How to find inverse laplace transform of $\frac{2\sqrt s}{2\sqrt s+1}$ How to find inverse laplace transform of
$$\dfrac{2\sqrt s}{2\sqrt s+1}$$
I tried to solve it, but I couldn't.
 A: Rewrite as
$$F(s) = \frac1{1+\frac12 s^{-1/2}} $$
Expand in a series about infinity:
$$F(s) = \sum_{k=0}^{\infty} (-1)^k 2^{-k} s^{-k/2} $$
Note that the inverse Laplace transform of $s^{-k}$ is $t^{k-1}/(k-1)!$ for $k \gt 0$, while the ILT of $s^{-k-1/2}$ is $t^{k-1/2}/\Gamma \left ( k+\frac12 \right )$.  Summing the series separately for these two cases, and treating the case $k=0$ still separately, we find that
$$\begin{align}f(t) &= \delta (t)-\frac{\sqrt{\pi } e^{t/4} \sqrt{t} \, \text{erf}\left(\frac{\sqrt{t}}{2}\right)+2}{4 \sqrt{\pi  t}}+\frac{1}{4} \exp \left(\frac{t}{4}\right) \\ &= \delta (t)- \frac1{2 \sqrt{\pi t}}+ \frac14 e^{t/4} \operatorname{erfc}{\left ( \frac{t}{4} \right )} \end{align}$$
Alternatively, rewrite as
$$F(s) = 1-\frac1{1+2 \sqrt{s}}$$
Then consider
$$\oint_C dz \frac{e^{z t}}{1+2 \sqrt{z}} $$
where $C$ is a keyhole contour about the negative real axis, as pictured below.

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 $C_1$.
The magnitude of the integral over $C_2$ vanishes as $R \to \infty$, as it is bounded by
$$\frac{R}{2\sqrt{R}-1} \int_{\pi/2}^{\pi} d\theta \, e^{R t \cos{\theta}} = \frac{R}{2\sqrt{R}-1} \int_{0}^{\pi/2} d\theta \, e^{-R t \sin{\theta}} \le \frac{R}{2\sqrt{R}-1} \int_{0}^{\pi/2} d\theta \, e^{-2 R t \theta/\pi} \le \frac{2 \pi}{2\sqrt{R}-1}$$
The integral over $C_6$ vanishes for a similar reason.  The integral over $C_4$ vanishes as $\epsilon \to 0$.
This leaves $C_1$, $C_3$, and $C_5$.  By Cauchy's theorem, as there are no poles inside $C$, we have
$$\int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{1+2 \sqrt{s}} + e^{i \pi} \int_{\infty}^0 dx \frac{e^{-x t}}{1+i 2 \sqrt{x}} + e^{-i \pi} \int_0^{\infty} dx \frac{e^{-x t}}{1-i 2 \sqrt{x}} = 0$$
so that we have
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{1+2 \sqrt{s}} = \frac{2}{\pi} \int_0^{\infty} dx \frac{\sqrt{x}}{1+4 x} e^{-x t}$$
The integral on the RHS may be evaluated by subbing $x=u^2$:
$$\frac1{2 \sqrt{\pi t}} - \frac1{\pi} \int_0^{\infty} dx \frac{e^{-t u^2}}{1+4 u^2} = \frac1{2 \sqrt{\pi t}}- \frac14 e^{t/4} \operatorname{erfc}{\left ( \frac{t}{4} \right )}$$
which checks out with the above result.
A: Since:
$$\forall\alpha>0,\quad\mathcal{L}^{-1}\left(\frac{1}{s^\alpha}\right)=\frac{t^{\alpha-1}}{\Gamma(\alpha)},$$
by exploiting the linearity of the Laplace (inverse) transform we have:
$$\begin{eqnarray*}\mathcal{L}^{-1}\left(\frac{2\sqrt{s}}{1+2\sqrt{s}}\right)&=&\mathcal{L}^{-1}\left(1-\frac{1}{2\sqrt{s}}+\frac{1}{4s}-\frac{1}{8s\sqrt{s}}+\ldots\right)\\&=&\delta_t+\sum_{k=1}^{+\infty}\frac{(-1)^k}{2^k}\mathcal{L}^{-1}\left(\frac{1}{s^{k/2}}\right)\\&=&\delta_t+\sum_{k=1}^{+\infty}\frac{(-1)^k}{2^k}\cdot\frac{t^{k/2-1}}{\Gamma(k/2)}\\&=&\color{red}{\delta_t-\frac{1}{\sqrt{4\pi t}}+\frac{1}{4}\,e^{\frac{t}{4}}\,\text{Erfc}\left(\frac{\sqrt{t}}{2}\right)},\end{eqnarray*}$$
where $\delta_t$ is the usual Dirac delta function. The last identity follows from:
$$\sum_{k=0}^{+\infty}\frac{(-1)^k t^{k}}{\Gamma(1+k/2)}= e^{t^2}\operatorname{Erfc}\left(t\right)\tag{1}$$
that can be proved in the following way:
$$\begin{eqnarray*}e^{t^2}\operatorname{Erfc}(t)&=&\frac{2}{\sqrt{\pi}}\int_{t}^{+\infty}e^{t^2-x^2}\,dx=\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}e^{-2tx}e^{-x^2}\,dx\\&=&\frac{2}{\sqrt{\pi}}\sum_{k=0}^{+\infty}\frac{(-1)^k (2t)^k}{k!}\int_{0}^{+\infty}x^k\,e^{-x^2}\,dx\tag{2}\end{eqnarray*}$$
by exploiting:
$$ \int_{0}^{+\infty}x^k\,e^{-x^2}\,dx = \frac{1}{2}\Gamma\left(\frac{k+1}{2}\right).\tag{3}$$
