# Fourier transform in Mathematica

When I calculate the Fourier transform of the function $$f(t) = \mathrm e^{-|t|/\tau} \text{ with } \tau >0$$ in Mathematica once via the function FourierTransform and once by hand, I get different results. And with "by hand" I mean letting Mathematica calculate the integral $$ft_2(\omega) = \int_\mathbb{R} f(t) \mathrm e^{-2 \pi \mathrm i \omega t} \, \mathrm dt$$

My input is:

f[t_] = Exp[-Abs[t]/\[Tau]];

ft1[\[Omega]_] = FourierTransform[f[t], t, \[Omega]]

ft2[\[Omega]_] =
Integrate[f[t]*Exp[-2*Pi*I*\[Omega]*t], {t, -Infinity, Infinity},
Assumptions -> {{\[Omega], \[Tau]} \[Element] Reals, \[Tau] > 0}]

\[Omega] = 0.123;
\[Tau] = 0.456;
ft1[\[Omega]]
ft2[\[Omega]]


And the generated output is:

$$ft_1(\omega) = \sqrt{\frac{2}{\pi}} \cdot \frac{\tau}{1+\tau^2\omega^2}$$ $$ft_2(\omega) = \frac{2 \tau}{(-\mathrm i + 2 \pi \tau \omega)(\mathrm i + 2 \pi \tau \omega)} = \frac{2 \tau}{1 + 4 \pi^2 \omega^2 \tau^2}$$ and

0.362694
0.811248+ 0. I


As you can see, we get different values for $\omega = 0.123$ and $\tau = 0.456$. I am most certain that there is some error with the integral, as the result from FourierTransform can also be obtained from rule 207 in the List from Wikipedia.

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You do know that FourierTransform[] has an unconventional normalization, don't you? Did you check the FourierParameters option? – J. M. Dec 3 '11 at 2:47

Does FourierTransform[Exp[-Abs[t]/\[Tau]], t, \[Omega], FourierParameters -> {0, -2 Pi}] work for you? It does for me...
@Vincent, one way you can do the reconciliation of normalizations by your own would be to look at the general definition for FourierTransform[] in Mathematica's docs and compare accordingly... – J. M. Apr 2 '13 at 13:16