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Laplace transformation

  1. Jan 14, 2006 #1
    laplace transform

    Could anyone should me how it is you arrive to [tex] \frac{n!}{(s - a)^{n+1}} [/tex] as being the laplace transform of [tex] t^n e^{at} [/tex] ? Or at least provide a link?
    I've done integration by parts but the term [tex]e^{(s-a)t}[/tex] will not go away.
     
    Last edited: Jan 14, 2006
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  3. Jan 14, 2006 #2

    benorin

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    If n is an integer the [tex]e^{(a-s)t}[/tex] term will go away after n iterations of integration by parts and one final integration of the term itself.
     
    Last edited: Jan 14, 2006
  4. Jan 14, 2006 #3
    So far I have [tex] \int t^n e^{(s-a)t} dt = \frac{t^n e^{(s-a)t}}{s-a} - \frac{n t^{n-1} e^{(s-a)t}}{(s-a)^{2}} + \frac{(n^2-n) t^{n-2} e^{(s-a)t}}{(s-a)^{3}} . . . . . . . [/tex]. I'm just not seeing how [tex] e^{(s-a)t} [/tex] goes away.
     
    Last edited: Jan 14, 2006
  5. Jan 14, 2006 #4
    I think the easiest way is to do it in pieces.

    prove L[ t^n ] = n!/s^(n+1).

    This is easy if you know
    [tex] n! = \Gamma ( n+1 ) = \int_0^{\infty} t^n exp(-t) dt [/tex]
    otherwise an inductive proof on n is pretty straight forward.

    Then i would show if L[f(t)] = F(s) then L[f(t)exp(at)] = F(s-a). Since this is just a relable-ing operation it should
    be easy to show.
     
    Last edited: Jan 14, 2006
  6. Jan 14, 2006 #5

    benorin

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    an inductive proof

    EDIT: changed the limits of integration to be from 0 to infinity (instead of -infinity to infinity).

    The Laplace transform of [itex] t^n e^{at} [/itex] is

    [tex]\int_{t=0}^{\infty} t^n e^{at} e^{-st}dt=\int_{t=0}^{\infty} t^n e^{(a-s)t} = \frac{n!}{(s - a)^{n+1}} [/tex]

    an inductive proof goes as follows:

    i. For n=0, [tex]\int_{t=0}^{\infty} t^0 e^{(a-s)t}= \frac{1}{s - a} \left[ e^{(a-s)t} \right]_{t=0}^{\infty} = \frac{0!}{(s - a)^{0+1}} [/tex]

    ii. Assume that

    [tex]\int_{t=0}^{\infty} t^n e^{(a-s)t} = \frac{n!}{(s - a)^{n+1}} [/tex]

    holds for some fixed integer n >0, then

    [tex]\int_{t=0}^{\infty} t^{n+1} e^{(a-s)t} dt= \left[ -\frac{t^{n+1}e^{(a-s)t}}{s-a}\right]_{t=0}^{\infty} + \frac{n+1}{s-a}\int_{t=0}^{\infty} t^{n} e^{(a-s)t} dt [/tex]

    which, by hypothesis,

    [tex]= \frac{n+1}{s-a}\frac{n!}{(s - a)^{n+1}} = \frac{(n+1)!}{(s - a)^{(n+1)+1}} [/tex]

    and so

    [tex]\int_{t=0}^{\infty} t^n e^{(a-s)t} = \frac{n!}{(s - a)^{n+1}} [/tex]

    holds for every non-negative integer n by induction.
     
    Last edited: Jan 15, 2006
  7. Jan 14, 2006 #6

    benorin

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    Remember that you differentiate the t^n term each iteration of integration by parts, so the exponent of t should decrese by one for successive terms in the above sum, after n such terms the exponent of t will be 0 leaving only the exponential to integrate.
     
  8. Jan 14, 2006 #7
    Nice advice, but I was instructed to derive the laplace transform from integration by parts, or else I would have used the property, If L{f(x)} = F(s), then L{(e^at)f(x)} = F(s-a).
     
  9. Jan 14, 2006 #8
    Sorry, that was a typo.
     
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