1. Not finding help here? Sign up for a free 30min 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!

Improper Integrals

  1. Mar 3, 2008 #1
    1. The problem statement, all variables and given/known data

    do the following integrals converge?

    i) [tex]\int_0^{1}\frac{dx}{x^{3/2}e^{x}}[/tex]

    ii) [tex]\int_0^{1}\frac{x}{\sqrt{1-x^{2}}}dx[/tex]


    3. The attempt at a solution

    looking at them i can guess that they both diverge - proving this is the hard part - this is what i have got but it doesn't prove anything...

    i) [tex]\frac{1}{x^{3/2}}[/tex][tex]\geq1[/tex]

    [tex]\frac{1}{x^{3/2}e^{x}}[/tex][tex]\geq{e^{-x}}[/tex]

    [tex]e^{-x}[/tex] converges



    [tex]\frac{1}{e^{x}}[/tex][tex]\leq1[/tex]

    [tex]\frac{1}{x^{3/2}e^{x}}[/tex][tex]\leq{x^{-3/2}[/tex]

    [tex]x^{-3/2}[/tex] divereges



    ii) [tex]x\leq1[/tex]

    [tex]\frac{x}{\sqrt{1-x^{2}}}[/tex][tex]\leq{\frac{1}{\sqrt{1-x^{2}}}[/tex]

    [tex]{\frac{1}{\sqrt{1-x^{2}}}[/tex] diverges (cannot prove it though)



    [tex]{\frac{1}{\sqrt{1-x^{2}}}\geq1[/tex]

    [tex]{\frac{x}{\sqrt{1-x^{2}}}[/tex][tex]\geq x[/tex]

    x converges

    any help would be much appreciated.
     
    Last edited: Mar 3, 2008
  2. jcsd
  3. Mar 3, 2008 #2
    I take your last statement above to mean that when you integrate both sides of the inequality with respect to x from 0 to 1, the right-hand side converges. That is true. Now what does that tell you about the left-hand side?

    Again, what does that tell you about the left-hand side? Ditto for ii).
     
  4. Mar 3, 2008 #3
    maybe i'm missing something but not much because the left hand side is greater than the right, the right converges, this does not mean the same for the left - it might diverge.

    The same for all the others - the comparisions don't tell us much.

    If all the inequalities were reversed then i wouldn’t have a problem.

    i think...?
     
    Last edited: Mar 3, 2008
  5. Mar 3, 2008 #4
    Yes. Exactly. You need to find an expression so that the inequalities are reversed.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Improper Integrals
  1. Improper Integral (Replies: 5)

  2. Improper Integrals (Replies: 3)

  3. Improper integrals (Replies: 5)

  4. Improper integrals (Replies: 2)

  5. Improper Integral (Replies: 7)

Loading...