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Math Help - Cauchy and fourier tranform

  1. #1
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    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
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  2. #2
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    The Fourier transform (without the constant) is F(\omega) = \int_{-\infty}^{\infty} e^{-i\omega t} \frac{1}{1+t^2} dt suppose for now that \omega > 0. And consider f(z) = e^{-i\omega z} (1+z)^{-2}. Let C be the semi-circular contour from -R to R transversed counterclockwise back to -R where R is a really big number. We will write, \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, \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, \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.

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  3. #3
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    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:
    \pi*e^{-2*\pi*|f|} <br />
( the FT of  \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
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