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...

  • $\begingroup$ I think your idea works, so asymptotic form of it simplify to $$\dfrac{2e^{-\alpha\beta}}{\pi\alpha\beta}\left(1+\sin \alpha\beta\right)$$ $\endgroup$ – Nosrati Jun 13 at 18:13

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

  • $\begingroup$ 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? $\endgroup$ – Lewis Proctor Jun 13 at 20:00
  • $\begingroup$ I don't know. As you know the final answer is asymptotic, not an exact. $\endgroup$ – Nosrati Jun 13 at 20:03
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    $\begingroup$ @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 $\beta$. $\endgroup$ – Maxim Jun 13 at 20:21

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