# Inverse Laplace Tranform of a function involving Bessel functions

I need to evaluate (if it exists) the inverse Laplace transform of the following complex function $$F(s)$$: $$F(s)=\sqrt{\frac{s}{a}} J_{1}(\sqrt{as})$$ where $$J_{1}(\cdot)$$ is the Bessel function of first kind and first order.

Does anyone have any suggestion?

I know that an inverse Laplace transform exists for a similar expression, i.e., $$L \left\{ \frac{1}{4t^2 } e^{-\frac{a}{4t}} \right\} = \sqrt{\frac{s}{a}} K_{1}(\sqrt{as})$$ where $$K_{1}(\cdot)$$ is the modified Bessel function of second kind and first order.

Thanks!

• Write the series expansion of $J_1$. Dec 21 '18 at 10:47
• But in this case I would obtain a series of distributions, wouldn't I? Dec 21 '18 at 16:30

I guess you're struggling with Voronoi's improvement of the Gauss circle problem. If you're interested, I am writing a section about it in my notes, it is not finished yet but it will probably be before the new year (2019). Anyway, you may consider Bessel's differential equation defining $$J_1$$, or directly the Maclaurin series of $$J_1$$, and discover that
$$\color{red}{\mathcal{L}}\left(\sum_{n\geq 0}\frac{(-1)^n a^{n} (s)^{n+1}}{ 2^{2n+1} n!(n+1)!}\right) = \frac{1}{2x^2} e^{-\frac{a}{4x}}$$ but the series is associated to an entire function, so its inverse Laplace transform is simply not defined, unless you're fine with a distributional identity $$\mathcal{L}^{-1}(s^n)(x)=\delta^{(n)}(x)$$.
• @MrCastozzo: the answer is no by Hamburger's theorem: the oscillations of $J_1$ violate log-convexity, so your expression cannot be the Laplace transform of a function defined on $\mathbb{R}^+$. On the other hand you already have your $\mathcal{L}^{-1}$: if it is a distribution and not a function, it is a distribution and not a function. Dec 21 '18 at 17:05