How can you get the asymptotics for the growth of Γ(n,n)?
Γ(n,n)=∫∞nxn−1exp(−x)dx
Answer
You can get it by a simple rescaling of the integral. Let x=nt; then
Γ(n,n)=nn∫∞1dttn−1e−nt=nn∫∞1dtten(logt−t)=nne−n∫∞0du1+uen[log(1+u)−u]
The contribution to this integral is dominated by that near u=0 as n→∞. We may then Taylor expand the term in the exponential and see that the leading asymptotic behavior of the integral is
Γ(n,n)∼nne−n∫∞0due−nu2/2=nne−n√π2n
We may find further terms in the asymptotic behavior by Taylor expanding the higher-order terms in the exponential and the term outside the exponential. The next higher order term is
(1+nu33)(1−u)∼1−u+nu33
Evaluating the integrals that result from this expansion, we find the next higher term in the expansion:
Γ(n,n)=nne−n[√π2n−13n+O(n−3/2)]
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