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Laplace Transform inversion

  1. Jan 3, 2010 #1
    Hi,

    I want to inverse this laplace transform, -(s^(1/2)), seems that the inverse is in complex plane. Where should i start to find this inverse....

    Thank you.
     
  2. jcsd
  3. Jan 3, 2010 #2

    HallsofIvy

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    According to this table, http://www.vibrationdata.com/Laplace.htm, [itex]1/s^k[/itex], where k can be any real number is the Laplace transform of [itex]t^{k-1}/\Gamma(k)[/itex] so the "inverse Laplace tranform" of [itex]-1/s^{1/2}[/itex] is [itex]-t^{-1/2}/\Gamma(1/2)= -1/\sqrt{\pi t}[/itex].
     
  4. Jan 3, 2010 #3
    Thank you for your reply. I'm sorry I should use latex, I'm looking inversion of [tex]-\sqrt{s}[/tex]
     
  5. Jan 4, 2010 #4
    In the first place does the inversion of [itex]-\sqrt{s}[/itex] exist????

    Let say it exist :surprised

    From the http://www.vibrationdata.com/Laplace.htm" [Broken]formula,
    [tex]L\{ \frac{df}{dt} \} = sF(s) - f(0^-)[/tex]

    Let [itex]f(0^-)=0[/itex] :surprised and choose F(s)=-s-1/2 so that [itex]f(t)=-1/\sqrt{\pi t}[/itex]

    Hence
    [tex] L^{-1} \{ -\sqrt{s} \} = \frac{d}{dt} (-1/\sqrt{\pi t}) = \frac{\pi}{2} (\pi t)^{-3/2} [/tex]


    So many contradiction!
     
    Last edited by a moderator: May 4, 2017
  6. Jan 4, 2010 #5

    Mute

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    If you choose f(t) to be t^{-1/2}, f(0-) is clearly not zero! You can't use the first derivative rule as you did because

    [tex]\mathcal L\left(\frac{d t^{-1/2}}{dt} = -\frac{1}{2}t^{-3/2}\right)[/tex]

    doesn't exist! The Laplace transform

    [tex]\int_{0^-}^\infty dt~t^q e^{-st}[/tex]

    exists only for [itex]\mbox{Re}(q) > -1[/itex], so saying

    [tex]\int_{0^-}^\infty dt~t^{-3/2} e^{-st} \propto \mathcal L[t^{-1/2}] + f(0^-)[/tex]

    is somewhat nonsensical (though you can argue that the LHS doesn't exist because the integral diverges, which is indicated by f(0-) being undefined).

    So, the transform of t^{-1/2} does exist, though you won't be able to see that from the first derivative rule.
     
    Last edited by a moderator: May 4, 2017
  7. Jan 4, 2010 #6
    Hi guys.

    Thank you for your reply. Actually I found one journal 'Multi-precision Laplace transform inversion' discussing about laplace inversion in complex plane named fixed talbot algorithm.

    this is the direct link to the said journal. http://www.pe.tamu.edu/valko/public_html/CV/ValkoPDF/2004AV_IJNME_Multi.pdf

    Talbot pioneered the approach of deforming the standard contour in the Bromwich integral

    [tex]f(t) = \frac{1}{2\pi i} \int_B \left exp(ts) \hat{f}(s) \right ds[/tex]

    [tex]B[/tex] in above equation is a vertical line defined by [tex]s=r+iy[/tex]. By Cauchy's theorem the deformed contour is valid where, line to a contour that ends in the left half plane (the integration from -infinity to infinity).

    by using this method ,the laplace inversion for

    [tex]- \sqrt{s} [/tex] is [tex]\frac{1}{2 \sqrt{\pi t^3}}[/tex]

    i still couldn't prove it manually, because couldn't understand a few concept involved. i need to discuss it with my 'sensei'
     
  8. Jan 5, 2010 #7
    I know my argument has a lot flaw. But I did get the right answer :wink: . I was hoping someone could improve on the method.

    This is a mystery because when I asked a friend to get the Laplace transform of [itex]t^{-3/2} [/itex] using Mathematica, it is possible to do it.
    [tex]L \{ \frac{1}{t^{3/2}} \} = -2\sqrt{\pi s}[/tex]

    One more question Mute. When you write the Laplace transform as

    [tex]\int_{0^-}^\infty dt~t^q e^{-st}[/tex]

    Is it not that t are positive. But why do we have the limit 0-. This also appears in the first derivative rule.
     
  9. Jan 5, 2010 #8

    Mute

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    I'm not sure. I would guess Mathematica is just giving that answer as a purely formal result based on the general formula in my last post. If you ask Mathematica to do the integral

    [tex]\int_0^\infty dt~\frac{e^{-st}}{t^{3/2}},[/tex]
    it will return the result "Integral does not converge".

    Approaching zero from the negative side is just so the laplace transform of the delta function gives 1 instead of zero. I don't think it makes a difference for other functions.
     
  10. Jan 6, 2010 #9
    Thanks for that info about the usage of 0-.
    I don't have access to Mathematica to experiment further. But if I'm to use http://integrals.wolfram.com/index.jsp?expr=Exp[-s*x]/x^(3/2)&random=false" for indefinite integral, it look like Mathematical can integrates it.

    [tex]\int dt~\frac{e^{-st}}{t^{3/2}} = -2\sqrt{\pi s} \mbox{ erf}(\sqrt{st}) - \frac{2e^{-st}}{\sqrt{t}} [/tex]

    It still looks like the function fail to exist when t tends to zero.

    Could it possible that Mathematica does not use Riemann integration? Lebesgue integration!! what's this?

    And where is https://www.physicsforums.com/showthread.php?t=355612"? I know he is an expert with that complex inversion formula. Can it be done with Bromwich integral ?
     
    Last edited by a moderator: Apr 24, 2017
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