Laplace transform of this Integral: $t \in [0, +\infty) \rightarrow \int_t^\infty \frac{e^{-s}}{\sqrt{s}} ds$

Question

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?

Answer 1

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

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

Answer 2

We can proceed directly using Fubini's Theorem. We have

$$\begin{align} \int_0^\infty e^{-st}\int_t^\infty \frac{e^{-x}}{\sqrt x}\,dx\,dt&=\int_0^\infty \frac{e^{-x}}{\sqrt x}\int_0^x e^{-st}\,dt\,dx\\\\ &=\frac1s \int_0^\infty \frac{e^{-x}-e^{-(s+1)x}}{\sqrt x}\,dx\\\\ &=\frac2s \int_0^\infty (e^{-x^2}-e^{-(s+1)x^2})\,dx\\\\ &=\frac1s \left(\sqrt \pi - \frac{\sqrt \pi }{\sqrt{s+1}}\right)\\\\ &=\frac{\sqrt{\pi}}{s}\left(1-\frac{1}{\sqrt{s+1}}\right) \end{align}$$

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