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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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closed as off-topic by Daniel Lichtblau, gpap, Johu, MarcoB, dr.blochwave Sep 26 '18 at 15:55

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  • "This question cannot be answered without additional information. Questions on problems in code must describe the specific problem and include valid code to reproduce it. Any data used for programming examples should be embedded in the question or code to generate the (fake) data must be included." – Daniel Lichtblau, gpap, MarcoB
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ If you want someone to explain you the mathematical equivalence and required assumptions, you should better ask it from Mathematics.SE. $\endgroup$ – Johu Sep 17 '18 at 15:14
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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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