Definite Exponential Integral evaluation 
What is the value of the following integral ?:
  $$
I=\int_0^{\infty}\exp\left(-\,{c \over x^{2} + a^{2}} - x^{2}\right)
\,\mathrm{d}x
$$

By change of variable $u = x^{2} + a^{2}$:
$$I
=
\mathrm{e}^{a^{2}}
\int_{a^{2}}^{\infty}{1 \over 2\,\sqrt{\, u - a^{2}\,}\,}\,
\exp\left(-\,{c \over u} - u\right)\,\mathrm{d}u
$$
I have worked on it a lot but nothing I found$\ldots$ Any thoughts ?.
 A: I will outline a possible approach. For simplicity, I will initially assume $a=c=1$, then provide a generalization for $a,c>0$. We have
$$ I(a,c)=\int_{0}^{+\infty}\exp\left(-\frac{1}{x^2+a}-x^2\right)\,dx = \int_{0}^{+\infty}\frac{e^{-x}}{2\sqrt{x}}\cdot\exp\left(-\frac{c}{x+a}\right)\,dx \tag{1}$$
where the Laplace transform of $\frac{e^{-x}}{2\sqrt{x}}$ is given by
$$ \mathcal{L}\left(\frac{e^{-x}}{2\sqrt{x}}\right)=\frac{\sqrt{\pi}}{2\sqrt{1+s}}\tag{2} $$
and the inverse Laplace transform of $\exp\left(-\frac{1}{x+1}\right)$ is given by:
$$ \mathcal{L}^{-1}\left(\exp\left(-\frac{1}{x+1}\right)\right) = \left(\delta(s)-\frac{J_1(2\sqrt{s})}{\sqrt{s}}\right)e^{-s}\tag{3} $$
so that:
$$\begin{eqnarray*}I(1,1)&=&\frac{\sqrt{\pi}}{2}-\sqrt{\pi}\int_{0}^{+\infty}\frac{J_1(2t)}{\sqrt{t^2+1}}e^{-t^2}\,dt\\&=&\frac{\sqrt{\pi}}{2}-\sqrt{\pi}\int_{0}^{+\infty}J_1(2\sinh\theta)\exp\left(-\sinh^2\theta\right)\,d\theta. \tag{4}\end{eqnarray*} $$
where the Bessel function $J_1(t)$ is an odd function, and for every odd $m\in\mathbb{N}$ the integral
$$ \int_{0}^{+\infty}\sinh(\theta)^{m}e^{-\sinh^2\theta}\,d\theta $$
is a linear combination with rational coefficients of $1$ and $e\sqrt{\pi}\,\text{Erfc}(1)$, due to integration by parts. Since the coefficient of $t^{2n+1}$ in $J_1(t)$ is given by $\frac{(-1)^n}{2^{2n+1}n!(n+1)!}$, the RHS of $(4)$ can be rearranged as a very fast-convergent series. The same principle holds if $a,c$ are positive but $\neq 1$.
