{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Statement\n\nLet $f \\colon [0, \\infty) \\to \\C$ be locally integrable, and let $F(s) = \\int_0^\\infty f(t) e^{-st}\\, dt$ be its [[Laplace transform|05XT5AVJBK]].\nLet $c \\in \\C$.\n\n1. (Behaviour at infinity.) If $\\frac{1}{T} \\int_0^T f(t)\\, dt \\to c$ as $T \\to \\infty$, then $F(s)$ converges for every real $s \u003e 0$, and $s F(s) \\to c$ as $s \\to 0^+$.\nIn particular this holds when $f(t) \\to c$ as $t \\to \\infty$.\n2. (Behaviour at zero.) If $\\int_0^\\infty \\abs{f(t)} e^{-s_0 t}\\, dt \u003c \\infty$ for some real $s_0$, and $f(t) \\to c$ as $t \\to 0^+$, then $s F(s) \\to c$ as $s \\to +\\infty$ through real values.\n\n## Remarks\n\nBoth parts compare $f$ with the constant $c$, using $s \\int_0^\\infty e^{-st}\\, dt = 1$ for $s \u003e 0$.\nFor part 2, write $sF(s) - c = s\\int_0^\\infty (f(t) - c) e^{-st}\\, dt$; the part of the integral over $[0, \\delta]$ is small because $f$ is close to $c$ there, and the rest tends to $0$ by [[dominated convergence|BYNV84PF86]].\nFor part 1, integrate by parts with $U(T) = \\int_0^T f$ to get $sF(s) = s^2 \\int_0^\\infty U(T) e^{-sT}\\, dT$, and use $s^2 \\int_0^\\infty T e^{-sT}\\, dT = 1$.\nExample: for $f(t) = 1 + e^{-t}$, $F(s) = 1/s + 1/(s + 1)$, so $sF(s) \\to 1 = \\lim_{t \\to \\infty} f(t)$ as $s \\to 0^+$, and $sF(s) \\to 2 = f(0)$ as $s \\to \\infty$.\nThe converses are false without extra conditions; they are the [[Tauberian theorems|NV54MWKT6T]].\n\n## Explanations\n\n### L1 human\n\nThese results say that the Laplace transform remembers the limits of $f$.\nThe long-run value of $f$ shows up in $sF(s)$ for small $s$, and the starting value of $f$ shows up in $sF(s)$ for large $s$.\nThis is the integral analogue of Abel's theorem on power series, which is where the name \"Abelian\" comes from.\n\n### L2 human\n\nIn engineering these are the final value theorem and the initial value theorem.\nThey read off the limits of a solution of a linear differential equation from its transform without inverting it.\nPart 1 is the Laplace analogue of the fact that Cesàro summability implies Abel summability.\nAnalogous Abelian theorems hold for other transforms, such as the Stieltjes and Mellin transforms: an asymptotic behaviour of the function gives a matching asymptotic behaviour of the transform.\n",
  "commit": "ff8a40749dc16545e3d2fc637f6b0196e81d0bfa",
  "front": {
    "about": [
      "0V49JEZW06",
      "D6XWK1NRY1"
    ],
    "kind": "statement",
    "msc": [
      "44A10",
      "40E05"
    ],
    "name": "Abelian theorems for integral transforms",
    "role": "theorem",
    "sources": [
      {
        "url": "https://en.wikipedia.org/wiki/Abelian_and_Tauberian_theorems"
      }
    ],
    "tag": "16W9PP8ET0",
    "uses": [
      "05XT5AVJBK",
      "D6XWK1NRY1",
      "BYNV84PF86"
    ]
  },
  "kind": "statement",
  "link": "https://tamnd.github.io/math2/node/16W9PP8ET0/",
  "links": [
    {
      "field": "about",
      "target": "0V49JEZW06"
    },
    {
      "field": "about",
      "target": "D6XWK1NRY1"
    },
    {
      "field": "uses",
      "target": "05XT5AVJBK"
    },
    {
      "field": "uses",
      "target": "D6XWK1NRY1"
    },
    {
      "field": "uses",
      "target": "BYNV84PF86"
    }
  ],
  "markdown": "https://tamnd.github.io/math2/node/16W9PP8ET0.md",
  "msc": [
    "44A10",
    "40E05"
  ],
  "name": "Abelian theorems for integral transforms",
  "prov": null,
  "role": "theorem",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/16W/16W9PP8ET0.md",
  "status": null,
  "tag": "16W9PP8ET0",
  "uri": "m2://16W9PP8ET0",
  "url": "https://tamnd.github.io/math2/wiki/abelian-theorems-for-integral-transforms/",
  "writing": {
    "tag": "16W9PP8ET0",
    "kind": "statement",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      },
      {
        "what": "a source with a locator",
        "have": 0,
        "want": 1
      }
    ]
  }
}