Integral of Gaussian

This is just a slick derivation of the definite integral of a Gaussian from minus infinity to infinity. With other limits, the integral cannot be done analytically but is tabulated. Functions are available in computer libraries to return this important integral.

The answer is

$$\bgroup\color{black}\int\limits_{-\infty}^{\infty}\; dx\; e^{-ax^2} = \sqrt{\pi\over a}.\egroup$$

Define

$$\bgroup\color{black}I=\int\limits_{-\infty}^{\infty}\; dx\; e^{-ax^2}.\egroup$$

Integrate over both $x$ and $y$ so that

$$\bgroup\color{black}I^2=\int\limits_{-\infty}^{\infty}\; dx\;... ...\int\limits_{-\infty}^{\infty}\; dxdy\; e^{-a(x^2+y^2)}.\egroup$$

Transform to polar coordinates.

$$\bgroup\color{black}I^2=2\pi\int\limits_{0}^{\infty}\;rdr\; e... ...i\left[-{1\over a}e^{-ar^2}\right]_0^\infty={\pi\over a}\egroup$$

Now just take the square root to get the answer above.

$$\bgroup\color{black}\int\limits_{-\infty}^{\infty}\; dx\; e^{-ax^2} = \sqrt{\pi\over a}.\egroup$$

Other forms can be obtained by differentiating with respect to $a$.

$$\bgroup\color{black}{\partial\over\partial a}\int\limits_{-\i... ... e^{-ax^2} = {\partial\over\partial a}\sqrt{\pi\over a}\egroup$$

$$\bgroup\color{black}\int\limits_{-\infty}^{\infty}\; dx\; x^2e^{-ax^2} = {1\over 2a}\sqrt{\pi\over a}\egroup$$