{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe integrate by parts on $[0, T]$ and let $T \\to \\infty$.\n\n1. Fix $s \\in \\mathbb{C}$ with $\\sigma = \\operatorname{Re} s \u003e a$, and write $s = \\sigma + i \\omega$ with $\\omega$ real. Let $E(t) = e^{-st}$ for $t \\ge 0$. Note that $M \\ge \\abs{f(0)} \\ge 0$.\n2. By the [[product of exponentials|0WM7DJCM6G]], $E(t) = e^{-\\sigma t} e^{-i \\omega t}$. By [[Euler's formula|F6PH4JEGS5]], $\\abs{e^{-i\\omega t}} = \\abs{\\cos(\\omega t) - i \\sin(\\omega t)} = 1$. So $\\abs{E(t)} = e^{-\\sigma t}$.\n3. For real $h \\ne 0$, $\\dfrac{E(t + h) - E(t)}{h} = E(t)\\, \\dfrac{e^{-sh} - 1}{h}$, by the product of exponentials. If $s \\ne 0$, this quotient tends to $-s E(t)$ as $h \\to 0$, by the [[derivative of the complex exponential|019H1Z6419]] at $0$. If $s = 0$, it is $0$. So $E'(t) = -s E(t)$.\n4. Let $g(t) = M e^{-(\\sigma - a) t}$. By step 2, $\\abs{f(t) E(t)} \\le M e^{at} e^{-\\sigma t} = g(t)$ for $t \\ge 0$.\n5. By the [[integral of the exponential function|06D0EWR2DV]] and [[integration by substitution|0D37QFPEZE]] with $x = -(\\sigma - a) t$, $\\int_0^T g(t)\\, dt = \\frac{M}{\\sigma - a}\\left( 1 - e^{-(\\sigma - a) T} \\right)$. This tends to $\\frac{M}{\\sigma - a}$ as $T \\to \\infty$, since $\\sigma - a \u003e 0$.\n6. Let $p$ be the real part of $t \\mapsto f(t) E(t)$. It is continuous, and $0 \\le p(t) + g(t) \\le 2 g(t)$ by step 4.\n7. By the [[comparison test for improper integrals|0FR1BDKK10]] and step 5, $\\int_0^\\infty (p + g)$ converges. So $\\int_0^T p = \\int_0^T (p + g) - \\int_0^T g$ has a limit as $T \\to \\infty$.\n8. The same argument applies to the imaginary part. So the limit $F(s) = \\lim_{T \\to \\infty} \\int_0^T f(t) E(t)\\, dt$ exists.\n9. The function $f'$ is continuous, so it is [[Riemann integrable|06SESV7GQY]] on each $[0, T]$. Thus its [[Laplace transform|05XT5AVJBK]] is defined as a limit, where the limit exists.\n10. By the [[product rule|0V9A43H7WV]], applied to real and imaginary parts, and step 3, $(f E)' = f' E - s f E$. This derivative is continuous on $[0, \\infty)$.\n11. By the [[second fundamental theorem of calculus|0XYAFCN0PT]], applied to the real and imaginary parts of $f E$ on $[0, T]$, we get $f(T) E(T) - f(0) = \\int_0^T f'(t) E(t)\\, dt - s \\int_0^T f(t) E(t)\\, dt$.\n12. So $\\int_0^T f'(t) e^{-st}\\, dt = f(T) e^{-sT} - f(0) + s \\int_0^T f(t) e^{-st}\\, dt$ for each $T \u003e 0$.\n13. By step 4, $\\abs{f(T) e^{-sT}} \\le M e^{-(\\sigma - a) T}$, and this tends to $0$ as $T \\to \\infty$.\n14. By steps 8 and 13, the right side of step 12 tends to $0 - f(0) + s F(s)$ as $T \\to \\infty$.\n15. Hence the Laplace transform of $f'$ converges at $s$, and $\\mathcal{L}\\{f'\\}(s) = s F(s) - f(0)$.\n\n## Idea\n\nIntegration by parts moves the derivative from $f$ onto $e^{-st}$, and differentiating $e^{-st}$ only multiplies it by $-s$. The boundary term at $0$ gives $-f(0)$. The boundary term at $T$ vanishes in the limit because $f$ grows at most like $e^{at}$, while $e^{-st}$ decays like $e^{-\\operatorname{Re}(s) t}$.\n",
  "commit": "68a8c3dc73fd55004b2796431c607c30960d1a85",
  "front": {
    "kind": "proof",
    "msc": [
      "44A10"
    ],
    "name": "Proof of the formula for the Laplace transform of a derivative by integration by parts",
    "proves": "0K090GVHVN",
    "sources": [
      {
        "locator": "Transform of a function's derivative",
        "url": "https://en.wikipedia.org/wiki/Laplace_transform"
      }
    ],
    "tag": "P6G7J7GK8N",
    "uses": [
      "05XT5AVJBK",
      "0WM7DJCM6G",
      "F6PH4JEGS5",
      "019H1Z6419",
      "0V9A43H7WV",
      "0XYAFCN0PT",
      "06SESV7GQY",
      "06D0EWR2DV",
      "0FR1BDKK10",
      "0D37QFPEZE"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/P6G7J7GK8N/",
  "links": [
    {
      "field": "proves",
      "target": "0K090GVHVN"
    },
    {
      "field": "uses",
      "target": "05XT5AVJBK"
    },
    {
      "field": "uses",
      "target": "0WM7DJCM6G"
    },
    {
      "field": "uses",
      "target": "F6PH4JEGS5"
    },
    {
      "field": "uses",
      "target": "019H1Z6419"
    },
    {
      "field": "uses",
      "target": "0V9A43H7WV"
    },
    {
      "field": "uses",
      "target": "0XYAFCN0PT"
    },
    {
      "field": "uses",
      "target": "06SESV7GQY"
    },
    {
      "field": "uses",
      "target": "06D0EWR2DV"
    },
    {
      "field": "uses",
      "target": "0FR1BDKK10"
    },
    {
      "field": "uses",
      "target": "0D37QFPEZE"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/P6G7J7GK8N.md",
  "msc": [
    "44A10"
  ],
  "name": "Proof of the formula for the Laplace transform of a derivative by integration by parts",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/P6G/P6G7J7GK8N.md",
  "status": null,
  "tag": "P6G7J7GK8N",
  "uri": "m2://P6G7J7GK8N",
  "url": "https://math2.tamnd.com/wiki/proof-of-the-formula-for-the-laplace-transform-of-a-derivative-by-integration-by-parts/",
  "writing": {
    "tag": "P6G7J7GK8N",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}