How to solve hard integral of Gaussian/ Normal distribution?

Question

How to find $$ {\displaystyle \int_{- \infty}^{\infty} A x^2 {\rm e}^{- x^2/2 \sigma^2} d x} $$ where A is a constant given that $$ {\displaystyle \int_{- \infty}^{\infty} A {\rm e}^{- x^2/2 \sigma^2} d x = 1} $$

The expression inside the second integral being the Gaussian or Normal Distribution.

Please help, i know by parts works but could someone please outline some main steps and an answer please as it has me really confused

Answer 1

Hint:

When you integrate by parts, you must choose a factor that you can integrate, and that later leads to a simplification of the integrand. Here you can split in three ways

• $1\cdot x^2e^{-x^2/2\sigma^2}$,

• $x\cdot xe^{-x^2/2\sigma^2}$,

• $x^2\cdot e^{-x^2/2\sigma^2}$.

Among these options you must investigate which factor you are able to integrate, and from this how the next integrand will evolve.

Sometimes integration requires to roll up your sleeves.

$$\int xe^{-x^2/2\sigma^2}\cdot x\,dx=-\sigma^2e^{-x^2/2\sigma^2}\cdot x+\sigma^2\int e^{-x^2/2\sigma^2}\,dx.$$

Answer 2

The expression is the standard normal distribution with mean $=0$ and variance $=\sigma^2$. Therefore $A=\frac{1}{\sqrt{2\pi\sigma^2}}$. The first integral $=\sigma^2$.

To integrate by parts use $u=x$ and $dv=xe^{-\frac{x^2}{2\sigma^2}}dx$ You will have an expression $B=C\int_{-\infty}^{\infty}e^{-\frac{x^2}{2\sigma^2}}dx$
Work with $B^2$. Convert to polar coordinates and you will get the final expression $B^2=\sigma^4$.

Details for use of polar coordinates:

$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{x^2+y^2}{2\sigma^2}}dxdy$$=\int_0^{2\pi}\int_0^\infty e^{-\frac{r^2}{2\sigma^2}}rdrd\theta$$=2\pi\int_0^\infty e^{-\frac{u}{\sigma^2}}du$$=2\pi\sigma^2$

Answer 3

We'll first prove, then use with $B:=\frac{1}{2\sigma^2}$, the fact that $\int_\Bbb R Ae^{-Bx^2}dx=A\sqrt{\pi}B^{-1/2}$.

Proof Call this integral $I$ so$$I^2=\int_{\Bbb R^2}A^2e^{-B(x^2+y^2)}dxdy.$$This is a double integral over the plane. Since $dxdy=rdrd\theta$,$$I^2=\int_0^{2\pi}\int_0^\infty A^2re^{-Br^2}drd\theta=2\pi A^2[-\tfrac{1}{2B}e^{-Br^2}]_0^\infty=\frac{\pi A^2}{B}.$$Taking the square root completes the proof, as clearly $I>0$.

Corollary Differentiating with respect to $-B$ gives$$\int_\Bbb R Ax^2e^{-Bx^2}dx=\frac12A\sqrt{\pi}B^{-3/2}\implies\int_\Bbb R Ax^2e^{-x^2/2\sigma^2}dx=A\sqrt{2\pi}\sigma^3.$$The choice $A=\frac{1}{\sigma\sqrt{2\pi}}$ verifies the famous $N(0,\,\sigma^2)$ PDF does, in fact, have variance $\sigma^2$.