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.
FourierTransform[Exp[I c t/Abs[k]], t, x, Assumptions -> Element[c, Reals]]
$\endgroup$ – bill s Nov 4 '16 at 16:42FourierTransform[Exp[I c t/Sqrt[k^2]], k, x, Assumptions -> Element[c, Reals]]
? I plotted the real and imaginary parts of this function forc == 1
andt == 1
, and they seemed to both be even functions. $\endgroup$ – march Nov 4 '16 at 18:53