Solving Bessel integration What would be the solution of the bessels equation,
$$b=k A(t)\int_0^{\infty} J_0 (k \rho) e^ \frac{-\rho^2}{R^2} \rho d \rho$$
Can I sove that by using this formulation?
$$c= \int_0^{\infty}j_0(t) e^{-pt} dt= \frac{1}{\sqrt{1+p^2}}$$
 A: No, the Laplace transform of a Bessel does not apply here.  Rather, what you are looking at is the Bessel transform of a Gaussian, which is really just the expression of the 2D Fourier transform of a 2D Gaussian:
$$\int_0^{\infty} d\rho \, \rho J_0(k \rho) \, e^{-\rho^2/R^2} = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dx \,  \int_{-\infty}^{\infty} dy \, e^{i (k_x x + k_y y)}\, e^{-(x^2+y^2)/R^2}$$
You may verify this using the relation
$$\int_0^{2 \pi} d\phi \, e^{i k \rho \cos{(\phi-\theta)}} = 2 \pi \, J_0(k \rho)$$
where $\rho = \sqrt{x^2+y^2}$ and $\phi-\theta$ is the angle between the vectors $(k_x,k_y)$ and $(x,y)$.
Can you compute each individual FT of the 1D Gaussians?  I get
$$\int_0^{\infty} d\rho \, \rho J_0(k \rho) \, e^{-\rho^2/R^2} = \frac{R^2}{2} e^{-R^2 k^2/4}$$
ADDENDUM
Here is the evaluation of the FT.  You can separate into $x$ and $y$ integrals:
$$ \frac{1}{2 \pi} \int_{-\infty}^{\infty} dx \,\int_{-\infty}^{\infty} dy \, e^{i (k_x x + k_y y)}\, e^{-(x^2+y^2)/R^2} = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dx \, e^{i k_x x} e^{-x^2/R^2} \, \int_{-\infty}^{\infty} dy \, e^{i k_y y} e^{-y^2/R^2}$$
The evaluation of each integral is identical in form, so let's look at the $x$ one.  The idea is to complete the square and use the fact that $\int_{-\infty}^{\infty} dx \, e^{-x^2} = \sqrt{\pi}$.  
$$\begin{align}\int_{-\infty}^{\infty} dx \, e^{i k_x x} e^{-x^2/R^2} &= \int_{-\infty}^{\infty} dx \, e^{-1/R^2 (x^2-i R^2 k_x x)}\\ &= \int_{-\infty}^{\infty} dx \, e^{-1/R^2 (x^2-i R^2 k_x x - R^4 k_x^2/4 + R^4 k_x^2/4)} \\ &= \int_{-\infty}^{\infty} dx \, e^{-1/R^2 (x^2-i R^2 k_x x - R^4 k_x^2/4)} e^{-R^4 k_x^2/(4 R^2)} \\ &= e^{-R^2 k_x^2/4} \int_{-\infty}^{\infty} dx \, e^{(x-i R^2 k_x/2)^2/R^2}\end{align}$$
Now, it turns out that this last integral is independent of the imaginary shifted term in the exponential, and therefore has value $\sqrt{\pi} R$.  Do the same for the $y$ integral, and the result follows.
A: According to Gradshteyn and Ryzhik, we have:
$$\int_0^{\infty}x^{\mu}\exp(-\alpha x^2)J_{\nu}(\beta x)dx = \frac{\beta^{\nu}\Gamma\left(\frac{1}{2}\nu+\frac{1}{2}\mu+\frac{1}{2}\right)}{2^{\nu+1}\alpha^{\frac{1}{2}(\mu+\nu+1)}\Gamma(\nu+1)}\mbox{}_1 F_1\left(\frac{\nu+\mu+1}{2};\mbox{ }\nu+1;\mbox{ }-\frac{\beta^2}{4\alpha}\right).$$
Here $\mbox{}_1F_1$ is a hypergeometric function. Inputting the proper values gives
$$\int_0^{\infty}\rho\exp\left(-\frac{\rho^2}{R^2}\right)J_0(k\rho)d\rho = \frac{\Gamma(1)}{2\frac{1}{R^2}\Gamma(1)}\mbox{}_1F_1\left(1;1;-\frac{k^2R^2}{4}\right).$$
Using a property of the hypergeometric function ($_1F_1(a;a;x) = \exp(x)$) we get:
$$\frac{R^2}{2}\exp\left(-\frac{k^2R^2}{4}\right).$$
