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  "body": "\n## Definition\n\nLet $f \\colon [0, \\infty) \\to \\C$ be integrable on each interval $[0, T]$ with $T \u003e 0$.\nThe *abscissa of absolute convergence* of the [[Laplace transform|05XT5AVJBK]] of $f$ is\n$$\\sigma_a = \\inf \\set{ \\sigma \\in \\R : \\int_0^\\infty \\abs{f(t)} e^{-\\sigma t}\\, dt \u003c \\infty } \\in [-\\infty, +\\infty],$$\nwith $\\inf \\emptyset = +\\infty$.\n\n## Remarks\n\nFor $\\operatorname{Re} s \u003e \\sigma_a$ the Laplace integral converges as a [[Lebesgue integral|XVEWR93EMV]], and for $\\operatorname{Re} s \u003c \\sigma_a$ it does not converge absolutely.\nAlways $\\sigma_c \\le \\sigma_a$, where $\\sigma_c$ is the [[abscissa of convergence|FTHFSJ30WJ]].\nExample: for $f(t) = e^{at}$ with $a \\in \\R$, $\\abs{f} = f$, so $\\sigma_a = \\sigma_c = a$.\nExample with $\\sigma_c \u003c \\sigma_a$: let $f(t) = e^t \\sin(e^t)$.\nThe substitution $u = e^t$ turns $\\int_0^\\infty f(t) e^{-\\sigma t}\\, dt$ into $\\int_1^\\infty \\sin(u)\\, u^{-\\sigma}\\, du$.\nThis converges for $\\sigma \u003e 0$ by the Dirichlet test and diverges for $\\sigma \\le 0$, so $\\sigma_c = 0$.\nIt converges absolutely exactly for $\\sigma \u003e 1$, since $\\int_1^\\infty \\abs{\\sin u}\\, u^{-\\sigma}\\, du \u003c \\infty$ if and only if $\\sigma \u003e 1$; so $\\sigma_a = 1$.\nFor a function of exponential order $a$ that is integrable on bounded intervals, $\\sigma_a \\le a$.\n",
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