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Indefinite integral and proving convergence

  1. Oct 2, 2015 #1
    1. The problem statement, all variables and given/known data

    okay so the equation goes:

    ∫(x*sin2(x))/(x3-1) over the terminals:
    b= ∞ and a = 2

    2. Relevant equations

    Various rules applying to the convergence or divergence of integrals such as the p-test, ratio test, squeeze test etc

    3. The attempt at a solution

    Okay so I have tried factorising the bottom, I have tried the squeeze test and the ratio test to no success..

    would method should I be applying, because I am unsure how to split up this integral...

    Thanks
     
  2. jcsd
  3. Oct 3, 2015 #2
    Here's a hint: first notice that [itex]\sin^2 x \leq 1[/itex] for all [itex]x[/itex]. Then you can compare your integral with another one whose integrand is always greater. You can then prove the convergence of that integral.
     
  4. Oct 3, 2015 #3
    Therefore using the comparison test should i compare this integral to 1/x3 over the same interval?
     
  5. Oct 3, 2015 #4

    Zondrina

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    Homework Helper

    Lets apply a useful theorem. If ##\int_a^b |f(x)| \space dx## converges, then ##\int_a^b f(x) \space dx## converges. Note the converse of this theorem is not true. So you should always determine the convergence of ##\int_a^b |f(x)| \space dx## to determine if ##\int_a^b f(x) \space dx## converges.

    Applying this theorem to your given problem:

    $$\int_{2}^{\infty} \left| \frac{x \sin^2(x)}{x^3 - 1} \right| \space dx = \int_{2}^{\infty} \frac{|x| |\sin^2(x)|}{|x^3 - 1|} \space dx \leq \int_{2}^{\infty} \frac{|x|}{|x - 1| |x^2 + x + 1|} \space dx$$

    For ##x \in [2, \infty)##, we can remove the absolute values in the expression because all of the terms would be positive anyway. So the new question is, does this integral converge:

    $$\int_{2}^{\infty} \frac{x}{(x - 1) (x^2 + x + 1)} \space dx$$

    Hint: Start with a partial fraction expansion, and then complete the square.
     
  6. Oct 3, 2015 #5
    Right now I understand. So the absolute of f(x) is always said to converge but the the opposite isn't always true ?

    Right do we are performing a comparison test to x/(x^3-1)
     
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