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Comparison test for basic integral

  1. Sep 25, 2011 #1
    integral 1/(1+x^2)dx from 0 to infinity

    I decided to compare that to 1/(1+x) (saying 1/(1+x) > (1/(1+x^2)) but this diverges when the original equation converges. Can someone explain why the integral 1/(1+x) was not a proper choice and what the process would be to find a correct comparison.
     
  2. jcsd
  3. Sep 25, 2011 #2

    Mark44

    Staff: Mentor

    A better choice is 1/x2, which is larger than 1/(1 + x2) for large x.

    1/(1 + x) was not a good choice if you believed that the original integral was convergent (which I think you did).

    When you use comparison for series or integrals, if you believe your series or integral is convergent, you want to find a series or integrand that 1) converges, and 2) is larger than the one you have.

    Iif you believe your series or integral is divergent, you want to find a series or integrand that 1) diverges, and 2) is smaller than the one you have.

    No other combinations are applicable.
     
  4. Sep 25, 2011 #3
    can you explain why 1/x^2 converges? The integral would be -1/x and from 0 to infinity that would be -1/(infinity)+1/0 wouldn't it? I know that 1/x^2 converges because I remember hearing that in class but i don't get why. Thank you very much for your help, I appreciate it a ton.
     
  5. Sep 25, 2011 #4
    Also, any general tips on comparison theorum would be awesome. My first test is in three days (calc II) and that is the only idea I don't fully understand.
     
  6. Sep 25, 2011 #5

    Mark44

    Staff: Mentor

    No, you can never substitute infinity in for a number as you have done.

    In the comparison I was thinking of, the limits are 1 and infinity. It shouldn't matter that the lower limit is different, since all we're doing is ignoring a small (and finite) part of the aream beneath the graph of y = 1/(1 + x2).

    To carry out the integral you asked about, you need to use limits.

    [tex]\int_1^{\infty}\frac{dx}{x^2} = \lim_{b \to \infty} \left.\frac{-1}{x}\right|_1^b[/tex]
    [tex]=\lim_{b \to \infty}-1/b + 1/1 = 1[/tex]
    As long as the lower limit of integration is positive,
    [tex]\int_a^{\infty}\frac{dx}{x^2}[/tex]
    converges. That's why I said that 1/x2 was a better choice "for large x."
     
  7. Sep 25, 2011 #6
    The limit is from 0 to infinity though, that is why I'm confused.
     
  8. Sep 25, 2011 #7

    Mark44

    Staff: Mentor

    Is the goal here to evaluate the integral or to just show that it is convergent? If all you need to do is show that the integral converges, then what I said earlier is applicable. It doesn't matter that the lower limit of integration is 1 instead of 0. For your function, the important thing is what happens as x gets large.
     
  9. Sep 25, 2011 #8
    Alright. So does 1/0 not mean that the integral diverges? and I am supposed to evaluate the integral which I understand how to do. I really can't see how 1/x^2 from 0 to infinity doesn't diverge. Thanks for bearing with me...
     
  10. Sep 25, 2011 #9

    Mark44

    Staff: Mentor

    1/0 is meaningless. If all you need to do is evaluate the integral, that's straightforward, but since this is an improper integral (because of the infinite limit of integration), you'll need to use a limit to do that. See post #5 for an example of this technique.

    You still seem to be struggling at understanding what I did. The fact that my integral runs from 1 to infinity instead of from 0 to infinity DOESN'T MATTER. As I said before, I am ignoring a small amount of area under the graph of y = 1/(1 + x2), and THAT DOESN'T MATTER, EITHER. The important thing is what happens when x gets very large, AND THAT DOES MATTER.
     
  11. Sep 25, 2011 #10
    I'll talk to my teacher tomorrow. You don't have to reply again, I don't want to annoy you with my inability to understand.

    [tex]\int_0^{\infty}\frac{dx}{x^2} = \lim_{c \to \infty} \left.\frac{-1}{x}\right|_0^c

    =\lim_{c \to \infty}-1/c + 1/0 = 0 + 1/0[/tex] ????
     
  12. Sep 25, 2011 #11
    What mark44 is trying to say, is that if you need to see if the TOTAL area converges or diverges then substituting 0 for 1 in your original function will create a finite area, then we can use that in the comparison instead of making it more complicated by going from 0 to inf when comparing it to 1/x^2.

    But like he said, are you just seeing if the integral converges or do you need to actually evaluate it...?
     
  13. Sep 25, 2011 #12

    Mark44

    Staff: Mentor

    I am evaluating this integral:

    [tex]\int_1^{\infty}\frac{dx}{x^2} = \lim_{c \to \infty} \left.\frac{-1}{x}\right|_1^c

    =\lim_{c \to \infty}-1/c + 1/1 = 1[/tex]
     
  14. Sep 25, 2011 #13
    My teacher never said we could change the limit for the comparison theorem. That is very helpful - thanks for your help.

    and romsofia, the actual limit was not my problem, although we were supposed to evaluate it.
     
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