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There appears to be a problem with version 10.1 compared to version 10.0 Your direct integration implies that your Fourier parameters are {-1, 1} $Version "10.0 for Mac OS X x86 (64-bit) (December 4, 2014)" F[w_] = 1/(1 - I*w); f[t_] = InverseFourierTransform[F[w], w, t, FourierParameters -> {-1, 1}] (2*Pi*HeavisideTheta[t])/E^t F[w] == ...


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The functions Fourier, FourierTransform, InverseFourier, and InverseFourierTransform (possibly others that I am not aware of) all accept a parameter called FourierParameters to determine the flavor of transform performed. By default, the value of FourierParameters is {0, 1}, which gives the results you've observed. To run a 'traditional' Fourier transform, ...


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InverseLaplaceTransform[(2 s^2 + s + 13)/((s - 1) ((s + 1)^2 + 4)), s, t] 2*E^t + (3/4)IE^((-1 - 2*I)t) (-1 + E^(4*I*t)) % // ComplexExpand // Simplify 2*E^t - (3*Cos[t]*Sin[t])/E^t % // TrigReduce ((1/2)*(4*E^(2*t) - 3*Sin[2*t]))/E^t In "one" step InverseLaplaceTransform[(2 s^2 + s + 13)/((s - 1) ((s + 1)^2 + 4)), s, t] // ...


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Update First I'd like to mention that after checking the definition of bilateral Laplace transform and Fourier transform carefully, I'm sure currently the formula for the relationship between them on the wikipedia page is wrong, the correct one should be: $$\mathcal{B}\{f(t)\}(s) = \sqrt{2\pi}\mathcal{F}\{f(t)\}(is)$$ Then we can use the internal function ...



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