Anyone can integrate $e^{-\frac{x^2}{3}}$ by hands? I just used wolfram integral calculator and the result is weird, there is something called error function.
$$
\int_{-\infty}^\infty e^{-\frac{x^2}{3}}\,\mathrm dx
$$
Hint says that change of variable might be helpful, but can't think of what to change and substitute. 
Anyone can do this by hands? 
 A: Let $$I=\int_{-\infty}^{\infty}\exp\left(-\frac{x^2}{3}\right)dx$$.
$$I^2=\int_{-\infty}^{\infty}\exp\left(-\frac{x^2}{3}\right)dx\int_{-\infty}^{\infty}\exp\left(-\frac{y^2}{3}\right)dy$$
$$=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\left(-\frac{x^2+y^2}{3}\right)dxdy$$
$$=\int_{0}^{2\pi}\int_{0}^{\infty}\exp\left(-\frac{r^2}{3}\right)rdrd\theta$$
$$=2\pi\frac{3}{2}\int_0^{\infty}\exp\left(-\frac{r^2}{3}\right)d\left(\frac{r^2}{3}\right)$$
$$=3\pi$$
$$I=\sqrt{3\pi}$$
A: hint: what's the density of a standard (normal) distribution? take $y := cx$ with appropriately chosen $c$ (and use the fact that integral of a density is $1$)
A: I believe that the problem assumes that you already know that
$$\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}$$
(or possibly a small variation of this) and your mission is to evaluate
$$\int_{-\infty}^{\infty} e^{-x^2/3}dx$$
by using a substitution to express a definite integral involving $e^{-x^2/3}$ into one involving $e^{-x^2}$.  Let's try.
First, note that
$$\int_{-\infty}^{\infty} e^{-x^2/3}dx = 2\lim_{b\rightarrow\infty} \int_0^b e^{-x^2/3}dx.$$
Thus, we'll focus on
$$\int_0^b e^{-x^2/3}dx.$$
Let us make the substitution $u=x/\sqrt{3}$.  Then $dx=\sqrt{3}du$ so
$$\int_0^b e^{-x^2/3}dx = \int_0^b e^{-(x/\sqrt{3})^2}dx = \sqrt{3}\int_0^{b/\sqrt{3}} e^{-u^2} du.$$
Taking a limit as $b\rightarrow\infty$, we get
$$\int_0^{\infty} e^{-x^2/3}dx = \sqrt{3}\int_0^{\infty} e^{-u^2} du = \frac{1}{2}\sqrt{3}\sqrt{\pi}.$$
The answer to your integral is twice this.
A few comments


*

*This is one of the first examples that most people see of a definite integral that can be computed in closed form without ever finding a closed form expression for the corresponding indefinite integral.

*The fact that
$$\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}$$
can be proved directly using the technique shown in Kyson's answer.

*An alternative formulation of the integral (equivalent by the same substitution technique is
$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-x^2/2}dx = 1.$$

*The previous integral is known as the Gaussian normal and is of tremendous importance in statistics.
