Should I trust Mathematica or numerous other sources on this Fourier transform 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?
 A: 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.
A: Different definitions
Note that Mathematica and your link use different definitions of the Fourier transform.
Mathematica uses (you can see this under "Details and Options" in the documentation):
$$\frac{1}{\sqrt{2\pi}} \int_{- \infty}^\infty f(t) e^{j \omega t} dt$$
and your link uses:
$$\int_{-\infty}^\infty f(t) e^{-j \omega t} dt$$
so the transforms differ by a factor of $\frac{1}{\sqrt{2 \pi}}$ and that Mathematica does not have the negative sign in the exponent.
In this case we can see that the only difference is the factor $\frac{1}{\sqrt{2 \pi}}$:
$$\frac{a\sqrt{\frac{2}{\pi}}}{a^2+\omega^2} \bigg/ {\frac{2a} {a^2+\omega^2}} = \frac{a\sqrt{\frac{2}{\pi}}(a^2 + \omega^2)}{2a(a^2 + \omega^2)} = \frac{\sqrt{\frac{2}{\pi}}}{2} = \frac{1}{\sqrt{2 \pi}}$$
so it works out.
Note that due to the fact of the missing minus sign in the exponent, we can get other differences as well, e.g. Mathematica says that the Fourier transform of $\operatorname{sign}(t)$ is
$$\frac{j \sqrt{\frac{2}{\pi }}}{\omega} = - \frac{1}{\sqrt{2\pi}} \underbrace{\frac{2}{j \omega}}_{:=A}$$
where $A$ is what your link says the transform is. The additional difference here is that we have a minus sign.
But why?
The definition that Mathematica uses has an advantage in that the inverse Fourier transform looks very similar to the Fourier transform:
$$\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty F(\omega) e^{-j \omega t} d\omega.$$
It is not only Mathematica that uses this definition, it is commonly used.
Getting Mathematica to do what you want
What you might want to do, to get the same behavior in Mathematica as in your link, is set FourierParameters to $\{1,-1\}$, so FourierTransform[Exp[-a Abs[t]], t, \[Omega], FourierParameters -> {1, -1}] returns
$$\frac{2 a}{a^2+\omega^2}$$
and FourierTransform[Sign[t], t, \[Omega], FourierParameters -> {1, -1}] returns
$$-\frac{2 i}{\omega}.$$
Note that FourierParameters can also be used with InverseFourierTransform. The details for what FourierParameters does can be found in the documentation for FourierTransform, under "Details and Options".
