$\int_{0}^{\infty}xe^{-x^2/2}dx= 1$?

Question

$X \sim N(0, 1)$

$$E(|X|) = \frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}|x|e^{-x^2/2}dx= \frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}xe^{-x^2/2}dx=\sqrt{\frac{2}{\pi}}$$

I don't understand how the last equality was arrived at. Why is it seemingly obvious that $\int_{0}^{\infty}xe^{-x^2/2}dx= 1$?

Is this some common identity?

Answer 1

Here are the steps that help make it obvious,
\[ \int_0^\infty xe^{\frac{-x^2}{2}} dx = \lim_{\beta \to \infty}\int_0^\beta xe^{\frac{-x^2}{2}} dx \] Let $u=\frac{-x^2}{2}$ then, \[ \frac{d}{dx}u = -x \Rightarrow du = -x\ dx \Rightarrow-du=x\ dx \] So now after adjusting the limits, we have \[ \lim_{\beta \to -\infty}-\int_0^\beta e^{u}\ du = \lim_{\beta \to -\infty}\int_\beta^0 e^{u}\ du = \lim_{\beta \to -\infty} e^u\Bigg|^0_\beta = \lim_{\beta \to -\infty}\left(e^0 - e^\beta\right) = 1 - 0 = 1 \] I hope this helps you understand the equation.

Answer 2

Isn't it obvious that

$$\int_0^\infty xe^{-x^2/2}=-e^{-x^2/2}\Bigg|_0^\infty=1\qquad?$$

Answer 3

I would substitute $u(x)=-\frac{1}{2}x^2$ $\Rightarrow \frac{du}{dx}=-x\Rightarrow -du=x \ dx$

$ -\int_0^{-\infty} e^{u} \ du $

The upper limit has been adjusted. The limits can be switched:

$ \int_{-\infty} ^0 e^{u} \ du $

Answer 4

$$ \int_0^\infty e^{-x^2/2} \Big( x\,dx\Big) = \int_0^\infty e^{-u}\,du. $$