Question: I need to calculate this integral of $\int\_{0}^{\infty} y^{2} \exp(-y^{2}/2) dy$

I started doing by parts: $u = y$, $v= - \exp(-y^{2}/2)$, $du=dv$ and $dv=-y \exp(- y^{2}/2)$

So, $-y \exp(-y^{2}/2) + \int\_{0}^{\infty} \exp(-y^{2}/2)$. $-y \exp(-y^{2}/2) =0$ when integrating in $(0, \infty)$ and I will have the integration of :$\int\_{0}^{\infty} \exp(-y^{2}/2)$

From now on I cant move.

Any help?

Many Thanks

Answer: 
The trick for these kinds of integral is to manipulate it into the form of the Gamma Function$$\Gamma(z)=\int\limits\_0^{\infty}\mathrm dt\, t^{z-1}e^{-t}$$To reduce your integral, label it as $\mathfrak{I}$ and let $t=\frac 12y^2$. Therefore $\mathrm dt=y\,\mathrm dy$ and $y=\sqrt{2t}$

$$\begin{align\*}\mathfrak{I} & =\int\limits\_0^{\infty}\mathrm dt\,\sqrt{2t}\, e^{-t}\\ & =\sqrt2\int\limits\_0^{\infty}\mathrm dt\, t^{1/2}e^{-t}\\ & =\sqrt2\Gamma\left(\frac 32\right)\end{align\*}$$

Since

$$\Gamma\left(\frac 32\right)=\frac {\sqrt\pi}2$$Thus$$\int\limits\_0^{\infty}\mathrm dt\, t^{z-1}e^{-t}\color{blue}{=\sqrt{\frac {\pi}2}}$$
the final integral you arrived at involves the limit for error function. you can either look for the properties of error function, or prove it by extending the integtal into the squareroot of a double integral and then transform into polar coordinates.