Stirling's formula for the gamma function

Remarks

The proof starts from Binet's formula, which writes $\operatorname{Log}\Gamma(z) - (z - \tfrac12)\operatorname{Log} z + z - \tfrac12\ln(2\pi)$ as an integral that is $O(1/\abs{z})$ for $\operatorname{Re} z > 0$; rotating the line of integration extends this bound to the whole sector. Another route is Laplace's method applied to Euler's integral $\Gamma(x + 1) = \int_0^\infty e^{x \ln t - t}\,dt$, whose integrand peaks at $t = x$. For real $x \to \infty$ it gives $\Gamma(x + 1) \sim \sqrt{2\pi x}\,(x/e)^x$; at integers this is Stirling's formula for $n!$. The sector condition is needed: on the negative real axis $\Gamma$ has poles, so no such estimate holds near it. The full asymptotic series for $\operatorname{Log}\Gamma(z)$ is Stirling's series.

Going deeper

For a student

The complex form gives the decay of $\Gamma$ on vertical lines: for fixed real $x$, $\abs{\Gamma(x + iy)} \sim \sqrt{2\pi}\,\abs{y}^{x - 1/2} e^{-\pi\abs{y}/2}$ as $\abs{y} \to \infty$. This decay is what makes Mellin–Barnes integrals and the inverse Mellin transform converge, and it is used to estimate the Riemann zeta function through its functional equation. Dividing two instances gives $\Gamma(x + a)/\Gamma(x) \sim x^a$ as $x \to \infty$.