I want to know how to explain $\left. \left(x e ^{-\frac{x^2}{2}} \right) \right|_{- \infty} ^{\infty}$ is zero?
Is it because the speed of exponentiation is greater than that of linear? How to prove? If using Bernoulli's rule, is there any other way to give an intuitive explanation?
Thanks!
$$...=\lim_{x\to\infty }xe^{-x^2/2}-\lim_{x\to-\infty }xe^{-x^2/2}=0-0=0$$
By l'Hopital Rules:
$$\lim_{x\to\infty }xe^{-x^2/2}=\lim_{x\to\infty }\frac{x}{e^{x^2/2}}=\lim_{x\to\infty }\frac{1}{xe^{x^2/2}}=\frac{1}{\infty }=0$$
By L'Hopital, $$\lim_{x \rightarrow \infty} xe^{-x^2/2} = \lim_{x \rightarrow \infty} \frac{x}{e^{x^2/2}} = \lim_{x \rightarrow \infty} \frac{1}{xe^{x^2/2}} = 0,$$ and it follows similarly that the limit is also $0$ as $x \rightarrow -\infty$.
$\exp -\frac{x^2}{2}$ is even.
$x$ is odd.
An odd function times an even function is odd.
The integral of an odd function over the reals is zero.
