Results 1 to 4 of 4

Thread: Improper integral

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    38

    Improper integral

    Evaluate $\displaystyle \int_0^1\sin(\ln x)\,dx$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by liyi View Post
    Evaluate $\displaystyle \int_0^1\sin(\ln x)\,dx$
    You can write, $\displaystyle \int_0^1 \sin (\ln x) dx = \int_0^1 x\sin (\ln x) \cdot (\ln x)' dx$

    Now let $\displaystyle t=\ln x$ and we get, $\displaystyle \int_{-\infty}^0 e^t \sin t ~dt$

    This integral is doable.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    You could try it this way:

    $\displaystyle \int_{0}^{1}sin(ln(x))dx$

    $\displaystyle \lim_{L\rightarrow{1}}\int_{0}^{L}sin(ln(x))dx$

    One way is you can use Integration by Parts for this:

    Let $\displaystyle u=sin(ln(x)), \;\ dv=dx, \;\ du=\frac{cos(ln(x))}{x}, \;\ v=x$

    Then you get $\displaystyle xsin(ln(x))-\int{cos(ln(x))}dx$

    Now, use parts again:

    $\displaystyle u=cos(ln(x)), \;\ dv=dx, \;\ du=\frac{-sin(ln(x))}{x}, \;\ v=x$

    Putting it together, we end up with:

    $\displaystyle \lim_{L\rightarrow{1}}\left[\frac{Lsin(ln(L))}{2}-\frac{Lcos(ln(L))}{2}\right]$

    Now, as you can see, as L approaches 1, we have:

    $\displaystyle 0-\frac{1}{2}=\frac{-1}{2}$
    Last edited by galactus; Jan 13th 2008 at 11:23 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    Quote Originally Posted by liyi View Post
    Evaluate $\displaystyle \int_0^1\sin(\ln x)\,dx$
    Double integration leads TPH's integral too.

    $\displaystyle \sin (\ln x) = \int_0^{\ln x} {\cos u\,du} .$ So

    $\displaystyle \int_0^1 {\sin (\ln x)\,dx} = - \int_0^1 {\int_{\ln x}^0 {\cos u\,du} \,dx} = - \int_{ - \infty }^0 {\underbrace {\int_0^{e^u } {dx} }_{e^u }\cos u\,du} .$

    Hence $\displaystyle \int_0^1 {\sin (\ln x)\,dx} = - \int_{ - \infty }^0 {e^u \cos u\,du} .$

    Now this is routine, substitute $\displaystyle u=-v,$

    $\displaystyle - \int_0^\infty {e^{ - v} \cos v\,dv} = - \text{Re} \int_0^\infty {e^{ - (1 - i)v} \,dv} = - \text{Re}\, \frac{1}
    {{1 - i}} = - \frac{1}
    {2}.$

    The rest follows.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. improper integral #5
    Posted in the Math Challenge Problems Forum
    Replies: 3
    Last Post: Jul 1st 2011, 01:29 PM
  2. improper integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Nov 18th 2009, 05:46 AM
  3. improper integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 16th 2009, 12:32 AM
  4. Improper Integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Mar 4th 2009, 01:52 PM
  5. Improper Integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jun 23rd 2008, 07:30 PM

Search Tags


/mathhelpforum @mathhelpforum