Complicated Inverse Laplace Transform I don't know how to show that the inverse Laplace transform of: 
$$k\cdot s^{-1/2} e^{-2\sqrt{s}}$$ is
$$K\cdot t^{-1/2}e^{-1/t}$$
where $k$ and  $K$ are constants.
In the standard tables of the Laplace transform I didn't see any property that justifies this. However I have found this in a deduction in a book and they write it without justifications. So It could be something trivial.
 A: Hints: Try to google the Error Function $\text{erf}(x)$ to find out the nature of this function. Then you will see that $$\mathcal{L}(\frac{1}{\sqrt{\pi t}}\text{e}^{-a^2/4t})=\frac{\text{e}^{-a\sqrt{s}}}{\sqrt{s}},~~ a>0$$
A: The difficulty with this ILT is that there is a branch point singularity at $s=0$ that a Bromwich contour must avoid.  That is, we are tasked with evaluating
$$\oint_C ds \: s^{-1/2} e^{-2 \sqrt{s}} e^{s t}$$
where $C$ is the following contour

We will define $\text{Arg}{z} \in (-\pi,\pi]$, so the branch is the negative real axis.  There are $6$ pieces to this contour, $C_k$, $k \in \{1,2,3,4,5,6\}$, as follows.
$C_1$ is the contour along the line $z \in [c-i R,c+i R]$ for some large value of $R$.
$C_2$ is the contour along a circular arc of radius $R$ from the top of $C_1$ to just above the negative real axis.
$C_3$ is the contour along a line just above the negative real axis between $[-R, -\epsilon]$ for some small $\epsilon$.
$C_4$ is the contour along a circular arc of radius $\epsilon$ about the origin.
$C_5$ is the contour along a line just below the negative real axis between $[-\epsilon,-R]$.
$C_6$ is the contour along the circular arc of radius $R$ from just below the negative real axis to the bottom of $C_1$.
We will show that the integral along $C_2$,$C_4$, and $C_6$ vanish in the limits of $R \rightarrow \infty$ and $\epsilon \rightarrow 0$.
On $C_2$, the real part of the argument of the exponential is
$$R t \cos{\theta} - 2 \sqrt{R} \cos{\frac{\theta}{2}}$$
where $\theta \in [\pi/2,\pi)$.  Clearly, $\cos{\theta} < 0$ and $\cos{\frac{\theta}{2}} > 0$, so that the integrand exponentially decays as $R \rightarrow \infty$ and therefore the integral vanishes along $C_2$.
On $C_6$, we have the same thing, but now $\theta \in (-\pi,-\pi/2]$.  This means that, due to the evenness of cosine, the integrand exponentially decays again as $R \rightarrow \infty$ and therefore the integral also vanishes along $C_6$.
On $C_4$, the integral vanishes in the limit as $\epsilon \rightarrow 0$.  Thus, we are left with the following by Cauchy's integral theorem (i.e., no poles inside $C$):
$$\left [ \int_{C_1} + \int_{C_3} + \int_{C_5}\right] ds \: s^{-1/2} e^{-\sqrt{s}} e^{s t} = 0$$
On $C_3$, we parametrize by $s=e^{i \pi} x$ and the integral along $C_3$ becomes
$$\int_{C_3} ds \: s^{-1/2} e^{-2\sqrt{s}} e^{s t} = -i e^{i \pi} \int_{\infty}^0 dx \:x^{-1/2}  e^{-i 2\sqrt{x}} e^{-x t}$$
On $C_5$, however, we parametrize by $s=e^{-i \pi} x$ and the integral along $C_5$ becomes
$$\int_{C_5} ds \: s^{-1/2} e^{-2\sqrt{s}} e^{s t} = i e^{-i \pi} \int_0^{\infty} dx \: x ^{-1/2} e^{i 2 \sqrt{x}} e^{-x t}$$
We may now write
$$\begin{align}\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: s^{-1/2} e^{-2\sqrt{s}} e^{s t} &= \frac{1}{\pi} \int_0^{\infty} dx \: x^{-1/2} \cos{(2 \sqrt{x})}\, e^{-x t}\\ &= \frac{1}{\pi} \underbrace{\int_{-\infty}^{\infty} du \: \cos{(2 u)} \, e^{-t u^2}}_{x=u^2} \\ &= \frac{1}{\pi} \Re{\left [\int_{-\infty}^{\infty} du \: e^{-t u^2} e^{i 2 u} \right ]}\\ &= \frac{e^{-1/t}}{\pi} \Re{\left [\int_{-\infty}^{\infty} du \: e^{-t (u-i/t)^2} \right ]}\\ &= \frac{e^{-1/t}}{\pi} \sqrt{\frac{\pi}{t}} \end{align}$$
Therefore the ILT is
$$\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: s^{-1/2} e^{-2\sqrt{s}} e^{s t} = (\pi t)^{-1/2} e^{-1/t}$$
