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  "body": "\n## Statement\n\nLet $f \\colon [0, \\infty) \\to \\C$ be integrable on each interval $[0, T]$, and let $F$ be its [[Laplace transform|05XT5AVJBK]].\nLet $a \\ge 0$ be real, and let $u$ be the [[Heaviside step function|09C4JP7Y0M]].\nLet $g(t) = u(t - a) f(t - a)$, that is, $g(t) = 0$ for $0 \\le t \u003c a$ and $g(t) = f(t - a)$ for $t \\ge a$.\nIf the Laplace integral of $f$ converges at the complex number $s$, then so does the Laplace integral of $g$, and\n$$\\mathcal{L}\\{g\\}(s) = e^{-a s} F(s).$$\n\n## Remarks\n\nThe substitution $t = \\tau + a$ gives $\\int_a^\\infty f(t - a) e^{-s t} \\, dt = e^{-a s} \\int_0^\\infty f(\\tau) e^{-s \\tau} \\, d\\tau$.\nExample: with $f = 1$ one gets $\\mathcal{L}\\{u(t - a)\\}(s) = e^{-a s} / s$ for $\\operatorname{Re} s \u003e 0$.\nA rectangular pulse $u(t - 1) - u(t - 2)$ has transform $(e^{-s} - e^{-2 s}) / s$.\nThe condition $a \\ge 0$ matters: for $a \u003c 0$ the shifted function would need values of $f$ at negative times, which the one-sided transform does not see.\n\n## Explanations\n\n### L1 human\n\nDelaying a signal by $a$ units of time multiplies its transform by $e^{-a s}$.\nThe step function makes sure that nothing happens before the delay is over.\n\n### L2 human\n\nThis rule is also called the second shifting theorem.\nIt is how one handles forcing terms that switch on and off in the [[Laplace method for initial value problems|0K3HFZJK5Q]].\nRead backwards, a factor $e^{-a s}$ in a transform means: invert the rest, then delay by $a$.\nCompare the [[frequency shift theorem|CD35MSBNE7]], where a factor $e^{a t}$ becomes a shift in $s$.\n",
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