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  "body": "\n## Definition\n\nLet $c \\ge 0$, and let $\\varphi$ be a holomorphic function on the half-plane $\\operatorname{Re} s \u003e c$.\nLet $\\gamma \u003e c$, and suppose that for each $t \u003e 0$ the integral below converges.\nHere the integral is the limit, as $R \\to \\infty$, of the integrals over the segment from $\\gamma - iR$ to $\\gamma + iR$.\nThe *Bromwich interpretation* of the Heaviside expression $\\varphi(p) H$ is the function\n$$x(t) = \\frac{1}{2\\pi i} \\int_{\\gamma - i\\infty}^{\\gamma + i\\infty} \\frac{\\varphi(s)}{s} \\, e^{st} \\, ds, \\qquad t \u003e 0,$$\na [[Bromwich integral|5KNAXNHRY6]], with $x(t) = 0$ for $t \u003c 0$.\n\n## Remarks\n\nUnder suitable growth conditions on $\\varphi$, the [[Laplace transform|0V49JEZW06]] of $x$ is $\\varphi(s)/s$, so this interpretation agrees with the Laplace transform reading of the [[Heaviside operational calculus|QQ98C2AXZP]].\nBy Cauchy's theorem, the value does not depend on $\\gamma \u003e c$.\nWhen $\\varphi$ is meromorphic and the line can be closed to the left, $x(t)$ is the sum of the [[residues|0PHN07KVQT]] of $\\varphi(s) e^{st} / s$.\n\nExample: $\\varphi(p) = p^{-n}$ with $n \\ge 0$.\nFor $t \u003e 0$, the line can be closed to the left, and the only pole of $e^{st}/s^{n+1}$ is at $0$, with residue $t^n / n!$.\nSo $x(t) = t^n/n!$, which is Heaviside's rule $p^{-n} H = t^n H / n!$.\n\nNon-example: $\\varphi(s) = e^{-s^2}$ is entire, but on the line $s = \\gamma + iy$ one has $\\abs{\\varphi(s)} = e^{y^2 - \\gamma^2}$.\nSo the integrand grows without bound, the integral diverges for every $t$, and $\\varphi(p) H$ has no Bromwich interpretation.\n",
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