Using $\int_{0}^{\infty}x^2e^{-x^2/2}\text{ d}x = \sqrt{\frac{\pi}{2}}$ to calculate $\int_{0}^{\infty}x^3e^{-x^2/\beta^2}\text{ d}x$, $\beta > 0$ Given $$\int_{0}^{\infty}x^2e^{-x^2/2}\text{ d}x = \sqrt{\dfrac{\pi}{2}}$$ how can I calculate $$\int_{0}^{\infty}x^3e^{-x^2/\beta^2}\text{ d}x\text{, }\beta > 0\text{?}$$
I am more interested in the thought process behind this problem - which isn't necessarily the solution.
My first thought was to attempt the substitution $\dfrac{u^2}{2} = \dfrac{x^2}{\beta^2}$ - but notice this makes 
$$x = \sqrt{\dfrac{\beta^2}{2}u^2} = \dfrac{\beta}{\sqrt{2}}u \implies x^3 = \dfrac{\beta^3}{2^{3/2}}u^3$$
so that $\text{d}x = \dfrac{\beta}{\sqrt{2}}\text{ d}u$,
giving me 
$$\int_{0}^{\infty}x^3e^{-x^2/\beta^2}\text{ d}x = \dfrac{\beta^3}{4}\int_{0}^{\infty}u^3e^{-u^2/2}\text{ d}u$$
and this isn't quite what I need.
Please do not make any other assumptions and don't use any special formulas other than what I've stated above.
 A: Hint
$$I(\alpha)= \int_{0}^{\infty}xe^{-\alpha x^2/2}\text{ d}x$$
we have
$$I'(\alpha)=-\frac{1}{2} \int_{0}^{\infty}x^3e^{-\alpha x^2/2}\text{ d}x$$
Note
$$I(\alpha)=\int_{0}^{\infty}xe^{-\alpha x^2/2}\text{ d}x=\frac{1}{\alpha}$$
A: This may not be exactly that you want  because it doesn't really use the given identity, but integrating by parts using $u = x^2, dv = xe^{-x^2/\beta^2} dx$, we find \begin{align*}\int_0^\infty x^3 e^{-x^2/\beta^2} dx &= \underbrace{\left[-\frac{\beta^2 x^2}{2}e^{-x^2/\beta^2}\right]^{x\to\infty}_{x=0}}_{\,\,\,\,\,\,\,\ = 0}  + \beta^2\int^\infty_0x e^{-x^2/\beta^2}dx\\
&= \left[-\frac{\beta^4}{2} e^{-x^2/\beta^2} \right]_{x=0}^{x\to\infty} = \frac{\beta^4} 2.
\end{align*}
A: $\int_{0}^{+\infty}x^3 e^{-x^2}\,dx $ is an elementary integral that can be computed by integration by parts:
$$ \int x^3 e^{-x^2}\,dx =-\frac{1+x^2}{2}e^{-x^2} $$
while $\int_{0}^{+\infty}x^2 e^{-x^2}\,dx $ is not, but anyway
$$ \int_{0}^{+\infty} x^k e^{-x^2}\,dx = \frac{1}{2}\int_{0}^{+\infty} z^{\frac{k-1}{2}}e^{-z}\,dz = \frac{1}{2}\cdot\Gamma\left(\frac{k-1}{2}\right). $$
In general, there is no way to compute $\Gamma\left(x+\frac{1}{2}\right)$ from $\Gamma(x)$ only with a simple formula, hence I guess there is some typo in your exercise, or maybe the intended one was to compute 
$$ \int_{0}^{+\infty} x^{\color{red}{4}} e^{-x^2/\beta^2}\,dx $$
from $\int_{0}^{+\infty}x^2 e^{-x^2}\,dx$. That makes sense since $\Gamma(x+1)=x\cdot\Gamma(x)$. The $\beta$ parameter is just syntactic sugar since it can be removed through the substitution $x=\beta z$.
