FourierTransform breaks function symmetry

Question

The fourier transform of a seemingly simple function gives me a complicated result: (I imaginary unit)

FourierTransform[Exp[I c t/Abs[k]], k, x, Assumptions -> Element[c, Reals]]

gives me a long expression of 15 terms, which can be simplified by applying FullSimplify to particular combinations of terms(*) to

Sqrt[2 π] DiracDelta[x]+

(-(I x + Abs[x]) KelvinKei[1, (2 Sqrt[Abs[x]])/Sqrt[I/(c t)]] + (-I x Abs[x]) KelvinKer[1, (2 Sqrt[Abs[x]])/Sqrt[I/(c t)]])/(Sqrt[π] Sqrt[I/(c t)] Abs[x]^(3/2))+

(c t (I MeijerG[{{}, {}}, {{0, 0, 1/2}, {-(1/2)}}, -(1/16)
   c^2 t^2 x^2] + MeijerG[{{}, {}}, {{-(1/2), 0, 1/2}, {0}}, -(1/16)
    c^2 t^2 x^2] Sign[x]))/(4 Sqrt[2 π]).

Now, as the original function to transform was even, its fourier transform must also be even. However, when plotting the output, the imaginary part looks even indeed, but the real part looks odd instead.

Where does the problem come from, and how can it be solved? By checking the original output, I saw that the problem was already there before the FullSimplify.

*) All Kelvin functions together, I tried simplifying all terms together but that one takes ages.

Answer 1

As pointed out by @march in a comment, instead of Abs[k] I should have used Sqrt[k^2], which is analytical and thus the only analytical continuation of absolute value to non-real numbers.*

The resulting expression gets much simpler and obeys the proper symmetry.

*This is not exactly true for the complex plane as a whole, but it allows to cut the complex plane in two parts where the function is analytical in each.