# 2d vs 1d FourierTransform

According to the documentation, the multidimensional Fourier transform is defined as $$\frac{1}{(2\pi)^{n/2}} \int_{-\infty}^\infty \int_{-\infty}^\infty \cdots f(t_1, t_2, \dots, t_n) \mathrm e^{\mathrm i(t_1 \omega_1 + t_2 \omega_2 + \dots + t_n \omega_n)} \mathrm d t_1 \mathrm d t_2 \dots \mathrm d t_n\text.$$ Define some assumptions:

$Assumptions = t > 0 && Element[a, Reals] && Element[b, Reals];  Now, I would expect if I do the 2D Fourier transform of$\mathrm e^{\frac{\mathrm i}{2} (x^2 + z^2) t}$, going$x \mapsto a$,$z \mapsto b$, that this factorizes and I could equally well do the product of the 1D Fourier transforms of$\mathrm e^{\frac{\mathrm i}{2} x^2}$and$\mathrm e^{\frac{\mathrm i}{2} z^2}$. Integrate confirms this: Simplify[ Integrate[Exp[I/2 (x^2 + z^2) t] Exp[I (x a + z b)], {x, -Infinity, Infinity}, {z, -Infinity, Infinity}] == Integrate[Exp[I/2 x^2 t] Exp[I x a], {x, -Infinity, Infinity}] Integrate[Exp[I/2 z^2 t] Exp[I z b], {z, -Infinity, Infinity}] ] > True  Now I try the same with FourierTransform: Simplify[ FourierTransform[Exp[I/2 (x^2 + z^2) t], {x, z}, {a, b}] == FourierTransform[Exp[I/2 x^2 t], x, a] FourierTransform[Exp[I/2 z^2 t], z, b] ] > 1 + Exp[I (a^2 + b^2)/t] == 0  How does this strange behavior arise? The factored version of FourierTransform agrees with the explicit integration (if you put FourierParameters -> {1, 1}) while the 2D version is basically the conjugate of the correct solution. Note that if I remove t, both versions agree. Removing just the assumption that t be positive leads to a result that contains$\sqrt{-t^2}\$ in the denominators, which then gives the wrong sign when the I is taken out.

• I filed this as a bug report on June 18. – Benjamin Desef Jul 5 '18 at 13:39
• According to WRI, this was fixed in Mathematica 12.0. – Benjamin Desef Dec 1 '19 at 14:16