How to Solve Complex Fourier Transform of Exponential Functions?

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SUMMARY

The discussion focuses on solving the Fourier transform of the function g(x) = (γ/2)e^{-γ|x|}, where γ > 0. The user successfully verified that the integral of g(x) over the entire real line equals 1 but encountered difficulties in calculating the Fourier transform, specifically in expressing it as \widetilde{g}(k) = a/(1 + (k²/s²)). The solution involves splitting the integral into two parts to handle the absolute value and combining the exponential terms, leading to the results a = 1/(2π) and s = γ, as confirmed by Wolfram Alpha.

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Homework Statement



I've been given the following function:

g(x) = \frac{\gamma}{2}e^{-\gamma \left|x\right|} with \gamma>0

First thing I needed to do was to prove \int^{-\infty}_{\infty}g(x)dx=1 which was simple enough.

I hit a problem when trying to find the Fourier transform \widetilde{g}(k) of g(x)

I'm asked to show the transform is of the form

\widetilde{g}(k)=a\frac{1}{1+\frac{k^{2}}{s^{2}}}

and find a and s

Homework Equations



\widetilde{g}(k) = \frac{1}{\sqrt{2\pi}}\int^{-\infty}_{\infty}e^{-ikx}g(x)dx

The Attempt at a Solution



I try to perform integration by parts on

\widetilde{g}(k) = \frac{1}{\sqrt{2\pi}}\int^{-\infty}_{\infty}e^{-ikx}\frac{\gamma}{2}e^{-\gamma \left|x\right|}dx

but whenever I try this I end up with uv equalling vdu so the whole expression turns out as zero. I think this is probably because I don't really know what I'm doing with the integration with those limits to infinity. I can split up g(x) because its an even function but not the e^{-ikx} part.

I've plugged this into wolfram alpha and it churned out a = \frac{1}{2\pi} and s = \gamma, I'm just not sure how to get there.

Any help appreciated!
 
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Divide integral into two parts to get rid of the abs(x) and combine the exponentials in the integrand. That way the integral is trivial. After some algebra you should get the desired result. Careful with the signs.
 
Aha thanks! Somehow I totally missed being able to combine the two powers of e, I guess I was staring at the problem for too long.
 

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