Evaluate $\int_0^{\infty} x^2 e^{-x^2}dx$ Evaluate $\int_0^{\infty} x^2 e^{-x^2}dx$
The original problem is : 
$$\text{Evaluate} \iint _R ye^{-x^2-y^2}dxdy$$
Where $R=\left\{ (x,y) \vert x\geq0,y\geq0\right\}$
I sub-ed $x=r\cos\theta, y=r\sin\theta$ and this changed to
$\left(\int_0^{\frac{\pi}{2}}\sin\theta d\theta \right)\left(\int_0^{\infty} r^2 e^{-r^2}dx\right)$
So I must calculate  $\int_0^{\infty} r^2 e^{-r^2}dr$, is there an easy way?
 A: Let $z=\sqrt{2}r$ so that $dz=\sqrt{2}dr$ and $r^2=\frac{1}{2}z^2$. Then:
$$
\int_0^\infty r^2e^{-r^2}dr=\int_0^\infty\frac{1}{2}z^2e^{-\frac{1}{2}z^2}\frac{dz}{\sqrt{2}}=\frac{1}{2\sqrt{2}}\int_0^{\infty}z^2e^{-z^2/2}dz=\frac{1}{4\sqrt{2}}\int_{-\infty}^\infty z^2e^{-z^2/2}dz.
$$
The last equality above uses symmetry. Next, we note:
$$
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty z^2e^{-z^2/2}dz=1
$$
because it is the second-moment of a standard normal distribution. And so:
$$
\int_0^\infty r^2e^{-r^2}dr=\frac{1}{4\sqrt{2}}\sqrt{2\pi}=\frac{\sqrt{\pi}}{4}.
$$
A: We can do :
\begin{align}
I  &=\int_{R^+}\int_{R^+}ye^{-x^2-y^2}dxdy\\
&=\int_{R^+}\left(\int_{R^+}ye^{-y^2}dy\right)e^{-x^2}dx\\
&=\frac12\int_{R^+}e^{-x^2}dx\\
&=\frac{\sqrt{\pi}}{4}
\end{align}
The last equality comes from the Gauss integral.
A: The answer here is slightly adapted from d'Aurizio's answer here. Substitute $u=x^2,\frac{du}{dx}=2x$:
$$\int_0^\infty x^2e^{-x^2}\,dx=\frac12\int_0^\infty u^{1/2}e^{-u}\,du$$
We recognise the integral now as $\Gamma\left(\frac32\right)=\frac12\sqrt\pi$. Therefore, the original integral is $\frac{\sqrt\pi}4$.
A: Let's define the integral $\int_0^{\infty} x^2 e^{-a x^2}dx$, which is equivalent to what be are about to evaluate when $a=1$. We'll use a classical approach of differentiation under the integral sign:
$$\int_0^{\infty} x^2 e^{-a x^2}dx=(-1)\frac{\partial}{\partial a}\int_0^{\infty} e^{-a x^2}dx=-\frac{1}{2}\frac{\partial}{\partial a} \sqrt{\frac{\pi}{a}}=\frac{\sqrt{\pi}}{4a^{3/2}}.$$
Where we used the standard Gaussian integral: $\int_0^{\infty} e^{-a x^2}dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}$
Setting $a=1$ we finally get the result:
$$\int_0^{\infty} x^2 e^{-x^2}dx=\frac{\sqrt{\pi}}{4}.$$
