Equivalent for the integral $\int_{\alpha n}^{\beta n}s^n\mathrm{e}^{-s}\mathrm{d}s$

Question

Let $0<\alpha<\beta$. I am looking for an equivalent for $\displaystyle I_n=\int_{\alpha n}^{\beta n}s^n\mathrm{e}^{-s}\mathrm{d}s$ when $n\longrightarrow+\infty$ : $I_n\underset{{n}\rightarrow{+\infty}}\sim ?$. I have tried integration by parts or use of asymptotics expansion for the Gamma function, I have changed variable, developed $\exp$... without success. Thank you.

Answer 1

If you make the change of integration variables $s = ne^t$, you find $$ \int_{\alpha n}^{\beta n} {s^n e^{ - s} ds} = n^{n + 1} e^{ - n} \int_{\log \alpha }^{\log \beta } {\exp ( - n(e^t - t - 1))e^t dt} . $$ Now $t=0$ is the sole saddle point of $e^t-t-1$ on the real line. The asymptotic form of the integral depends on whether this saddle is inside the interval of integration or not. See https://dlmf.nist.gov/2.3#iii and https://dlmf.nist.gov/2.4#iv

Alternatively, \begin{align*} & \int_{\alpha n}^{ + \infty } {s^n e^{ - s} ds} - \int_{\beta n}^{ + \infty } {s^n e^{ - s} ds} = \Gamma (n + 1,\alpha n) - \Gamma (n + 1,\beta n) \\ & = n\Gamma (n,\alpha n) - n\Gamma (n,\beta n) + (\alpha n)^n e^{ - \alpha n} - (\beta n)^n e^{ - \beta n} . \end{align*} The asymptotics of the incomplete gamma functions will depend on the sizes of $\alpha$ and $\beta$. See http://dlmf.nist.gov/8.11.iii You can also obtain a uniform asymptotic approximation where you do not need to distinguish any cases, see http://dlmf.nist.gov/8.12 with $Q(a,z)=\Gamma(a,z)/\Gamma(a)$.