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Math Help - integrals

  1. #1
    Super Member Random Variable's Avatar
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    integrals

    1)  \int_{0}^{\infty} e^{-ax^{2}} \sin \Big( \frac{b}{x^{2}} \Big) \ dx


    2)  \int_{0}^{\infty} e^{-ax^{2}} \cos \Big(\frac{b}{x^{2}} \Big) \ dx


    3)  \int_{0}^{\infty} \cos ax^{2} \cos 2bx \ dx


    4)  \int_{0}^{\infty} \sin ax^{2}} \cos 2bx \ dx


    I can evaluate all of these integrals, but using a method that's just too difficult to justify. Namely evaluating an integral with real-valued parameters and assuming the solution is valid when one of the parameters is imaginary.
    Last edited by Random Variable; January 5th 2012 at 07:53 PM.
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  2. #2
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    Re: integrals

    Quote Originally Posted by Random Variable View Post
    1)  \int_{0}^{\infty} e^{-ax^{2}} \sin \Big( \frac{b}{x^{2}} \Big) \ dx
    I am out of my depth here, but have an idea.

    The solution is clearly real.

    What if we consider the complex integral:
     \int_{0}^{\infty} e^{-az^{2}} \sin \Big( \frac{b}{z^{2}} \Big) \ dx ; where I assume a,b are real and z = x + iy

    Now
    Re(\sin \Big( \frac{b}{z^{2}} \Big)) = 0 when x = y.

    also
    Re(\sin \Big( \frac{b}{z^{2}} \Big)) tends to zero, when x is sufficiently large

    So if we integrate along the path from the origin out to (R + iR) and then around the arc (centred on the origin) from R+iR to \sqrt2R we find, after taking R to the limit of infinity, that the solution is zero.
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