Proof that Normal Distribution is Normalized

Question

How do we know that $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{(-x^2/2)}dx$ = 1.

Or how do we know that the normal distribution is normalized?

Or how do we know $erf(\infty) = 1$ ?

Answer 1

If you consider$\left(\int_{-\infty}^\infty e^{-x^2} dx \right) \left(\int_{-\infty}^\infty e^{-y^2} dy\right) =\int_{-\infty}^\infty \int_{\infty}^\infty e^{-(x^2+y^2)} dx dy$, you can evaluate the integral on the right using change of coordinates to polar coordinates.

Then, $\int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\int_{-\infty}^\infty \int_{\infty}^\infty e^{-(x^2+y^2)} dx dy}$.