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Math Help - Calculate Laplace Transform{t}

  1. #1
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    Calculate Laplace Transform{t}

    I need help completing (or have already made a mistake) calculating the Laplace Transform of g(t)=t. My work (1st equal: def of Lapl xform; 2nd equal: integration by parts):

    L \{ t \} = \int_0^{\infty} t e^{-st} dt = \frac{t e^{-st}}{-s} - \int_0^{\infty} \frac{e^{-st}}{-s} dt = \left[ \frac{t e^{-st}}{-s} - \frac{e^{-st}}{s^2} \right]_0^\infty

    = lim_{b \rightarrow \infty} \left[ \frac{b e^{-sb}}{-s} - \frac{e^{-sb}}{s^2} \right] - \left[ \frac{0 \cdot e^{-s0}}{-s} - \frac{e^{-s \cdot0}}{s^2} \right] = \left[ \frac{\infty \cdot e^{- \infty}}{-s} - 0 \right] - \left[ 0 - \frac1{s^2} \right]

    = \frac{\infty \cdot 0}{-s} + \frac1{s^2}

    P.S. How do you make an euler's constant "e" so it doesn't look like a variable e?
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  2. #2
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    Re: Calculate Laplace Transform{t}

    I have the following:

    \mathcal{L}[t]=\int_{0}^{\infty}t\,e^{-st}\,dt=\frac{t\,e^{-st}}{-s}\Bigg|_{0}^{\infty}-\int_{0}^{\infty}\frac{e^{-st}}{-s}\,dt=\left[\lim_{t\to\infty}\frac{t\,e^{-st}}{-s}\right]-\frac{e^{-st}}{s^{2}}\Bigg|_{0}^{\infty}.

    To evaluate the remaining limits, you must assume that \text{Re}(s)\ge 0. If that's true, then you can use L'Hopital's Rule to finish them off.
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  3. #3
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    Re: Calculate Laplace Transform{t}

    Can't use l'Hospital's rule. The first derivative of the bottom of \frac{te^{-st}}{-s} is zero.
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    Re: Calculate Laplace Transform{t}

    Quote Originally Posted by MSUMathStdnt View Post
    Can't use l'Hospital's rule. The first derivative of the bottom of \frac{te^{-st}}{-s} is zero.
    Ah, but you can. Put the exponential in the denominator and flip the sign of the exponent.
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  5. #5
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    Re: Calculate Laplace Transform{t}

    Quote Originally Posted by Ackbeet View Post
    Ah, but you can. Put the exponential in the denominator and flip the sign of the exponent.
    Clearly, I must review l'Hospital's. Thanks.
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  6. #6
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    Re: Calculate Laplace Transform{t}

    Quote Originally Posted by Ackbeet View Post
    I have the following:

    \mathcal{L}[t]=\int_{0}^{\infty}t\,e^{-st}\,dt=\frac{t\,e^{-st}}{-s}\Bigg|_{0}^{\infty}-\int_{0}^{\infty}\frac{e^{-st}}{-s}\,dt=\left[\lim_{t\to\infty}\frac{t\,e^{-st}}{-s}\right]-\frac{e^{-st}}{s^{2}}\Bigg|_{0}^{\infty}.

    To evaluate the remaining limits, you must assume that \text{Re}(s)\ge 0. If that's true, then you can use L'Hopital's Rule to finish them off.
    Just cleaning up some old posts. I think the answer to the limit is

    \lim_{t\to\infty}\frac{t\,e^{-st}}{-s} = \lim_{t\to\infty}\frac{-t}{s\,e^{st}} = \text{l'Hospital's rule} = \lim_{t\to\infty}\frac{-1}{s^2\,e^{st}} = 0.
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  7. #7
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    Re: Calculate Laplace Transform{t}

    Quote Originally Posted by MSUMathStdnt View Post
    Just cleaning up some old posts. I think the answer to the limit is

    \lim_{t\to\infty}\frac{t\,e^{-st}}{-s} = \lim_{t\to\infty}\frac{-t}{s\,e^{st}} = \text{l'Hospital's rule} = \lim_{t\to\infty}\frac{-1}{s^2\,e^{st}} = 0.
    Looks correct to me.
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