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Abs Value of Sine Integral

  1. Apr 9, 2005 #1
    i know that the sine integral converges to pi/2. But what about the abs value of the sine integral. It seems to me that it would have value oo. But I'm having trouble coming up with a lower bound that diverges.
     
  2. jcsd
  3. Apr 9, 2005 #2

    dextercioby

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    Do you mean
    the absolute value of Si(x).

    [tex] \left|Si(x)\right|=:\left|\int_{0}^{x} \frac{\sin t}{t} \ dt \right| [/tex]

    or the abolute value of sinc(x)

    [tex]\tilde{Si}(x)=:\int_{0}^{x} \left|\frac{\sin t}{t}\right| \ dt [/tex]

    Daniel.
     
  4. Apr 9, 2005 #3
    The latter. Sorry for the confusion.
     
  5. Apr 9, 2005 #4

    dextercioby

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    The graph is deceiving.My computer wouldn't compute the intagral.I dunno whether it's finite or not...

    Daniel.
     
  6. Apr 9, 2005 #5

    shmoe

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    Break it up into intervals over the period of |sin(x)|

    [tex]\int_{k*\pi}^{(k+1)*\pi}\left|\frac{\sin{t}}{t}\right|dt\geq \int_{k*\pi}^{(k+1)*\pi}\frac{|\sin{t}|}{(k+1)*\pi}dt[/tex]

    Then sum over k=0,1,..,whatevers appropriate. There will be a little left over if x is not a multiple of pi, but this won't matter (you're bounding from below and your integrand is positive).
     
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