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Laplace Transform Definition

  1. Jun 17, 2013 #1
    In the attachment that I added I highlighted the portion I am questioning.

    I will define L[f(t)](s) to be the laplace transform of the function f(t).

    f(t) = e^t

    L[f(t)](s) = 1/(s-1). The laplace transform is defined for all values s≠1.

    L[f(t)](2) = 1.

    Question: "What do they mean by sufficiently large vales of s" because I don't consider s = 2 a large value.

    Attached Files:

    • Lap.jpg
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  2. jcsd
  3. Jun 17, 2013 #2

    Ray Vickson

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    Sufficiently large means s ≥ a (or maybe s > a) for some finite number 'a'. Although 1/(s-1) exists for all s ≠ 1, it really only represents the Laplace integral for s > 1. Here, '1' is a sufficiently large number. You are right: '0' is not sufficiently large.
  4. Jun 17, 2013 #3
    I think it would be better if they said s can be any value so long the integral (in the attachment) converges.


    L[e^(at)](s) = ∫e-steatdt [from 0 to ∞] = e[t(a-s)]/(a-s) [from 0 to ∞] = 0 + 1/(s-a)

    where the term e[R(a-s)]/(a-s) tends to 0 as R→∞ IF s>a
  5. Jun 17, 2013 #4
    Sorry I noticed that the laplace didn't converge for s = 0 so I changed it to s = 2
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