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I have to use InverseFourierSinTransForm in Mathematika for the function u[ω,t] but infortunately it does not work.It gives back the same! I tried it without the assumptions but it does not work again! Any help please?How can I write InverseFourierSinTransform to find the result? $$u(x,t)=\frac{2}{π}\int^{\infty}_{0}\hat{u}(\omega,t)sin(\omega x)d\omega$$

$$\hat{u}(\omega,t)= \sqrt{\frac{2}{π}}\frac{e^{-\omega^2 t}\omega}{\omega^2+a^2}$$

u[x_, t_] = InverseFourierSinTransform[v[ω, t], ω, x, Assumptions -> α > 0]

where $\hat{u} \to $ v[ω,t]

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  • $\begingroup$ It would depend on the (missing) definition of v. $\endgroup$ Commented May 4, 2023 at 19:28
  • $\begingroup$ @DanielLichtblau the ``` v[ω,t]``` is defined as the $\hat{u}$.I did n't give all the code because I think that it does not matter.Have you any idea what I have to add to run this? $\endgroup$
    – george
    Commented May 4, 2023 at 20:37
  • $\begingroup$ It would require that definition for v. $\endgroup$ Commented May 4, 2023 at 23:25

2 Answers 2

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In principle, the integral can be reduced to finding

FourierTransform[Exp[- t omega^2]/(omega +- I a), omega, s]

because by partial fraction decomposition and diffentiating wrt to t

(1/(2*I*a*\[Omega]))*D[(E^(-t*\[Omega]^2)*(1/(\[Omega] + I*a)) - 1/(\[Omega] - I*a)), t]

Mathematica does not know, but searching the tables could be done.

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Looks like: $$\text{InverseFourierSinTransform}\left[\frac{\exp \left(-\omega ^2 t\right) \omega }{\omega ^2+a^2},\omega ,x\right]=\\\int_0^{\infty } \frac{\sqrt{\frac{2}{\pi }} \sin (\omega x) \left(\omega \exp \left(-\omega ^2 t\right)\right)}{\omega ^2+a^2} \, d\omega =-\frac{e^{a^2 t} \pi \left(e^{-a x} \text{erf}\left(a \sqrt{t}-\frac{x}{2 \sqrt{t}}\right)-e^{a x} \text{erf}\left(a \sqrt{t}+\frac{x}{2 \sqrt{t}}\right)+2 \sinh (a x)\right)}{2 \sqrt{2 \pi }}$$

  HoldForm[
  InverseFourierSinTransform[
  Exp[-\[Omega]^2*t]/(\[Omega]^2 + a^2)*\[Omega], \[Omega], x] == 
  Integrate[
  Sqrt[2/Pi]*
  Sin[\[Omega] x]*(\[Omega]*
  Exp[-\[Omega]^2*t]/(\[Omega]^2 + a^2)), {\[Omega], 0, 
  Infinity}] == -(1/(2 Sqrt[2 \[Pi]]))*
  E^(a^2 t) \[Pi] (E^(-a x) Erf[a Sqrt[t] - x/(2 Sqrt[t])] - 
  E^(a x) Erf[a Sqrt[t] + x/(2 Sqrt[t])] + 2 Sinh[a x])]

Example 3.954.1 on page 535 from book: "Table of Integrals, Series, and Products Eighth Edition-I. S. Gradshteyn and I. M. Ryzhik"

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