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.

  • $\begingroup$ It looks like you are using k as your time variable? Then what is t? I suspect you mean: FourierTransform[Exp[I c t/Abs[k]], t, x, Assumptions -> Element[c, Reals]] $\endgroup$
    – bill s
    Nov 4 '16 at 16:42
  • 1
    $\begingroup$ @bills I am transforming from k to x (wavenumber to position). I know this is commonly called inverse fourier-transformation, but according to the documentation FourierTransform is the one without a minus-sign in the exponential. t is a parameter that denotes time, but it remains invariant under the transformation. Real t is sufficient in fact. $\endgroup$
    – Wouter
    Nov 4 '16 at 18:29
  • 2
    $\begingroup$ Did you try instead, FourierTransform[Exp[I c t/Sqrt[k^2]], k, x, Assumptions -> Element[c, Reals]]? I plotted the real and imaginary parts of this function for c == 1 and t == 1, and they seemed to both be even functions. $\endgroup$
    – march
    Nov 4 '16 at 18:53
  • $\begingroup$ @march thanks, Seems to work indeed, with an expression that is much simpler. Will mark as solved on monday, if I can compare it with my previous results. Strangely, I'd think Abs should be the least ambiguous regarding branch cuts etc. $\endgroup$
    – Wouter
    Nov 4 '16 at 21:00

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.


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