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How to calculate the following integral? $\int_{- \infty}^{\infty} \mathrm{e}^{- \frac{x^2}{2}} \mathrm{d} x$

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marked as duplicate by Marvis, vonbrand, Micah, muzzlator, Amzoti Apr 8 '13 at 17:37

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The standard method here is to multiply the integral by itself, in the form $\int_{-\infty}^{\infty}e^{\frac{-y^2}{2}}dy$ and then integrate over the plane using polar coordinates. Remember to take the square root at the end. – Jared Apr 8 '13 at 17:08
up vote 0 down vote accepted

Hint: Let $y=\frac{x}{\sqrt{2}}$ and make a change of variables, then use the formula $$\int_{-\infty}^\infty e^{-x^2}\,dx=\sqrt{\pi}.$$

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Do you mean we should learn this integral? I have been told by my teacher that indefinite integral of this doesn't exist. – ABC Apr 8 '13 at 17:07
How do u get $\int_{-\infty}^\infty e^{-x^2}\,dx=\sqrt{\pi}$? – N Zhang Apr 8 '13 at 17:09
@NZhang: It would appear there are plenty of answers to help guide you along the proof of the formula. The main idea is using polar coordinates and getting the extra $r$ to integrate. – Clayton Apr 8 '13 at 17:10
@exploringnet, the indefinite integral does exist, it just can't be expressed in terms of elementary functions (algebraic, exponentials, trigonometric functions and their inverses). – vonbrand Apr 8 '13 at 17:17

There's a well known trick for this integral: \begin{align*} \left(\int_{-\infty}^\infty e^{-\frac{x^2}{2}}\,\mathrm{d}x\right)^2 &= \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{x^2+y^2}{2}}\,\mathrm{d}x\mathrm{d}y \\ &= \int_0^\infty\int_0^{2\pi} e^{-\frac{r^2}{2}}r\,\mathrm{d}\theta\mathrm{d}r \\ &= 2\pi \left[-e^{-\frac{r^2}{2}}\right]_{0}^\infty \\ &= 2\pi \end{align*} Hence, $\int_{-\infty}^\infty e^{-\frac{x^2}{2}}\,\mathrm{d}x = \sqrt{2\pi}$

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$$S:=\int\limits_{-\infty}^\infty e^{-x^2/2}dx\implies S^2=\int\limits_{-\infty}^\infty e^{-x^2/2}dx\int\limits_{-\infty}^\infty e^{-y^2/2}dy=\int\limits_{-\infty}^\infty e^{-\frac{x^2+y^2}{2}}dx\,dy=$$

$$=\int\limits_0^\infty\int_0^{2\pi}re^{-\frac{r^2}{2}}d\theta\,dr=\left.-2\pi e^{-\frac{r^2}{2}}\right|_0^\infty=\ldots$$

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