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Improper Integral Sinx/x^2 and similar with sinx

  1. Jun 9, 2015 #1
    1. The problem statement, all variables and given/known data

    This is the improper integral of which I have to study the convergence.

    ∫[1,+∞] sinx/x2 dx

    3. The attempt at a solution

    I have tried to use the absolute convergence.

    ∫f(x)dx converges ⇔ ∫|f(x)|dx converges

    but after i have observed that x^2 is always positive and the absolute value is only for the sinx, I do not how to move!

    How to see the convergence or the divergence of sinx when you can not to approximate to x for x->0?

    Thank you for the help!!:biggrin:
     
  2. jcsd
  3. Jun 9, 2015 #2

    bhobba

    Staff: Mentor

    Try the p test, then the comparison test for absolute convergence.

    Thanks
    Bill
     
  4. Jun 9, 2015 #3
    Try starting with the Squeeze Theorem.
     
  5. Jun 9, 2015 #4
    Try starting with the Squeeze Theorem.
     
  6. Jun 9, 2015 #5
    I tried the Squeeze Theorem.

    -1≤sinx≤+1 ⇒ -1/x2≤sinx/x2≤+1/x2

    the two functions converge so my function converge!

    Thank you very much!
     
  7. Jun 9, 2015 #6

    bhobba

    Staff: Mentor

    I gave you a like because you are on the right track - but that is wrong - the squeeze test applied to functions that converge to the same thing.

    Like I said use the p test on 1/x^2, note that the absolute value of your function is less than or equal to 1/x^2. Thus its absolute value converges by the comparison test, then apply the absolute convergence test.

    Thanks
    Bill
     
  8. Jun 9, 2015 #7

    Zondrina

    User Avatar
    Homework Helper

    You have already pointed this out:

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