Integrating ${e}^{-{x}^{2}}{x}^{2}$

Show that $\mathcal{L}\{{t}^{1/2}\}=\sqrt{\pi}/(2{s}^{3/2}),\phantom{\rule{mediummathspace}{0ex}}s>0$

By the definition of Laplace transform we get:

$$\mathcal{L}\{{t}^{1/2}\}={\int}_{0}^{\mathrm{\infty}}{t}^{1/2}{e}^{-st}\phantom{\rule{thinmathspace}{0ex}}dt=\{x=\sqrt{st}\}={\displaystyle \frac{2}{{s}^{3/2}}}{\int}_{0}^{\mathrm{\infty}}{e}^{-{x}^{2}}{x}^{2}\phantom{\rule{thinmathspace}{0ex}}dx.$$

Show that $\mathcal{L}\{{t}^{1/2}\}=\sqrt{\pi}/(2{s}^{3/2}),\phantom{\rule{mediummathspace}{0ex}}s>0$

By the definition of Laplace transform we get:

$$\mathcal{L}\{{t}^{1/2}\}={\int}_{0}^{\mathrm{\infty}}{t}^{1/2}{e}^{-st}\phantom{\rule{thinmathspace}{0ex}}dt=\{x=\sqrt{st}\}={\displaystyle \frac{2}{{s}^{3/2}}}{\int}_{0}^{\mathrm{\infty}}{e}^{-{x}^{2}}{x}^{2}\phantom{\rule{thinmathspace}{0ex}}dx.$$