# Gaussian Integral Evaluation

The evaluation of the Gaussian integral $\displaystyle I = \int_{-\infty}^{+\infty} e^{-x^2} dx$ using the double integrals and the polar coordinates is presented.



## From Double Integrals in Rectangular Coordinates to Polar Coordinates

Changing a double integrals from rectangualr to polar coordinates is done as follows [1] $\iint_R f(x,y) \;dy \;dx = \int_{\theta_1}^{\theta_2} \int_{r_1(\theta)}^{r_2(\theta)} f(r,\theta) r \;dr \;d\theta \qquad (I)$ with the relationships between the rectangular coordinates $x$ and $y$; and the polar coordinates $r$ and $\theta$ given by [3]
$x = r \cos \theta$ , $y = r \sin \theta$ , $r^2 = x^2 + y^2$

## Evaluate the Gaussian Integral

The Gaussian integral is defined as follows $I = \int_{-\infty}^{+\infty} e^{-x^2} dx$ and we need to evaluate $I$.
We first note that the integrals $\displaystyle \int_{-\infty}^{+\infty} e^{-x^2} dx$ and $\displaystyle \int_{-\infty}^{+\infty} e^{-y^2} dy$ have equal values. We therefore may write that

$I^2 = \displaystyle \left(\int_{-\infty}^{+\infty} e^{-x^2} dx \right) \left(\int_{-\infty}^{+\infty} e^{-y^2} dy \right)$

The above may be written as a double integral as follows
$I^2 = \displaystyle \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-(x^2+y^2)} \; dx \; dy$

Let $x = r \cos \theta$ , $y = r \sin \theta$ , $r^2 = x^2 + y^2$ be the relationships between the rectangular and polar coordinates and use the change of integral from rectangular to polar coordinates (I) given above.
$I^2 = \displaystyle \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-(x^2+y^2)} \; dx \; dy = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r \;dr \;d\theta$

Note that $\displaystyle \int_{0}^{\infty} e^{-r^2} r \;dr = -(1/2) e^{-r^2}$, which gives

$I^2 = \displaystyle \int_{0}^{2\pi} \left[-(1/2) e^{-r^2}\right]_0^{\infty} \;d\theta$
which may be written as
$I^2 = \displaystyle \int_{0}^{2\pi} (-1/2) \left[e^{-\infty}-e^0 \right] \;d\theta$

Note that using limits, $e^{-\infty} = \lim_{a\to\infty} e^{-a} = 0$, we now evaluate
$I^2 = \displaystyle \int_{0}^{2\pi} \dfrac{1}{2} \;d\theta$

Evaluate the above integral
$I^2 = \dfrac{1}{2} \left[\theta\right]_0^{2\pi}$
$= \dfrac{1}{2} \left[2\pi - 0 \right]$
$= \pi$

So far we have calculated $I^2 = \pi$ and therefore taking the square root, we have $I = \int_{-\infty}^{+\infty} e^{-x^2} dx = \sqrt {\pi}$