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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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1 Answer 1

up vote 4 down vote accepted

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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