Integrand is sharply peaked

Question

I have a gamma function, $$n!=\int_0^\infty e^{-x}x^n \mathrm{d}x$$ and if I take the derivative of the logarithm of the integrand, I get $$\dfrac{\mathrm{d}}{\mathrm{dx}}(n\ln x -x)=\dfrac{n}{x}-1.$$ But why I can argument that the integrand is sharply peaked with the contribution important only near $x=n$?

Answer 1

The gamma function is $$ \Gamma(n+1) = \int_0^\infty dz e^{-z}z^n.$$ Rewrite the integrand as $$ \Gamma(n+1) = \int_0^\infty dz \exp\big[-z + n \ln z\big].$$ The integrand goes to $0$ as $z\rightarrow 0$ or $z \rightarrow \infty. $ Between these extremes, the function somewhere finds a maximum. When $n$ is large, the decay toward $0$ on either side of this maximum becomes very rapid (as you could see by analyzing the first derivative). Therefore, for large $n$, the area very near the integrand's maximum becomes the primary contribution to the integral.

This maximum occurs where $$ 0 = \frac{d}{dz}\exp\big[-z + n \ln z\big] = \big(-1 + \frac{n}{z}\big) \exp \big[-z + n \ln z\big],$$ which means $z=n$. Typically one would now Taylor expand the argument of the exponential around $z=n$ to second order and then integrate in what's called a Saddle-Point Expansion. This yields the so-called Stirling's Formula for $n!$.

Answer 2

$e^{-x} x^n/\Gamma(n+1)$ for $x > 0$ is the probability density function of the Gamma distribution with scale parameter $1$ and shape parameter $n+1$. When $n$ is a positive integer, this is the distribution of the sum $X$ of $n+1$ independent random variables $X_n$, each with exponential distribution with parameter $1$. The Central Limit Theorem lets you approximate this with a normal distribution of mean $n+1$ and standard deviation $\sqrt{n+1}$. This is "strongly peaked" because $\sqrt{n+1}$ is small compared to $n+1$ when $n$ is large. The Weak Law of Large Numbers says that for any $\epsilon > 0$, $$ \lim_{n \to \infty} \mathbb P(1-\epsilon < X/(n+1) < 1+\epsilon) = 1$$