# Simplest way to integrate this expression : $\int_{-\infty}^{+\infty} e^{-x^2/2} dx$ [duplicate]

I'm toying around with statistics and calculus for a project of mine and I'm trying to find the simplest/fastest way to integrate this formula :

$$\int_{-\infty}^{+\infty} e^{-x^2/2} dx$$

• I do not want to use a table.
• I'm taking this opportunity to get more practice with my new calculus skills
• It seems that a Taylor series approx is the only way to go

Best Regards

• en.wikipedia.org/wiki/Gaussian_integral – icurays1 Jun 29 '15 at 19:20
• Good question but surely a duplicate. – Michael Hardy Jun 29 '15 at 19:32
• Taylor series might be involved if it were an indefinite integral or had finite bounds other than $0$, but for the integral over the whole line, one can prove that it's $\sqrt{2\pi\ {}}$ by any of several methods not involving Taylor series. ${}\qquad{}$ – Michael Hardy Jun 29 '15 at 19:33
• Try googling "Gaussian Integral Trick" – John Joy Jun 29 '15 at 23:05

If we set $$I := \int_{\mathbb{R}} \exp \left(- \frac{x^2}{2} \right) \, dx,$$ then

$$I^2 = \int_{\mathbb{R}} \int_{\mathbb{R}} \exp \left( - \frac{x^2+y^2}{2} \right) \, dx \, dy.$$

Introducting polar coordinates, i.e.

$$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} r \cos \varphi \\ r \sin \varphi \end{pmatrix},$$

yields

$$I^2 = \int_{r=0}^{\infty} \int_{\varphi=0}^{2\pi} e^{-r^2/2} r \, dr \, d\varphi = \left( \int_{0}^{\infty} r e^{-r^2/2} \, dr \right) \left( \int_{\varphi=0}^{2\pi} d \varphi \right).$$

This expression can be easily calculated.

• Thanks, exactly what I needed. – keeepkool Jun 30 '15 at 10:47

The "easiest" way is to use a change of variables to change your integral into a multiple of

$$\int_{-\infty}^{+\infty}e^{-u^2}\,du$$

and use the famous fact that that last integral equals $\sqrt{\pi}$.

• The value of $\int_{-\infty}^\infty e^{-x^2/2}\; dx$ is just as famous. – Robert Israel Jun 29 '15 at 20:09

Hint: Change variables into the integral $\displaystyle\int_{-\infty}^{+\infty} e^{-x^2}\,dx$, square that, then transform into a polar integration.