Another way to evaluate $\int_0^{\infty} x^2 e^{-x^2}dx$ In Stewart's Calculus book I came across the following Gaussian integral.

Using $\int_{-\infty}^{\infty}\exp{(-x^2)}dx = \frac{\sqrt{\pi}}{2}$ evaluate 
  $$
\int_0^{\infty} x^2 e^{-x^2}dx
$$

I read in this pdf that
$$
 \int_{-\infty}^{\infty} e^{-ax^2}dx =\sqrt{ \frac{\pi}{a}}
$$
and how to use differentiation under the integral sign to evaluate it (recreated below for convenience).
$$
\begin{align*}
I(a) &= \int_{-\infty}^{\infty} e^{-ax^2} dx =\sqrt{ \frac{\pi}{a}} \\ I'(a)&=  -\int_{-\infty}^{\infty} x^2  e^{-ax^2}dx = -\frac{1}{2}\sqrt{\pi} a^{-3/2} \\I'(1) &= \frac{\sqrt{\pi}}{2} 
\end{align*}
$$
Using the results above (and Wolfram Alpha), I was able to conclude that 
$$
\int_0^{\infty} x^2 e^{-x^2}dx = \frac{\sqrt{\pi}}{4}
$$
however, I was wondering if there is some substitution or an another way to evaluate the aforementioned integral seeing as Leibniz's rule is not mentioned anywhere in the chapter.
 A: Observe that $$\left(e^{-x^2}\right)'=-2xe^{-x^2} \implies xe^{-x^2}=-\frac12\left(e^{-x^2}\right)'$$ so that the integral can be written as $$-\frac12\int_{0}^{+\infty}x\left(e^{-x^2}\right)'dx$$ and integration by parts gives $$-\frac12\int_{0}^{+\infty}x\left(e^{-x^2}\right)'dx=-\frac12\left[xe^{-x^2}\right]_{0}^{+\infty}du+\frac12\int_{0}^{+\infty}e^{-x^2}dx$$ (notice the abuse of notation in the upper limit. It should be $\lim_{t\to \infty}$ instead of $\infty$). It remains to show that $\lim_{x\to \infty}xe^{-x^2}=0$ which can be done by L'Hopitals rule, since $$\lim_{x\to \infty}xe^{-x^2}=\lim_{x\to\infty}\frac{x}{e^{x^2}}\overset{\frac{\infty}{\infty}}=\lim_{x\to\infty}\frac{1}{2xe^{x^2}}=0$$
A: Alternatively, you can use hueristics that stem from "probabilistic" results, i.e., note that if $X\sim \mathcal{N}(0,1/2)$, then its density function is given by 
$$
f_X(x) = \frac{1}{\sqrt{\pi}}e^{-x^2}, \quad x\in \mathbb{R}
$$
hence, $\mathbb{V}(X)=\mathbb{E}(X^2) = \int_{-\infty}^{\infty}\frac{x^2}{\sqrt{\pi}}e^{-x^2}dx=1/2 $, thus by using symmetry of $f_X(x)$ w.r.t. $0$,
\begin{align}
 \int_{-\infty}^{\infty}\frac{x^2}{\sqrt{\pi}}e^{-x^2}dx&= \int_{-\infty}^{0}\frac{x^2}{\sqrt{\pi}}e^{-x^2}dx + \int_{0}^{\infty}\frac{x^2}{\sqrt{\pi}}e^{-x^2}dx\\
 &=\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}x^2e^{-x^2}dx = 1/2
\end{align}
rearranging the equation we get,
$$
\int_{0}^{\infty}x^2e^{-x^2}dx = \frac{\sqrt{\pi}}{4}\, .
$$
A: For $$\int\limits_{0}^{\infty} x^{2} \mathrm{e}^{-x^{2}} \mathrm{d} x$$ let $y = x^{2}$
\begin{equation}
\int\limits_{0}^{\infty} x^{2} \mathrm{e}^{-x^{2}} \mathrm{d} x = 
\frac{1}{2} \int\limits_{0}^{\infty} \mathrm{e}^{-y} y^{\frac{1}{2}} \mathrm{d} y
 = \frac{1}{2} \Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{4}
\end{equation}
