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Math Help - Laplace Transform

  1. #1
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    Laplace Transform

    I've been working on a problem, this one in particular:

    f(t) = cosh(bt)

    where (e^{bt} + e^{-bt})/2 is equal to cosh bt


    I attempt to take the Laplace transform:

    \frac {1} {2} \{L\}(e^bt) + \frac {1} {2} \{\L\}(e^-{bt})

    = \frac {1} {2} \int e^{-st} e^{bt} + \frac {1} {2} \int e^{-st} e^{-bt}

    = \lim_{A\to\infty} \frac {1} {2} \displaystyle\int^A_0 e^{t(b-s)}\,dt + \frac {1} {2} \lim_{A\to\infty} \displaystyle\int^A_0 e^{-t(b+s)}\,dt

    However, this is where I'm stuck. I'm not sure what do when I have a non-negative "t" (cause if it's negative, I can assume it's going to equal zero as A goes towards infinity). Can anyone show me where I went wrong, or what step is next? Thank you!
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  2. #2
    MHF Contributor
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    Just out of curiosity is the answer

    \frac{1}{2}\big[\frac{1}{s-b}+\frac{1}{s+b}\big]?
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  3. #3
    MHF Contributor
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    It is actually an advantage to having negative ts.

    Here is what the integration will look like.

    \frac{1}{2}\bigg[\frac{-1}{e^{t(s+b)}(s+b)}-\frac{1}{e^{t(s+b)}(s-b)}\bigg]_{0}^{\infty}

    Can you see how this is good for the bounds integration?
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