Integrating $e^{-x^2}x^2$ 
Show that $\mathcal{L}\{t^{1/2}\}=\sqrt{\pi}/(2s^{3/2}), \: s>0$

By the definition of Laplace transform we get:
$$\mathcal{L}\{t^{1/2}\} = \int_0^\infty t^{1/2}e^{-st} \, dt = \{x = \sqrt{st} \} = \dfrac{2}{s^{3/2}} \int_0^\infty e^{-x^2} x^2 \, dx. $$
A known and easily proved result is $\int_0^\infty e^{-x^2} \, dx = \dfrac{\sqrt{\pi}}{2}$. How (if possible) can I use this result to determine the integral above? I was thinking either integration by parts or perhaps a suitable coordinate transformation (e.g. polar coordinates) but my attempts failed. 
 A: Rewrite the integral as $$\int\limits_0^\infty-\frac{x}{2} \left(-2xe^{-x^2}\right)dx$$ and use integration by parts with $f(x) = -\frac{x}{2}$ and $g(x) = e^{-x^2}$ to obtain $$\left. -\frac x 2 e^{-x^2}\right|_0^\infty + \frac{1}{2} \int\limits_0^\infty e^{-x^2}dx = \frac{\sqrt{\pi}}{4}$$
A: $$
\int_0^\infty x^2 e^{-x^2} \, dx = \int_0^\infty \frac x 2 e^{-x^2} \Big( 2x\, dx \Big) = \int_0^\infty \frac {\sqrt u} 2  e^{-u} \, du = \frac 1 2 \Gamma\left( \frac 3 2 \right).
$$
A: Hint: Consider integration by parts on
$$
\left(
x\,e^{-x^2}
\right)
\cdot x
$$
A: Start off with the substitution $u=x^2$ such that$$I=\frac 12\int\limits_0^{\infty}du\,\sqrt ue^{-u}$$The latter integral is just simply the factorial function.$$n!=\int\limits_0^{\infty}dt\,t^ne^{-t}$$So$$I=\frac 12\left(\frac 12\right)!=\frac {\sqrt{\pi}}4$$
A: You can try integration by parts with  $u=x$ and $dv=xe^{-x^2}dx$ to evaluate$$  \int_0^\infty e^{-x^2} x^2 \, dx.$$ Note that,$du=dx$ and $v=-(1/2)e^{-x^2}.$ Therefore; $$\int e^{-x^2} x^2 \, dx=-(1/2)xe^{-x^2} +1/2\int e^{-x^2}\,dx.$$ Upon evaluating from $0$ to $\infty$,the first term vanishes and you get $$\int_0^\infty e^{-x^2} x^2 \, dx=1/2\int_0^\infty e^{-x^2}\,dx=\frac{\sqrt{\pi}}4.$$
