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Convergence of an integral.

  1. Jun 13, 2006 #1
    i need to verify my work, i need to evalute the next integral, and decide if it converges or diverges:
    [tex]\int_{1}^{\infty}cos(\frac{2}{x})dx[/tex]

    here what i did:
    2/t=x (-2/t^2)dt=dx
    [tex]\int_{1}^{\infty}-2cos(t)/t^2dt[/tex]

    cos(t)>=-1
    -2cos(t)/t^2<=2/t^2
    the integral of 1/t^2 between 1 and infinity converges to 1, and therefore also the orginal integral with cosine converges.

    is this line of reasoning correct?
     
  2. jcsd
  3. Jun 13, 2006 #2

    shmoe

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    When you substituted 2/x=t, you didn't change the endpoints.

    You might want to take a moment to think about what the graph of cos(2/x) looks like on [1,infinity)
     
  4. Jun 13, 2006 #3
    ok i understand it diverges.
    so i get:
    [tex]\int_{2}^{0}\frac{-2cos(t)}{t^2}dt=-\int_{0}^{2}\frac{-2cos(t)}{t^2}dt[/tex]

    when
    -2cos(t)/t^2>=-2/t^2

    and -2/t^2 diverges on those limits.
     
  5. Jun 13, 2006 #4

    shmoe

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    -2cos(t)/t^2>=-2/t^2 won't help show that diverges. You're bounding it from the wrong side, your bound is 0<=2cos(t)/t^2<=2/t^2, on [0,1] say. Bounding something from above by something divergent tells you nothing.

    You can fix this, but why bother with a change of variables at all? You know what cos(2/x) is doing as x->infinity right?
     
  6. Jun 13, 2006 #5
    lim cos(2/x)=1
    2/x->0
    but how do i find the integral of cos(2/x)?
     
  7. Jun 13, 2006 #6

    shmoe

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    What do you mean "find the integral"? I thought you already agreed it diverged?
     
  8. Jun 13, 2006 #7
    then how do i prove it?
     
  9. Jun 13, 2006 #8

    shmoe

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    cos(2/x)->1 as x->infinity, so the integral over [1,infinity) diverges.

    You've surely got a result to this effect, if not just show you can make the integral over [1,N] arbitrarily large by taking N large enough.
     
  10. Jun 14, 2006 #9
    but surely the integral of cos(2/x) is different than cos(2/x), so how the limit of cos(2/x) helps us to prove the integral of cos(2/x) diverges?
     
  11. Jun 14, 2006 #10

    siddharth

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    Here's one way of looking at it. Consider the function f(x) = cos(2/x). Now as [itex] x \rightarrow \infty [/itex], f(x) tends to 1. So, when you calculate the area under the graph from 1 to infinity, can you see that the area blows up to infinity?
     
    Last edited: Jun 14, 2006
  12. Jun 14, 2006 #11
    i uderstand that it diverges but how do i prove it rigourosly?
     
  13. Jun 14, 2006 #12

    arildno

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    In order to prove this rigorously, use some version of the simple ordering property of the integral:
    If the integral of a function f (strictly greater than your own integrand) diverges, then your integral diverges as well ("bigger integrand means bigger integral").
     
  14. Jun 14, 2006 #13
    Why not expand it out as an infinite series, then integrate?
    It can be shown that the first term diverges, and the remainder converges.
     
    Last edited: Jun 14, 2006
  15. Jun 14, 2006 #14

    StatusX

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    Another way would be to write:

    [tex]\int_1^\infty \cos(2/x)dx = \sum_{n=1}^\infty \int_n^{n+1} \cos(2/x)dx[/tex]

    And then use what you know about the convergence of infinite series.
     
  16. Jun 14, 2006 #15

    shmoe

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    The limit of cos(2/x) is 1, so there is an M where cos(2/x)>1/2 if x>M. How big can you then say the integral of cos(2/x) is on the interval [M,N]?
     
  17. Jun 14, 2006 #16

    VietDao29

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    Ok, what does it mean if someone says that:
    [tex]\lim_{x \rightarrow \infty} \cos \left( \frac{2}{x} \right) = 1[/tex]?
    It means that, for every [tex]\varepsilon > 0[/tex], there exists a [tex]\delta > 0[/tex], such that [tex]\forall x > \delta \ : \ \left| \cos \left( \frac{2}{x} \right) - 1 \right| < \varepsilon[/tex]
    As x increases without bound, [tex]\cos \left( \frac{2}{x} \right) < 1[/tex], so:
    [tex]\cos \left( \frac{2}{x} \right) - 1 < 0[/tex], breaking the absolute value, we have:
    [tex]\forall x > \delta \ : \ - \cos \left( \frac{2}{x} \right) + 1 < \varepsilon[/tex]
    [tex]\forall x > \delta \ : \ \cos \left( \frac{2}{x} \right) > 1 - \varepsilon[/tex]
    Now, you can choose [tex]\varepsilon = \frac{1}{2}[/tex] as shmoe.
    So we have:
    [tex]\forall x > \delta \ : \ \cos \left( \frac{2}{x} \right) > 1 - \frac{1}{2} = \frac{1}{2}[/tex]
    So we have:
    [tex]\int_{\delta} ^ \infty \cos \left( \frac{2}{x} \right) dx > \int_\delta ^ \infty \frac{dx}{2}[/tex].
    Does the integral [tex]\int_{\delta} ^ \infty \cos \left( \frac{2}{x} \right) dx[/tex] converge or diverge? :)
    ----------------------
    Ok, it works just like series, we know that if the series:
    [tex]\sum_{n = \alpha} ^ {\infty} a_n[/tex] converges then [tex]\lim_{n \rightarrow \infty} a_n = 0 , \ n \in \mathbb{N}[/tex].
    So if: [tex]\int_{\alpha} ^ \infty f(x) dx[/tex] converges, then [tex]\lim_{x \rightarrow \infty} f(x) = 0[/tex]. :)
    Note that the reverse is, however not true.
     
    Last edited: Jun 14, 2006
  18. Jun 14, 2006 #17

    arildno

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    Why bother with FOTC at all???
     
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