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?
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?
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 $I = \int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$.
We had a nice discussion of different ways to prove this over on Tim Gowers' blog a few years ago.
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 ..............