TITLE: Another way to evaluate $\int_0^{\infty} x^2 e^{-x^2}dx$
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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}