Inverse Laplace transform of $1/(1+\sqrt{1+s^2})$

Question

A seemingly simple function,

$F(s)=\frac{1}{1+\sqrt{1+s^2}}$.

But what is its inverse Laplace transform?

If there is not a closed form, can I get some long/short time asymptotic?

It must have something to do with the Bessel J function.

Answer 1

Hint

Let the inverse Laplace be $f(t)$ . Then what is the Laplace of $\frac{f(t)}{t}$

Answer 2

We can use the trick of writing $$\frac 1 {1 + \sqrt {1 + s^2}} = \frac 1 s \sqrt {1 + \frac 1 {s^2}} - \frac 1 {s^2},$$ which holds for $s$ in the right half-plane. Then $$\mathcal L^{-1} \!{\left[ \frac 1 {1 + \sqrt {1 + s^2}} \right]} = -\mathcal L^{-1}[s^{-2}] + \sum_{k \geq 0} \binom {1/2} k \mathcal L^{-1}[s^{-2 k - 1}] = \\ -t + \sum_{k \geq 0} \binom {1/2} k \frac {t^{2 k}} {\Gamma(2 k + 1)} = {_1\hspace{-2px}F_2} {\left( -\frac 1 2; \frac 1 2, 1; -\frac {t^2} 4 \right)} - t.$$ The hypergeometric function can be converted to a combination of Bessel and Struve functions.