Editing Gaussian integral (section)

=== By [[Laplace's method]] ===

In Laplace approximation, we deal only with up to second-order terms in Taylor expansion, so we consider <math>e^{-x^2}\approx 1-x^2 \approx (1+x^2)^{-1}</math>.  

In fact, since <math>(1+t)e^{-t} \leq 1</math> for all <math>t</math>, we have the exact bounds:<math display="block">1-x^2 \leq e^{-x^2} \leq (1+x^2)^{-1}</math>Then we can do the bound at Laplace approximation limit:<math display="block">\int_{[-1, 1]}(1-x^2)^n dx \leq \int_{[-1, 1]}e^{-nx^2} dx \leq \int_{[-1, 1]}(1+x^2)^{-n} dx</math>  

That is,
<math display="block">2\sqrt n\int_{[0, 1]}(1-x^2)^n dx \leq \int_{[-\sqrt n, \sqrt n]}e^{-x^2} dx \leq 2\sqrt n\int_{[0, 1]}(1+x^2)^{-n} dx</math>  

By trigonometric substitution, we exactly compute those two bounds: <math>2\sqrt n(2n)!!/(2n+1)!!</math> and <math>2\sqrt n (\pi/2)(2n-3)!!/(2n-2)!!</math>  

By taking the square root of the [[Wallis formula]], <math display="block">\frac \pi 2 = \prod_{n=1} \frac{(2n)^2}{(2n-1)(2n+1)}</math>we have <math>\sqrt \pi = 2 \lim_{n\to \infty} \sqrt{n} \frac{(2n)!!}{(2n+1)!!}</math>, the desired lower bound limit. Similarly we can get the desired upper bound limit.
Conversely, if we first compute the integral with one of the other methods above, we would obtain a proof of the Wallis formula.