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.


  • $\begingroup$ Write the series expansion of $J_1$. $\endgroup$
    – Nosrati
    Dec 21 '18 at 10:47
  • $\begingroup$ But in this case I would obtain a series of distributions, wouldn't I? $\endgroup$
    – MrCastozzo
    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 $\mathcal{L}^{-1}(s^n)(x)=\delta^{(n)}(x)$.

  • $\begingroup$ For some reasons, I would need to express the propagator (advanced, retarted, or Feynmann propagator) as a Laplace transform of some quantity. $\endgroup$
    – MrCastozzo
    Dec 21 '18 at 16:48
  • $\begingroup$ And unfortunately the expression of the sought inverse Laplace transform as a series of distributions is not useful for my purposes. I need a closed form expression to be further manipulated. In any case, I also wonder if such an inverse Laplace transform exists in the sense of functions... $\endgroup$
    – MrCastozzo
    Dec 21 '18 at 16:57
  • 1
    $\begingroup$ @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. $\endgroup$ Dec 21 '18 at 17:05
  • $\begingroup$ Dear Jack, thanks for your answer and explanation. $\endgroup$
    – MrCastozzo
    Dec 21 '18 at 17:45

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