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Evaluating Improper Integral with limits and comparison theorem

  1. Mar 24, 2013 #1
    1. The problem statement, all variables and given/known data

    evaluate the integral 1/(u^2 -36) from 0 to 6

    does the integral converge?

    2. Relevant equations
    3. The attempt at a solution

    integral 1/(u^2 -36)
    integral 1/((u-6)(u+6))
    Partial fraction decomposition
    1/((u-6)(u+6)) = A/(u-6) + B/(u+6)
    1=A(u+6) + B(u-6)
    1=(A+B)u +(6A-6B)
    A+B=0
    A=-B
    6A -6B=1
    -12B=1
    B=-1/12
    A=1/12
    1/12 int 1/(y-6) - 1/12 int 1/(y+6)
    1/12 ln|y-6| - 1/12 ln|y+6|

    I'm being told however by wolframalpha that it should be 1/12 ln|6-y| - 1/12 ln|y+6|
    How did that happen?

    I also wanted to try to use the comparison theorem to see if it converges or not.
    I use the function g(x)=1/y^2 and i know that does not converge from 0 to 6, and since it is larger than f(x) then f(x) also does not converge. Did i go about that correctly?
     
    Last edited: Mar 24, 2013
  2. jcsd
  3. Mar 24, 2013 #2

    Dick

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    Try and stick with one variable. I don't know why you changed from u to y, but there is no difference between |6-y| and |y-6|. And to show it diverges by a comparision test you want a function that's less than 1/((y-6)(y+6)) that diverges. 1/y^2 isn't less than 1/((y-6)(y+6)).
     
  4. Mar 24, 2013 #3
    Hmm, not sure why I changed the variables myself. Oh right, I forgot I had to take the absolute value. It also seems that I have the entire comparison theorem backwards..that's not good haha. I can't think of any obvious functions less than 1/((y-6)(y+6)) that diverges. Could you point me toward the right direction of finding one? Thanks for your help I really appreciate it. :smile:
     
  5. Mar 24, 2013 #4

    Dick

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    Does 1/(y-6) diverge on [0,6]? The 1/(y+6) factor is between 1/12 and 1/6 on [0,6]. Use that info to make a comparison function.
     
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