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(Comparison Theorem) Why is x/(x^3+1) convergent on interval 0 to infinity?

  1. Jan 20, 2012 #1
    ∫0->∞ x/(x^3 + 1) dx. Use comparison theorem to determine whether the integral is convergent of divergent.



    2. Relevant equations

    None.

    3. The attempt at a solution

    ∫0->∞ x/(x^3 + 1) dx

    = ∫0->∞ x/(x^3) dx
    = ∫0->∞ 1/(x^2) dx

    From my class I learned that
    ∫1->∞ 1/(x^2) dx , is convergent

    But now that the interval begins from 0 to infinity
    ∫0->∞ 1/(x^2) dx is divergent!

    Although my professor, and as well as the back of the book, tells me that ∫0->∞ x/(x^3+1) dx is convergent.

    This must mean that I did something wrong... what would that be?
     
  2. jcsd
  3. Jan 20, 2012 #2

    micromass

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    Uuuh, why?
     
  4. Jan 20, 2012 #3
    1/(x+1)

    if x = ∞
    1/(∞+1) ≈ 0
    0 ≈ 1/(∞+1) ≈ 1/(∞) ≈ 0
     
  5. Jan 20, 2012 #4

    HallsofIvy

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    You can compare [itex]x/(x^3+ 1)< x/x^3= 1/x^2[/itex] to show that the integral from 1 to infinity is finite. And now, because the original integrand is finite on the interval from 0 to 1, that entire integral is convergent.
     
  6. Jan 21, 2012 #5
    For the original integrand, we know that it is finite on the interval from 0 to 1 by plugging in 0 and 1? It seems like the original integrand does not have an asymptote like 1/x^2 does.
     
  7. Jan 21, 2012 #6

    SammyS

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    That is not enough to show that it's finite on the whole interval [0, 1].
     
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