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Improper integral comparison test

  1. Jan 3, 2015 #1
    • Member warned about not using the homework template
    The question asks whether the following converges or diverges.

    [tex]\int_{0}^{\infty } \frac{\left | sinx \right |}{x^2} dx[/tex]

    Now I think there might be a trick with the domain of sine function but I couldn't make up my mind on this.
    I tried to compare it with 1/x^2, (sinx)/x, and sinx. I actually expected that I would get something good with 1/x^2, but as the lower limit of the integral is zero, it ended up with infinity on (0, inf) and since 1/x^2 is greater than (sinx)/x^2, and is divergent as we just found, we cannot say whether the given function diverges or converges. So I'm wondering what is the right track on this question?
     
    Last edited: Jan 3, 2015
  2. jcsd
  3. Jan 3, 2015 #2

    Stephen Tashi

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    Science Advisor

    [tex]\int_{0}^{\infty } \frac{\left | sinx \right |}{x^2} dx = [/tex]

    [tex] \int_{0}^{ \frac{\pi}{2} } \frac{\left | sinx \right |}{x^2} dx + \int_{ \frac{\pi}{2} }^{\infty} \frac{\left | sinx \right |}{x^2} dx [/tex]

    Deal with the two integrals separately.

    In the interval [itex] [0,\frac{\pi}{2}] [/itex] you can establish an inequality between [itex] |\sin(x)| [/itex] and a non-periodic function. Try comparing it to [itex] f(x) = x [/itex].
     
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