Integral of Exponent of Half Square over Reals
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Theorem
- $\ds \int_{\mathop \to -\infty}^{\mathop \to +\infty} e^{- x^2 / 2} \rd x = \sqrt {2 \pi}$
Proof
Let $t = \dfrac {x^2} 2$.
Then:
\(\ds \int_0^{\mathop \to +\infty} e^{- x^2 / 2} \rd x\) | \(=\) | \(\ds \int_0^{\mathop \to +\infty} \paren {2 t} e^{-t} \rd t\) | Integration by Substitution | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\sqrt 2} \map \Gamma {\frac 1 2}\) | Definition of Gamma Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\sqrt 2} \sqrt \pi\) | Gamma Function of One Half | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sqrt {2 \pi} } 2\) | multiplying top and bottom by $\sqrt 2$ |
We have that $e^{- x^2 / 2}$ is an even function.
From Definite Integral of Even Function: Corollary:
- $\ds \int_{\mathop \to -\infty}^{\mathop \to +\infty} e^{- x^2 / 2} \rd x = 2 \int_0^{\mathop \to +\infty} e^{- x^2 / 2} \rd x$
Hence the result.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 17.7 \ (6)$