The Gaussian Integral is given by $$ \int_{-\infty}^{\infty} e^{-x^2} dx $$ This is a beautiful integral with many important applications to probability and statistics, namely normal distributions. Here, we will evaluate this integral using techniques from Calculus 3. First, if we define \(I=\int_{-\infty}^{\infty} e^{-x^2} dx\) we have that $$ I^2=\int_{-\infty}^{\infty} e^{-x^2} dx \int_{-\infty}^{\infty} e^{-y^2} dy= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2} dy dx= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)} dy dx $$ From here, we will preform a change into polar coordinates, with \((x,y) \rightarrow (t, \theta)\) and \(x=r\cos\theta\) , \(y=r\sin\theta\) . From this, since \(r^2=x^2+y^2\) , \(r \in (0,\infty)\) and \(\theta \in (0, 2\pi)\) we have that our integral is now $$ I^2=\int_{0}^{\infty} \int_{0}^{2\pi} re^{-r^2} d\theta dr $$ after this change of coordinates. By Fubini’s theorem, we have that $$ I^2=\int_{0}^{\infty} \int_{0}^{2\pi} re^{-r^2} d\theta dr = \int_{0}^{2\pi} d\theta \int_{0}^{\infty} re^{-r^2} dr = 2\pi \int_{0}^{\infty} re^{-r^2} dr $$ Now evaulating this integral of one dimension with the substitution $$ t=r^2 \implies \sqrt{t}=r \implies dt=2rdr \implies rdr=\frac{dt}{2} $$ we have that
$$ \int_{0}^{\infty} re^{-r^2} dr = \int_{0}^{\infty} \frac{e^{-t}}{2} dt = \frac{1}{2} (-e^{-t})|_{0}^{\infty}) = \frac{1}{2} (0-(-1)) = \frac{1}{2} $$
Finally, we have that
$$ I^2 = 2\pi \int_{0}^{\infty} re^{-r^2} dr = 2\pi (\frac{1}{2}) = \pi $$
With
$$ I^2=\pi $$
we have that
$$ I=\sqrt{\pi} \implies \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi} $$