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Quite often when I ask W|A to solve something it gives me an answer in terms of $\Gamma(n+1,n)$ or exponential integral $\operatorname{E}_{-n}(n)$. Looking up the definition of the incomplete gamma function and the exponential integral I can get a formal definition but I have no feeling for how these function behaves.

What is the asymptotic behaviour of these two functions?

Sasha
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Start with the definition: $$ \Gamma(n+1,n) = \int_n^\infty x^n \exp(-x) \mathrm{d}x $$ Changing $x = n (1+t)$ we get: $$ \begin{eqnarray} \Gamma(n+1,n) &=& n^{n+1} \exp(-n) \int_0^\infty (1+t)^n \exp(-n t) \mathrm{d}t \\ &=& n^{n+1} \exp(-n) \int_0^\infty \exp\left(-n \left(t-\log(1+t) \right) \right) \mathrm{d}t \end{eqnarray} $$ The integrand is maximal at $t=0$. Expanding $t-\log(1+t)$ in the series around $t=0$ and using Laplace's method: $$ \int_0^\infty \exp\left(-n \left(t-\log(1+t) \right) \right) \mathrm{d}t = \int_0^\infty \exp\left(-n \frac{t^2}{2}\right) \left(1 - \frac{n}{3} t^3 + n \frac{t^4}{4} + \mathcal{o}(t^5) \right) \mathrm{d} t $$ giving $$ \Gamma(n+1,n) = n^{n+\frac{1}{2}} \exp(-n) \sqrt{\frac{\pi}{2}} \left( 1 + \frac{1}{\sqrt{n}} \frac{2}{3} \sqrt{\frac{2}{\pi}} + \frac{3}{4n} + \mathcal{o} \left(n^{-1}\right) \right) $$

Sasha
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    $\Gamma(n+1,n) e^{n} n^{-n} \sim \sqrt{\pi n/2}$ . Is it obvious how to get the $\sqrt{\pi/2}$ from your solution? –  Mar 11 '13 at 17:18
  • Thanks! Would you mind adding some more details or a reference to somewhere with them please? It's not obvious how to do the "Expand in series and using Laplace's method" part. Or could there a simpler solution by some change of variables? http://math.stackexchange.com/questions/314220/variance-of-time-to-find-first-duplicate has a related integral solved that way. –  Mar 11 '13 at 17:51
  • Is it obvious how $\operatorname{E}_{-n}(n)$ and $\Gamma(n+1,n)$ are related from their definitions? –  Mar 11 '13 at 18:51
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    After a change of variables:$$ \Gamma(n+1,n) = \int_n^\infty x^n \mathrm{e}^{-x} \mathrm{d}x = n^{n+1} \int_1^\infty t^n \mathrm{e}^{-n t} \mathrm{d}x = n^{n+1} \operatorname{E}_{-n}(n)$$ – Sasha Mar 11 '13 at 18:53
  • Thanks (accepted answer). Do you know if there are any texts online or in print that cover asymptotics of the sort you showed? –  Mar 11 '13 at 19:04
  • The book by Olver is good. – Sasha Mar 11 '13 at 19:21