$\displaystyle \int_{0}^{\infty}x^2e^{-x^2}$
If so, how can you integrate it?
Also, what about $\displaystyle \int_0^\infty e^{-x^2}$
Parts won't work this time, my apologies
Wolfram Mathematica Online Integrator
this can be done.
Choose
$\displaystyle u=x$
$\displaystyle dv=xe^{-x^2}dx$
Notice that then v=$\displaystyle \frac{-e^{-x^2}}{2}$
You will then notice that the uv term of the integration by parts is 0 (remember that for definite integrals, this term gets evaluated at the main integral's bounds, i.e., $\displaystyle \infty$ and 0.) What you're left with is
$\displaystyle \frac{1}{2}\int_0^\infty e^{-x^2}$
A rather cool method for that one will use polar coordinates. here's a link:
Gaussian integral - Wikipedia, the free encyclopedia
(note the limits of integration for the Gaussian integral: You will need to divide by 2 (justified as the integrand is an even function)).
Use the fact that $\displaystyle \int_{0}^{\infty} e^{-x^{2}} \ dx = \frac{\sqrt{\pi}}{2} $
Make the substitution $\displaystyle x^{2}= au^{2} $
which implies that $\displaystyle 2xdx=2audu $
so $\displaystyle dx=\frac{au}{x} dx = \frac{ax}{\sqrt{a}x}dx = \sqrt{a} dx $
then $\displaystyle \int^{\infty}_{0} e^{-x^{2}} \ dx = \sqrt{a} \int^{\infty}_{0} e^{-au^{2}} du = \frac{\sqrt{\pi}}{2} $
and $\displaystyle \int^{\infty}_{0} e^{-au^{2}} \ du =$ $\displaystyle \frac{1}{2} \sqrt{\frac{\pi}{a}}$
notice that $\displaystyle - \int^{\infty}_{0} \frac{\partial}{\partial a} \ e^{-au^{2}} \ du =$$\displaystyle \int^{\infty}_{0} u^{2} e^{-au^{2}} \ du $
now swtich the order of integration and differentiation (which in this case is allowed)
$\displaystyle - \frac{\partial}{\partial a} \int^{\infty}_{0} e^{-au^{2}} \ du = - \frac{\partial}{\partial a} \ \frac{1}{2} \sqrt{\frac{\pi}{a}} = $$\displaystyle \ \frac{\sqrt{\pi}}{4}a^{-3/2}$
so $\displaystyle \int^{\infty}_{0} u^{2} e^{-au^{2}} \ du = \frac{\sqrt{\pi}}{4}a^{-3/2} $
letting a=1
$\displaystyle \int^{\infty}_{0} u^{2} e^{-u^{2}} \ du = \frac{\sqrt{\pi}}{4} $