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  "body": "\n## Example\n\nLet $P \u003e 0$ and let $f \\colon [0, \\infty) \\to \\C$ be [[periodic|089KWVYD69]] with period $P$, that is $f(t + P) = f(t)$ for $t \\ge 0$, and integrable on $[0, P]$.\nIts [[Laplace transform|05XT5AVJBK]] is\n$$F(s) = \\frac{1}{1 - e^{-Ps}} \\int_0^P f(t) e^{-st}\\, dt \\quad (\\operatorname{Re} s \u003e 0).$$\nCheck: for $\\operatorname{Re} s \u003e 0$ the integral converges absolutely, since $\\int_0^\\infty \\abs{f} e^{-\\sigma t}\\, dt = \\sum_{k \\ge 0} e^{-k P \\sigma} \\int_0^P \\abs{f(t)} e^{-\\sigma t}\\, dt \u003c \\infty$ for $\\sigma = \\operatorname{Re} s$.\nSplit $[0, \\infty)$ into the intervals $[kP, (k + 1)P]$ and substitute $t = u + kP$:\n$$F(s) = \\sum_{k \\ge 0} e^{-kPs} \\int_0^P f(u) e^{-su}\\, du,$$\na geometric series with ratio $e^{-Ps}$, of modulus less than $1$.\nInstance: $f(t) = \\abs{\\sin t}$ has period $\\pi$, and $\\int_0^\\pi \\sin t\\, e^{-st}\\, dt = \\frac{1 + e^{-\\pi s}}{s^2 + 1}$.\nSo\n$$\\mathcal{L}\\{\\abs{\\sin t}\\}(s) = \\frac{1 + e^{-\\pi s}}{(s^2 + 1)(1 - e^{-\\pi s})} = \\frac{\\coth(\\pi s / 2)}{s^2 + 1}.$$\n\n## Explanations\n\n### L1 human\n\nA periodic function repeats the same piece forever.\nEach copy of the piece is the first copy delayed by a multiple of the period, and a delay by $kP$ multiplies the transform by $e^{-kPs}$.\nAdding up all copies gives a geometric series, so only the transform of one period is needed.\n",
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