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Inverse laplace of 1

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data

    anyone know how to find inverse laplace of 1?
    that is

    3. The attempt at a solution

    can we use
    L-11=s ?

  2. jcsd
  3. Nov 12, 2008 #2


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    Staff Emeritus
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    Normally, the Laplace tranform of a function of x is written as a function of s. You seem to be confusing the two. The Laplace transform of f(x)= x is
    [tex]\int_0^\infty xe^{-sx}dx= \frac{1}{s^2}[/tex]
    by integration by parts, not 1. And you certainly cannot just multiply a Laplace transform you already know by the variable to get another Laplace transform!

    The inverse Laplace transform of the constant 1 is the Dirac delta function [itex]\delta(x)[/itex]:
    [tex]\int_0^\infty e^{-sx}\delta(x)dx= e^{-s(0)}= 1[/tex]
    since, by definition, [itex]\int_S f(x)\delta(x) dx= f(0)[/itex] as long as the region of integration, S, includes 0.

    Here's a good table of Laplace and inverse Laplace transforms:
  4. Nov 12, 2008 #3
    thank you:smile:
  5. Nov 12, 2008 #4
    regarding this topic , using the Laplace transform properties would it be valid that

    [tex] \mathcal L^{-1} (s^{k})= D^{k}\delta (t) [/tex] ??

    where k >0 any real number (at least this property seems to work with Fourier transforms)

    the case k <0 would involve integration , but the integrals of the Dirac delta (with k integer) are well defined for t >0 (except perhaps at the point t=0 )
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