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I searched for the inverse fourier transformation of
$$ \mathcal{F(\omega)} = \frac{2}{(1+i\cdot \omega)^2 +4} \rightarrow i^2=-1 $$

My solution (compliant with the solution from my textbook): $$ \mathcal{F}^{-1}\{\mathcal{F}\{\omega\}\} = e^{-t}\cdot \sin{(2t)}\cdot \sigma {(t)}$$
Wolfram|Alpha: $$ \mathcal{F}^{-1}\{\mathcal{F}\{\omega\}\} = i \sqrt{\frac{\pi }{2}} e^{(1-2 i) t} \left(-1+e^{4 i t}\right) \theta (-t) $$
Where does this difference come from and how can I make the output equal to my solution?

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  • $\begingroup$ If you want someone to explain you the mathematical equivalence and required assumptions, you should better ask it from Mathematics.SE. $\endgroup$
    – Johu
    Commented Sep 17, 2018 at 15:14

1 Answer 1

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Use FourierParameters-> {1, -1}:

InverseFourierTransform[2/((1 + I ω)^2 + 4), ω, t, FourierParameters -> {1, -1}] // FullSimplify

$$e^{-t} \theta (t) \sin (2 t)$$

More info about FourierParameters in the documentation.

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