TLDR: How do I calculate the Fourier inverse transform (in terms of $\alpha$) of
$\tilde{T}(r,\theta,\alpha) = \frac{A\sigma}{J_0(\alpha R)\sqrt{4\pi \ln{2}}}\textrm{exp}{\left(-\frac{\alpha^2\sigma^2}{4\ln{2}}\right)}J_0(\alpha r)$?
I probably have to use the asymptotic behavior of $J_n(r)$ which says $J_n(r)\to \left(\frac{2}{\pi r}\right)^{1/2}\cos{\left(r-\frac{n\pi}{2}-\frac{\pi}{4}\right)}$ as $r\to \infty $ but I'm not sure how to proceed.
I was given the following problem.
Steady state temperature distribution in a medium of constant heat conductivity is governed by the Laplace equation $\nabla^2 T=0$. In cylindrical coordinates $(r,\theta, x)$, the Laplace's equation is;
$T_{rr}+\frac{1}{r}T_r +\frac{1}{r^2}T_{\theta \theta}+T_{xx}=0$
Find the temperature distribution inside an infinitely long cylinder of radius $R$, if the surface temperature is kept at $T(R,\theta,x)=A\exp{\left[-\ln{2}\left(\frac{x}{\sigma}\right)^2\right]}$
We use the Fourier transform/inverse Fourier transform with the following definition.
$\displaystyle \tilde{f} (\alpha) = \frac{1}{2\pi}\int_{-\infty}^{\infty} f(x)e^{-i\alpha x}dx$
$\displaystyle f(x) = \int_{-\infty}^{\infty} \tilde{f}(\alpha)e^{i\alpha x}d\alpha$
My attempt We apply the Fourier transform in terms of $x$ on both sides of the equation. By the fact $\tilde{f_x}=i\alpha \tilde{f}$, we obtain
$\displaystyle \tilde{T}_{rr}+\frac{1}{r}T_{r}+\frac{1}{r^2}T_{\theta\theta}-\alpha^2 T=0$
We apply the separation of variable $\tilde{T}(r,\theta,\alpha)=\tilde{S}(r,\alpha)\tilde{\Theta}(\theta,\alpha)$ and we obtain two equations:
$\displaystyle \begin{cases} \tilde{S}_{rr}+\frac{1}{r}\tilde{S}_{r}-\left(\frac{C}{r^2}+\alpha^2\right)\tilde{S}=0 \\ \tilde{\Theta}_{\theta\theta}=C\tilde{\Theta} \end{cases}$
where $C=C(\alpha)$ is a function of $\alpha$. We solve the second ODE using the periodic boundary condition(since we are using the cylindrical coordinate), and we obtain eigenfunctions:
$\Theta_{n}(\theta,\alpha)=D(\alpha)\cos{n\theta}+E(\alpha)\sin{n\theta}$
for an eigenvalue $C=-n^2$ with $n=0,1,2,\cdots$.
Using $C=-n^2$, the ODE for $S$ can be written as
$\tilde{S}_{rr}+\frac{1}{r}\tilde{S}_{r}+\left(\alpha^2-\frac{n^2}{r^2}\right)\tilde{S}=0$
Now, apply the change of variable $\rho =\alpha r$. Then, we have $\tilde{S}_r = \frac{\partial \rho}{\partial r}\frac{\partial \tilde{S}}{\partial \rho}=\alpha \tilde{S}_{\rho}$ and $\tilde{S}_{rr}=\alpha^2 \tilde{S}_{\rho\rho}$, and therefore the equation can be written as
$\alpha^2 \tilde{S}_{\rho\rho}+\frac{\alpha^2}{\rho}\tilde{S}_{\rho}+\left(\alpha^2-\alpha^2\frac{n^2}{\rho^2}\right)\tilde{S}=0$
or
$\tilde{S}_{\rho\rho}+\frac{1}{\rho}\tilde{S}_{\rho}+\left(1-\frac{n^2}{\rho^2}\right)\tilde{S}=0$
This is Bessel's equation. Therefore, the general solution is
$\tilde{S}(\rho,\alpha) = F(\alpha)J_n (\rho)+G(\alpha)Y_n(\rho)$
where $F,G$ are functions of $\alpha$. Using $\rho = \alpha r$, we have
$\tilde{S}(r,\alpha) = F(\alpha)J_n (r\alpha)+G(\alpha)Y_n(r\alpha)$
Due to its physical meaning, we want a bounded solution as $r\to 0$, but $Y_n$ diverges as $r\to 0$. Therefore $G=0$. Therefore our solution corresponding to the eigenvalue $-n^2$ is
$\tilde{S}_{n}(r,\alpha)=F(\alpha)J_{n}(\alpha r)$
Now, multiplying together, we have
$\tilde{T}_{n}(r,\theta,\alpha)=FJ_n(\alpha r)(D\cos{n\theta}+E\sin{n\theta})$
We define $D_n(\alpha)=F(\alpha)D(\alpha),E_n(\alpha)=F(\alpha)E(\alpha)$. Then
$\tilde{T}_{n}(r,\theta,\alpha)=J_n(\alpha r)(D_n\cos{n\theta}+E_n\sin{n\theta})$
Thus, the general solution for $\tilde{T}$ is
$\tilde{T} = \sum_{n=0}^{\infty}\tilde{T}_n = \sum_{n=0}^{\infty}J_{n}(\alpha r)(D_n\cos{n\theta}+E_n\sin{n\theta})$
To apply the boundary condition at $r=R$, we calculate the Fourier transform of $T(R,\theta,x)=A\exp{\left[-\ln{2}\left(\frac{x}{\sigma}\right)^2\right]}$.
After a long calculation, we obtain
$\tilde{T}(R,\theta,\alpha) = \frac{A\sigma}{\sqrt{4\pi \ln{2}}}\textrm{exp}{\left(-\frac{\alpha^2\sigma^2}{4\ln{2}}\right)}$
Thus, we have
$\sum_{n=0}^{\infty}J_{n}(\alpha R)(D_n\cos{n\theta}+E_n\sin{n\theta}) = \frac{A\sigma}{\sqrt{4\pi \ln{2}}}\textrm{exp}{\left(-\frac{\alpha^2\sigma^2}{4\ln{2}}\right)}$
Since the right hand side does not depend on $\theta$, we have $D_n=E_n=0$ for $n\geq 1$. So we have
$J_0(\alpha R)D_0 = \frac{A\sigma}{\sqrt{4\pi \ln{2}}}\textrm{exp}{\left(-\frac{\alpha^2\sigma^2}{4\ln{2}}\right)}$
Therefore we have
$D_0 = \frac{A\sigma}{J_0(\alpha R)\sqrt{4\pi \ln{2}}}\textrm{exp}{\left(-\frac{\alpha^2\sigma^2}{4\ln{2}}\right)}$
This means that the solution in Fourier space is
$\tilde{T}(r,\theta,\alpha) = \frac{A\sigma}{J_0(\alpha R)\sqrt{4\pi \ln{2}}}\textrm{exp}{\left(-\frac{\alpha^2\sigma^2}{4\ln{2}}\right)}J_0(\alpha r)$
Now I want to find the inverse Fourier transform, but I don't know how to deal with it, especially $J_0(\alpha r)/J_0(\alpha R)$ part.
How do I calculate the Fourier inverse transform?