Results 1 to 2 of 2

Thread: Help with integration

  1. #1
    Junior Member
    Joined
    Dec 2009
    Posts
    39
    Thanks
    1

    Help with integration

    Could you please help me with this:

    show that for -T<t<T, the function $\displaystyle f_{t} : \mathbb{R} \rightarrow \mathbb{R} $ defined by
    $\displaystyle f_{t} (x) = \frac{e^{-x} (1-cos(tx))}{x^2} \bold{1}_{ [0, \infty) } (x) $

    is Lebesgue integrable and that

    $\displaystyle \int_{0}^{\infty} \frac{e^{-x} (1-cos(tx))}{x^2} dx = t .tan^{-1} t - \frac{1}{2} log (1+t^{2}) $

    We are allowed to use results from Lebesgue integration.

    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by Mimi89 View Post
    Could you please help me with this:

    show that for -T<t<T, the function $\displaystyle f_{t} : \mathbb{R} \rightarrow \mathbb{R} $ defined by
    $\displaystyle f_{t} (x) = \frac{e^{-x} (1-cos(tx))}{x^2} \bold{1}_{ [0, \infty) } (x) $

    is Lebesgue integrable and that

    $\displaystyle \int_{0}^{\infty} \frac{e^{-x} (1-cos(tx))}{x^2} dx = t .tan^{-1} t - \frac{1}{2} log (1+t^{2}) $

    We are allowed to use results from Lebesgue integration.
    One way would be to use Fubini's theorem. Start with the fact that $\displaystyle \int_0^\infty e^{-x}\cos(tx)\,dx = \frac1{1+t^2}$. The justification for that is that $\displaystyle e^{-x}\cos(tx)$ is dominated by $\displaystyle e^{-x}$ and therefore integrable over $\displaystyle [0,\infty)$; and the integral is equal to $\displaystyle \text{Re}\!\int_0^\infty e^{-(1+it)x}dx = \text{Re}\left.\frac{e^{-(1+it)x}}{-(1+it)}\right|_0^\infty = \frac1{1+t^2}.$

    Integrate with respect to t to see that $\displaystyle \int_0^T\!\!\!\int_0^\infty e^{-x}\cos(tx)\,dxdt = \int_0^T\frac1{1+t^2}dt$. Now use Fubini to change the order of integration and deduce that $\displaystyle \int_0^\infty\frac{e^{-x}\sin(tx)}xdx = \tan^{-1}t$ (after replacing T by t).

    Now repeat the process: integrate from t=0 to T again, use Fubini, and conclude that $\displaystyle \int_{0}^{\infty} \frac{e^{-x} (1-cos(tx))}{x^2} dx = \int_0^t\tan^{-1}u\,du$. Finally, integrate by parts to see that $\displaystyle \int_0^t\tan^{-1}u\,du = t \tan^{-1} t - \tfrac{1}{2}\log (1+t^{2}) $.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Nov 3rd 2010, 12:54 AM
  2. Replies: 2
    Last Post: Nov 2nd 2010, 04:57 AM
  3. Replies: 8
    Last Post: Sep 2nd 2010, 12:27 PM
  4. Replies: 2
    Last Post: Feb 19th 2010, 10:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 06:58 AM

Search Tags


/mathhelpforum @mathhelpforum