Show that: $\int_{0}^{\infty}{x^2e^{-x^2}}{dx} = \frac{1}{2}\int_{0}^{\infty}{e^{-x^2}}{dx}$ I am fully uncertain of how to approach this problem:
Show that: $$\int_{0}^{\infty}{x^2e^{-x^2}}{dx} = \frac{1}{2}\int_{0}^{\infty}{e^{-x^2}}{dx}$$
We've just completed the section on improper integrals in our book, and it seems that I haven't the necessary knowledge required to give a proof for this. The most elementary approach to this proof (considering that the furthest I've gotten in maths is improper integrals, thus far) would be ideal. Any insight would be greatly appreciated.
 A: Approach 1:
Write
$$I(a)=\int_0^{\infty}e^{-ax^2}dx =\frac12 \sqrt{\frac{\pi}{a}}\tag1$$
Then, take the derivative of $I$ to find
$$I'(a)=-\int_0^{\infty}x^2e^{-ax^2}dx=-\frac14 \sqrt{\pi}a^{-3/2}$$
And we are done after setting $a=1$.

Approach 2:
Use integration by parts to show
$$\int_0^{\infty}x^2\,e^{-x^2}dx=\left. xe^{-x^2}\right|_0^{\infty}+\frac12 \int_0^{\infty}e^{-x^2}dx=\frac12 \int_0^{\infty}e^{-x^2}dx$$
as expected!
A: You may integrate by parts
$$
\begin{align}
\int_{0}^{\infty}{x^2e^{-x^2}}{dx} &=\left. x\times \frac{-e^{-x^2}}{2}\right|_0^{\infty}+\frac{1}{2}\int_0^{\infty}e^{-x^2}\:dx\\\\
&=0+\frac{1}{2}\int_0^{\infty}e^{-x^2}\:dx\\\\
&=\frac{1}{2}\int_0^{\infty}e^{-x^2}\:dx\\\\
&=\color{blue}{\frac{\sqrt{\pi }}{4}}
\end{align}
$$ where we have used a well known gaussian result.
A: Approach 3:
Consider the integral:
$$I(\epsilon) = \int_0^{\infty}\exp\left[-(1-\epsilon)x^2\right]dx$$
If you expand this in powers of $\epsilon$ then you see that the first order term will be the desired integral. To compute that term, you can do a change of variables by putting:
$$x = \frac{t}{\sqrt{1-\epsilon}}$$
This yields:
$$I(\epsilon) = \frac{1}{\sqrt{1-\epsilon}}\int_0^{\infty}\exp\left(-t^2\right)dt$$
We then see that
$$I(\epsilon) = I(0)\left[1 + \frac{\epsilon}{2}+\mathcal{O}\left(\epsilon^2\right) \right]$$
The integral is thus $\frac{1}{2}I(0)$, because that is the coefficient of $\epsilon$ in $I(\epsilon)$.
