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I dont get this convergent/divergent problem

  1. Feb 8, 2006 #1
    please help. its 4 am and im still stuck on my math hw

    find the values of p for which the interval converges and evalaute the integral for those values of p

    frankly, i dont really get this convergent/divergent fooey

    here is the question

    48. interval of 1/(x(lnx)^p) dx from e to infinity

    ill draw it in a second

    Last edited: Feb 8, 2006
  2. jcsd
  3. Feb 8, 2006 #2
    lol nice sketch

    use compartive p test to check answers
  4. Feb 8, 2006 #3
    \int_{e}^{\infty} x dx = \frac{1}{xln(x)^p}

    Will converge for p>1
    Diverge for p less or equal 1

    based on

    \int_{1}^{\infty} x dx = \frac{1}{x^p}
  5. Feb 8, 2006 #4
    thanks so much
  6. Feb 8, 2006 #5


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    Homework Helper

    An improper integral is convergent if it is equal to a finite number, and divergent otherwise.

    In your integral (by the way: nice sketch) put u = ln x so that du = 1/x dx and hence

    [tex]\int_{x=e}^{\infty} \frac{1}{x(lnx)^{p}} dx = \int_{u=1}^{\infty} \frac{1}{u^{p}} du[/tex]

    Can you get it from there?
  7. Feb 8, 2006 #6
    i got it now. thanks

  8. Feb 8, 2006 #7
    sorry to bother you guys, but im still a little confused

    so all i do is find the integral of xlnx^-p?


    would that be all i have to do since i said that it has to be greater than 1?

    and its still divergent by the way, even though i made p = 2
    Last edited: Feb 8, 2006
  9. Feb 8, 2006 #8
    \int_{e}^{\infty} \frac{1}{xln(x)^p}
    Will converge to 1
    when p=2
  10. Feb 8, 2006 #9
    Like benorin said since you can use substitution
    with u=ln(x) and du=dx/x
    you will get the
    Substitute in the bounds with u
    Lower: u(e)=1
    Higher: u(infinity)=infinity
    So from
    \int_{e}^{\infty} \frac{1}{xln(x)^p}

    you get
    \int_{1}^{\infty} \frac{1}{x^p}du
    which is a definition of a ptest which defines that
    \int_{e}^{\infty} \frac{1}{x)^p}dx
    will Converge for p>1 and Diverge for p less than or equal to 1

    Btw another way you can test when p=2 is the
    \int_{1}^{\infty} \frac{1}{x^2}

    [tex] \lim_{B->\infty}[ \frac{-1}{x} ]_{1}^{B}

    -1/B- (-1/1)
    the (-1/B) goes to zero so you have
    Last edited: Feb 8, 2006
  11. Feb 8, 2006 #10
    thanks sooooo much. you guys just made my day :cool:
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