Gaussian Integral

Theorem

The Gaussian Integral is the integral over $\R$ of the function $f(x) = e^{-x^2}$.

Its value is $\sqrt{\pi}$.

That is:

$\displaystyle \int_{-\infty}^{+\infty} e^{-x^2} \ \mathrm d x = \sqrt \pi$

Proof

One of the most famous proofs of this result uses the trick of calculating instead a two-dimensional integral in polar coordinates, as follows.

Notice that, as $e^{-x^2 -y^2} = e^{-x^2} e^{-y^2}$:

$\displaystyle (1) \qquad \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-x^2 - y^2} \ \mathrm d x \ \mathrm d y = \left({\int_{-\infty}^{+\infty} e^{-x^2} \ \mathrm d x}\right)^2$

so if we calculate this two-dimensional integral, we will also have the value we want to find.

Changing to polar coordinates:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-x^2 -y^2} \ \mathrm d x \ \mathrm d y\)	\(=\)	\(\displaystyle \int_0^{2\pi} \int_0^{+\infty} r e^{-r^2} \ \mathrm d r \ \mathrm d \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 2 \pi \left[-\frac 1 2 e^{-r^2} \right]_0^{+\infty}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \pi\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

The result follows from $(1)$.

$\blacksquare$

Source of Name

This entry was named for Carl Friedrich Gauss.