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Laplace transform

  1. Oct 17, 2005 #1
    how do i find the laplace transform of log[x]
     
  2. jcsd
  3. Oct 17, 2005 #2

    Tide

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    The Laplace transform of [itex]\log t[/itex] is

    [tex]\int_0^{\infty}e^{-st} \log t dt[/tex]
     
  4. Oct 17, 2005 #3
    i know that but i tink the final answer is infinity,thats ridiculus,so i need confirmation
     
  5. Oct 17, 2005 #4

    Tide

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    I don't think there is a "closed form" expression for the integral but the integral should be finite since log x integrates to x log x - x which goes to 0 as x -> 0.
     
  6. Oct 17, 2005 #5

    Tom Mattson

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    I don't think that this Laplace transform exists. A necessary condition for the existence of the Laplace transform of [itex]f(t)[/itex] is that [itex]f[/itex] be continuous on [itex]0 \leq t < \infty[/itex], but [itex]\log(t)[/itex] isn't even defined at [itex]t=0[/itex].
     
  7. Oct 17, 2005 #6

    Tide

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    The condition for the existence of the Laplace Transform is that function must be piecewise continuous and of exponential order. In short, it has to be integrable.
     
  8. Oct 17, 2005 #7

    Tom Mattson

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    Right, but the interval has to include [itex]t=0[/itex], where the integrand has a vertical asymptote. Doesn't that screw things up?
     
    Last edited: Oct 17, 2005
  9. Oct 17, 2005 #8

    Tide

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    Yes, the fact that the discontinuity occurs at t = 0 poses a problem but you can define the Laplace transform by setting the lower limit to [itex]\epsilon > 0[/itex] and passing to the limit 0.

    In fact, we know that the integral

    [tex]\int_{0}^{\infty}\ln x e^{-x} dx = -\gamma[/tex]

    is just the Euler-Mascheroni constant. We can use this result to evaluate the Laplace transform:

    [tex]\int_{0}^{\infty} \ln t e^{-st} dt = \int_{0}^{\infty} \frac{-\ln s +\ln x}{s}e^{-x}dx[/tex]

    with the result

    [tex]\int_{0}^{\infty} \ln t e^{-st} dt = - \frac {\ln s + \gamma}{s}[/tex]

    which wasn't as bad as I first thought it would be!
     
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