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Integration by Parts / Laplace Transform watch

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    I'm not even sure that what I'm about to ask is possible. Just a pre-warning.

    Because I live such a rivoting life I've spent some time this Friday evening trying to find the Laplace Transform of t^n, where n is a positive integer constant.

    Why am I doing this? Well, the transform of t is easy to find, requiring just one iteration of IBP. t^2 takes a bit more effort, but is also relatively easy. Anything else is somewhat overkill. Plus I'm bored.

    Now, the Laplace Transform of a function f(t) is given by \displaystyle\int^\infty_0 f(t)e^{-st} dt

    i.e. for this function, we're considering \displaystyle\int^\infty_0 t^ne^{-st} dt

    IBP:  \displaystyle\int^\infty_0 u\frac{dv}{dt} dt =\left[ uv \right]_0 ^{\infty} - \displaystyle\int^\infty_0 v\frac{du}{dt}dt

    Letting

     u = t^n so \dfrac{du}{dt} = nt^{n-1} and

    \dfrac{dv}{dt} = e^{-st} so that v = -\dfrac{1}{s}e^{-st}

    which leads us to consider

    \displaystyle\int^\infty_0 t^ne^{-st} dt =\left[ -\dfrac{t^n}{s}e^{-st} \right]_0 ^{\infty} - \displaystyle\int^\infty_0 -\dfrac{1}{s}nt^{n-1}e^{-st} dt

    which simplifies down to

    \displaystyle\int^\infty_0 t^ne^{-st} dt = -\dfrac{n}{s} \displaystyle\int^\infty_0 t^{n-1}e^{-st} dt

    Which also requires integration by parts. Using the same steps as above, we come to

    \displaystyle\int^\infty_0 t^ne^{-st} dt = -\dfrac{n}{s} . -\dfrac{n-1}{s} \displaystyle\int^\infty_0 t^{n-2}e^{-st} dt = \dfrac{n(n-1)}{s^2}\displaystyle\int^\infty  _0 t^{n-2}e^{-st} dt

    I would imagine, but haven't checked that the next stage would come out as -\dfrac{n(n-1)(n-2)}{s^3}\displaystyle\int^\infty  _0 t^{n-3}e^{-st} dt

    Obviously this continues n times.

    My question is this - how do you go about actually evaluating this integral? It's probably something blindingly obvious but my brain has stopped working. Or is it just not doable using elementary functions?
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    Well let I_n(s) = \displaystyle \int_0^{\infty} t^ne^{-st}\, dt. You've found that I_n(s) = -\dfrac{n}{s}I_{n-1}(s), and so you can apply this formula recursively (putting n \leftarrow n-1 and so on) until you get I_n(s) = (\text{something}) \times I_0(s), but I_0(s) = \displaystyle \int_0^{\infty} e^{-st}\, dt is something you can evaluate.
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    (Original post by nuodai)
    Well let I_n(s) = \displaystyle \int_0^{\infty} t^ne^{-st}\, dt. You've found that I_n(s) = -\dfrac{n}{s}e^{-st}I_{n-1}(s), and so you can apply this formula recursively (putting n \leftarrow n-1 and so on) until you get I_n(s) = (\text{something}) \times I_0(s), but I_0(s) = \displaystyle \int_0^{\infty} e^{-st}\, dt is something you can evaluate.
    Not sure if it's a typo on your part, or obtuseness on mine, but how did a factor of e^-st get into the RHS? Surely that's contained within I_{n-1}(s).

    EDIT: lol, speaking of typos... *sigh* [/irony]
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    (Original post by EEngWillow)
    Not sure it it's a typo on your part, or obtuseness on mine, but how did a factor of e^-st get into the RHS? Surely that's contained within I_{n-1}(s).
    Woops yeah, that was definitely a typo.
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    (Original post by nuodai)
    Woops yeah, that was definitely a typo.
    Anyway, just using this recursively is giving out the results I'd expect. Thanks! :-)
    (I knew it was something obvious! xD)
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    You might also want to look up the gamma function.
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    (Original post by DFranklin)
    You might also want to look up the gamma function.
    Thanks for the suggestion. Will give it some looking at later, right now I have a cake to make.
    (Does the gamma function make this easier or more difficult than nuodai's suggestion?)
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    The gamma function is essentially the same as the integral you're wanting to find. (So you'll see lots of possible approaches for evaluating it.
 
 
 
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