In the defining integral $$\mathcal{L}\left\{t^r\right\} = \int_0^\infty e^{-st}t^r\,dt$$ of the Laplace transform of the power function $t \mapsto t^r$ , we make the substitution $u := st$ : $$\mathcal{L}\left\{t^r\right\} = \int_0^\infty e^{-u}\left(\frac{u}{s}\right)^r\frac{du}{s} = \frac{1}{s^{n+1}}\int_0^\infty e^{-u}u^{r+1-1}\,du$$ Here we have assumed that $r > -1$ and $s > 0$ . According to the definition of the gamma function, the last integral is equal to $\Gamma(r\!+\!1)$ . Thus we obtain

(1)

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

(2)