/2} = \\pi\\, x^{-1/2}$.\nDirectly, the substitution $t = x u^2$ gives $\\int_0^\\infty \\frac{t^{-1/2}}{x + t}\\, dt = \\frac{2}{\\sqrt x} \\int_0^\\infty \\frac{du}{1 + u^2} = \\frac{\\pi}{\\sqrt x}$.\n\n## Explanations\n\n### L1 human\n\nApplying the Laplace transform twice gives another well-known transform, the Stieltjes transform.\nThe reason is the elementary integral $\\int_0^\\infty e^{-(x+t)s}\\, ds = 1/(x + t)$.\n\n### L2 human\n\nThis identity lets one move facts between the two transforms.\nProperties of the Stieltjes transform, such as analyticity in $x$ and its behaviour as $x \\to 0^+$ or $x \\to \\infty$, follow from the corresponding properties of the Laplace transform applied twice.\nConversely, inversion formulas for the [[Laplace transform|0V49JEZW06]], such as [[Post's inversion formula|D6QKPBS2EN]], can be applied twice to recover $f$ from its Stieltjes transform.\n",
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