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Math Help - Integration tips

  1. #1
    Member mabruka's Avatar
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    Integration tips

    Hi

    I need to compute the next integral:

     \int_{0}^{\infty} \frac{cos{(tx)}(1- cos{(x)})}{x^2}dx
    =\int_{0}^{\infty} \frac{cos(tx)}{ x^2} dx - \int_{0}^{\infty} \frac{cos(xt)\cos(x)}{x^2}dx


    but i dont know even where to start ! Any hint how to do this?{


    thank you
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  2. #2
    Member mabruka's Avatar
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    I tried using integration by parts on the first one, and everything seems fine except for the limits...



    I want to avoid integrating ont he complex plane but... what do you think?
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mabruka View Post
    Hi

    I need to compute the next integral:

     \int_{0}^{\infty} \frac{cos{(tx)}(1- cos{(x)})}{x^2}dx
    =\int_{0}^{\infty} \frac{cos(tx)}{ x^2} dx - \int_{0}^{\infty} \frac{cos(xt)\cos(x)}{x^2}dx


    but i dont know even where to start ! Any hint how to do this?{


    thank you
    Hint:

    Spoiler:
    For the second one use the product-to-sum identity to make it into two similar cases of the first integral. For that one use the fact that \int_0^{\infty}y\cdot e^{-y\cdot x}\text{ }dy
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  4. #4
    Member mabruka's Avatar
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    Thank you, im gonna work in it right now
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  5. #5
    Member mabruka's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Hint:

    For the second one use the product-to-sum identity to make it into two similar cases of the first integral. For that one use the fact that \int_0^{\infty}y\cdot e^{-y\cdot x}\text{ }dy
    Sorry, i dont understand the fact. What fact about that integral? The only think i can think of is its relation with the gamma function

    <br />
\int_0^{\infty}y\cdot e^{-y\cdot x}\text{ }dy = \frac{1}{x^2}\int_0^{\infty}u\cdot e^{-u}\text{ }du = \frac{ \Gamma (2)}{x^2} = \frac{ 1}{x^2}
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  6. #6
    Member mabruka's Avatar
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    Guess im gonna go over the complex field to integrate that last integral!

    Hope it works just fine


    Thnx for the help
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mabruka View Post
    Sorry, i dont understand the fact. What fact about that integral? The only think i can think of is its relation with the gamma function

    <br />
\int_0^{\infty}y\cdot e^{-y\cdot x}\text{ }dy = \frac{1}{x^2}\int_0^{\infty}u\cdot e^{-u}\text{ }du = \frac{ \Gamma (2)}{x^2} = \frac{ 1}{x^2}
    \int_0^{\infty}\frac{\cos(tx)}{x^2}dx=\int_0^{\inf  ty}\int_0^{\infty}ye^{-yx}\cos(tx)dydx switch the order of integration.
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  8. #8
    Math Engineering Student
    Krizalid's Avatar
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    Nice try but that integral diverges.

    And you'll need a justification to swap the integration order.
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  9. #9
    Member mabruka's Avatar
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    After working with it a bit i found out

    \int_0^{\infty}\frac{\cos(tx)}{x^2}dx=\int_0^{\inf  ty}\int_0^{\infty}ye^{-yx}\cos(tx)dydx = \int_{0}^{ \infty}\frac{y^2}{y^2+t^2} dy

    which as Krizalid pointed out it is not finite.

    The problem here is we can not use Fubini's theorem because the function

    f(x,y)= ye^{-yx}\cos(tx) is not integrable.

    So the only way i can think of doing it is over the complex field.


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