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!

Homework Help: Use comparison theorem to show if integral is convergent or divergent

  1. Mar 21, 2010 #1
    1. The problem statement, all variables and given/known data

    int (e^-x)/(x)dx from 0 to infinity

    Determine if integral is convergent or divergent


    2. The attempt at a solution
    I assume because the bottom limit is 0 and there is an x in the bottom of the integral that this is going to be divergent but I still have to use the comparison theorem. I'm trying to find a function less then (e^-x)/(x) that also diverges to show that (e^-x)/(x) will diverge but I'm drawing a blank. I've tried messing around with 1/x^2 or just e^-x but both of those are still larger then the original (e^-x)/(x). Any suggestions?
     
  2. jcsd
  3. Mar 21, 2010 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    1/x is an interesting function in that it diverges both ways, the integral from 0 to 1 of 1/x is divergent, as is the integral from 1 to infinity (compare this to [tex]x^{1/2}[/tex] and [tex] \frac{1}{x^2}[/tex]). So when trying to do comparison with a 1/x term, it often helps to break up the integral into one from 0 to 1 and one from 1 to infinity, and see on each of those whether it converges or diverges
     
  4. Mar 21, 2010 #3
    Ok, so if I can break the function into two parts, as long as one part (in this case 1/x) converges or diverges then the entire function does?

    I thought I could only do these type of integrals by picking a function that was larger or smaller and then comparing.
     
  5. Mar 21, 2010 #4
    Since this is an improper integral at both limits of integration, that's why you should consider the two intervals (0, 1] and [1, ∞) separately. Try finding whether the integral from (0, 1] converges first.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook