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Proof for this Laplace transform

  1. Nov 26, 2006 #1
    Can someone help me out with the proof of the Laplace transform of the function tn?

    I did have a go at this one.
    [tex]L[t^n] = \int_0^{\infty} t^n e^{-st}dt[/tex]

    [tex]= t^n\frac{-e^{-st}}{s}\vert_0^{\infty} + {1\over s}\int_0^{\infty} nt^{n-1}e^{-st}dt[/tex]

    [tex]={n\over s}\int_0^{\infty} t^{n-1}e^{-st}dt[/tex]

    [tex]={n\over s}L[t^{n-1}][/tex]

    I am supposed to arrive at the result:
    [tex]L[t^n] = \frac{(n+1)!}{s^{n+1}}[/tex]
  2. jcsd
  3. Nov 26, 2006 #2
    Just an addendum, should I prove this using induction?
  4. Nov 26, 2006 #3
    [tex]L(t^{n}) = \int_{0}^{\infty} t^{n}e^{-st}dt[/tex]

    Let [tex]v=st[/tex] to give [tex]L(t^{n}) = \frac{1}{s^{n+1}}\int_{0}^{\infty}v^{n}e^{-v}dv = \frac{1}{s^{n+1}}\Gamma(n+1)[/tex]

    [tex]\Gamma(n+1) = n![/tex], giving the result [tex]L(t^{n}) = \frac{n!}{s^{n+1}}[/tex]. This isn't what you've got, but check by using n=1 to see whose correct.

    [tex]L(t) = \int_{0}^{\infty}te^{-st}dt =\left[ \frac{te^{-st}}{-s}\right]_{0}^{\infty} - \int_{0}^{\infty}\frac{1}{-s}e^{-st}dt = \frac{1}{s^{2}}[/tex]

    Hence [tex]L(t^{1}) = \frac{1!}{s^{2}}[/tex], so the formula you were trying to derive isn't correct.
    Last edited: Nov 26, 2006
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