Results 1 to 14 of 14

Math Help - Integration of unbounded function.

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Integration of unbounded function.

    For which values of p the integral \int^2_1 \frac{x}{(x-1)^p}dx converges?

    Thank you...
    Last edited by Also sprach Zarathustra; June 24th 2010 at 08:22 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    What have you tried so far?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Nothing actually...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Well, I would try computing the integral. Do your normal Calc II stuff and see what happens.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    I solved this!

    \int^2_1\frac{x}{(x-1)^p}dx<\int^2_1\frac{2}{(x-1)^p}dx

    Now, \int\frac{2}{(x-1)^p}dx=\frac{1}{1-p}\frac{1}{(x-1)^{p-1}}

    By some theorem...
    If p-1<1 or p<2 then the above integral converges.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Well, I would have used u = x-1 on the original integral.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by Also sprach Zarathustra View Post
    \int^2_1\frac{x}{(x-1)^p}dx<\int^2_1\frac{2}{(x-1)^p}dx

    Now, \int\frac{2}{(x-1)^p}dx=\frac{1}{1-p}\frac{1}{(x-1)^{p-1}}

    By some theorem...
    If p-1<1 or p<2 then the above integral converges.
    Check your work, the answer is  p<1 .
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by chiph588@ View Post
    Check your work, the answer is  p<1 .
    I disagree!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    I agree with chiph588. Try u substitution! The result pops right out.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by Also sprach Zarathustra View Post
    I disagree!
    Let  p=1 :

     \int \frac{x}{x-1}dx = x+\ln|x-1| . What happens when you plug in  x=1 ?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    I still don't get it!

    ...with substetution x-1=u.

    \int\frac{x}{(x-1)^p}dx=\int\frac{u+1}{(u)^p}du=\int\frac{1}{u^{p-1}}du+\int\frac{1}{u^p}du=\frac{1}{u^{p-2}}\frac{1}{2-p}+\frac{1}{u^{p-1}}\frac{1}{1-p}

    ...then:

    p-2<1 and p-1<1

    or:


    p<3 and p<2

    Therefore:

    p<2
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by Also sprach Zarathustra View Post
    ...with substetution x-1=u.

    \int\frac{x}{(x-1)^p}dx=\int\frac{u+1}{(u)^p}du=\int\frac{1}{u^{p-1}}du+\int\frac{1}{u^p}du=\frac{1}{u^{p-2}}\frac{1}{2-p}+\frac{1}{u^{p-1}}\frac{1}{1-p}

    ...then:

    p-2< 0 and p-1< 0

    or:


    p<3 and p<2

    Therefore:

    p<2
    .
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Why is that???
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    I have got it!!!! silly me!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. range of unbounded function
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 11th 2012, 03:27 AM
  2. Holomorphic function on a unbounded domain
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 5th 2011, 12:46 AM
  3. Continuous and Unbounded Function
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 25th 2010, 01:06 AM
  4. Continous, unbounded function in metric spaces
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 17th 2009, 05:28 AM
  5. The UnBounded Function
    Posted in the Calculus Forum
    Replies: 9
    Last Post: April 18th 2007, 08:21 PM

Search Tags


/mathhelpforum @mathhelpforum