How to calculate the Laplace transform $$ h: t \in[0,+\infty[  \to \int_{t}^\infty \frac{1}{e^s\sqrt{s}}ds$$ 
I have to calculate the Laplace transform of $h$ in $0$
I know that $L[\int_{0}^\infty f(t)dt](p)= \frac{1}{p}L[f](p)$
but i don't understand how to procede next.
 A: We have:
$$\begin{eqnarray*}(\mathcal{L}h)(0^+)&=&\lim_{s\to 0^+}\int_{0}^{+\infty}\int_{t}^{+\infty}\frac{1}{e^u \sqrt{u}}\cdot e^{-st}\,du\,dt\\ (u=tv)\quad&=&\lim_{s\to 0^+}\int_{0}^{+\infty}\int_{1}^{+\infty}\frac{t}{e^{tv} \sqrt{tv}}\cdot e^{-st}\,dv\,dt\\(\text{Fubini})\quad&=&\lim_{s\to 0^+}\int_{1}^{+\infty}\int_{0}^{+\infty}\sqrt{\frac{t}{v}}\,e^{-(s+v)t}\,dt\,dv\\&=&\lim_{s\to 0^+}\int_{1}^{+\infty}\frac{\sqrt{\pi}}{2(s+v)^{3/2}\sqrt{v}}\,dv\\&=&\lim_{s\to 0^+}\frac{\sqrt{2\pi}}{s}\left(1-\frac{1}{\sqrt{1+s}}\right)=\color{red}{\frac{\sqrt{\pi}}{2}}.\end{eqnarray*}$$ 
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\color{#f00}{%
\int_{0}^{\infty}\pars{\int_{t}^{\infty}{\dd s \over \expo{s}\root{s}}}\dd t} =
\int_{0}^{\infty}{1 \over \expo{s}\root{s}}\int_{0}^{s}\dd t\,\dd s =
\int_{0}^{\infty}s^{1/2}\expo{-s}\,\dd s = \Gamma\pars{3 \over 2}
\\[3mm] = &\
\half\,\ \overbrace{\Gamma\pars{\half}}^{\ds{\root{\pi}}} =
\color{#f00}{{\root{\pi} \over 2}}
\end{align}
