Take Laplace Transform of the integral J_0 I was just wondering how to use tables from Spiegal to solve $\int_0^\infty J_0(2\sqrt{ut}) J_0(u) du$ At the moment, I see similar transforms on page 244, but I don't actually know how to combine the laplace transforms of the first $J_0$ and the second $J_0$
Any help is appreciated,
Thank you!
 A: Well, I have a way to do it without referring to the tables, but instead using two well-known representations of a Bessel function.  First write
$$J_0(u) = \frac{1}{\pi} \int_0^{\pi} d\theta \, e^{i u \cos{\theta}}$$
Then assuming we may interchange the order of integration, the integral is equal to
$$\begin{align}\int_0^{\infty} du \, J_0 \left ( 2 \sqrt{u t}\right ) J_0(u) &= \frac{1}{\pi}\int_0^{\pi} d\theta \, \int_0^{\infty} du \, e^{i u \cos{\theta}} J_0 \left ( 2 \sqrt{u t}\right )\\ &= \frac{2}{\pi}\int_0^{\pi} d\theta \, \int_0^{\infty} dv \, v e^{i \cos{\theta}\; v^2}J_0 \left ( 2 \sqrt{t} v\right ) \end{align}$$
Here we use the well-known relation 
$$\int_0^{\infty} dv \, v \, e^{i a v^2} J_0(b v) = \frac{i}{2 a} e^{-i b^2/(4 a)}$$
so that we get that
$$\begin{align}\int_0^{\infty} du \, J_0 \left ( 2 \sqrt{u t}\right ) J_0(u) &= \frac{i}{\pi} \int_0^{\pi} d\theta \, \sec{\theta} \, e^{-i t \, \sec{\theta}}\\ &= \frac{1}{\pi} \int_{-1}^1 dy \, \left (1-y^2\right )^{-1/2} e^{-i t y} \end{align}$$
This last integral is simply another well-known representation of a Bessel.  Therefore,
$$\int_0^{\infty} du \, J_0 \left ( 2 \sqrt{u t}\right ) J_0(u) = J_0(t)$$
