Is there a way to speed up this Fourier transform and optimise the results?

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 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 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 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 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 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 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 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 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.

– Q.P.
Jul 25 '19 at 14:43
• For $m=1$ Maple 2019.1 performs a simpler result in terms of the erf function. Jul 25 '19 at 15:12
• @user64494 thanks. Unfortunately I don't have Maple!
– Q.P.
Jul 25 '19 at 15:13
• May present the Maple result through Dropbox on demand. Jul 25 '19 at 15:39
• Here dropbox.com/s/8m89hyzmf05uddg/FT.pdf?dl=0 it is. It should be noticed that the definition of Fourier transform in Maple (see maplesoft.com/support/help/Maple/view.aspx?path=examples/… ) differs from the Mathematica one by a constant multiplier. Jul 25 '19 at 16:13

The explicit integral doesn't take long and is quite simple (compared to what you have):

Assuming[τ > 0 && m > 0 && Element[ω0, Reals] && Element[ω, Reals],
1/Sqrt[2 π] Integrate[Exp[-t/τ] Cos[(m t + ω0) t] E^(I ω t),
{t, 0, ∞}] // FullSimplify]

$$\frac{(-1)^{1/4} \left(e^{\frac{2 \tau \omega +i}{2 m \tau ^2}} \left(1+\text{erf}\left(\frac{(-1)^{3/4} (\tau (\omega +\omega_0)+i)}{2 \sqrt{m} \tau }\right)\right)-i e^{\frac{i \left(\omega ^2+\omega_0^2\right)}{2 m}} \left(1+\text{erf}\left(\frac{(-1)^{1/4} (\tau (\omega -\omega_0)+i)}{2 \sqrt{m} \tau }\right)\right)\right) \exp \left(-\frac{i \tau ^2 (\omega +\omega_0)^2+2 \tau (\omega -\omega_0)+i}{4 m \tau ^2}\right)}{4 \sqrt{2m}}$$

In case it's not correct to assume $$m>0$$ as I did, then you can replace that assumption by $$m\in\mathbb{R}$$ and get a slightly different answer.

Make a plot:

With[{ω0 = 1, τ = 100, m = 0.001},
Plot[Abs[((-1)^(1/4) E^(-((I + 2 τ (ω - ω0) + I τ^2 (ω + ω0)^2)/(4 m τ^2))) (-I E^((I (ω^2 + ω0^2))/(2 m)) (1 + Erf[((-1)^(1/4) (I + τ (ω - ω0)))/(2 Sqrt[m] τ)]) + E^((I + 2 τ ω)/(2 m τ^2)) (1 + Erf[((-1)^(3/4) (I + τ (ω + ω0)))/(2 Sqrt[m] τ)])))/(4 Sqrt Sqrt[m])]^2,
{ω, 0, 2}, PlotRange -> All]] • really thanks a lot for this. One additional question, if I further complicate this integral by adding say a Hanning Window, do you have any hint as to how to make this execute more efficiently? I'm playing with your example and adding HannWindow[t] significantly increases computation time. At the time of writng it is still running...
– Q.P.
Jul 27 '19 at 19:11
• Usually you would convolve with a window function, which would be a multiplication in Fourier space: just multiply the already found Fourier transform with the Fourier transform of the Hann function. Or did I misunderstand and you want to use the Hann function in time instead of the UnitStep[t] factor in the original problem? Jul 27 '19 at 20:41
• For a Hann window of duration $\tau$, maybe this could work: Assuming[τ > 0 && m > 0 && ω0 > 0, 1/Sqrt[2 π] Integrate[Sin[(π t)/τ]^2 Cos[(m t + ω0) t] E^(I ω t), {t, 0, τ}]]. Maybe add a FullSimplify inside the Assuming if you can bear the slowness. Jul 27 '19 at 22:22
• I'm trying to get a line shape analytically, that resembles the FFT of some data. And as I understand it the windows should always be multiplied in, in the time domain. So yeah a straight swap of UnitStep. I only ever add unit step, as I usually use FourierTransform rather than the direct integration, as without UnitStep FourierTransform returns strange results even for simple transforms.
– Q.P.
Jul 28 '19 at 8:43

What version of Mathematica were you using? Using MMA Version 11.2.0 for Linux I get, using

ft=FourierTransform[UnitStep[t] Exp[-t/\[Tau]] Cos[(m t + \[Omega]0) t], t, \[Omega]] ,

the result

(((1 + I*Erfi[(I + \[Tau]*\[Omega] - \[Tau]*\[Omega]0)/(2*Sqrt[I*m]*\[Tau])])*(Cos[(I + \[Tau]*\[Omega] - \[Tau]*\[Omega]0)^2/(4*m*\[Tau]^2)] + I*Sin[(I + \[Tau]*\[Omega] - \[Tau]*\[Omega]0)^2/(4*m*\[Tau]^2)]))/Sqrt[I*m] + ((1 + Erf[(m*(I + \[Tau]*\[Omega] + \[Tau]*\[Omega]0))/(2*((-I)*m)^(3/2)*\[Tau])])*(Cos[(I + \[Tau]*\[Omega] + \[Tau]*\[Omega]0)^2/(4*m*\[Tau]^2)] - I*Sin[(I + \[Tau]*\[Omega] + \[Tau]*\[Omega]0)^2/(4*m*\[Tau]^2)]))/Sqrt[(-I)*m])/(4*Sqrt)

which, displayed in TeXForm, is $$\frac{\frac{\left(1+\text{erf}\left(\frac{m (\tau \omega +\tau \text{\omega 0}+i)}{2 (-i m)^{3/2} \tau }\right)\right) \left(\cos \left(\frac{(\tau \omega +\tau \text{\omega 0}+i)^2}{4 m \tau ^2}\right)-i \sin \left(\frac{(\tau \omega +\tau \text{\omega 0}+i)^2}{4 m \tau ^2}\right)\right)}{\sqrt{-i m}}+\frac{\left(1+i \text{erfi}\left(\frac{\tau \omega -\tau \text{\omega 0}+i}{2 \sqrt{i m} \tau }\right)\right) \left(i \sin \left(\frac{(\tau \omega -\tau \text{\omega 0}+i)^2}{4 m \tau ^2}\right)+\cos \left(\frac{(\tau \omega -\tau \text{\omega 0}+i)^2}{4 m \tau ^2}\right)\right)}{\sqrt{i m}}}{4 \sqrt{2}}$$.

I'm not sure what a sensible range of parameters is, but I tried a plotting a few cases, e.g., Plot[Abs[ft/.{m->2,\[Tau]->4/3,\[Omega]0->1/5}],{\[Omega],-20,20}], and obtained, I guess, plausible looking results. (But forgot how to include graphics here.)

• Interesting, I'm using Mathematica 11.0....Perhaps time for an upgrade!
– Q.P.
Jul 25 '19 at 20:15
• Can't reproduce it in version 12.0 on Windows 10 32-bit. I obtain the same result as in the question. Jul 26 '19 at 3:56
• @user64494 Does it agree with your Maple results? Now, I don’t have Maple, nor have I installed MMA version 12. Jul 26 '19 at 4:02
• @QuantumPenguin if you only care about additional stuff like this, the online version of Mathematica (in the Wolfram Cloud) can do stuff like this and doesn't cost you anything more. Jul 28 '19 at 17:21

Using MMA Version 12.0 for Windows I get:

FT1 = Assuming[{m > 0, τ > 0, ω0 > 0}, FourierTransform[
UnitStep[t] Exp[-t/τ] Cos[(m t + ω0) t], t, ω]] // FullSimplify;
ComplexExpand[Abs[FT1], TargetFunctions -> {Abs, Arg}] // FullSimplify

(* ((E^((2 (ω + ω0))/(m τ)))^(1/4)
Abs[1 + Erf[((-1)^(3/4) (I + τ (ω + ω0)))/(
2 Sqrt[m] τ)] -
I E^((I ((I + τ ω)^2/τ^2 + ω0^2))/(2 m))
Erfc[-(((-1)^(1/4) (I + τ (ω - ω0)))/(
2 Sqrt[m] τ))]])/(4 Sqrt (m^2)^(1/4)) *)

which, displayed in TeXForm, is:

$$\frac{\sqrt{e^{\frac{2 (\omega +\text{\omega 0})}{m \tau }}} \left|\text{erf}\left(\frac{(-1)^{3/4} (\tau (\omega +\text{\omega 0})+i)}{2 \sqrt{m} \tau }\right)-i e^{\frac{i \left(\frac{(\tau \omega +i)^2}{\tau ^2}+\text{\omega 0}^2\right)}{2 m}} \text{erfc}\left(-\frac{\sqrt{-1} (\tau (\omega -\text{\omega 0})+i)}{2 \sqrt{m} \tau }\right)+1\right|}{4 \sqrt{2} \sqrt{m^2}}$$