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A difficult integration (antiderivative) problem

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Decide whether f(x)=[tex]\int[/tex] (1-cos(x))/x^2 is improperly integrable on (0, infinity).


    2. Relevant equations



    3. The attempt at a solution
    I understand the concept of improper integration, but I don't see how to take the antiderivative -- I tried substitution and by parts, but I can't get there.
     
  2. jcsd
  3. Nov 16, 2009 #2

    Hurkyl

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    Have you tried an approach that doesn't rely on finding an elementary anti-derivative?
     
  4. Nov 16, 2009 #3
    Hmm -- what do you mean? I did graph the function, and it seems apparant that it is indeed improperly integrable -- but I don't think stating as much is a sufficient response.
     
  5. Nov 16, 2009 #4

    Mark44

    Staff: Mentor

    Do you know about Maclaurin series? E.g., the Maclaurin series for cos x is 1 - x2/2! + x4/4! - x2/6! ...
     
  6. Nov 16, 2009 #5

    Hurkyl

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    Can you articulate why it is apparent?


    P.S. is it really apparent from graphing the functions that
    [tex]\int_1^{+\infty} \frac{dx}{x^2}[/tex]​
    exists, but
    [tex]\int_1^{+\infty} \frac{dx}{\sqrt{x}}[/tex]​
    does not? They look pretty much the same to me.
     
  7. Nov 16, 2009 #6
    Hmm, yes, I didn't think of that. But in that case, I get that the value of the entire integral approaches infinity as x-->infinity: all the terms go to zero except that second term [(x^2)/2!] / x^2, which is 1/2; the antiderivative is therefore x/2, which approaches infinity as x-> infinity?
     
  8. Nov 16, 2009 #7

    lanedance

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    the way i read the question, you don't have to give a value, but just show the integral either exists or doesn't

    so, wth that in mind, how about splitting it up into two integrals over (0,1) and (1,infinity), consider each part separately and have another look at Hurkyl's & Mark's comments
     
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