I was looking at the documentation here. In the "Details and Options" section they mention that $f(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} F(\omega)e^{-i \omega t} d\omega$. I interpreted this to mean

InverseFourierTransform[expr,$\omega$,t] = $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} $ expr $ e^{-i \omega t} d\omega$

Is this correct? I tried to type in the following two expressions to confirm this but I was not getting the same results.

     InverseFourierTransform[1/(1 + I*\[Omega]), \[Omega], -t]
    (1/Sqrt[(2*Pi)]) Integrate[(1/(1 + I*\[Omega]))*
   Exp[-I*\[Omega]*t], {\[Omega], -Infinity, Infinity}]

What I was expecting was for both of the above expressions to give me E^-t Sqrt[2 \[Pi]] HeavisideTheta[t] but that isn't what happened. I tried switching Integrate[] to NIntegrate[] but then the integral won't evaluate at all.

Am I misinterpreting what InverseFourierTransform[] does? Or is there an error in my other integral? Thank you very much in advance.

  • 2
    $\begingroup$ why would you use -t as the time variable? Try InverseFourierTransform[1/(1 + I*w), w, t] The mapping is from w (frequency) to t (time) and not from w to -t. The Fourier transforms maps t to w and not t to -w. Try to stick to the definitions unless you want results which will not make too much sense. $\endgroup$
    – Nasser
    Dec 8, 2022 at 2:23
  • $\begingroup$ @Nasser I was doing this to try to match the definition of the bilateral Laplace transform so I needed the -t. I didn't do it perfectly though I'm now seeing some other errors that I need to fix $\endgroup$
    – k12345
    Dec 9, 2022 at 1:24

1 Answer 1


Indeed, the results of

InverseFourierTransform[1/(1 + I*\[Omega]), \[Omega], t]

E^t Sqrt[2 \[Pi]] HeavisideTheta[-t]

N[E^t Sqrt[2 \[Pi]] HeavisideTheta[-t] /. t -> -1]


and (pay you attention to 1/Sqrt[2*Pi], not 1/(2*Pi))

Integrate[ 1/(1 + I*\[Omega])*Exp[-I*\[Omega]*t], {\[Omega], -Infinity, 
Infinity}, Assumptions -> t > -Infinity]/Sqrt[2*Pi]

-(1/(4 \[Pi])) I ((1/Sqrt[\[Pi]])(MeijerG[{{1/2, 1, 1}, {}}, {{1}, {}}, (2 I)/ Abs[t], 1/2] - I MeijerG[{{1/2, 1/2, 1}, {}}, {{1/2}, {}}, (2 I)/Abs[t], 1/ 2] Sign[t]) + 2 CosIntegral[ I Abs[t]] (Cosh[Abs[t]] + Sign[t] Sinh[Abs[t]]) - (Cosh[Abs[t]] Sign[t] + Sinh[Abs[t]]) (I \[Pi] + 2 SinhIntegral[Abs[t]]))

% /. t -> -1.0000000

0.922137 + 2.21457*10^-17 I

differ symbolically, but coincide numerically for t==-1.

  • $\begingroup$ It should be noticed that the above improper integral converges only conditionally. Because of this reason have to handle it carefully and carefully. $\endgroup$
    – user64494
    Dec 8, 2022 at 8:32
  • $\begingroup$ Thank you very much for your help. Sorry about the typos. I've fixed them. $\endgroup$
    – k12345
    Dec 9, 2022 at 1:29

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