Why is my result different from what the teacher does when I used Inverse Fourier Transformer?
InversFouiere Transform.nb (14.8 KB)
This should give you an idea. The output simplified, used as input, gives your orig. function except with rules. Fourier tables in books are a little infamous for not all being in the same form (and some being incorrect “in ways”).
In: FourierTransform[(
E^((a x)/(2 \[Pi])) HeavisideTheta[-x Sign[Re[a]]] Sign[Re[a]])/
Sqrt[2 \[Pi]], x, w] // Simplify
Out: 1/(a + 2 I \[Pi] w) /; Re[a] != 0
There is an option FourierPrameters which controls which convention is in use for the FT and IFT. The one in the book differs from the default behavior of InverseFourierTransform. Also there is a need to make an assumption about the parameter a in order to have convergence of the underlying integral.
InverseFourierTransform[1/(a + 2 I f \[Pi]), f, t,
FourierParameters -> {0, -2 Pi}, Assumptions -> Re[a] > 0]
(* Out[13]= E^(-a t) HeavisideTheta[2 \[Pi] t] *)
Note that this is equivalent to the textbook result since the HeavisideTheta function makes its transition at the origin for either result.