Does the following infinite series converge?
\begin{align} \label{eq:series} \sum_{j=1}^{\infty} \mathrm{e}^{-\lambda_{j}^{2} \,t}, \end{align}
where $t\in R$, $t>0$ and $\lambda_{j}$ is the j-th zero of the Bessel function of the first kind $J_{3/2}(x)$. If it does converge, how to obtain a closed-form expression of it?
The following are what I have tried.
By using argument principle, one can convert the infinite series into contour integral on complex plane
\begin{align} \sum_{j=1}^{\infty} \mathrm{e}^{-\lambda_{j}^{2} \,t} = \frac{1}{2\pi\mathrm{i}} \oint_{C}\mathrm{e}^{-z^{2}\,t} \frac{J'_{3/2}(z)}{J_{3/2}(z)} \mathrm{d} z, \end{align}
where the contour curve $C$ should be carefully chosen. I do not know how to choose the contour curve properly so that the contour integral can be calculated explicitly.
I am also thinking about simplifying the infinite series by Laplace transform. Denote
\begin{align} f(t) \equiv \sum_{j=1}^{\infty} \mathrm{e}^{-\lambda_{j}^{2}\,t}. \end{align}
The Laplace transformed form is
\begin{align} F(s) = \int_{0}^{\infty} f(t) \mathrm{e}^{-s\,t}\mathrm{d} t = \int_{0}^{\infty} \sum_{j=1}^{\infty} \mathrm{e}^{-\lambda_{j}^{2}\,t} \mathrm{e}^{-s\,t}\mathrm{d} t = \sum_{j=1}^{\infty} \int_{0}^{\infty} \mathrm{e}^{-\lambda_{j}^{2}\,t} \mathrm{e}^{-s\,t}\mathrm{d} t = \sum_{j=1}^{\infty} \frac{1}{s+\lambda_{j}^{2}}. \end{align}
Is it legitimate to swap the integral and summation here given that $s\in \mathcal{C}$? If it is, then by argument principle, there should be
\begin{align} F(s) = \sum_{j=1}^{\infty} \frac{1}{s+\lambda_{j}^{2}} = -\frac{1}{2\mathrm{i}\sqrt{s}} \frac{J'_{3/2}(\mathrm{i} \sqrt{s})}{J_{3/2}(\mathrm{i} \sqrt{s})} = \frac{1}{2\sqrt{s}} \frac{I'_{3/2}(\sqrt{s})}{I_{3/2}(\sqrt{s})}. \end{align}
But how to calculate the inverse Laplace transform
\begin{align} \frac{1}{2\pi \mathrm{i}} \lim_{T\to\infty} \int_{\gamma-\mathrm{i}\, T}^{\gamma + \mathrm{i}\, T} \mathrm{e}^{s\,t} \frac{1}{2\sqrt{s}} \frac{I'_{3/2}(\sqrt{s})}{I_{3/2}(\sqrt{s})} \mathrm{d} s , \end{align}
where $\gamma$ is real and is greater than the real part of all the singularities of $F(s)$ according to Wikipedia: inverse Laplace transform.
Any information is greatly appreciated.
