# $\int e^{-x^2}dx$ [duplicate]

Possible Duplicate:
Proving $\int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$

How does one integrate $\int e^{-x^2}\,dx$? I read somewhere to use polar coordinates.

How is this done? What is the easiest way?

## marked as duplicate by Mike Spivey, t.b., Américo Tavares, Ross Millikan, Willie WongApr 23 '11 at 16:13

• As an indefinite integral, this is not expressible by elementary functions. I guess you mean a definite integral? – Raeder Apr 23 '11 at 12:00
• This is an interesting integral. There is an expository paper by K Conrad dealing with this. – user230452 Jul 2 '16 at 5:44
• One of most elementary ways to prove it is using limits and arctangent : math.stackexchange.com/questions/9286/… Side note : On the same page are other solutions as well so check them out :) – Machinato Jul 15 '16 at 0:00

You can integrate the function $e^{-x^2}$ only as a definite integral, as you mention you can do it using polar coordinates as follows:
Let $I = \int_{-\infty}^{\infty} e^{-x^2}\,dx$ then multiply it by $I = \int_{-\infty}^{\infty} e^{-y^2}\,dy$ so we have $I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2}\,dx\right) \left(\int_{-\infty}^{\infty} e^{-y^2}\,dy\right) = \left(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)}\,dx\,dy\right)$ now we can change this integral to polar coordinates using $\rho^2 = x^2 + y^2$ and now $I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-\rho^2} \rho \, d\rho \, d\theta = \pi$ and finally $I = \int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$.
The function you have mentioned is called the gaussian integral.... Read more about it on wiki. It has explanation regarding solving it by using the polar co-ordinates too ..............