<s>[INST] Laplace inverse of $\frac{e^{-\sqrt{s+2}}}{s}$ I want to find out $$\mathcal{L^{-1}}\{\frac{e^{-\sqrt{s+2}}}{s}\}$$
How do you find the inverse Laplace? 
thanks
 [/INST] Thank you for the interesting question.  Here is a rather brute force solution, which may add a few steps to Jan Erland's solution.
First, let us recall that, if
$$
\mathcal L(f) = \int_0^\infty f(t) \, e^{-st} \, dt =  F(s)
$$
Then
$$
\mathcal L(f') = \int_0^\infty f'(t) \, e^{-st} \, dt = s \, F(s) - f(0).
$$
In our case, $F(s) = e^{-\sqrt{s+2}}/s$, so, we shall seek a function $f'(t)$ whose Laplace transform is $sF(s) = e^{-\sqrt{s+2}}$.  Then $f(t)$ can be found from integrating $f'(t)$ from $t = 0$, where $f(0) = 0$.
Using the inverse formula, we have
$$
f'(t)
= \frac{1}{2\, \pi \, i}\int_{\gamma-i\infty}^{\gamma+i\infty} e^{-\sqrt{s+2} + st} \, ds,
$$
where $\gamma$ is a large positive number, such that all poles, if any, lie on the left side of the line $y = \gamma$.
For our case, there is no pole, but only a branch cut at $s = -2$. We shall let the branch cut extend to $-\infty$ from $s = -2$, and wrap the contour around the branch cut from $-\infty + 0^- \, i$ to $-2 + 0^- \, i$ (lower half) and then from $-2 + 0^+ \, i$ to $-\infty + 0^+ \, i$ (higher half).
Let $s = -2 + A\,e^{-i\pi}$ with $A \ge 0$, then
$$
\begin{aligned}
f'(t)
&= \frac{e^{-2t}}{\pi}
\int_0^{+\infty} \sin(\sqrt{A}) \, e^{- A t} dA \\
&= \frac{e^{-2t}}{\pi}
\int_{-\infty}^{+\infty} u \, \sin(u) \, e^{-u^2 t} du \\
&= \frac{e^{-2t}}{\pi}
\mathrm{Im} \left(\frac{\partial}{\partial v}
\int_{-\infty}^{+\infty} e^{-u^2 t + v u} d u \right)_{v = i} \\
&= \frac{e^{-2t}}{\pi}
\mathrm{Im} \left[\frac{\partial}{\partial v}\left(
\sqrt{\frac{\pi}{t}} \, \exp{\frac{v^2}{4t}} \right)\right]_{v = i} \\
&= \frac{1}{2 \, \sqrt{\pi \, t^3}} \, \exp\left(-2 \, t-\frac{1}{4\,t}\right).
\end{aligned}
$$
Here, we have used the fact that $\mathrm{Im} \, e^{iu} = \sin(u)$,
Finally, let us integrate $f'(t)$.
$$
\begin{aligned}
f(t)
&= \int_0^t f'(\tau) \, d\tau \\
&=\frac{1}{\sqrt\pi}
  \int_{1/\sqrt{t}}^\infty
    \exp\left(-\frac{x^2}{4}-\frac{2}{x^2}\right) \, dx \\
&=\frac{1}{\sqrt\pi}
\int_{1/\sqrt{t}}^\infty
  \exp\left(-\frac{x^2}{4}-\frac{2}{x^2}\right) \,
    d\left( \frac{x}{2}+\frac{\sqrt{2}}{x} \right) \\
&\quad+
\frac{1}{\sqrt\pi}
\int_{1/\sqrt{t}}^\infty
  \exp\left(-\frac{x^2}{4}-\frac{2}{x^2}\right) \,
    d\left( \frac{x}{2}-\frac{\sqrt{2}}{x} \right)
 \\
&=\frac{1}{\sqrt\pi}
\left[
\int_{\frac{1}{2\sqrt{t}}+\sqrt{2t}}^\infty
  \exp\left(\sqrt{2} -y^2\right) \,  dy
+
\int_{\frac{1}{2\sqrt{t}}-\sqrt{2t}}^\infty
  \exp\left(-\sqrt{2} -z^2\right) \, dz
\right] \\
&=
  \frac{e^{\sqrt 2}}{2} \, \mathrm{erfc}\left(
    \frac{1}{2\sqrt{t}}+\sqrt{2\, t} \right)
 +\frac{e^{-\sqrt 2}}{2} \, \mathrm{erfc}\left(
    \frac{1}{2\sqrt{t}}-\sqrt{2\, t} \right).
\end{aligned}
$$
Here, we have changed variables
$$
\begin{aligned}
x &\equiv \frac{1}{\sqrt{\tau}}, \\
y &\equiv \frac{x}{2} + \frac{\sqrt{2}}{x}, \\
z &\equiv \frac{x}{2} - \frac{\sqrt{2}}{x}.
\end{aligned}
$$
Our result agrees with Jan Erland's.
</s>