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Math Help - related integrals

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

     \int_{0}^{\infty} \cos \Big(x + \frac{1}{x} \Big) \frac{dx}{\sqrt{x}}  = 2 \int^{\infty}_{0} \cos \Big(x^{2}+ \frac{1}{x^{2}} \Big) dx = 2 \ \Re \int^{\infty}_{0} e^{-i(x^{2}+\frac{1}{x^{2}})} dx


    and  \int_{0}^{\infty} \sin \Big(x + \frac{1}{x} \Big) \frac{dx}{\sqrt{x}}  = 2 \int^{\infty}_{0} \sin \Big(x^{2}+ \frac{1}{x^{2}} \Big) dx = -2 \ \Im \int^{\infty}_{0} e^{-i(x^{2}+\frac{1}{x^{2}})} dx


    Could I use  \int^{\infty}_{0} e^{-a(x^{2}+\frac{1}{x^{2}})} dx = \frac{\sqrt{\frac{\pi}{a}}}{2e^{2a}} and substitute i for a?
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  2. #2
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    Re: related integrals

    Good question! Any kind of substitution would have to be justified. It boils down to whether the last result extends to complex a's. I'll need to think about this a while.
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  3. #3
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    Re: related integrals

    The answer to your question is yes you can. You can extend the identity to complex values of a, in a suitable neighborhood, because of the uniqueness principle in complex analysis.
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  4. #4
    Super Member Random Variable's Avatar
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    Re: related integrals

    I'm not familiar with a uniqueness principle. All I know is that when a is real valued and a \le 0, the integral diverges. I don't know for what imaginary or complex values the integral is convergent. Substituting  a= i does lead to the correct solution, though.

    EDIT: It should be OK if I can show that  \int^{\infty}_{0} \cos \Big( x^{2}+\frac{1}{x^{2}} \Big) \ dx and  \int^{\infty}_{0} \sin \Big( x^{2} + \frac{1}{x^{2}} \Big) \ dx actually converge.
    Last edited by Random Variable; June 26th 2011 at 06:57 PM.
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  5. #5
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    Re: related integrals

    The question of whether an identity involving a real parameter can extended to complex values comes up a lot in complex analysis. There's more to it than just showing that certain integrals converge. A reference for the uniqueness principle is Gamelin's book on complex analysis. He addresses questions very similiar to the one you're considering.
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