Stirling's formula for Gamma

Taking logarithms in [L1] on the chosen sector gives [L1, given, algebra] $\log \mathrm{\Gamma }(z)=-\log z-\gamma z+{\sum }_{n\ge 1}(\frac{z}{n}-\mathrm{Log}(1+\frac{z}{n})),$ where $\mathrm{Log}$ denotes the principal logarithm.

For an integer $N\ge 1$, define [step 1.1, algebra] $I_N(z):={\int }_0^N\frac{\lfloor u\rfloor -u+1\mathrm{/}2}{u+z}du\mathrm{.}$ On each interval $[n,n+1]$ with $0\le n\le N-1$, one has ${\int }_n^{n+1}\frac{n-u+1\mathrm{/}2}{u+z}du=(n+\frac{1}{2}+z)\mathrm{Log}\frac{n+1+z}{n+z}-1.$ Summing these equalities and telescoping the logarithms yields $I_N(z)=(N+z-\frac{1}{2})\mathrm{Log}(N+z)-(z+\frac{1}{2})\mathrm{Log}z-N-\log (N-1)!-z{H}_{N-1}+{\sum }_{n=1}^{N-1}(\frac{z}{n}-\mathrm{Log}(1+\frac{z}{n}))\mathrm{.}$

For fixed $z$ in the sector, [L2] and [L3] give [L2, L3, step 1.1, step 2.1, algebra] $H_{N-1}=\log N+\gamma +o(1),\log (N-1)!=(N-\frac{1}{2})\log N-N+\frac{1}{2}\log (2\pi )+o(1)\mathrm{.}$ The logarithm needs its first-order term: $\mathrm{Log}(N+z)=\log N+\mathrm{Log}(1+z\mathrm{/}N)=\log N+\frac{z}{N}+O_z(N^{-2}),$ whence $(N+z-\frac{1}{2})\mathrm{Log}(N+z)=(N+z-\frac{1}{2})\log N+z+o(1)\mathrm{.}$ Substituting these three expansions into step 2.1, the $N$, $\log N$, and $z\log N$ terms cancel. The remaining finite sum converges by step 1.1, so the limit gives the Binet-type formula $\log \mathrm{\Gamma }(z)=(z-\frac{1}{2})\mathrm{Log}z-z+\frac{1}{2}\log (2\pi )+{\int }_0^{\mathrm{\infty }}\frac{\lfloor u\rfloor -u+1\mathrm{/}2}{u+z}du\mathrm{.}$

Let [step 3.1, algebra] $\mathrm{\Phi }(u):={\int }_0^u(\lfloor v\rfloor -v+1\mathrm{/}2)dv\mathrm{.}$ Then $\mathrm{\Phi }$ is $1$-periodic and therefore bounded. Integrating by parts in step 3.1 gives ${\int }_0^{\mathrm{\infty }}\frac{\lfloor u\rfloor -u+1\mathrm{/}2}{u+z}du={\int }_0^{\mathrm{\infty }}\frac{\mathrm{\Phi }(u)}{(u+z{)}^2}du\mathrm{.}$ On the closed sector $\mathrm{\mid }\arg z\mathrm{\mid }\le \pi -\delta$, one has $\mathrm{\mid }u+z\mathrm{\mid }\ge c_{\delta }(u+\mathrm{\mid }z\mathrm{\mid })$ for a constant $c_{\delta }>0$, so the integral above is $O_{\delta }(\mathrm{\mid }z{\mathrm{\mid }}^{-1})$ uniformly as $\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty }$. Hence $\log \mathrm{\Gamma }(z)=(z-\frac{1}{2})\mathrm{Log}z-z+\frac{1}{2}\log (2\pi )+O_{\delta }(\mathrm{\mid }z{\mathrm{\mid }}^{-1}),$ and exponentiating yields $\mathrm{\Gamma }(z)=\sqrt{2\pi }{z}^{z-1\mathrm{/}2}{e}^{-z}(1+O_{\delta }(\mathrm{\mid }z{\mathrm{\mid }}^{-1}))\mathrm{.}$