Integral of Exponent of Half Square over Reals

Theorem

 * $\displaystyle \int_{\mathop \to -\infty}^{\mathop \to +\infty} e^{- x^2 / 2} \rd x = \sqrt {2 \pi}$

Proof
Let $t = \dfrac {x^2} 2$.

Then:

We have that $e^{- x^2 / 2}$ is an even function.

From Definite Integral of Even Function: Corollary:
 * $\displaystyle \int_{\mathop \to -\infty}^{\mathop \to +\infty} e^{- x^2 / 2} \rd x = 2 \int_0^{\mathop \to +\infty} e^{- x^2 / 2} \rd x$

Hence the result.