Page 1 of 2 12 LastLast
Results 1 to 15 of 17

Math Help - Improper Integrals

  1. #1
    Member
    Joined
    Feb 2008
    Posts
    102

    Improper Integrals

    Show that  \int_{0}^{1} cos(x)dx converges.

    A little bit puzzled here.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by larson View Post
    Show that  \int_{0}^{1} cos(x)dx converges.

    A little bit puzzled here.
    It is defined on [0,1] so just do it \int_0^{1}cos(x)dx=sin(x)\bigg|_0^{1}=sin(1)-0=sin(1) therefore it converges
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2008
    Posts
    102
    Quote Originally Posted by Mathstud28 View Post
    It is defined on [0,1] so just do it \int_0^{1}cos(x)dx=sin(x)\bigg|_0^{1}=1-0=1 therefore it converges
    crap... i wrote the problem wrong... it's ...  \int_{0}^{1} \frac {cos(x)}{x^2}dx show that it CONVERGES.
    Last edited by larson; April 19th 2008 at 06:29 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by larson View Post
    Show that  \int_{0}^{1} cos(x)dx converges.

    A little bit puzzled here.
    This seems a little weird. Cosine is continous for all real numbers and has an antiderivative.

    This type of question is usuall asked when you have a vertical asymptote or somthing.

    It would converge to \sin(1)

    Maybe I missed the point?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by larson View Post
    crap... i wrote the problem wrong... it's ...  \int_{0}^{1} \frac {cos(x)}{x}dx
    ahh the plot thickens

    express cosine as its power series \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}

    so then

    \frac{\cos(x)}{x}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n+1)!}

    integrate term by term and evaluate.

    This will converges by the AST.

    Good luck.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Quote Originally Posted by larson View Post
    crap... i wrote the problem wrong... it's ...  \int_{0}^{1} \frac {cos(x)}{x}dx
    That integral diverges, note that \frac{\cos(x)}{x}\sim{\frac{1}{x}} as x\rightarrow{0}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Feb 2008
    Posts
    102
    Quote Originally Posted by TheEmptySet View Post
    ahh the plot thickens

    express cosine as its power series \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}

    so then

    \frac{\cos(x)}{x}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n+1)!}

    integrate term by term and evaluate.

    This will converges by the AST.

    Good luck.
    I'm a little confused here by this... \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!} is that just general knowledge?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by larson View Post
    crap... i wrote the problem wrong... it's ...  \int_{0}^{1} \frac {cos(x)}{x}dx
    The integral diverges?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    My above post is wrong I use the sine power series

    express cosine as its power series \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}

    so then

    \frac{\cos(x)}{x}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n-1}}{(2n)!}

    integrate term by term and evaluate and the sum diverges.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by larson View Post
    I'm a little confused here by this... \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!} is that just general knowledge?
    I used the wrong power series. sorry
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Feb 2008
    Posts
    102
    Quote Originally Posted by Mathstud28 View Post
    The integral diverges?
    I'm so sorry, I'm making this much much more confusing that what I intended it to be. It converges...
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by larson View Post
    I'm so sorry, I'm making this much much more confusing that what I intended it to be. It converges...
    This integral \int_0^{1}\frac{cos(x)}{x}dx diverges?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member
    Joined
    Feb 2008
    Posts
    102
    Quote Originally Posted by Mathstud28 View Post
    This integral \int_0^{1}\frac{cos(x)}{x}dx diverges?
    Sorry, I wrote it down wrong. The problem is...

     \int_{0}^{1} \frac{cos(x)}{x^2}dx explain how this CONVERGES.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Quote Originally Posted by larson View Post
    I'm so sorry, I'm making this much much more confusing that what I intended it to be. It converges...
    Maybe it is \int_0^1\frac{1-\cos(x)}{x}dx or \int_0^1\frac{1-\cos(x)}{x^2}dx
    (both converge since 1-\cos(x)\sim{\frac{x^2}{2}} as x\rightarrow{0})
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by larson View Post
    Sorry, I wrote it down wrong. The problem is...

     \int_{0}^{1} \frac{cos(x)}{x^2}dx explain how this CONVERGES.
    ok \int_0^{1}\frac{cos(x)}{x^2}dx=\int_0^{1}\sum_{n=1  }^{\infty}\frac{(-1)^{n}x^{2n-2}}{(2n)!}dx=\sum_{n=1}^{\infty}\frac{(-1)^{n}x^{2n-1}}{(2n-1)\cdot{(2n)!}} which diverges from 0 to 1...it still doesn't converge
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Improper Integrals
    Posted in the Differential Geometry Forum
    Replies: 9
    Last Post: February 15th 2011, 12:09 AM
  2. improper integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 7th 2009, 05:06 AM
  3. Improper Integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 17th 2009, 08:29 PM
  4. Improper Integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 5th 2008, 11:57 AM
  5. Improper Integrals
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 24th 2007, 05:05 PM

Search Tags


/mathhelpforum @mathhelpforum