Laplace transform of power function

In the defining integral (http://planetmath.org/ImproperIntegral)

$$ℒ\left\{t^r\right\}={\int }_0^{\mathrm{\infty }}{e}^{-st}{t}^r𝑑t$$

of the Laplace transform of the power function  $t\mapsto t^r$,  we make the substitution (http://planetmath.org/SubstitutionForIntegration)  $u:=st$:

$$ℒ\left\{t^r\right\}={\int }_0^{\mathrm{\infty }}{e}^{-u}{\left(\frac{u}{s}\right)}^r\frac{du}{s}=\frac{1}{s^{n+1}}{\int }_0^{\mathrm{\infty }}{e}^{-u}{u}^{r+1-1}𝑑u$$

Here we have assumed that  $r>-1$  and $s>0$.  According to the definition of the gamma function, the last integral is equal to$\mathrm{\Gamma }(r+1)$.  Thus we obtain

$ℒ\left\{t^r\right\}={\displaystyle \frac{\mathrm{\Gamma }(r+1)}{s^{r+1}}}.$(1)

The special case  $r=-\frac{1}{2}$  gives the result

$ℒ\left\{{\displaystyle \frac{1}{\sqrt{t}}}\right\}=\sqrt{{\displaystyle \frac{\pi }{s}}}.$(2)