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Math Help - Finding asymptotic expansion using Laplace's method

  1. #1
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    Finding asymptotic expansion using Laplace's method

    Consider the integral I defined by
    \displaystyle I = \int^1_0 e^{-xt} t \cos t dt.

    Use Laplace's method to obtain the first term of the asymptotic expansion of the integral as x \to \infty.

    How can I start? Can someone point me to somewhere with a similar example?

    Thank you.
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  2. #2
    Grand Panjandrum
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    Re: Finding asymptotic expansion using Laplace's method

    Quote Originally Posted by math2011 View Post
    Consider the integral I defined by
    \displaystyle I = \int^1_0 e^{-xt} t \cos t dt.

    Use Laplace's method to obtain the first term of the asymptotic expansion of the integral as x \to \infty.

    How can I start? Can someone point me to somewhere with a similar example?

    Thank you.
    I don't see that this is Laplace's method but:

    For large x the function e^{-xt} is small everywhere except close to t=0, so we expand the other term in the integrand about t=0:

    \int_0^1 e^{-xt}t \cos(t)\; dt = \int_0^1 e^{-xt}t\ [1-t^2/2+...]\; dt

    Truncating after the first term gives:

    \int_0^1 e^{-xt}t \cos(t)\; dt \sim \int_0^1 e^{-xt}t\; dt=x^{-2}[1-(x+1)e^{-x}]

    and you can probably replace the square brackets by 1.

    CB
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  3. #3
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    Re: Finding asymptotic expansion using Laplace's method

    Thanks! I can see that the approximated solution is similar to the solution. But this question is definitely a Laplace method question. How can I apply the Laplace method to the question?
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  4. #4
    Grand Panjandrum
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    Re: Finding asymptotic expansion using Laplace's method

    Quote Originally Posted by math2011 View Post
    Thanks! I can see that the approximated solution is similar to the solution. But this question is definitely a Laplace method question. How can I apply the Laplace method to the question?
    Well that was the version of Laplace's method from the first part of this document, except the maximum is of \phi(x) is at an end point of the interval so not quadratic, and so this is covered by the middle paragraph of the further notes section.


    CB
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  5. #5
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    Re: Finding asymptotic expansion using Laplace's method

    According to the tutorial 2.8 referred to by the middle paragraph of the further notes section, the higher boundary 1 should be taken to \infty. This is what I got after doing that.

    \int^1_0 e^{-xt} t \cos(t) dt =& \int^1_0 e^{-xt} t \left(1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \cdots \right) dt \\ \sim & \int^\infty_0 e^{-xt} t dt \\ =& -\frac{1}{x} \left[ e^{-xt} t \right]^\infty_0 - \int^\infty_0 \left(-\frac{1}{x}\right) e^{-xt} dt \\ =& -\frac{1}{x} \left[ e^{-xt} t \right]^\infty_0 - \frac{1}{x^2} \left[ e^{-xt} \right]^\infty_0 \\ =& \frac{1}{x^2}

    Thanks for helping me getting into this topic.
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