Editing Stirling's approximation (section)

=== Higher orders ===
In fact, further corrections can also be obtained using Laplace's method. From previous result, we know that <math>\Gamma(x) \sim x^x e^{-x}</math>, so we "peel off" this dominant term, then perform two changes of variables, to obtain:<math display="block">x^{-x}e^x\Gamma(x) = \int_\R e^{x(1+t-e^t)}dt</math>To verify this: <math>\int_\R e^{x(1+t-e^t)}dt \overset{t \mapsto \ln t}{=} e^x \int_0^\infty t^{x-1} e^{-xt} dt \overset{t \mapsto t/x}{=} x^{-x} e^x \int_0^\infty e^{-t} t^{x-1} dt = x^{-x} e^x \Gamma(x)</math>.

Now the function <math>t \mapsto 1+t - e^t</math> is unimodal, with maximum value zero. Locally around zero, it looks like <math>-t^2/2</math>, which is why we are able to perform Laplace's method. In order to extend Laplace's method to higher orders, we perform another change of variables by <math>1+t-e^t = -\tau^2/2</math>. This equation cannot be solved in closed form, but it can be solved by serial expansion, which gives us <math>t = \tau - \tau^2/6 + \tau^3/36 + a_4 \tau^4 + O(\tau^5) </math>. Now plug back to the equation to obtain<math display="block">x^{-x}e^x\Gamma(x) = \int_\R e^{-x\tau^2/2}(1-\tau/3 + \tau^2/12 + 4a_4 \tau^3 + O(\tau^4)) d\tau = \sqrt{2\pi}(x^{-1/2} + x^{-3/2}/12) + O(x^{-5/2})</math>notice how we don't need to actually find <math>a_4</math>, since it is cancelled out by the integral. Higher orders can be achieved by computing more terms in <math>t = \tau + \cdots</math>, which can be obtained programmatically.{{NoteTag|note=For example, a program in Mathematica: <syntaxhighlight lang="mathematica">
series = tau - tau^2/6 + tau^3/36 + tau^4*a + tau^5*b;
(*pick the right a,b to make the series equal 0 at higher orders*)
Series[tau^2/2 + 1 + t - Exp[t] /. t -> series, {tau, 0, 8}]

(*now do the integral*)
integral = Integrate[Exp[-x*tau^2/2] * D[series /. a -> 0 /. b -> 0, tau], {tau, -Infinity, Infinity}];
Simplify[integral/Sqrt[2*Pi]*Sqrt[x]]

</syntaxhighlight>|name=mathematica-program|content=content|text=text}}

Thus we get Stirling's formula to two orders:<math display="block"> n! = \sqrt{2\pi n}\left(\frac{n}{e}\right)^n \left(1 + \frac{1}{12 n}+O\left(\frac{1}{n^2}\right) \right).
</math>