TITLE: Should I trust Mathematica or numerous other sources on this Fourier transform
QUESTION [5 upvotes]: Assume $a>0$
So Mathematica claims
$$F\{e^{-a|t|}\}(\omega) = \frac{a\sqrt{\frac{2}{\pi}}}{a^2+\omega^2}$$
However, I've read about another transform pair (page 3):
$$F\{e^{-a|t|}\}(\omega) = \frac{2a}{a^2+\omega^2}$$
They clearly are not equal, when you let $a=2$, mathematica claims the transform is $\frac{1.60}{4+\omega^2}$, and the other source claims it's $\frac{4}{4+\omega^2}$. Which one is right, and why do they differ?

REPLY [3 votes]: There are different definitions of the Fourier Transform. The confusion comes from Mathematica and your reference using different definitions for the Fourier Transform. They are equivalent up to a constant. Most people, and Mathematica use the definition: $$F[f](\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-ix\omega}dx.$$
Others use the definition 
$$F[f](\omega)=\int_{-\infty}^\infty f(x)e^{-ix\omega}dx.$$
The reason for the difference comes from the inverse transform. If we use the definition, the inverse transform is: $$F^{-1}[f](x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(\omega)e^{ix\omega}d\omega.$$
The second definition yields an inverse transform of $$F^{-1}[f](\omega)=\frac{1}{2\pi}\int_{-\infty}^\infty f(\omega)e^{ix\omega}d\omega.$$
So notice that both definitions have a total coefficient of $1/(2\pi)$. Different people have different preferences as to how do balance the coefficient between the transform and the inverse transform.