How are Fourier transformations calculated?

Question

Lets say my function is $$f(t)=\left\{\begin{matrix} 1 &, -T_p/2\leq t \leq T_p/2\\ 0& , |t|>T_p/2 \end{matrix}\right.$$ than a Fourier transformation is calculated $$F(j\omega)=\int_{-\infty}^\infty f(t)e^{-j\omega t}dt$$ for $j$ being an imaginary unit. The integral above can be simplified to $$F(j\omega)=\int_{-T_p/2}^{T_p/2}e^{-j\omega t}dt$$ which evaluates to $$F(j\omega)=\frac{1}{j\omega}(e^{j\omega T_p/2}-e^{-j\omega T_p/2})$$ and written in a more compact form $$F(j\omega )=Tp\frac{\sin(\frac{\omega T_p}{2})}{\frac{\omega T_p}{2}}$$ Correct me if I am wrong. Because what Mathematica calculates using

f[t_] := Piecewise[{{1, -Tp/2 <= t <= Tp/2}, {0, t > Tp/2}, {0, 
    t < -Tp/2}}]
FourierTransform[f[t], t, \[Omega], Assumptions -> Tp > 0]

is $$F(j\omega)=\frac{\sqrt{\frac{2}{\pi }} \sin \left(\frac{\text{Tp} \omega }{2}\right)}{\omega }$$

which is not and never will be the same result.

So how does Mathematica calculate the Fourier Transformations? Where does that factor $\sqrt{\frac 2 \pi}$ come from? Etc.. Just wondering which method is more correct and why they don't match.

Answer 1

The documentation indicates the following:

$$\sqrt{\frac{|b|}{(2\pi)^{1-a}}}\int_{-\infty}^\infty f(t)e^{ib\omega t}\,dt$$

with FourierParameters -> {a, b}.

f[t_] = Piecewise[{{1, -Tp/2 <= t <= Tp/2}, {0, t > Tp/2}, {0, t < -Tp/2}}];
 FourierTransform[f[t], t, \[Omega], FourierParameters -> {1, 1}, Assumptions -> Tp > 0]