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  "body": "\n## Statement\n\nLet $f \\colon [0, \\infty) \\to \\C$ be integrable on each interval $[0, T]$ with $T \u003e 0$.\nAssume that the [[Laplace integral|05XT5AVJBK]] $\\int_0^\\infty f(t) e^{-s_0 t}\\, dt$ converges, as an improper integral, at some $s_0 \\in \\C$.\nLet $\\sigma_c$ be the [[abscissa of convergence|FTHFSJ30WJ]] of $f$.\nThen:\n\n1. The integral $F(s) = \\int_0^\\infty f(t) e^{-st}\\, dt$ converges for every $s \\in \\C$ with $\\operatorname{Re} s \u003e \\operatorname{Re} s_0$.\n2. The function $F$ is [[holomorphic|0DG0Z72CQD]] on the half-plane $\\set{s \\in \\C : \\operatorname{Re} s \u003e \\sigma_c}$.\n3. For every $n \\in \\N$ and every $s$ with $\\operatorname{Re} s \u003e \\sigma_c$,\n$$F^{(n)}(s) = \\int_0^\\infty (-t)^n f(t) e^{-st}\\, dt,$$\nand this integral converges.\n\n## Remarks\n\nFor part 1, let $\\varphi(t) = \\int_0^t f(u) e^{-s_0 u}\\, du$; it is continuous and bounded on $[0, \\infty)$ because it has a finite limit.\nIntegration by parts gives $\\int_0^T f(t) e^{-st}\\, dt = \\varphi(T) e^{-(s - s_0)T} + (s - s_0) \\int_0^T \\varphi(t) e^{-(s - s_0)t}\\, dt$, and both terms converge when $\\operatorname{Re}(s - s_0) \u003e 0$.\nThe same formula shows that the truncated integrals converge locally uniformly on the half-plane $\\operatorname{Re} s \u003e \\operatorname{Re} s_0$.\nBy part 1 the integral converges at real points $\\sigma$ arbitrarily close to $\\sigma_c$ from the right, so these half-planes cover $\\operatorname{Re} s \u003e \\sigma_c$.\nA locally uniform limit of holomorphic functions is holomorphic, which gives part 2.\nPart 3 follows by differentiating under the integral sign on $[0, T]$ and passing to the limit.\nExample: for $f = 1$, $F(s) = 1/s$ is holomorphic on $\\operatorname{Re} s \u003e 0$, and $F'(s) = -1/s^2 = \\int_0^\\infty (-t) e^{-st}\\, dt$.\nThe half-plane is in general not the largest domain of holomorphy: $1/s$ extends to $\\C \\setminus \\set{0}$, but the integral diverges for $\\operatorname{Re} s \\le 0$.\n\n## Explanations\n\n### L1 human\n\nThe Laplace transform of a function is a very smooth object: wherever the integral converges (off the boundary line), the transform is a holomorphic function of the complex variable $s$.\nThe region of convergence is always a right half-plane.\nDerivatives of the transform are transforms of $f$ multiplied by powers of $-t$.\n\n### L2 human\n\nThis theorem is why complex analysis applies to Laplace transforms.\nResidues, contour shifts and analytic continuation are used to invert transforms, as in the Bromwich integral, and to read off the asymptotic behaviour of $f$ from the singularities of $F$.\nPart 3 is the rule \"multiplication by $-t$ corresponds to differentiation in $s$\".\nIt is the analogue of the fact that a power series converges on a disc and is holomorphic inside it; Dirichlet series obey the same pattern.\n",
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