Result of $\int \limits_{-\infty}^{+\infty}x^2\times\exp\left(\dfrac{-x^2}{2}\right)\mathrm{d}x$ I would like to read a very thorough and explained calculation process for a couple of integrals. For the life of me I just can't figure out the result on my own, and no resource on the web were able to help me.
First, a small question: Is a primitive of $\exp\left(\frac{-x^2}{2}\right)$, $-x\times\exp\left(\frac{-x^2}{2}\right)$ ? If so, what's the derivative ?
Now the real question. I would like to find the result of: 
$$\int \limits_{-\infty}^{+\infty}\exp\left(\dfrac{-x^2}{2}\right)\mathrm{d}x$$
And then, the result of:
$$\int \limits_{-\infty}^{+\infty}x^2\times\exp\left(\dfrac{-x^2}{2}\right)\mathrm{d}x$$
Supposedly both are equal to $\sqrt{2\pi}$, but there's no way I can get there on my own.
 A: $e^{-x^2/2}$ has no elementary antiderivative. To compute its definite integral over the line, we perform the trick:
$$\left ( \int_{-\infty}^\infty e^{-x^2/2} dx \right )^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-x^2/2-y^2/2} \, dx \, dy.$$
Then we convert to polar coordinates, obtaining
$$\int_0^{2 \pi} \int_0^\infty e^{-r^2/2} r \, dr \, d \theta.$$
The inner integral can be done by a simple substitution; the key is that to substitute $u=r^2/2$ you need a factor of $r$, but the polar coordinate transformation "gives us" this factor of $r$. Then the outer integral is trivial (there is no dependence on $\theta$). Finally you finish by taking the square root of the result.
With the $x^2$, you integrate by parts with $u=x$ with $dv$ absorbing everything else, and then the problem reduces to the preceding integral.
And yes, this is a rather famous problem that would've been fairly easy to find on MSE and on Google. I went ahead and wrote up this answer largely so I could easily find it in my own history when this problem comes up again.
A: Hint: Let $~I(a)=\displaystyle\int_{-\infty}^\infty e^{-ax^2}~dx,~$ and then evaluate $I'\bigg(\dfrac12\bigg)$.
