Fourier transform of DawsonF not recovered by using Erfi

Question

Bug introduced in 8 or earlier and fixed in 12.2

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I wish to compute the Fourier transform related to the Dawson function:

FourierTransform[1/u DawsonF[1/u], u, x]

This gives Hypergeometric functions.

(1/2 \[Pi]^(3/2)
   HypergeometricPFQ[{}, {1/2, 1}, x^2/4] - \[Pi] Abs[
   x] HypergeometricPFQ[{}, {3/2, 3/2}, x^2/4])/Sqrt[2 \[Pi]]

However, the Dawson integral can also be written using Erfi:

$$D(x) = \frac{\sqrt{\pi}}{2}e^{-x^2}Erfi(x)$$

So the same solution should be obtained using

FourierTransform[1/u Sqrt[Pi]/2 Exp[-1/u^2] Erfi[1/u], u, x]

This gives a MeijerG function, which has the same real part, but an additional imaginary part.

MeijerG[{{1/2}, {}}, {{0, 1/2, 1/2}, {0}}, -(x^2/4)]/(2 Sqrt[2 \[Pi]])

So am I missing something or is this a bug? And which of the two is correct, if any?

Update: So this was a bug, but it is fixed now in version 12.2

Answer 1

Clear["Global`*"]

ft1 = FourierTransform[1/u DawsonF[1/u], u, x]

(* (1/Sqrt[2 π])(1/2 π^(3/2)
    HypergeometricPFQ[{}, {1/2, 1}, x^2/4] - π Abs[
    x] HypergeometricPFQ[{}, {3/2, 3/2}, x^2/4]) *)

ft2 = FourierTransform[FunctionExpand[1/u DawsonF[1/u]], u, x]

(* (1/(4 Sqrt[
 2 π]))(MeijerG[{{1/2}, {}}, {{0, 1/2, 1/2}, {0}}, -((I x)/2), 1/2] + 
  MeijerG[{{1/2}, {}}, {{0, 1/2, 1/2}, {0}}, (I x)/2, 1/2]) *)

{ft1, ft2} /. x -> 1.0`50

(* {0.2941039674327754399621322037300034824681806340199, 
 0.2941039674327754399621322037300034824681806340199 + 0.*10^-50 I} *)

The negligible imaginary part of ft2 is an artifact of using finite precision.

Graphically,

Plot[{ft1, ft2}, {x, -5, 5},
 PlotStyle -> {Automatic, Dashed}]