Inverse Laplace transform of the Bessel function

Question

I want to calculate the inverse Laplace transform of Bessel function: $$J_{as}(x)=\sum_{m=0}^\infty\frac{(-1)^{m}(\frac{x}{2})^{2m+as}}{m!Γ(m+as+1)}=\sum_{m=0}^\infty\frac{(-1)^{m}(\frac{x}{2})^{2m}(\frac{x}{2})^{as}}{m!Γ(m+as+1)}=\sum_{m=0}^\infty\frac{(-1)^{m}(\frac{x}{2})^{2m}e^{asln(\frac{x}{2})}Γ(-m-as)sin\pi(1+m+as)}{m!\pi}$$ if I know Laplace's reversal of the multiplication of the exponential function and the sin function, I also know that Laplace's reversal of the gamma function, thus I can solve with convolution theorem my problem .your question this is, what is the inverse Laplace transform of gamma function that is raised in question?

Answer 1

$$J_\alpha(x) = \frac{1}{\pi} \int_0^\pi \cos(\alpha\tau - x \sin\tau)\,d\tau - \frac{\sin \alpha\pi}{\pi} \int_0^\infty e^{-x \sinh t - \alpha t} \, dt$$

Apply the change of variables $u = \sin \tau, v = \sinh \tau$ to obtain that $J_\alpha(x)$ is a Laplace transform, whose inverse Laplace transform is thus obvious by the inversion theorem.

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As $e^{i \pi s}$ and $\sin(\pi s)$, $\frac{1}{\Gamma(s)}$ grows way too fast on vertical lines to be the Laplace transform of a function.

Its regularized inverse Laplace transform is an analytic functional (which tell us an expression for $\mathcal{L}^{-1}[\frac{F(s)}{\Gamma(s)}]$ in term of $\mathcal{L}^{-1}[F(s)]$ assuming both are well-defined). See how it works for $\lim_{n \to \infty} \mathcal{L}^{-1}[e^{i \pi s}e^{s^2/n^2}]$