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!

Classify the following improper integral with the parameter

  1. Jul 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Discuss for alpha the convergence of the following improper integral:
    [tex] \displaystyle \int\limits_{0}^{3}{\frac{{{x}^{3\alpha }}}{{{\left( 9-{{x}^{2}} \right)}^{\alpha }}}}[/tex]



    2. Relevant equations



    3. The attempt at a solution
    Well, my attempt was to simplify the integral to:

    [tex] \displaystyle \int\limits_{0}^{3}{{{\left( \frac{{{x}^{3}}}{9-{{x}^{2}}} \right)}^{\alpha }}}[/tex]

    After that my idea was to bound the function above by [tex] \displaystyle {{x}^{3a}}[/tex] but I don't know if that's correct, because [tex] \displaystyle {{x}^{3a}}[/tex] is greater than the original function but it could happen that it's not greater for all x, so how can I find a function I can use to bound the function in the interval of integration [0,3]?

    Thanks for your help.
     
  2. jcsd
  3. Jul 4, 2012 #2
    ANy help?
     
  4. Jul 4, 2012 #3

    Mute

    User Avatar
    Homework Helper

    For positive [itex]\alpha[/itex] your integrand diverges at x = 3, where the denominator is zero. How are you going to find a function which bounds that? I think you need a revised strategy to figure out the convergence properties of this integral. You might try choosing a function (or functions) which is greater than your integrand but has the same divergence, and which you can evaluate the integral of.

    Similarly, for negative [itex]\alpha[/itex] your integrand diverges at x = 0.
     
  5. Jul 4, 2012 #4
    Well, I did find two functions which bounds that:


    [tex] \dfrac{4x}{9-x^2}\leq \dfrac{x^3}{9-x^2}\leq \dfrac{9x}{9-x^2}[/tex]

    Why you say I cannot bound it?

    Now what can I do with that? It's still tough for me
     
  6. Jul 4, 2012 #5

    Mute

    User Avatar
    Homework Helper

    Perhaps I misunderstood your original post. It appeared that you were trying to bound the function with x3a rather than x3a/(9-x2).

    In any event, I'm not sure if just using bounds alone are going to be the easiest way to show the result you want, at least for positive alpha. You might want to try changing variables first, and then finding a bound.

    Another piece of knowledge that will be useful is the answer to this question: for what values of [itex]\alpha[/itex] does

    [tex]\int_0^c du~\frac{1}{u^\alpha}[/tex]

    converge, where the upper limit c is a finite constant?
     
  7. Jul 4, 2012 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    BTW:

    I have this pet peeve about leaving the differential symbol out of integrals, dx, in this case.

    [itex] \displaystyle \int\limits_{0}^{3}{\frac{{{x}^{3\alpha }}}{{{\left( 9-{{x}^{2}} \right)}^{\alpha }}}}\,dx[/itex]
     
    Last edited: Jul 4, 2012
  8. Jul 4, 2012 #7

    Mute

    User Avatar
    Homework Helper

    I think the entire point of the OP's problem is to find that out! You probably shouldn't just give it away! =P
     
  9. Jul 4, 2012 #8

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Oops ! I edited that out.

    You might want to also.
     
  10. Jul 5, 2012 #9

    Mute

    User Avatar
    Homework Helper

    Too late to completely edit it out the fact that it diverges for some values of alpha; fortunately I had the foresight to remove the value from the quoted post when I posted.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Classify the following improper integral with the parameter
  1. Improper integrals (Replies: 5)

  2. Improper integrals (Replies: 2)

  3. Improper Integral (Replies: 7)

Loading...