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Another test for convergence

  1. Sep 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Test the following integral for convergence
    [tex] \int^{2}_{0}\frac{dx}{1-x^{2}} [/tex]


    2. Relevant equations



    3. The attempt at a solution

    So far I have brought it down to

    [tex] \int^{2}_{0}\frac{1}{1-x}+\frac{1}{1+x} dx[/tex]

    However, it seems that this integral produces a natural logarithm, which then must be evaluated with a negative logerand, making it a complex number.
    My textbook is introductory calculus, so it deals strictly with real numbers, so I am puzzled.

    BiP
     
  2. jcsd
  3. Sep 17, 2012 #2

    LCKurtz

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    Have you noticed there is a problem at ##x=1##? Also, for what it's worth,$$
    \int \frac 1 u\, du = \ln|u| + C$$which works when the argument is negative.
     
  4. Sep 17, 2012 #3

    Dick

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    For one thing, the integrand is singular at x=1. You don't want to integrate directly from 0 to 2. That's dangerous. Try the integral from 0 to 1 first. And for another, you can take the integral of 1/(1-x) to be -log(|1-x|). Note the absolute value.
     
  5. Sep 17, 2012 #4
    Thanks!

    The problem has been reduced to

    [tex] \lim_{x→1-}ln(|1-x|)-\lim_{x→1+}ln(|1-x|) [/tex]

    I can't fathom how I would compute this though. I am not very good with one-sided limits being subtracted from each other.

    BiP
     
  6. Sep 17, 2012 #5

    Dick

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    Don't do that. Just work on the interval [0,1) first. If it doesn't converge, then the whole integral can't converge.
     
  7. Sep 17, 2012 #6

    Zondrina

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    Consider when x→1-. So you have the possibility of all values less than one being considered here, including the negative ones.

    As for x→1+, note that ln|-x| = ln(x). You can apply the abs before taking the ln I believe.
     
  8. Sep 17, 2012 #7

    Dick

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    Both limits are negative infinity. That's what's confusing Bipolarity. The difference is quite undefined.
     
  9. Sep 17, 2012 #8

    Zondrina

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    I was just stating that they could be computed :P, but yeah what you said would actually help him get the answer he wants.
     
  10. Sep 18, 2012 #9
    I get how to do the problem but can that limit actually be computed? If so, how?

    BiP
     
  11. Sep 18, 2012 #10

    Dick

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    The difference of the two terms can't be computed. You are clear on that, right?
     
  12. Sep 18, 2012 #11
    I see. Is it always the case that the difference between two limits, each of which does not by itself exist, is incalculable?

    BiP
     
  13. Sep 18, 2012 #12

    Dick

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    Almost. The form (+infinity)-(+infinity) is not defined. On the other hand you can define (+infinity)+(+infinity) to be +infinity. So you can define (+infinity)-(-infinity) to be +infinity. Do you see how that applies to your example when you have the other sign on the infinite limits?
     
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