# Thread: Cauchy and fourier tranform

1. ## 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

2. 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.

Is that what you are looking for?

3. 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|}
$
( 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