Laplace transform of this Integral: $t \in [0, +\infty) \rightarrow \int_t^\infty \frac{e^{-s}}{\sqrt{s}} ds$ I have to calculate the Laplace transform of this integral
$$ t \in [0, +\infty) \rightarrow \int_t^\infty \frac{e^{-s}}{\sqrt{s}} ds $$
I know that I can write the Laplace Transform of $ \int_0^t f(s)ds $ as $ \frac{L[f(t)](z)}{z} $ but, honestly, I have no idea how to manage that thing.
Can somebody please explain me how to do that?
 A: Hint. One may also integrate by parts and use a gaussian result,
$$
\begin{align}
\mathcal{L}\left[\int_t^\infty \frac{e^{-u}}{\sqrt{u}} du\right](s)&=\int_0^\infty e^{-st}\left[\int_t^\infty \frac{e^{-u}}{\sqrt{u}} du\right]dt
\\\\&=\left[\frac{e^{-st}}{-s}\cdot\int_t^\infty \frac{e^{-u}}{\sqrt{u}} du\right]_0^\infty -\frac{1}{s}\int_0^\infty e^{-st}\cdot \frac{e^{-t}}{\sqrt{t}} \:dt
\\\\&=\frac{1}{s}\cdot\int_0^\infty \frac{e^{-u}}{\sqrt{u}} du -\frac{1}{s}\int_0^\infty  \frac{e^{-(s+1)t}}{\sqrt{t}} \:dt
\\\\&=\frac{\sqrt{\pi}}{s} -\frac{\sqrt{\pi}}{s\sqrt{s+1}}, \qquad s>0,
\end{align}
$$ thus

$$
\mathcal{L}\left[\int_t^\infty \frac{e^{-u}}{\sqrt{u}} du\right](s)=\frac{\sqrt{\pi}}{s+1+\sqrt{s+1}}, \qquad s>0.
$$

A: We can proceed directly using Fubini's Theorem.  We have
$$\begin{align}
\int_0^\infty e^{-st}\int_t^\infty \frac{e^{-x}}{\sqrt x}\,dx\,dt&=\int_0^\infty \frac{e^{-x}}{\sqrt x}\int_0^x e^{-st}\,dt\,dx\\\\
&=\frac1s \int_0^\infty \frac{e^{-x}-e^{-(s+1)x}}{\sqrt x}\,dx\\\\
&=\frac2s \int_0^\infty (e^{-x^2}-e^{-(s+1)x^2})\,dx\\\\
&=\frac1s \left(\sqrt \pi - \frac{\sqrt \pi }{\sqrt{s+1}}\right)\\\\
&=\frac{\sqrt{\pi}}{s}\left(1-\frac{1}{\sqrt{s+1}}\right)
\end{align}$$

