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Math Help - Integration by Parts (with infinity)

  1. #1
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    SOLVED Integration by Parts (with infinity)

    F=\int_0^\infty e^{-st}t^2 dt

    My unsuccessful attempt:
    Integration by parts: \int fg'=fg-\int f'g
    Using "liate":
    f=t^2  f'=2t
    g=- \frac{1}{s} e^{-st}  g'=e^{-st}

    F=- \frac{1}{s} e^{-st} \cdot t^2 |^\infty_0 - \int _0^\infty 2t\cdot - \frac{1}{s} e^{-st} dt

    F=  \lim_{A \to \infty} (-\frac{1}{s} e^{-sA} \cdot A^2) - \int _0^\infty 2t\cdot - \frac{1}{s} e^{-st} dt

    Using L'Hopital on the limit I get that the limit is equal to 0. So...
    F=  \frac{2}{s} \int _0^\infty t\cdot e^{-st} dt

    I use integration by parts again and I get...
    F=  \frac{2}{s}  [-\frac{1}{s}e^{-st}\cdot t|^\infty _0 -\int _0^\infty e^{-st}dt ]
    F=  \frac{2}{s}  [\lim_{A\to\infty} (0) +  \frac{1}{s} \lim_{A\to\infty} (-1)]

    F=  \frac{-2}{s^2}

    The answer is supposed to be F=\frac{2!}{s^3}

    Where did I go wrong?
    Last edited by Carbon; May 24th 2011 at 12:12 AM. Reason: solved!
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  2. #2
    MHF Contributor chisigma's Avatar
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    With the substitution s t= \tau You obtain...

    \int _{0}^{\infty} t^{2}\ e^{-s t}\ dt = \frac{1}{s^{3}}\ \int_{0}^{\infty} \tau^{2} e^{-\tau}\ d \tau (1)

    ... where the s variable is 'pulled out' the integral so that integration [by parts...] is more confortable...

    Kind regards

    \chi \sigma
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  3. #3
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    I'm probably making a really dumb mistake but I don't see where the s^3 comes from.

    \int _{0}^{\infty} t^{2}\ e^{-s t}\ dt =\int _{0}^{\infty} \frac{\tau ^2}{s^{2}}e^{-\tau}\ d \tau= \frac{1}{s^{2}}\int _{0}^{\infty} \tau ^2 e^{-\tau} d\tau
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Carbon View Post
    I'm probably making a really dumb mistake but I don't see where the s^3 comes from.

    \int _{0}^{\infty} t^{2}\ e^{-s t}\ dt =\int _{0}^{\infty} \frac{\tau ^2}{s^{2}}e^{-\tau}\ d \tau= \frac{1}{s^{2}}\int _{0}^{\infty} \tau ^2 e^{-\tau} d\tau
    ... that is because s t= \tau \implies dt= \frac{d \tau}{s}...

    Kind regards

    \chi \sigma
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  5. #5
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    There it is. :P

    Thanks! Do you happen to know if this tau substitution works for most cases involving computing laplace transforms?
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  6. #6
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Carbon View Post
    There it is. :P

    Thanks! Do you happen to know if this tau substitution works for most cases involving computing laplace transforms?
    ... pratically the substitution s t=\tau\ ,\ dt=\frac{d \tau}{s} always works...

    Kind regards

    \chi \sigma
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  7. #7
    Ted
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    Your integral = Laplace transform of t^2 = \dfrac{2}{s^3}.
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