How does the integral of e^(-x^2), between +/-infinity, equal pi?

Question

Or did I get that wrong?

Answer 1

It's actually √π

If you want to see "how this was done", here's a link to a Wolfram article on it.

Answer 2

Set

I= \int e^(-x^2) dx. Now,

I^2=\int e^(-x^2) dx \int e^(-y^2) dy= \int \int e^(-(x^2+y^2)) dx dy

Use polar coordinates:

I^2=\int \int e^(-r^2) r dr d\theta = 1/2 \int \int e^-u du d\theta,
where u=r^2, and runs from 0 to infinity. \theta goes from 0 to 2\pi.

So I^2= 1/2 * 2\pi *\int e^-u du=\pi which means that

I=\sqrt{\pi}.

Answer 3

You must use integration in the complex plane