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Calculus II - Improper Integral Problem

  1. Dec 1, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral: ∫(0 to ∞) [dv/((1+v^2)(1+tan^-1(v))]

    2. Relevant equations

    U-substitution, taking limit to evaluate improper integrals


    3. The attempt at a solution

    http://imgur.com/CjkRF
    As you can see in the image, I try u-substitution and then take the integral. I end up with ln(0), though, because arctan(0) = 0. The correct answer is ln(1 + ∏/2), but I'm not sure how to get there.
     
    Last edited: Dec 1, 2012
  2. jcsd
  3. Dec 1, 2012 #2

    LCKurtz

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    With your substitution the denominator is ##1+u##, not just ##u##. It works either way, but I would suggest the substitution ##u=1+\arctan v## in the first place.
     
  4. Dec 1, 2012 #3
    Sorry, I intended to write u = 1 + arctan(v), not u = arctan(v). The 1 goes to zero either way though, so I still have the same problem of getting ln(0).
     
  5. Dec 1, 2012 #4
    Nevermind, I see what you're saying, that would change the bounds. Thanks!
     
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