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To be more exact, I have a function F[w_,s_], where $w$ is the Fourier transform of $x$ and $s$ is the Laplace transform of $t$.

Now I want to perform the double inverse transforms $s\to t$ and $w\to x$.

Thanks!

If it is not convenient to share the program, can you help me to perform? My parameters are

t0 = 0.1;
alpha = 1.5;
beta = alpha;
mx = alpha*t0/(alpha - 1);
a = 1;
\[Sigma] = 1;
balpha = t0^alpha*Abs[Gamma[1 - alpha]];

My function is

F[w_,s_]:= (1/(s + (I*a*w)/mx + ((\[Sigma]^2 + a^2)*w^2)/(2*mx) + 
 balpha/mx^2*s^(alpha - 1)*(I*w*a + 1/2*(\[Sigma]^2 + a^2)*w^2)))

I want to get F[x,t] in real space with $t=1000$, where $x$ is rang from 1500 to 4500.

Thanks!

PS: Here I added a figure, I have tried some programs. The red solid line is the theory I expected and the other lines are obtained from numerical inverse transforms. enter image description here

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  • 1
    $\begingroup$ You define tt, but don't use it. $\endgroup$ – user64494 Aug 2 at 7:09
  • $\begingroup$ @user64494 Please ignore it since the above are the total parameters. So some formulas may do not use it. $\endgroup$ – Blueka Aug 2 at 7:16
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Too long for a comment.

First, your F[w_,s_]:= (1/(s + (I*a*w)/mx + ((\[Sigma]^2 + a^2)*w^2)/(2*mx) + balpha/mx^2*s^(alpha - 1)*(I*w*a + 1/2*(\[Sigma]^2 + a^2)*w^2))) is a complex valued function. I don't see any reason why its inverse Fourier transform should be a real valued function.

Second, up to the definition used by Mathematica, its inverse Fourier transform equals

Integrate[(1/(s + (I*a*w)/mx + ((\[Sigma]^2 + a^2)*w^2)/(2*mx) + 
  balpha/mx^2*s^(alpha - 1)*(I*w*a + 1/2*(\[Sigma]^2 + a^2)*w^2)))*
  Exp[-I*x*w - I*s*t], {w, -Infinity, Infinity}, {s, -Infinity,  Infinity}]/(2*Pi)

I have doubts whether the above double improper integral can be expressed in closed form. The one can be calculated numerically for $t=1000$ as

t0 = 1/10;alpha = 3/2;beta = alpha;mx = alpha*t0/(alpha - 1);a = 1;\[Sigma] = 1;
balpha = t0^alpha*RealAbs[Gamma[1 - alpha]];
f[x_] := NIntegrate[(1/(s + (I*a*w)/
     mx + ((\[Sigma]^2 + a^2)*w^2)/(2*mx) + 
    balpha/mx^2*
     s^(alpha - 1)*(I*w*a + 1/2*(\[Sigma]^2 + a^2)*w^2)))*
Exp[-I*x*w - I*s*1000], {w, -Infinity, Infinity}, {s, -Infinity, 
Infinity}, AccuracyGoal -> 5, PrecisionGoal -> 5,  WorkingPrecision -> 15]/(2*Pi)
f[3600] // AbsoluteTiming
(*{1874.91, -0.000207174182415258 - 0.000187617006989998 I}*)

Third, the integrand is very oscillating in both variables. I don't know good numeric methods to calculate such sort integrals. Asymptotic methods are known to this end (see M. Fedoryuk. The saddle-point method. Moskva: ”Nauka”. 368 p. (1977) (in Russian) https://zbmath.org/?q=ai%3Afedoryuk.mikhail-v+py%3A1977 ).

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  • $\begingroup$ I'd like to add that F[w,s] has an integrable singularity at the origin in view of Simplify[ComplexExpand[ Abs[(1/(s + (I*a*w)/mx + ((\[Sigma]^2 + a^2)*w^2)/(2*mx) + balpha/mx^2* s^(alpha - 1)*(I*w*a + 1/2*(\[Sigma]^2 + a^2)*w^2)))]], Assumptions -> s < 0 && w > 0] which results in $$\frac{9}{\sqrt{-4 s w \left(9 \sqrt{10 \pi } \sqrt[4]{s^2}+10 \pi w^3+5 (2 \pi -27) w\right)+81 s^2+900 \left(w^4+w^2\right)}} $$ and similar expansions in other quadrants. $\endgroup$ – user64494 Aug 2 at 19:27
  • $\begingroup$ The addition of the Exclusions -> {{s, w} == {0, 0}} option changes nothing but time: nowf[3600] // AbsoluteTiming produces {9986.51, -0.000207174182415258 - 0.000187617006989998 I}. $\endgroup$ – user64494 Aug 3 at 9:59
  • $\begingroup$ thanks! Usuaslly it is not easy to solve such a problem. $\endgroup$ – Blueka Aug 5 at 11:42

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