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In my asymptotic analysis and combinatorics class I was asked this question:

We first remember the definition f the Gamma function $ \Gamma(n+1) = n! = \int_{0}^{\infty} t^{n} e^{-t} dt $ and using this definition we are to prove Stirling's approximation formula for very large n

$ n! \sim (\frac{n}{e})^n \sqrt{2 \pi n} $

I realize the idea is to show the limit at $ n \to \infty $ of the quotient is 1 but since n is discrete then l'Hopital's rule is gone out the window and I do not see how to use the Gamma function definition to derive this even after giving it some thought so I am asking here in the hope of finding help. Thanks to all helpers.

Community
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Croc2Alpha
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    http://math.stackexchange.com/a/1409131/44121 – Jack D'Aurizio Mar 25 '16 at 18:17
  • After much observation of other authors' shifting and/or scaling of the variable of integration, I hit on a change-of-variable that moves the lower limit of integration to $-\infty$ and almost moves the peak of the integrand to the origin: https://math.stackexchange.com/a/4212957/307861. – Gavin R. Putland Aug 02 '21 at 09:02

1 Answers1

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Note that the Gamma Function has the integral representation

$$\Gamma(z+1)=\int_0^\infty t^ze^{-t}\,dt \tag 1$$

for $\text{Re}(z)>0$.

Enforcing the substitution $t=zs$ yields

$$\begin{align} \Gamma(z+1)&=z^{z+1}\int_0^\infty t^ze^{-zt}\,dt \\\\ &=z^{z+1}\int_0^\infty e^{z(\log(t)-t)}\,dt \tag 2 \end{align}$$

Using Laplace's Method in $(2)$ with $M=z$ and $f(t)=\log(t)-t$, we obtain

$$\begin{align} \Gamma(z+1)&\sim \sqrt{\frac{2\pi}{z}}e^{-z}z^{z+1}\\\\ &=\sqrt{2\pi z}\left(\frac{z}{e}\right)^z \tag 3 \end{align}$$

Finally, setting $z=n$ in $(3)$ yields

$$\bbox[5px,border:2px solid #C0A000]{\Gamma(n+1)=n!\sim \sqrt{2\pi n}\left(\frac ne\right)^n}$$

as was to be shown!

Mark Viola
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  • My solution at https://math.stackexchange.com/a/4212957/307861 can be understood as a variant of Laplace's method, allowing a slowly-varying second factor in the integrand. – Gavin R. Putland Aug 02 '21 at 09:09
  • @GavinR.Putland Hi Gavin. I posted This solution, which provides error bounds on Stirling's formula, on the same page you posted yours. My EE PhD was in electromagnetic field theory. What is your specialty? – Mark Viola Aug 02 '21 at 14:49
  • My PhD was on electric-acoustic analogs. Later I helped to produce a book on time-frequency analysis. Then I spent a decade working for an economic NGO.... I'm not sure if my approach to Stirling's approximation is conducive to working out error bounds or additional terms, but my aims were to avoid complexity and minimize reliance on prior knowledge. – Gavin R. Putland Aug 02 '21 at 16:39
  • Interesting specialty ... Laplace's Method does admit higher-order asymptotic approximations. – Mark Viola Aug 02 '21 at 16:43