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  "body": "\n## Summary\n\nResidue calculus for Laplace inversion evaluates the [[Bromwich integral|5KNAXNHRY6]] with the [[residue theorem|032HKYS329]].\nOne closes the vertical line of integration with a large arc to the left.\nWhen $F$ decays on such arcs, the [[inverse Laplace transform|07WBTX1MHA]] is the sum of the [[residues|0PHN07KVQT]] of $e^{s t} F(s)$ at the poles of $F$.\nIt turns inversion into a finite or infinite sum of explicit terms, one for each pole.\n\n## Explanations\n\n### L1 human\n\nEach pole of the transform $F$ produces one piece of the original function of time.\nA simple pole at $s = a$ produces a multiple of $e^{a t}$; a double pole produces terms like $t e^{a t}$.\nFor $F(s) = 1/(s^2 + 1)$, the poles $\\pm i$ give $\\frac{e^{i t}}{2 i} - \\frac{e^{-i t}}{2 i} = \\sin t$.\nThe method needs $F$ to be small on large arcs in the left half-plane; this fails for factors such as $e^{-s}$, which encode a time delay.\n",
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