How could you show that the normal density integrates to 1?
$$ \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi \sigma^2}} e^{-(x+\mu)^2 / \sigma^2} dx = 1 $$
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2See Gaussian integral and How to prove $ \int_0^\infty e^{-x^2} ; dx = \frac{\sqrt{\pi}}2$ without changing into polar coordinates? – Martin Sleziak Dec 01 '12 at 14:19
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The function $f(x) = \exp(-x^2)$ is a so called Schwartz-function for which you can use Fubini's theorem in 2-dimension.
It holds:
$$\int_{-\infty}^{+\infty} \exp(-x^2)dx\cdot\int_{-\infty}^{+\infty}\exp(-x^2)dx = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\exp(-x^2-y^2)dxdy$$
This can be solved in polar-coordinates easilly and its value is $\pi$.
Because of $\exp(-x^2)\gt0\,\forall\,x$ you have $\int_{-\infty}^{+\infty}\exp(-x^2)\gt0$ and finally $\int_{-\infty}^{+\infty}\exp(-x^2) = \sqrt{\pi}$
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