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Math Help - Improper Integral

  1. #1
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    Improper Integral

    Hi,

    Im trying to solve the following, Integrate[x^2*Cos[w*x], x], lower limit = 0, upper limit = infinity,

    Im getting (2*x*Cos[w*x])/w^2 + ((-2 + w^2*x^2)*Sin[w*x])/w^3

    I need to evaluate this now between the lower limit of 0 and upper limit of infinity, but am having trouble,

    any thoughts?

    Thanks!!.....
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  2. #2
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    Before taking the antiderivative, change the upper limit to "t". Then, take your antiderivative (answer will have t's in it). Perform the normal subtraction for integrals. Finally, take the limit as t approaches infinity.


    can you repost using latex code? just to make sure i have it right
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  3. #3
    Forum Admin topsquark's Avatar
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    Here's the LaTeX version:

    Quote Originally Posted by citbquinn View Post
    Hi,

    Im trying to solve the following,
    \displaystyle \int_0^{\infty} x^2~cos(wx) ~ dx

    Im getting
    \displaystyle \frac{2x~cos(wx)}{w^2} + \frac{(-2 + w^2x^2)~sin(wx)}{w^3}

    I need to evaluate this now between the lower limit of 0 and upper limit of infinity, but am having trouble,

    any thoughts?

    Thanks!!.....
    -Dan
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  4. #4
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    Thanks Dan that looks good.

    citbquinn,

    simply evaluate your antiderivative from 0 to t. your answer will have t's in it. then take the limit of that as t approaches infinity. that's your answer
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  5. #5
    Forum Admin topsquark's Avatar
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    Your integral looks good. Now what happens when you take the limit as x goes to infinity?

    -Dan
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  6. #6
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    evaluating from 0 to t, gives,

    =\displaystyle\lim_{t\to \infty}\dfrac{(2tcos(wt))}{w^2}   +  \dfrac{((-2+w^2t^2)sinwt)}{w^3}

    so is this pretty much my answer??

    Thank you!!
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  7. #7
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    Quote Originally Posted by citbquinn View Post
    evaluating from 0 to t, gives,

    =\displaystyle\lim_{t\to \infty}\dfrac{(2tcos(wt))}{w^2}   +  \dfrac{((-2+w^2t^2)sinwt)}{w^3}

    so is this pretty much my answer??

    Thank you!!
    Now you actually have to evaluate the limit ><
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  8. #8
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    Quote Originally Posted by Prove It View Post
    Now you actually have to evaluate the limit ><
    Yes,

    this is where my problem is, because coswt for example is periodic as t-->inf,
    so there is no limiting value??
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by citbquinn View Post
    Yes,

    this is where my problem is, because coswt for example is periodic as t-->inf,
    so there is no limiting value??
    Since -1 \leq cos(wt) \leq 1 what can you say about the values of t \cdot cos(wt)?

    -Dan
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  10. #10
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    Quote Originally Posted by topsquark View Post
    Since -1 \leq cos(wt) \leq 1 what can you say about the values of t \cdot cos(wt)?

    -Dan
    yes....??

    -t \leq cos(wt) \leq t
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by citbquinn View Post
    yes....??

    -t \leq cos(wt) \leq t
    Let me be more specific. You want
    \displaystyle \lim_{t \to \infty} t \cdot cos(wt) = \left ( \lim_{t \to \infty} t \right ) \cdot \left ( \lim_{t \to \infty} cos(wt) \right )

    We know that \displaystyle -1 \leq cos(wt) \leq 1 so cos(wt) is bounded by -1 and 1 so even though the limit is undefined, it is not infinite.

    What about the limit of t?

    -Dan
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