Integral over a Gaussian function

Question

I want to determine the following integral:

$$\int_\mathbb{R}\left|x-\dfrac{\sigma}{\sigma_{n}}x\right|^{2}e^{\frac{-x^2}{2\sigma^2}}\,dx$$ Thanks for your help!!

Answer 1

Hint. One may write $$ \begin{align} \int_\mathbb{R}\left|x-\dfrac{\sigma}{\sigma_{n}}x\right|^2e^{\frac{-x^{2}}{2\sigma^{2}}}\:dx&=\left|1-\dfrac{\sigma}{\sigma_{n}}\right|^2\int_{-\infty}^\infty\,|x|^2\,e^{\frac{-x^{2}}{2\sigma^{2}}}\,dx \\\\&=2\left|1-\dfrac{\sigma}{\sigma_{n}}\right|^2\int_{0}^\infty\,x^2\,e^{-x^2/2\sigma_{n}^{2}}\,dx \\\\&=\left|1-\dfrac{\sigma}{\sigma_{n}}\right|^2\cdot \sigma^{3}\sqrt{2\pi} \end{align} $$ where we have used the standard gaussian result $$ \int_{0}^\infty\,x^2\,e^{-x^2/2a^2}\,dx=\sqrt{\frac \pi2}\:a^3,\quad a>0. $$

Answer 2

$$\int_\mathbb{R}|x-\dfrac{\sigma}{\sigma_{n}}x|^{2}e^{\frac{-x^{2}}{2\sigma^{2}}}dx = \left(1-\dfrac{\sigma}{\sigma_{n}}\right)^2\int_\mathbb{R}x^{2}\exp\left(\frac{-x^{2}}{2\sigma^{2}}\right)dx$$

Integrating by parts, one has $$ \left(1-\dfrac{\sigma}{\sigma_{n}}\right)^2\int_\mathbb{-\infty}^{\infty}x^{2}\exp\left(\frac{-x^{2}}{2\sigma^{2}}\right)dx = \left(1-\dfrac{\sigma}{\sigma_{n}}\right)^2(-\sigma^2)\int_\mathbb{-\infty}^{\infty}x\left(\exp\left(\frac{-x^{2}}{2\sigma^{2}}\right)\right)^\prime dx = \sigma^2\left(1-\dfrac{\sigma}{\sigma_{n}}\right)^2\int_\mathbb{-\infty}^{\infty}\exp\left(\frac{-x^{2}}{2\sigma^{2}}\right) dx $$

Last step is to use $$\int_\mathbb{-\infty}^{\infty}\exp\left(\frac{-x^{2}}{2\sigma^{2}}\right) = \sqrt{2\pi\sigma^2}$$ and get $$\int_\mathbb{R}|x-\dfrac{\sigma}{\sigma_{n}}x|^{2}e^{\frac{-x^{2}}{2\sigma^{2}}}dx = \sqrt{2\pi}\sigma^3\left(1-\dfrac{\sigma}{\sigma_{n}}\right)^2$$