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.
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
\[ I = \int_{-\infty}^{+\infty} e^{-x^2} dx = \sqrt {\pi}\]