The FourierTransform
doesn't seem to work on the following function:
fun[x_, y_, t_] =
1/((x - 4 y) (x + 4 y) (x^2 + 2 y^2))
E^(-(1/4)
t (3 x + Sqrt[
x^2 - 16 y^2])) (-E^(1/4 t (3 x + Sqrt[x^2 - 16 y^2]))
x^2 (x - 4 y) (x + 4 y) + x^3 Sqrt[x^2 - 16 y^2] -
E^(1/2 t Sqrt[x^2 - 16 y^2]) x^3 Sqrt[x^2 - 16 y^2] +
5 x y^2 Sqrt[x^2 - 16 y^2] -
5 E^(1/2 t Sqrt[x^2 - 16 y^2]) x y^2 Sqrt[
x^2 - 16 y^2] + (1 + E^(1/2 t Sqrt[x^2 - 16 y^2])) (x -
4 y) (x + 4 y) (x^2 + y^2));
Doing FourierTransform[fun[x, y, t], t, \[Omega]]
simply returns the same thing (i.e., FourierTransform[expression of the function]. How can I calculate the FourierTransform of this function and plot it?
Expand
beforehand might work. $\endgroup$x
andy
. If its imaginary part is non-zero, you will need to convert it to real numbers (e.g. usingAbs
,Re
,Im
orReIm
) $\endgroup$x
andy
thatFourierTransform
simply does not exist. For example, putting{x -> 1, y -> 1}]
, we otain a meaningless result-8 I Sqrt[(2 \[Pi])/15] DiracDelta[-3 I + Sqrt[15] - 4 s] + 8/3 Sqrt[2 \[Pi]] DiracDelta[-3 I + Sqrt[15] - 4 s] - 1/3 Sqrt[2 \[Pi]] DiracDelta[s] + 8 I Sqrt[(2 \[Pi])/15] DiracDelta[3 I + Sqrt[15] + 4 s] + 8/3 Sqrt[2 \[Pi]] DiracDelta[3 I + Sqrt[15] + 4 s]
. Heres
instead of\[Omega]
is used. Putting{x -> 5, y -> 1}
, one obtains-(25/27) + 76/27 E^(-9 t/2) - (8 E^(-3 t))/9
. Its FT does not exist. $\endgroup$