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  "body": "\n## Definition\n\nLet $a$ be a real number, and let $F$ be a [[holomorphic function|0DG0Z72CQD]] on the half-plane $\\{s \\in \\C : \\operatorname{Re} s \u003e a\\}$.\nLet $\\gamma \u003e a$ be real and let $t$ be real.\nThe *Bromwich integral* of $F$ along the line $\\operatorname{Re} s = \\gamma$, at time $t$, is\n$$\\frac{1}{2 \\pi i} \\int_{\\gamma - i \\infty}^{\\gamma + i \\infty} e^{s t} F(s) \\, ds = \\frac{1}{2 \\pi i} \\lim_{T \\to \\infty} \\int_{\\gamma - i T}^{\\gamma + i T} e^{s t} F(s) \\, ds.$$\nEach integral on the right is the [[contour integral|00FWNP9TXY]] along the vertical segment from $\\gamma - i T$ to $\\gamma + i T$.\nThe Bromwich integral exists at $t$ when this limit exists.\n\n## Remarks\n\nExample: let $F(s) = 1/s$ on $\\operatorname{Re} s \u003e 0$, and $\\gamma \u003e 0$.\nFor $t \u003e 0$ one closes the segment by a large arc to the left, whose contribution tends to $0$.\nThe residue of $e^{s t}/s$ at $0$ is $1$, so the integral is $1$.\nFor $t \u003c 0$ one closes to the right, where there are no poles, and gets $0$; for $t = 0$ the symmetric limit is $\\frac{1}{2 \\pi i} \\lim_{T \\to \\infty} \\left( \\log(\\gamma + i T) - \\log(\\gamma - i T) \\right) = \\tfrac{1}{2}$.\nNon-example: for $F(s) = 1$ and $t \\ne 0$, the integral over the segment equals $e^{\\gamma t} \\sin(T t) / (\\pi t)$, which has no limit as $T \\to \\infty$; so the Bromwich integral does not exist.\nThe limit is a symmetric one; it can exist even when $e^{s t} F(s)$ is not absolutely integrable on the line.\nThe integral is also called the Fourier–Mellin integral.\nWhen $F$ is the [[Laplace transform|05XT5AVJBK]] of a suitable function $f$, the integral equals $f(t)$ at each $t \u003e 0$ where $f$ is continuous; this is Mellin's inverse formula.\n",
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