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Integrability of sin(x)*x^(-a)

  1. Jan 10, 2009 #1
    Hello, question: does the integral [tex]\int_0^\infty \frac{\sin(x)}{x^a}[/tex] converge (in the sense of Lebesgue principal value) for all [tex]a \in (0;2)[/tex]? For a=1/2, it's the Fresnel integral, but other than that, I'm not sure how to approach this.
  2. jcsd
  3. Jan 11, 2009 #2
    It can be shown with some clever maneuvering and the use of the Gamma function that:


    Gamma is undefined at 0, so one can see that a=2 leads to Gamma(0) and a=0 gives 1.

    Of course, if a=1/2, then we have [tex]\sqrt{\pi}[/tex], which is the solution of the Fresnel integral.

    Remember that [tex]{\Gamma}(\frac{1}{2})=\sqrt{\pi}[/tex].
  4. Jan 11, 2009 #3
    this trick works all the time!!!

    [tex]\int_0^\infty \frac{\sin(x)}{x^a}dx=\int_0^\infty \int_0^\infty \frac{t^{a-1}}{\Gamma(a)} \sin(x)e^{-xt} dxdt[/tex]
  5. Jan 12, 2009 #4
    Thanks for the replies.

    Fredoniahead: thanks for the formula (Maple says it needs an extra [tex]2^{-a}[/tex] factor). It's curious/didactic that the actual integral is undefined for a=0, while the formula is perfectly well-behaved there.

    tim_lou: neat trick, I'll remember it.
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