Gaussian integral evaluation

Asked a question to evaluate the Gaussian Integral,

$$\dfrac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty x^2 \exp(-x^2/2) dx $$

using the the following approximation,

$J=\Bbb E[X^2] \sim J_N = 1/N \sum_1^NX_n^2 $

where $x_n \sim N(0, 1)$

Examin $J_N$ for different values of $N$ and plot an error graph (difference between exact and numerical values)

I can simulate and generate the approximations but I am stuck at getting the exact value to compare to. As far as I am aware there are no analytical solutions for gaussian integrals. The other way I can generate comparable numbers is to go use numerical integration techniques but those are approximate values too.

Any help will be greatly appreciated!

Solutions Collecting From Web of "Gaussian integral evaluation"

$$I(a)=\int_{-\infty}^\infty e^{-ax^2}dx\iff-I'(a)=\int_{-\infty}^\infty x^2\cdot e^{-ax^2}dx$$ Now, since we already know that $I(1)=\sqrt\pi$, it is trivial to show, by a simple change in variable, that $I(a)=\sqrt{\dfrac\pi a}$ , in which case our integral is simply $-\dfrac{I’\Big(\tfrac12\Big)}{\sqrt{2\pi}}$ . For more information on this topic, see here.

The object that you have there is essentially the definition of the variance of a standard normal, which equals one.

Hint: integrate by parts:$$
x^2\exp (-x^2/2) = x\times x\exp (-x^2/2)
$$

details:
You get
$$
\int_{-\infty}^\infty x^2\exp (-x^2/2) dx =
[x\times -\exp (-x^2/2)]_{-\infty}^\infty
– \int_{-\infty}^\infty 1\times -\exp (-x^2/2) dx \\
= \int_{-\infty}^\infty \exp (-x^2/2) dx
=\sqrt{2\pi}
$$


More generally,
$$
\int_{-\infty}^\infty x^{s+2}\exp (-x^2/2) dx =
(s+1)\int_{-\infty}^\infty x^{s}\exp (-x^2/2) dx
$$