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?
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 $$ \Gamma(n+1,n) = n^{n+\frac{1}{2}} \exp(-n) \sqrt{\frac{\pi}{2}} \left( 1 + \frac{1}{\sqrt{n}} \frac{2}{3} \sqrt{\frac{2}{\pi}} + \frac{3}{4n} + \mathcal{o} \left(n^{-1}\right) \right) $$