Asymptotic behaviour for an integral with parameter $n$

Question

I am trying to find an asymptotic development up to the precision $o(1)$ for the integral $\displaystyle\int_0^{+\infty} ne^{-nt}\left (\sum_{k=0}^{n-1}\frac{t^k}{k!}\right)^n\text{d}t$ as $n\longrightarrow +\infty$... but I do not manage either to find an equivalent. .. Many thanks if you have some ideas.

Answer 1

Let $g_n(t)=e^{-t}\sum_{k=0}^{n-1}\frac{t^k}{k!}$: the given integral is $\int_{0}^{+\infty}n g_n(t)^n\,dt $ and

$$ g_n(t) = \sum_{k=0}^{n-1}\sum_{j\geq 0}\frac{(-1)^j t^{j+k}}{j! k!}=1-\frac{t^n}{n!}+\frac{t^{n+1}}{(n+1)(n-1)!}+O(t^{n+2}). $$ $g_n(t)$ is related to a Poisson distribution: it is well-known that the law of large numbers implies $\lim_{n\to +\infty}g_n(n)=\frac{1}{2}$, and in general we have $$ g_n(t) \approx \frac{1}{2}\left[1-\operatorname{Erf}\left(\frac{t-n}{\sqrt{2n}}\right)\right]=\frac{1}{2}\,\operatorname{Erfc}\left(\frac{t-n}{\sqrt{2n}}\right)$$ so $\int_{0}^{+\infty}n g_n(t)^n\,dt$ roughly behaves like $\frac{2n^2}{e}$ for moderately large values of $n$, but in order to derive the asymptotic expansion up to the $o(1)$ term we need a pretty heavy machinery, like the Berry-Esseen theorem and refined versions of the Laplace method. What is the purpose of finding such asymptotic expansion?