<s>[INST] integrations by parts. Show $\int_0^\infty 2y^2e^{-y^2} dy = .886$
$2\int_0^\infty y^2e^{-y^2} dy$
Then by parts; 
$f=y^2$     $dg=e^{-y^2}$
$df=2y$     $g=-2ye^{-y^2}$
$2\int_0^\infty y^2e^{-y^2} dy$ = $-2y^3e^{-y^2} -\int_0^\infty -4y^2e^{-y^2}$=$-2y^3e^{-y^2} +4\int_0^\infty y^2e^{-y^2}$
equivalently; $-2\int_0^\infty y^2e^{-y^2} dy = -2y^3e^{-y^2}$
or $2\int_0^\infty y^2e^{-y^2} dy = 2y^3e^{-y^2}$
$\left(2(\infty)^3e^{-\infty^2}\right)$-$\left(2(0)^3e^{-0^2}\right)=0-0=0$
Where have I gone wrong? 
 [/INST] $$\int y^2e^{-y^2}dy=\int \underbrace{y}_{=u}\cdot\underbrace{ ye^{-y^2}}_{=v'}dy=[-\frac{1}{2}ye^{-y^2}]+\frac{1}{2}\int_0^\infty  e^{-y^2}dy$$
The integral $\int_0^\infty  e^{-y^2}dy$ is not that obvious. If you set $I=\int_0^\infty  e^{-y^2}dy$, you can compute $I^2$ using polar coordinate. You can also know that $$\int_{\mathbb R}e^{-x^2}dx=\sqrt\pi$$
or make the substitution $u=y^2$ to get$$\int_0^\infty  e^{-y^2}dy=\int_0^\infty  e^{-y^2}dy=\frac{1}{2}\int_0^\infty u^{\frac{1}{2}-1}e^{-u}du=\frac{1}{2}\Gamma\left(\frac{1}{2}\right)=\frac{\sqrt{\pi}}{2}.$$
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