# Are there some programs about inverse Fourier and Laplace transfroms?

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.

• You define tt, but don't use it. Aug 2 '20 at 7:09
• @user64494 Please ignore it since the above are the total parameters. So some formulas may do not use it. Aug 2 '20 at 7:16

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 ).

• 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. Aug 2 '20 at 19:27
• 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}. Aug 3 '20 at 9:59
• thanks! Usuaslly it is not easy to solve such a problem. Aug 5 '20 at 11:42