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  "body": "\n## Example\n\nLet $H$ be the [[Heaviside step function|09C4JP7Y0M]], with $H(t) = 0$ for $t \u003c 0$ and $H(t) = 1$ for $t \u003e 0$, and let $a \\ge 0$.\nThe delayed step $f(t) = H(t - a)$ is $0$ on $[0, a)$ and $1$ for $t \u003e a$.\nIts [[Laplace transform|05XT5AVJBK]] is\n$$\\mathcal{L}\\{H(t - a)\\}(s) = \\frac{e^{-as}}{s} \\quad (\\operatorname{Re} s \u003e 0).$$\nCheck: for $\\operatorname{Re} s \u003e 0$,\n$$\\int_0^\\infty H(t - a) e^{-st}\\, dt = \\int_a^\\infty e^{-st}\\, dt = \\lim_{T \\to \\infty} \\frac{e^{-as} - e^{-sT}}{s} = \\frac{e^{-as}}{s}.$$\nThe value of $H$ at the single point $0$ does not change the integral.\nFor $a = 0$ this is $\\mathcal{L}\\{1\\} = 1/s$.\nA rectangular pulse $H(t - a) - H(t - b)$ with $0 \\le a \u003c b$ has transform $(e^{-as} - e^{-bs})/s$, by linearity.\n\n## Explanations\n\n### L1 human\n\nThe step function models a switch turned on at time $a$.\nDelaying a signal by $a$ multiplies its transform by $e^{-as}$, and here the undelayed signal is the constant $1$ with transform $1/s$.\nSums of steps describe piecewise constant inputs, and their transforms are sums of terms $e^{-as}/s$.\n",
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