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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?

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  • $\begingroup$ Welcome to mathSE! I'm sorry but your integral don't have a closed form. $\endgroup$ – rlartiga Jan 27 '15 at 17:00
  • $\begingroup$ You sure you didn't misread the question? $\endgroup$ – Akiva Weinberger Jan 27 '15 at 17:01
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    $\begingroup$ The reason why I put statistic tag is that the result of this integral is widely used in the field of statistic, that's what my professor said. $\endgroup$ – kik Jan 27 '15 at 17:05
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    $\begingroup$ yes I am sure, and I found out that integration of e^(-x^2/2) over the same boundary is square root of 2pi, would that be helpful? $\endgroup$ – kik Jan 27 '15 at 17:22
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    $\begingroup$ So you want $u$ such that $\frac{u^2}{2}=\frac{x^2}{3}.$ It's just a constant of proportionality, easiest substitution in the book. $\endgroup$ – David K Jan 28 '15 at 1:26
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Let $$I=\int_{-\infty}^{\infty}\exp\left(-\frac{x^2}{3}\right)dx$$.

$$I^2=\int_{-\infty}^{\infty}\exp\left(-\frac{x^2}{3}\right)dx\int_{-\infty}^{\infty}\exp\left(-\frac{y^2}{3}\right)dy$$

$$=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\left(-\frac{x^2+y^2}{3}\right)dxdy$$ $$=\int_{0}^{2\pi}\int_{0}^{\infty}\exp\left(-\frac{r^2}{3}\right)rdrd\theta$$ $$=2\pi\frac{3}{2}\int_0^{\infty}\exp\left(-\frac{r^2}{3}\right)d\left(\frac{r^2}{3}\right)$$ $$=3\pi$$

$$I=\sqrt{3\pi}$$

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  • $\begingroup$ Basically re-deriving the constant factor in the standard normal distribution function. Or you can simply cite the well-known result as mm-aops did and save three or four lines of calculations. $\endgroup$ – David K Jan 28 '15 at 1:14
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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 $1$)

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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.

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