I'm trying to find the inverse Laplace transform of $$\mathrm{e}^{-\beta\alpha}J_0(\frac{\beta\alpha}{2})^2$$ where $\alpha=const$. Would an idea be because since I'm looking for the $\alpha\rightarrow\infty$, to calculate the asymptotic form of $J_0(x)$ i.e., $\sqrt{\pi/x}\cos(x-\pi/4)$ and try it that way?
But I can't seem to solve it that way either...
I think your idea works, so asymptotic form of it is $$\dfrac{2e^{-\alpha\beta}}{\pi\alpha\beta}\left(1+\sin \alpha\beta\right)\tag{1}$$ also $\dfrac{1+\sin \alpha\beta}{\alpha\beta}$ is bounded and $$\dfrac{e^{-\alpha\beta}}{\alpha\beta}\left(1+\sin \alpha\beta\right)\to0$$ as $\alpha\to\infty$ where $\beta$ is a constant, then $(1)$ has the inverse Laplace transform. With $f(u)=\dfrac{e^{-u}}{u}\left(1+\sin u\right)$, $u=0$ is a simple pole and the residue of $$f(u) e^{t u}$$ is $1$, sum od residues is $1$, then the final answer should be $\dfrac{2}{\pi}$.