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A Definite integral where solution. involves infinity - infinity

  1. Mar 18, 2009 #1
    Last edited: Mar 18, 2009
  2. jcsd
  3. Mar 18, 2009 #2

    Mark44

    Staff: Mentor

    After using partial fractions decomposition, I get this:
    [tex]
    \int_0^B \frac{-1/2}{x + 1} + (1/2)\frac{x + 1}{x^2 + 1}dx
    [/tex]

    I'm not sure this is what you got. After integrating, the antiderivative will be in terms of B. Take the limit of this expression as B approaches infinity.
     
  4. Mar 18, 2009 #3

    Dick

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    Combine the parts that go to infinity into one term. ln((x+1)^(1/2)/(x^2+1)^(1/4)). You want to figure out the limit of that as x->infinity. Can you see where to go from there?
     
  5. Mar 18, 2009 #4
    Thank you very much Mark44 and Dick.
    I got the same integral as Mark44.
    I understand what Dick is saying but I'm not allowed to use L hospital's rule at school. I can't figure out any other way.
    Is there an alternate way to evaluate the above integral?
     
  6. Mar 18, 2009 #5

    Dick

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    Factor an x^(1/2) out of the numerator and denominator of (x+1)^(1/2)/(x^2+1)^(1/4).
     
  7. Mar 18, 2009 #6
    Correct me if i am wrong.
    ln [x^(1/2) { 1+ 1/ x^(1/2) } / x^(1/2) { 1+ 1/ x^(1/2) }]

    I would get ln 1.

    which is equal to zero and hence the required answer.
    Thanks a lot Mark 44 and Dick.
     
  8. Mar 18, 2009 #7

    Dick

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    The conclusion is right, but your algebra leaves a lot to be desired. If you mean
    (x+1)^(1/2)=(x)^(1/2)*(1+1/x)^(1/2), that's ok. If you mean (x+1)^(1/2)=x^(1/2)+1/(x^(1/2)) or x^(1/2)+(1/x)^(1/2), both of those are wrong. And what happened with the 1/4 power in the denominator? Try to use enough parentheses to make it clear what you mean.
     
  9. Mar 18, 2009 #8
    oops!
    {x^(1/2){1+1/x}^(1/2)} / x ^(1/2){1+1/(x^2)}^(1/4) .
    i got this in the notebook, forgot to use paranthesis and forgot to edit the second term that i copy pasted from the first.
     
  10. Mar 18, 2009 #9

    Dick

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    Perfect.
     
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