Large $n$ asymptotic of $\int_0^\infty \left( 1 + x/n\right)^{n-1} \exp(-x) \, \mathrm{d} x$

With the change of variables $x\to(n-1)t-1$, we get $$ \begin{align} &\int_0^\infty\left(1+\frac{x}{n}\right)^{n-1}e^{-x}\;\mathrm{d}x\\ &=ne\left(1-\frac{1}{n}\right)^n\int_\frac{1}{n-1}^\infty e^{-(n-1)(t-\log(1+t))}\;\mathrm{d}t\tag{1}\\ \end{align} $$ Since $$ 1+n\log\left(1-\frac{1}{n}\right)=-\frac{1}{2n}-\frac{1}{3n^2}+O\left(\frac{1}{n^3}\right) $$ exponentiating and multiplying by $n$, we get $$ \begin{align} ne\left(1-\frac{1}{n}\right)^n &=ne^{-\frac{1}{2n}-\frac{1}{3n^2}+O\left(\frac{1}{n^3}\right)}\\ &=n-\frac{1}{2}-\frac{5}{24n}+O\left(\frac{1}{n^2}\right)\tag{2} \end{align} $$ Note that $$ \begin{align} \int_0^\frac{1}{n-1} e^{-(n-1)(t-\log(1+t))}\;\mathrm{d}t &=\int_0^\frac{1}{n-1}\left(1-\tfrac{n-1}{2}t^2+\tfrac{n-1}{3}t^3+\tfrac{(n-1)^2}{8}t^4\right)\;\mathrm{d}t+O\left(\tfrac{1}{n^4}\right)\\ &=\frac{1}{n-1}-\frac{1}{6(n-1)^2}+\frac{13}{120(n-1)^3}+O\left(\frac{1}{n^4}\right)\\ &=\frac{1}{n}+\frac{5}{6n^2}+\frac{31}{40n^3}+O\left(\frac{1}{n^4}\right)\tag{3} \end{align} $$ Finally, setting $\frac{u^2}{2}=t-\log(1+t)$, so that $t=u+\frac{u^2}{3}+\frac{u^3}{36}-\frac{u^4}{270}+\frac{u^5}{4320}+\frac{u^6}{17010}+O(u^7)$, we get $$ \begin{align} &\int_0^\infty e^{-(n-1)(t-\log(1+t))}\;\mathrm{d}t\\ &=\int_0^\infty e^{-(n-1)u^2/2}\;\mathrm{d}t\\ &=\int_0^\infty e^{-(n-1)u^2/2}\;(1+\frac{2u}{3}+\frac{u^2}{12}-\frac{2u^3}{135}+\frac{u^4}{864}+\frac{u^5}{2835}+O(u^6))\;\mathrm{d}u\\ &=\sqrt{\tfrac{\pi}{2(n-1)}}+\tfrac{2}{3(n-1)}+\sqrt{\tfrac{\pi}{288(n-1)^3}}-\tfrac{4}{135(n-1)^2}+\sqrt{\tfrac{\pi}{165888(n-1)^5}}+\tfrac{8}{2835(n-1)^3}+O\left(\tfrac{1}{n^{7/2}}\right)\\ &=\sqrt{\tfrac{\pi}{2n}}\left(1+\tfrac{7}{12n}+\tfrac{145}{288n^2}\right)+\left(\tfrac{2}{3n}+\tfrac{86}{135n^2}+\tfrac{346}{567n^3}\right)+O\left(\tfrac{1}{n^{7/2}}\right)\tag{4} \end{align} $$ Combining $(3)$ and $(4)$, we get $$ \int_\frac{1}{n-1}^\infty e^{-(n-1)(t-\log(1+t))}\;\mathrm{d}t=\sqrt{\tfrac{\pi}{2n}}\left(1+\tfrac{7}{12n}+\tfrac{145}{288n^2}\right)-\left(\tfrac{1}{3n}+\tfrac{53}{270n^2}+\tfrac{3737}{22680n^3}\right)+O\left(\tfrac{1}{n^{7/2}}\right) $$ Including $(2)$, yields $$ \int_0^\infty\left(1+\frac{x}{n}\right)^{n-1}e^{-x}\;\mathrm{d}x=\sqrt{\tfrac{n\pi}{2}}\left(1+\tfrac{1}{12n}+\tfrac{1}{288n^2}\right)-\left(\tfrac{1}{3}+\tfrac{4}{135n}-\tfrac{8}{2835n^2}\right)+O\left(\tfrac{1}{n^{5/2}}\right) $$

A related result was given in the problems column of the American Mathematical Monthly not too long ago. This is problem 11353 whose solution was published in the January 2010 issue.

Let $$g(s)=\int_0^\infty \left(1+{x\over s}\right)^se^{-x}\, dx-\sqrt{s\pi\over 2}.$$ Show that $g(s)$ decreases from $1$ to $2/3$ as $s$ ranges from $0$ to $\infty$.

Note that the exponent in the integral is $s$, not $s-1$.

Interesting. I've got a representation $$ \mathcal{I}_n = n e^n \int_1^\infty t^{n-1} e^{- nt}\, dt $$ which can be obtained from yours by the change of variables $t=1+\frac xn$. After some fiddling one can get $$ 2\mathcal{I}_n= n e^n \int_0^\infty t^{n-1} e^{- nt}\, dt+o(\mathcal{I}_n)= n^{-n} e^n \Gamma(n+1)+\ldots=\sqrt{2\pi n}+\ldots. $$