TITLE: A problem in calculating integral
QUESTION [0 upvotes]: I have a problem in calculating the integral:
$I=\displaystyle\int\limits_{-\infty}^{\infty} \dfrac{xe^{-x^2/2}}{\sqrt{2\pi}}
dx$.
My teacher said that $\dfrac{xe^{-x^2/2}}{\sqrt{2\pi}}$ is an odd function, thus $I=0$. 
This is my solution, but I wonder if it's right.
$I=\displaystyle\int\limits_{-\infty}^{\infty} \dfrac{xe^{-x^2/2}}{\sqrt{2\pi}}
dx=\dfrac{1}{\sqrt{2\pi}}\displaystyle\int\limits_{-\infty}^{\infty} e^{-x^2/2}d(\dfrac{x^2}{2})=0$ 
Is the statement $\displaystyle\int\limits_{-\infty}^{\infty} \dfrac{xe^{-x^2/2}}{\sqrt{2\pi}}
dx=\dfrac{1}{\sqrt{2\pi}}\displaystyle\int\limits_{-\infty}^{\infty} e^{-x^2/2}d(\dfrac{x^2}{2})$ right? 
My teacher said that $f(x)=\dfrac{xe^{-x^2/2}}{\sqrt{2\pi}}$ needs to be injective so that we can write as above, but I can't find any theorem confirm that. Anyone help me please?

REPLY [0 votes]: Well, substitute $\text{u}=-\frac{x^2}{2}$ and $\text{d}\text{u}=-x\space\text{d}x$:
$$\mathcal{I}\left(x\right)=\frac{1}{\sqrt{2\pi}}\int x\exp\left(-\frac{x^2}{2}\right)\space\text{d}x=-\frac{1}{\sqrt{2\pi}}\int\exp\left(\text{u}\right)\space\text{d}\text{u}=\text{C}-\frac{\exp\left(-\frac{x^2}{2}\right)}{\sqrt{2\pi}}$$
Now:

$$\lim_{x\to-\infty}\left\{-\frac{\exp\left(-\frac{x^2}{2}\right)}{\sqrt{2\pi}}\right\}=-\frac{1}{\sqrt{2\pi}}\cdot\lim_{x\to-\infty}\frac{1}{\exp\left(\frac{x^2}{2}\right)}=0$$
$$\lim_{x\to\infty}\left\{-\frac{\exp\left(-\frac{x^2}{2}\right)}{\sqrt{2\pi}}\right\}=-\frac{1}{\sqrt{2\pi}}\cdot\lim_{x\to\infty}\frac{1}{\exp\left(\frac{x^2}{2}\right)}=0$$
{"set_name": "stack_exchange", "score": 0, "question_id": 2006198}