1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding Where an Improper Integral Converges

  1. Oct 17, 2012 #1
    Hi all,

    This is a case of a book answer going against Wolfram's and my answer.

    The problem is ∫e(ln(x)/x)dx

    The book claims the answer is ∞.

    I would think it is a case of ∞/∞ and use L'Hospital's Rule. Wolfram has the same solution.

    *= lima->∞(1/x)/1
    = 0

    Which would be correct?
     
  2. jcsd
  3. Oct 17, 2012 #2

    pwsnafu

    User Avatar
    Science Advisor

  4. Oct 17, 2012 #3
    Good call. I must have typed it in incorrectly.

    Is it correct that the integral diverges because lim as x approaches infinity of lnx/x is an indeterminate?

    Thanks!
     
  5. Oct 17, 2012 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    No. Being indeterminate has nothing to do with it.

    It's true that [itex]\displaystyle \lim_{x\,\to\,\infty} \frac{\ln(x)}{x}=0\,,\ [/itex] as well as [itex]\displaystyle \lim_{x\,\to\,\infty} \frac{1}{x}=0\ .[/itex]

    Those results are necessary conditions (but not sufficient) for [itex]\displaystyle \int_{e}^{\infty} \frac{\ln(x)}{x}\,dx\ [/itex] to converge.

    Does [itex]\displaystyle \int_{e}^{\infty} \frac{1}{x}\,dx\ [/itex] converge?
     
  6. Oct 19, 2012 #5
    No, but I'm not sure why.


    EDIT:

    It's because ∫1/x dx = ln(x)

    and ln(∞) = ∞

    So since

    ∫ln(x)/x = ln(x)2/2, and ln(∞)2/2= ∞,

    the integral diverges.

    Correct?
     
    Last edited: Oct 19, 2012
  7. Oct 20, 2012 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    We know that [itex]\displaystyle \int_{e}^{\infty} \frac{1}{x}\,dx[/itex] diverges, because we know that [itex]\displaystyle \sum_{n=3}^{\infty} \frac{1}{n}[/itex] diverges.
     
  8. Oct 21, 2012 #7
    But 1/infinity goes to zero, correct?
     
  9. Oct 21, 2012 #8

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes 1/n goes to zero as n goes to ∞ . That's basically looking at the sequence [1/n] .

    However, the infinite series, [itex]\displaystyle \sum \frac{1}{n}\ ,[/itex] diverges.
     
  10. Oct 21, 2012 #9
    Thanks for clarifying that; I need to practice these.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook