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  "body": "\n## Definition\n\nLet $a$ be a real number and let $F$ be a complex function on the half-plane $\\{s \\in \\mathbb{C} : \\operatorname{Re} s \u003e a\\}$.\nAn inverse Laplace transform of $F$ is a piecewise continuous function $f : [0, \\infty) \\to \\mathbb{C}$ with two properties. First, there are constants $M, b$ with $|f(t)| \\le M e^{b t}$ for $t \\ge 0$. Second, the [[Laplace transform|05XT5AVJBK]] of $f$ equals $F(s)$ for each $s$ with $\\operatorname{Re} s \u003e \\max(a, b)$.\nThe notation is $f = \\mathcal{L}^{-1}\\{F\\}$.\n\n## Remarks\n\nBy Lerch's theorem, two such functions with the same transform agree at each point where both are continuous. So $\\mathcal{L}^{-1}\\{F\\}$ is unique in practice, and unique among continuous functions.\nLet $\\gamma$ be a real number larger than the real parts of all singularities of $F$. Under suitable conditions, the Bromwich integral recovers $f$: $f(t) = \\frac{1}{2 \\pi i} \\lim_{T \\to \\infty} \\int_{\\gamma - iT}^{\\gamma + iT} e^{s t} F(s)\\, ds$.\nIn practice, one splits $F$ into partial fractions and reads the inverse of each term from a table. For example, $\\mathcal{L}^{-1}\\{1/(s - a)\\} = e^{a t}$.\n\n## Explanations\n\n### L1 human\n\nThe inverse transform goes back from the $s$ side to the time side.\nAfter we solve a problem in terms of $F(s)$, the inverse transform gives the solution as a function of time.\n",
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