Laplace transform of Bessel function I am stuck in a question, and don't know where to start. I have to obtain the Laplace transform of $J_0(t)$, I have to let:
$$a_n=\int_{0}^{\pi}(\sin \theta)^{2n}d\theta$$
And now wish to show that:
$$a_n= \frac{(2n)!}{2^{2n}(n!)^2}\pi$$
My idea was:
I know that:
$$J_0(t)=\sum_{n=0}^{\infty}\frac{(-1)^n t^{2n}}{(n!)^2 2^{2n}}$$
The Laplace transform is represented by:
$$\mathcal{L}(f)=\int_{0}^\infty e^{-st}f(t)dt$$
But can I just plug in the first $a_n$? I don't think so. But where to start now?

It is possible to solve this using the DE-definition of Bessel function as well as some basic properties of Laplace transforms. 
Recall that the definition of the zeroth order Bessel function $J_0(x)$ is that it satisfies 
$$xJ_0''+J_0'+xJ_0=0.$$
Take the Laplace transform, and using a property about derivative of Laplace transform:
$$\widehat{f'}=p\widehat{f}-f(0)$$
$$\widehat{f''}=p^2\widehat{f}-pf(0)-f'(0)$$
and a property about derivative of a Laplace transform:
$$\widehat{tg}=-\widehat{g}\ '$$
We get the following equation 
$$(p^2+1)\widehat{J_0} \ ' + p\widehat{J_0} = 0$$
and thus $\widehat{J_0} = A/\sqrt{1+p^2}$. We can further find $A=1$ using $J_0(0)=1$ and the property that 
$$p\widehat{f} \rightarrow f(0) \ \textrm{as} \ p\rightarrow \infty.$$
