Fourier Transform of exp(-a|x-.5|)

Question

So I've been working on the fourier transform of $exp(-a|x-\frac{1}{2}|)$ (with $a>0$) and keep getting:
$\left(e^{-\pi i}\right)\left(\frac{2a}{a^+4\pi^2x^2}\right)$.
A research partner keeps getting something much more complicated - am I doing something wrong?

Answer 1

The Fourier transform of an integrable function is bounded, so having $x^2$ in the denominator is out of question. I used Wolfram Alpha to remind myself that the transform of $\exp(-|x|)$ is the Cauchy distribution: $$\mathcal F\left(e^{-|x|}\right) = \frac{2}{\pi}\frac{1}{\omega^2+1}$$ By the scaling property, $$\mathcal F\left(e^{-a|x|}\right) = \frac{2}{\pi}\frac{a}{\omega^2+a^2}$$ And then shift by one half will produce the factor of $e^{-2\pi i \omega (1/2)}$.

I can't promise that my conventions for $\mathcal F$ agree with yours.