1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Proof for this Laplace transform
  1. Laplace Transform (Replies: 3)