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Math Help - Laplace Transform

  1. #1
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    Laplace Transform

    Hello MHF!

    I've been learning Laplace Transform and I understand the method but I don't understand the boundary.

    For example, Laplace transform on 1 would be \frac{1}{s} for s>0. Why only for s>0, I don't understand the boundary.
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  2. #2
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    Quote Originally Posted by Simplicity View Post
    Hello MHF!

    I've been learning Laplace Transform and I understand the method but I don't understand the boundary.

    For example, Laplace transform on 1 would be \frac{1}{s} for s>0. Why only for s>0, I don't understand the boundary.
    Then you know that the Laplace transform of f(x) is \int_0^\infty e^{-st}f(t)dt. In particular, the Laplace transform of f(x)= 1 is \int_0^\infty e^{-st}dt. If s were negative, that integral would not exist. For example, if s= -1, that integral is \int_0^\infty e^t dt which would be \lim_{t\to\infty} e^t -1 and that limit does not exist.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Then you know that the Laplace transform of f(x) is \int_0^\infty e^{-st}f(t)dt. In particular, the Laplace transform of f(x)= 1 is \int_0^\infty e^{-st}dt. If s were negative, that integral would not exist. For example, if s= -1, that integral is \int_0^\infty e^t dt which would be \lim_{t\to\infty} e^t -1 and that limit does not exist.
    But that limit would work. It would just be 0-1=-1? Why is that wrong. It's not like we are getting an infinite answer.
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  4. #4
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    Quote Originally Posted by Simplicity View Post
    But that limit would work. It would just be 0-1=-1? Why is that wrong. It's not like we are getting an infinite answer.
    Look at it again: \lim_{t \to \infty} e^t = e^{\infty} = \infty so the function diverges. There is no limit.
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