How can you get the asymptotics for the growth of $\Gamma(n,n)$?

$$ \Gamma(n,n) = \int_n^\infty x^{n-1} \exp(-x) \mathrm{d}x $$

  • $\begingroup$ Asymptotics to what order? $\endgroup$ – cardinal Mar 11 '13 at 23:35
  • $\begingroup$ By thinking about this in terms of a Poisson process (or doing integration by parts), one can obtain the explicit solution $\Gamma(n,n) = (n-1)! \sum_{k=0}^{n-1} \frac{e^{-n} n^k}{k!}$. $\endgroup$ – cardinal Mar 11 '13 at 23:50
  • $\begingroup$ @cardinal Could you add that as an answer? I would like the first few lower order terms which rlgordonma gave but your explicit solution looks perfect. $\endgroup$ – user66307 Mar 12 '13 at 8:08
  • $\begingroup$ @Lembik: you did ask for an asymptotic expansion, which cardinal's solution is not. These are two separate things. One is for computing numbers for any value of $n$, the other is for understanding how a function behaves in a certain limit. Perhaps you could ask for the explicit solution in another question, but if it was the explicit solution that you wanted, you should have asked for it to begin with. $\endgroup$ – Ron Gordon Mar 12 '13 at 17:53
  • $\begingroup$ @RonGordon You are right. It's simply that I know how to compute the asymptotics of cardinal's sum immediately. $\endgroup$ – user66307 Mar 12 '13 at 18:42

You can get it by a simple rescaling of the integral. Let $x=n t$; then

$$\begin{align}\Gamma(n,n) &= n^n \int_1^{\infty} dt \: t^{n-1} e^{-n t}\\ &= n^n \int_1^{\infty} \frac{dt}{t} e^{n (\log{t}-t)} \\ &= n^n e^{-n} \int_0^{\infty} \frac{du}{1+u} e^{n [\log{(1+u)}-u]} \end{align}$$

The contribution to this integral is dominated by that near $u=0$ as $n \rightarrow \infty$. We may then Taylor expand the term in the exponential and see that the leading asymptotic behavior of the integral is

$$\Gamma(n,n) \sim n^n e^{-n} \int_0^{\infty} du \: e^{-n u^2/2} = n^n e^{-n} \sqrt{\frac{\pi}{2 n}}$$

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

$$\left ( 1 + \frac{n u^3}{3} \right ) (1-u) \sim 1-u +\frac{n u^3}{3} $$

Evaluating the integrals that result from this expansion, we find the next higher term in the expansion:

$$\Gamma(n,n) = n^n e^{-n} \left [ \sqrt{\frac{\pi}{2 n}} - \frac{1}{3 n} + O\left ( n^{-3/2} \right ) \right ]$$


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