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Improper integral

  1. Dec 10, 2007 #1
    Regarding the integral from -infinity to 0

    of 6/(5x-2)


    --> I arrive at 6/5*ln(u), is this the right thing?

    How would I evaluate something like ln(-2), or do I just assume it is divergent since it the other limit will come out to neg infinity?
     
  2. jcsd
  3. Dec 10, 2007 #2

    quasar987

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    You have made the change of variables

    [tex]\int_{-\infty}^{0}\frac{6}{5x-2}dx = \frac{6}{5}\int_{-\infty}^{-2}\frac{du}{u}[/tex]

    But obviously, it's not true that ln(u) is an antiderivative of 1/u on the domain (-\infinity,-2) simply because ln(u) is not defined there. So your change of variable was not so useful. Try u=2-5x.
     
  4. Dec 10, 2007 #3
    wait, do you mean 5x - 2?
     
  5. Dec 10, 2007 #4

    Hurkyl

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    ln(u) is not the antiderivative of 1/u over the nonzero reals. ln(u) is a simplification that is only valid for positive u.
     
  6. Dec 10, 2007 #5
    So, what substitution should I use?
     
  7. Dec 10, 2007 #6

    Dick

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    Hurkyl is trying to get you to recall ln(|u|) is also an antiderivative for 1/u.
     
  8. Dec 10, 2007 #7
    I am still unsure wth I should do -- u-subs or what?

    Thank you.
     
  9. Dec 10, 2007 #8

    Dick

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    You already did the u-subs correctly. The antiderivative is (6/5)*ln(|5x-2|). Now you just have to think what happens if you put the limits in.
     
  10. Dec 10, 2007 #9

    quasar987

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    No, I mean 2-5x. Try it.
     
  11. Dec 10, 2007 #10
    where did that come from, though?
     
  12. Dec 10, 2007 #11

    quasar987

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    it comes from choosing a change of coordinate such that the bounds of the integral become positive, so that now ln(u) is an antiderivative to the integrand.
     
  13. Dec 10, 2007 #12
    But how was that derived, can anyone explain?
     
  14. Dec 10, 2007 #13

    quasar987

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    Because with u(x)=5x-2, the upper bound of the intregral became -2. So using u(x) = -(5x-2), instead, it becomes -(-2)=2
     
  15. Dec 11, 2007 #14

    HallsofIvy

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    If u is negative, |u|= -u. Since, on this range, 5x-2 is negative, |5x-2|= 2- 5x.
     
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