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Natural antiderivative is defined as follows, using Fourier transform:

$$f^{(-1)}(x)=\frac{i}{2\pi}\int_{-\infty}^{+\infty} \frac{e^{- i \omega x}}{\omega} \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$

Particularly, if $x=0$,

$$f^{(-1)}(0)=\frac{i}{2\pi}\int_{-\infty}^{+\infty} \frac{1}{\omega} \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$

Expressed as Mathematica code:

(I/Sqrt[2*Pi])*Integrate[(1/w)*FourierTransform[f[t], t, w], {w, -Infinity, Plus[Infinity]}]

That said, I wonder if Mathematica always adheres to the rule of providing the antiderivative which would satisfy this condition.

For instance, Integrate[Sin[x], x] gives -Cos[x], which satisfies the condition as -Cos[0] == -1.

I wonder how much this rule is adhered to in the output of Integrate.

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Your equation has a sign error for the inverse Fourier transform. A simpler prescription is

I InverseFourierTransform[FourierTransform[f[t], t, w]/w, w, x]

Although there is a lrge range of function for which this works, it will of course fail for all those functions for which the Fourier transform can't be calculated. Anyway, it does work for polynomials, e.g. as follows:

 Integrate[x^n, x] == 
  I InverseFourierTransform[FourierTransform[t^n, t, w]/w, w, x], {n, 
  1, 10}]

(* ==> {True, True, True, True, True, True, True, True, True, True} *)

But an example for which Integrate works and the Fourier approach doesn't is this simple case:

I InverseFourierTransform[FourierTransform[t^(1/2), t, w]/w, 
  w, x]

==> I InverseFourierTransform[FourierTransform[Sqrt[t], t, w]/w, 
  w, x]
share|improve this answer
Are there cases when both integrate works, and the natural integral formula works but they give different results? – Anixx Aug 24 '14 at 6:39
I have tried to correct the error. Is it correct now? – Anixx Aug 24 '14 at 7:06
I found a function where there is difference by Eurer's Gamma constant: $\frac1{x-1}$ – Anixx Aug 24 '14 at 12:16
$$\left( \begin{array}{ccc} \sin x & -(\cos x) & -(\cos x) \\ \cos x & \sin x & \sin x \\ 1 & x & x \\ x & \frac{x^2}{2} & \frac{x^2}{2} \\ x^2 & \frac{x^3}{3} & \frac{x^3}{3} \\ e^{-x^2} & \frac{1}{2} \sqrt{\pi } (\text{erf} x) & \frac{1}{2} \sqrt{\pi } (\text{erf} x) \\ e^{-x^2} x & -\frac{e^{-x^2}}{2} & -\frac{e^{-x^2}}{2} \\ \sin ^3 x & \frac{1}{12} (\cos (3 x)-9 (\cos x)) & \frac{1}{12} (\cos (3 x)-9 (\cos x)) \\ \frac{1}{x-1} & \log (x-1) & \log \left| x-1\right| +\gamma \\ \end{array} \right)$$ – Anixx Aug 24 '14 at 12:17
But I think it is an error in FurierTransform function: taking the outer Furier transform by hand, gives zero in x=0, so normal integral function is more correct here. The same problem is with function 1/(x+1) – Anixx Aug 24 '14 at 12:49

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