How to calculate the Laplace transform

Question

$$ h: t \in[0,+\infty[ \to \int_{t}^\infty \frac{1}{e^s\sqrt{s}}ds$$

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.

Answer 1

We have:

$$\begin{eqnarray*}(\mathcal{L}h)(0^+)&=&\lim_{s\to 0^+}\int_{0}^{+\infty}\int_{t}^{+\infty}\frac{1}{e^u \sqrt{u}}\cdot e^{-st}\,du\,dt\\ (u=tv)\quad&=&\lim_{s\to 0^+}\int_{0}^{+\infty}\int_{1}^{+\infty}\frac{t}{e^{tv} \sqrt{tv}}\cdot e^{-st}\,dv\,dt\\(\text{Fubini})\quad&=&\lim_{s\to 0^+}\int_{1}^{+\infty}\int_{0}^{+\infty}\sqrt{\frac{t}{v}}\,e^{-(s+v)t}\,dt\,dv\\&=&\lim_{s\to 0^+}\int_{1}^{+\infty}\frac{\sqrt{\pi}}{2(s+v)^{3/2}\sqrt{v}}\,dv\\&=&\lim_{s\to 0^+}\frac{\sqrt{2\pi}}{s}\left(1-\frac{1}{\sqrt{1+s}}\right)=\color{red}{\frac{\sqrt{\pi}}{2}}.\end{eqnarray*}$$

Answer 2

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\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}