Can anyone suggest a method of speeding up the evaluation of the following Fourier transform?
FourierTransform[UnitStep[t] Exp[-t/τ] Cos[(m t + ω0 ) t], t, ω]
I'm surprised by the amount of time this takes to evaluate, I have added in a linear chirp to the frequency so the variable $t$ in the argument of the $\cos$ now enters quadratically; so I suppose this is the culprit.
This takes quite some time to evaluate and when it does I am left with
(1/(24 (m^2)^(7/4) Sqrt[
2 π] τ^3))(6 m (m^2)^(
3/4) τ^2 (I + τ ω - τ ω0) \
HypergeometricPFQ[{1}, {3/4, 5/
4}, -((I + τ ω - τ ω0)^4/(
64 m^2 τ^4))] -
6 m (m^2)^(
3/4) τ^2 (I + τ ω + τ ω0) \
HypergeometricPFQ[{1}, {3/4, 5/
4}, -((I + τ ω + τ ω0)^4/(
64 m^2 τ^4))] +
I (m^2)^(3/
4) (I + τ ω - τ ω0)^3 \
HypergeometricPFQ[{1}, {5/4, 7/
4}, -((I + τ ω - τ ω0)^4/(
64 m^2 τ^4))] +
I (m^2)^(3/
4) (I + τ ω + τ ω0)^3 \
HypergeometricPFQ[{1}, {5/4, 7/
4}, -((I + τ ω + τ ω0)^4/(
64 m^2 τ^4))] +
3 (-I m^3 + (m^2)^(3/2)) Sqrt[
2 π] τ^3 (Cos[(I + τ ω - τ ω0)^2/(
4 m τ^2)] +
I Sin[(I + τ ω - τ ω0)^2/(
4 m τ^2)]) +
3 (I m^3 + (m^2)^(3/2)) Sqrt[
2 π] τ^3 (Cos[(I + τ ω + τ ω0)^2/(
4 m τ^2)] -
I Sin[(I + τ ω + τ ω0)^2/(
4 m τ^2)]))
This I wouldn't mind, but I don't know how to deal with the HypergeometricPFQ in the answers.
My goal is to eventually have a lovely peak-like lineshape at the end I can play with.
By adding some assumptions I can get something a little nicer
FT1 = FourierTransform[UnitStep[t] Exp[-t/τ] Cos[(m t + ω0 ) t], t, ω, Assumptions-> t > 0 && τ > 0 && ω > 0 && ω0 > 0]
However when I take the absolute value using ComplexExpand[Abs[FT1]]
I get a similarly as ugly
√((-(((m^2)^(1/4)
Cos[(-1 + (τ ω - τ ω0)^2)/(
4 m τ^2)] Cos[
1/2 Arg[I m]] Cosh[(τ ω - τ ω0)/(
2 m τ^2)])/(2 Sqrt[2] m)) + ((m^2)^(1/4)
Cos[(-1 + (τ ω + τ ω0)^2)/(
4 m τ^2)] Cosh[(τ ω + τ ω0)/(
2 m τ^2)])/(4 m) - (
Cosh[(τ ω + τ ω0)/(
2 m τ^2)] Sin[(-1 + (τ ω + τ \
ω0)^2)/(4 m τ^2)])/(
4 (m^2)^(1/4)) + ((m^2)^(1/4)
Cosh[(τ ω - τ ω0)/(
2 m τ^2)] Sin[(-1 + (τ ω - τ \
ω0)^2)/(4 m τ^2)] Sin[1/2 Arg[I m]])/(
2 Sqrt[2] m) + ((m^2)^(1/4)
Cos[(-1 + (τ ω - τ ω0)^2)/(
4 m τ^2)] Cos[
1/2 Arg[I m]] Sinh[(τ ω - τ ω0)/(
2 m τ^2)])/(
2 Sqrt[2] m) - ((m^2)^(1/4)
Sin[(-1 + (τ ω - τ ω0)^2)/(
4 m τ^2)] Sin[
1/2 Arg[I m]] Sinh[(τ ω - τ ω0)/(
2 m τ^2)])/(
2 Sqrt[2] m) + ((m^2)^(1/4)
Cos[(-1 + (τ ω + τ ω0)^2)/(
4 m τ^2)] Sinh[(τ ω + τ ω0)/(
2 m τ^2)])/(4 m) - (
Sin[(-1 + (τ ω + τ ω0)^2)/(
4 m τ^2)] Sinh[(τ ω + τ ω0)/(
2 m τ^2)])/(4 (m^2)^(1/4)))^2 + ((
Cos[(-1 + (τ ω + τ ω0)^2)/(
4 m τ^2)] Cosh[(τ ω + τ ω0)/(
2 m τ^2)])/(
4 (m^2)^(1/4)) + ((m^2)^(1/4)
Cos[1/2 Arg[I m]] Cosh[(τ ω - τ ω0)/(
2 m τ^2)] Sin[(-1 + (τ ω - τ \
ω0)^2)/(4 m τ^2)])/(
2 Sqrt[2] m) + ((m^2)^(1/4)
Cosh[(τ ω + τ ω0)/(
2 m τ^2)] Sin[(-1 + (τ ω + τ \
ω0)^2)/(4 m τ^2)])/(
4 m) + ((m^2)^(1/4)
Cos[(-1 + (τ ω - τ ω0)^2)/(
4 m τ^2)] Cosh[(τ ω - τ ω0)/(
2 m τ^2)] Sin[1/2 Arg[I m]])/(
2 Sqrt[2] m) - ((m^2)^(1/4)
Cos[1/2 Arg[
I m]] Sin[(-1 + (τ ω - τ ω0)^2)/(
4 m τ^2)] Sinh[(τ ω - τ ω0)/(
2 m τ^2)])/(
2 Sqrt[2] m) - ((m^2)^(1/4)
Cos[(-1 + (τ ω - τ ω0)^2)/(
4 m τ^2)] Sin[
1/2 Arg[I m]] Sinh[(τ ω - τ ω0)/(
2 m τ^2)])/(2 Sqrt[2] m) + (
Cos[(-1 + (τ ω + τ ω0)^2)/(
4 m τ^2)] Sinh[(τ ω + τ ω0)/(
2 m τ^2)])/(
4 (m^2)^(1/4)) + ((m^2)^(1/4)
Sin[(-1 + (τ ω + τ ω0)^2)/(
4 m τ^2)] Sinh[(τ ω + τ ω0)/(
2 m τ^2)])/(4 m))^2)
Seeing as I am posting a bounty on this question I think it is prudent to state what I am looking for. I am looking for a way to perform Fourier transforms of this form, what would be helpful is not only an answer to this problem, but also a more general guide when dealing with complicated time transients that one wants to transform. This is a damped wave with a linear frequency chirp, which is well understood.
I'd like to be able to figure out how I can get Mathematica to perform such computations so I can play around a look at my resultant line-shapes.
