Calculation of $\frac{1}{\sqrt{2\pi}\sigma(x)}\int_{-\infty}^{\infty}|u|\exp\left(-\frac{u^2}{2\sigma{^2}(x)}\right)\ du$

Question

How to show that $$\dfrac{1}{\sqrt{2\pi}\sigma(x)}\int_{-\infty}^{\infty}|u|\operatorname{exp}\left(-\dfrac{u^2}{2\sigma{^2}(x)}\right)\mathop{du}=\sqrt{\dfrac{2}{\pi}}\sigma(x)$$

I think it has to be shown through polar coordinates and then $u$ susbstituion, but I am not sure.

Answer 1

Exploit the evenness of the integrand to write

$$\frac{2}{\sqrt{2\pi}\sigma(x)}\int_{0}^{\infty}u\operatorname{exp}\left(-\dfrac{u^2}{2\sigma{^2}(x)}\right)\ du$$

Then let $ z= \frac{u^2}{2\sigma{^2}(x)}$ so $\sigma{^2}(x)\ dz = u\ du$. Note that $\sigma$ has no functional dependence on $u$ and can be brought in front of the integral.

$$\begin{align*} =& \frac{\sqrt{2}}{\sqrt{\pi}\sigma(x)}\int_{0}^{\infty}e^{-z}\sigma{^2}(x) dz \\ \\ =& \sqrt{\frac{2}{\pi}}\sigma(x)\int_{0}^{\infty}e^{-z}\ dz\\ \\ =& \sqrt{\frac{2}{\pi}}\sigma(x) \end{align*}$$

Answer 2

Observe that you are doing integration of an even function and $\sigma(x)$ is nothing but a constant. $$ \int_{0}^{\infty}u\exp\left(-\dfrac{u^2}{2\sigma{^2}}\right)\ du = \int_{0}^{\infty}\frac{\sqrt{2}\sigma}{2}\exp\left(-\dfrac{u^2}{2\sigma{^2}}\right)\ d\left(\frac{u^2}{\sqrt{2}\sigma}\right)\\ =\frac{\sqrt{2}\sigma}{2} \int_0^\infty\exp(-v)\ dv $$ with $\sigma:=\sigma(x)$.

Answer 3

$$\dfrac{1}{\sqrt{2\pi}\sigma(x)}\int_{-\infty}^{\infty}|u|\operatorname{exp}\left(-\dfrac{u^2}{2\sigma{^2}(x)}\right)\mathop{du} =\dfrac{1}{\sqrt{2\pi}\sigma(x)}\left(\int_{0}^{\infty}u\operatorname{exp}\left(-\dfrac{u^2}{2\sigma{^2}(x)}\right)\mathop{du}- \int_{-\infty}^{0}u\operatorname{exp}\left(-\dfrac{u^2}{2\sigma{^2}(x)}\right)\mathop{du}\right)=\\ \dfrac{2}{\sqrt{2\pi}\sigma(x)}\int_{0}^{\infty}u\operatorname{exp}\left(-\dfrac{u^2}{2\sigma{^2}(x)}\right)\mathop{du}=\\ \dfrac{2\sigma (x)^2}{\sqrt{2\pi}\sigma(x)} \left(-e^{-\frac{u^2}{2 \sigma (x)^2}}\right)\Bigg|_0^\infty =\sqrt{\dfrac{2}{\pi}}\sigma(x) $$

Answer 4

I will let $\sigma(x)$ be $\sigma$. Using symmetry, \begin{align*} \frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^\infty|u|\exp\left\{-\frac{1}{2}\frac{u^2}{\sigma^2}\right\} du &= \frac{2}{\sqrt{2\pi}\sigma}\int_{0}^\infty u \exp\left\{-\frac{1}{2}\frac{u^2}{\sigma^2}\right\}du\\ &= \frac{2}{\sqrt{2\pi}\sigma}\sigma^2\int_{0}^\infty \frac{u}{\sigma^2} \exp\left\{-\frac{1}{2}\frac{u^2}{\sigma^2}\right\}du\tag{1}\\ &=\sqrt{\frac{2}{\pi}}\sigma, \end{align*} where $(1)$ is a Rayleigh distribution.