Results 1 to 3 of 3

Thread: Cauchy and fourier tranform

  1. #1
    Member
    Joined
    Nov 2005
    Posts
    111

    Cauchy and fourier tranform

    Hi,
    I have this problem and I dont know how to finish it:

    Using the Cauchy Theorem, prove that the fourier tranform of 1/(1+t^2) is
    (pi)*e^(-2*pi*|f|) .( you must show the intergration contour) Stetch the power spectrum.

    I applied the fourier transform formula but then tried to break down the
    1/(1+t^2) but I get stuck to apply the Cauchy theorem.
    Please can I have some help?

    Thank you
    B
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    The Fourier transform (without the constant) is $\displaystyle F(\omega) = \int_{-\infty}^{\infty} e^{-i\omega t} \frac{1}{1+t^2} dt$ suppose for now that $\displaystyle \omega > 0$. And consider $\displaystyle f(z) = e^{-i\omega z} (1+z)^{-2}$. Let $\displaystyle C$ be the semi-circular contour from $\displaystyle -R$ to $\displaystyle R$ transversed counterclockwise back to $\displaystyle -R$ where $\displaystyle R$ is a really big number. We will write, $\displaystyle \frac{e^{-i\omega z}}{z^2+1} = \frac{\frac{1}{2i}e^{-i\omega z}}{z+i} - \frac{\frac{1}{2i}e^{-i\omega z}}{z-i}$. Thus, $\displaystyle \oint_C \frac{e^{-i\omega z}}{z^2+1} = \frac{\frac{1}{2i}e^{-i\omega z}}{z+i} - \frac{\frac{1}{2i}e^{-i\omega z}}{z-i} dz = \oint_C \frac{\frac{1}{2i}e^{-i\omega t}}{z-i} dz$ because the first fraction is holomorphic in some open set containg the curve ands its boundary so by Cauchy's closed curve theorem the integral is zero. But, $\displaystyle \oint_C \frac{\frac{1}{2i}e^{-i\omega z}}{z-i} dz = \frac{1}{2i}\cdot (2\pi i)\cdot [e^{-i\omega z}]'_{z=i} = -i\pi \omega e^{\omega} $ by Cauchy's integral theorem.

    Is that what you are looking for?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2005
    Posts
    111
    Hi ThePerfectHacker,

    thank you for the help. However it is not the same expression we need to prove. Basically I think that after the integration, you were supposed to find:
    $\displaystyle \pi*e^{-2*\pi*|f|}
    $ ( the FT of $\displaystyle \frac{1}{1+t^2}$)

    What you did seems correct, and I got the idea how I need to proceed. I will try to work on it to see if I can find the right expression.
    I will let you know if I find it.
    B
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Subsequence of a Cauchy Sequence is Cauchy
    Posted in the Differential Geometry Forum
    Replies: 9
    Last Post: Sep 30th 2010, 01:29 AM
  2. How to tranform 2x=3y?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Apr 2nd 2010, 05:21 PM
  3. how to tranform between functions in integral
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: Mar 5th 2010, 12:58 AM
  4. Laplace Tranform ?!?!
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jun 26th 2009, 11:14 PM
  5. from fourier transform to fourier series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 1st 2008, 06:35 AM

Search Tags


/mathhelpforum @mathhelpforum