Math 2.0

Integral transforms, operational calculus Definition

Laplace transform

Builds on Function

Definition

Let f:[0,∞)→ℂf : [0, \infty) \to \mathbb{C} be a function that is integrable on each interval [0,T][0, T] with T>0T > 0. The Laplace transform of ff is the function F=ℒ{f}F = \mathcal{L}\{f\} given by F(s)=∫0∞f(t) e−st dt=lim⁡T→∞∫0Tf(t) e−st dt,F(s) = \int_0^\infty f(t)\, e^{-s t}\, dt = \lim_{T \to \infty} \int_0^T f(t)\, e^{-s t}\, dt, for each complex number ss for which the limit exists.

Remarks

Let ff be of exponential order: there are constants M,aM, a with ∣f(t)∣≤Meat|f(t)| \le M e^{a t} for all t≥0t \ge 0. Then the integral converges absolutely for Re⁡s>a\operatorname{Re} s > a. Some transforms: ℒ{1}(s)=1/s\mathcal{L}\{1\}(s) = 1/s and ℒ{eat}(s)=1/(s−a)\mathcal{L}\{e^{a t}\}(s) = 1/(s - a) for Re⁡s>Re⁡a\operatorname{Re} s > \operatorname{Re} a, and ℒ{sin⁡ωt}(s)=ω/(s2+ω2)\mathcal{L}\{\sin \omega t\}(s) = \omega / (s^2 + \omega^2) for Re⁡s>0\operatorname{Re} s > 0 and real ω\omega. The transform is linear: ℒ{cf+g}=cℒ{f}+ℒ{g}\mathcal{L}\{c f + g\} = c \mathcal{L}\{f\} + \mathcal{L}\{g\} where both sides converge. It turns derivatives into multiplication by ss. See the Laplace transform of a derivative.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Laplace_transform, Formal definition
  2. Web https://en.wikipedia.org/wiki/Laplace_transform, Region of convergence
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