{
  "aliases": null,
  "backlinks": [
    {
      "field": "uses",
      "target": "DBM13YSJAM"
    },
    {
      "field": "uses",
      "target": "Z22Y9N8EHN"
    }
  ],
  "body": "\n## Statement\n\nLet $f$ be a locally integrable function on $[0, \\infty)$.\nLet $h$ be the constant function $h(t) = 1$ on $[0, \\infty)$, the restriction of the [[Heaviside step function|09C4JP7Y0M]].\nWrite $f * g$ for the [[convolution on the half-line|FCKX8KFVGN]], $(f * g)(t) = \\int_0^t f(\\tau) \\, g(t - \\tau) \\, d\\tau$.\nThen:\n\n1. $(h * f)(t) = \\int_0^t f(\\tau) \\, d\\tau$ for all $t \\ge 0$.\n2. For every integer $n \\ge 1$, the $n$-fold convolution power of $h$ is $h^{*n}(t) = \\dfrac{t^{n-1}}{(n-1)!}$.\n3. For every integer $n \\ge 1$, $(h^{*n} * f)(t) = \\displaystyle \\int_0^t \\frac{(t - \\tau)^{n-1}}{(n-1)!} f(\\tau) \\, d\\tau$, and this is the $n$-fold iterated integral of $f$ from $0$ to $t$.\n\n## Remarks\n\nClause 1 is the definition of convolution with $h(t - \\tau) = 1$ for $0 \\le \\tau \\le t$.\nClause 2 follows by induction, since $\\int_0^t \\tau^{n-1}/(n-1)! \\, d\\tau = t^n / n!$.\nClause 3 follows from clauses 1 and 2 and the associativity of convolution; it is the Cauchy formula for repeated integration.\nFor example, with $f(t) = \\cos t$, one gets $(h * f)(t) = \\sin t$ and $(h * h * f)(t) = 1 - \\cos t$.\nIn the [[Mikusiński operational calculus|TD5XK5AHNV]] this result lets one treat integration as multiplication by the element $h$ of a commutative ring.\n\n## Explanations\n\n### L1 human\n\nIntegrating a function from $0$ to $t$ is the same as convolving it with the function that is constantly $1$.\nSo the operation \"integrate\" becomes multiplication by a fixed object in a ring whose product is convolution.\nIntegrating twice is multiplication by $h * h$, which is the function $t$.\n\n### L2 human\n\nThis is the starting point of Mikusiński's calculus.\nThe convolution ring of functions on $[0, \\infty)$ has no zero divisors by the Titchmarsh convolution theorem, so it has a field of fractions.\nIn that field $h$ has an inverse $s$, which plays the role of differentiation.\nUnder the Laplace transform, $h$ corresponds to $1/s$, in line with the rule that the transform of $\\int_0^t f$ is $F(s)/s$.\n",
  "commit": "ff8a40749dc16545e3d2fc637f6b0196e81d0bfa",
  "front": {
    "about": [
      "TD5XK5AHNV",
      "FCKX8KFVGN"
    ],
    "kind": "statement",
    "msc": [
      "44A40",
      "44A35"
    ],
    "name": "Integration operator as convolution with the unit step",
    "role": "proposition",
    "sources": [
      {
        "url": "https://en.wikipedia.org/wiki/Operational_calculus"
      }
    ],
    "tag": "BDBDJSP6DF",
    "uses": [
      "FCKX8KFVGN",
      "09C4JP7Y0M"
    ]
  },
  "kind": "statement",
  "link": "https://tamnd.github.io/math2/node/BDBDJSP6DF/",
  "links": [
    {
      "field": "about",
      "target": "TD5XK5AHNV"
    },
    {
      "field": "about",
      "target": "FCKX8KFVGN"
    },
    {
      "field": "uses",
      "target": "FCKX8KFVGN"
    },
    {
      "field": "uses",
      "target": "09C4JP7Y0M"
    }
  ],
  "markdown": "https://tamnd.github.io/math2/node/BDBDJSP6DF.md",
  "msc": [
    "44A40",
    "44A35"
  ],
  "name": "Integration operator as convolution with the unit step",
  "prov": null,
  "role": "proposition",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/BDB/BDBDJSP6DF.md",
  "status": null,
  "tag": "BDBDJSP6DF",
  "uri": "m2://BDBDJSP6DF",
  "url": "https://tamnd.github.io/math2/wiki/integration-operator-as-convolution-with-the-unit-step/",
  "writing": {
    "tag": "BDBDJSP6DF",
    "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
      }
    ]
  }
}