# Inverse Laplace transformation of Bessel and exponential

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 simplify to $$\dfrac{2e^{-\alpha\beta}}{\pi\alpha\beta}\left(1+\sin \alpha\beta\right)$$ – Nosrati Jun 13 at 18:13

## 1 Answer

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 $$\dfrac{2}{\pi}$$.

• Thanks this is great! I was just refreshing myself on residue calculus and (for your $f(u)$) I get the same answer. I've been looking into keeping the $\alpha$ term, and looking at the Laurent series, and $\mathrm{e}^{tu-au}J_0(au/2)^2$ doesn't have any poles. How would one do this? – Lewis Proctor Jun 13 at 20:00
• I don't know. As you know the final answer is asymptotic, not an exact. – Nosrati Jun 13 at 20:03
• @LewisProctor If $\beta$ is real, (1) as a function of $\alpha$ is not bounded on a vertical line. In fact, it grows exponentially and the inverse transform doesn't exist as an ordinary function and doesn't exist even as a tempered distribution. Same if $\alpha$ is a real constant and (1) is a function of $\beta$. – Maxim Jun 13 at 20:21