Integral of Exponent of Half Square over Reals

Theorem

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

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

Then:

$\ds \int_0^{\mathop \to +\infty} e^{- x^2 / 2} \rd x$

$=$

$\ds \int_0^{\mathop \to +\infty} \paren {2 t} e^{-t} \rd t$

$\ds $

$=$

$\ds \frac 1 {\sqrt 2} \map \Gamma {\frac 1 2}$

Definition of Gamma Function

$\ds $

$=$

$\ds \frac 1 {\sqrt 2} \sqrt \pi$

$\ds $

$=$

$\ds \frac {\sqrt {2 \pi} } 2$

multiplying top and bottom by $\sqrt 2$

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

$\ds \int_{\mathop \to -\infty}^{\mathop \to +\infty} e^{- x^2 / 2} \rd x = 2 \int_0^{\mathop \to +\infty} e^{- x^2 / 2} \rd x$

Hence the result.

$\blacksquare$