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Definite Integral (Si)

  1. Jul 13, 2009 #1

    zcd

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    1. The problem statement, all variables and given/known data
    [tex]\int_{0}^{\infty} \frac{\sin^{3}(x)}{x^{3}}\,dx[/tex]


    2. Relevant equations
    [tex]e^{ix}=\sin(x)+i\cos(x)[/tex]
    [tex]\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}[/tex]

    3. The attempt at a solution
    [tex]\int_{0}^{\infty} \frac{\sin^{3}(x)}{x^{3}}\,dx=\int_{0}^{\infty} \frac{1}{x^{3}} (\frac{e^{ix}-e^{-ix}}{2i})^3\,dx[/tex]

    reduces eventually to
    [tex]\int_{0}^{\infty} \frac{\sin(3x)-3\sin(x)}{4x^{3}}\,dx[/tex]

    by parts
    [tex]=\frac{\sin(3x)-3\sin(x)}{-8x^{2}}+\int_{0}^{\infty} \frac{3\cos(3x)-3\cos(x)}{2x^{2}}\,dx[/tex]

    by parts again
    [tex]=\frac{\sin(3x)-3\sin(x)}{-8x^{2}}-\frac{3\cos(3x)-3\cos(x)}{8x}-\int_{0}^{\infty} \frac{9\sin(3x)-3\sin(x)}{8x}\,dx}[/tex]

    reduces to
    [tex]\lim_{t \to \infty} \frac{3}{8} {\rm Si}(t)-\frac{9}{8} {\rm Si}(3t) + \left[ \frac{3\sin(x)}{8x^{2}}-\frac{\sin(3x)}{8x^{2}}+\frac{3\cos(x)}{8x}-\frac{3\cos(3x)}{8x}\right]_{0}^{t}[/tex]

    I'm not sure how to evaluate at last step. Did I make some error before? or is it supposed to be divergent because none of the terms inside the bracket can handle the 0 limit.
     
    Last edited: Jul 14, 2009
  2. jcsd
  3. Jul 13, 2009 #2

    Dick

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    Is the problem just to decide whether the integral is convergent or divergent? You can do that without finding the indefinite integral. Use a comparison test. The quantity inside the brackets goes to zero at both limits once you correct 9*cos(3x)/8x to 3*cos(3x)/8x.
     
    Last edited: Jul 13, 2009
  4. Jul 13, 2009 #3

    zcd

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    I know that [tex]-\frac{1}{x^{3}}\leq\frac{\sin^{3}(x)}{x^{3}}\leq{\frac{1}{x^{3}}(\forall x \in \mathbb{R})[/tex], and [tex]\pm\int_{0}^{\infty}\frac{1}{x^{3}}\,dx[/tex] are both convergent, so [tex]\int_{0}^{\infty}\frac{\sin^{3}(x)}{x^{3}}\,dx[/tex] must be convergent.
     
    Last edited: Jul 13, 2009
  5. Jul 13, 2009 #4

    Dick

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    Not quite, the integral of 1/x^3 diverges near 0. On the other hand your original integrand sin(x)^3/x^3 doesn't have this problem. Split the integration region into [0,1] and [1,infinity) and use the comparison test on the second one.
     
  6. Jul 13, 2009 #5

    Dick

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    sin(x)^3/x^3 has a finite limit as x->0. It's bounded on (0,1].
     
  7. Jul 13, 2009 #6

    zcd

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    I just noticed the mistake. But that does leave [tex]\int_{1}^{\infty}\frac{\sin^{3}(x)}{x^{3}}\,dx[/tex] convergent

    Would this second statement be valid?

    Because [tex]\lim_{x \to 0}\frac{\sin(x)}{x}=1[/tex], then [tex]\lim_{x \to 0}(\frac{\sin(x)}{x})^{3}=1[/tex], and [tex]\int_{0}^{1}\frac{\sin^{3}(x)}{x^{3}}\,dx[/tex] would converge. If [tex]\int_{0}^{1}\frac{\sin^{3}(x)}{x^{3}}\,dx[/tex] and [tex]\int_{1}^{\infty}\frac{\sin^{3}(x)}{x^{3}}\,dx[/tex] are both convergent, then [tex]\int_{0}^{\infty}\frac{\sin^{3}(x)}{x^{3}}\,dx[/tex] is convergent.

    Edit: sorry I just restated what you said while I was working on the latex code :D
     
  8. Jul 13, 2009 #7

    Dick

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    The integral over [1,infinity) is finite by the comparison test since |sin(x)^3/x^3|<=1/x^3 which converges. Your integrand is not really defined at x=0. But it has a limit as x->0 and it's continuous for x>0. Isn't that enough to say the improper integral is also well defined on [0,1]?
     
  9. Jul 13, 2009 #8

    zcd

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    Yes, that is what I was asking the validity of. But now that I know it's convergent, how do I solve the actual limits? Do I try to reduce sin(3x) and cos(3x) back to a function of sin(x) and cos(x) and hope that it reduces the denominator?
     
  10. Jul 13, 2009 #9

    Dick

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    If you actually want to find the value of the integral, you know Si(t) and Si(3t) have a limit as t->infinity, right? The limit of the terms inside the bracket go to 0 as t->infinity, also right? Now you just want to show they also go to zero as t->0. That's clear for the sin terms, it's not true for the cos terms because you have an error in one of the coefficients. If you fix it, they will go to 0 as t->0 as well. Use the series expansion of cos(x) and cos(3x) around x=0.
     
  11. Jul 13, 2009 #10

    zcd

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    Ah, never thought of using taylor series to eliminate cos(x) and cos(3x) at x = 0. Intuitively I can guess that Si(t) has convergent limit as t->infinity, but I don't know how to actually calculate that limit.
     
  12. Jul 13, 2009 #11

    Dick

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    I haven't been paying much attention to actually calculating the exact limit, because I was guessing the idea was just to show convergence. Finding Si(infinity) uses complex analysis (stuff like contour integrals). I'd just make sure you are clear on why your integral converges using elementary arguments without worrying about the value it converges to.
     
  13. Jul 13, 2009 #12

    zcd

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    The question was asking for the exact value, and wikipedia gives the answer [tex]\lim_{t \to \infty} {\rm Si}(t) = \frac{\pi}{2}[/tex], so I'll just go with [tex]\frac{3\pi}{8}[/tex] for now. Thanks for the help!
     
    Last edited: Jul 14, 2009
  14. Jul 13, 2009 #13

    Dick

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    3*pi/8 it is. You're welcome.
     
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