Prove that $\sum_{n=0}^{\infty} \frac{L_n(t)}{n!} = eJ_0(2\sqrt{t}) $ using Laplace transform $$\sum_{n=0}^{\infty} \frac{L_n(t)}{n!} = eJ_0(2\sqrt{t}) $$
where $L_n(t)$ are the Laguerre polynomials and $J_0(t)$ the Bessel function of first kind
I have calculated the Laplace transform of $L_n(t)$ which is $\frac{(s-1)^n}{s^{n+1}}$
The sum after a Laplace transform will evaluate to $\frac{1}{s} e^{1-\frac{1}{s}}$ by using the definition.
But $e^{-\frac{1}{s}}$ has no inverse Laplace transform so I guess that's not how to approach this problem.
 A: As I said in the comments, just take the explicit representation of a Laguerre polynomial:
$$L_n(t) = \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{t^k}{k!} $$
The sum in question is then
$$\begin{align} \sum_{n=0}^{\infty} \frac{L_n(t)}{n!} &= \sum_{n=0}^{\infty} \frac1{n!} \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{t^k}{k!} \\ &= \sum_{k=0}^{\infty} (-1)^k \frac{t^k}{k!} \sum_{n=k}^{\infty} \frac1{n!} \binom{n}{k} \\ &= 
\sum_{k=0}^{\infty} (-1)^k \frac{t^k}{k!} \sum_{m=0}^{\infty} \frac1{m! k!} \\&= e  \sum_{k=0}^{\infty} (-1)^k \frac{t^k}{k!^2} \\ &= e J_0 \left ( 2 \sqrt{t} \right ) \end{align}$$
A: Your approach also works.
We just need to show that the Laplace transform of $ J_{0}(2 \sqrt{t})$ is $\frac{e^{-1/s}}{s}$.
That is, we need to show that$$\int_{0}^{\infty}  e^{-st} J_{0}(2 \sqrt{t}) \,  \mathrm dt = \frac{e^{-1/s}}{s} $$
Using the series definition of the Bessel function of the first kind, we get
$$\begin{align} \int_{0}^{\infty}  e^{-st} J_{0}(2 \sqrt{t}) \, \mathrm dt &= \int_{0}^{\infty} e^{-st} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(n!)^{2}} \, t^{n} \, \mathrm dt \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(n!)^{2}} \, \int_{0}^{\infty}e^{-st} \, t^{n} \, dt \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(n!)^{2}} \, \frac{\Gamma(n+1)}{s^{n+1}} \\ &= \frac{1}{s} \sum_{n=0}^{\infty} \frac{1}{n!} \left(-\frac{1}{s} \right)^{n} \\ &= \frac{1}{s} \, e^{-1/s} \end{align}$$
