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I would like to compute the following limit of an integral $$I=\lim_{n\to+\infty}\int_0^n\frac{x^n\text{e}^{-x}}{n!}\text{d}x$$ which I encountered in a statistics problem. I can easily do this using Gamma distribution and Central Limit Theorem:

Consider a random variable $X_n$ with PDF $f_n(x)=\frac{x^n\text{e}^{-x}}{n!}$. From the properties of Gamma distribution, it can be decomposed into exponential distributions, $X_n=\sum_{k=1}^n X_k$ where $X_k\sim\text{Exp}(1)$. Now let $F_n(x)=\text{P}(\frac{X_n-n}{\sqrt n}\leq x)$; from Central Limit Theorem, $$\lim_{n\to+\infty}F_n(x)=\lim_{n\to+\infty}\text{P}\left(\frac{\sum_{k=1}^n X_k -n}{\sqrt n}\leq x\right)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^x\text{e}^{\frac{-t^2}{2}}\text{d}t$$ Plugging $x=0$ in gives $I=\frac 1 2$.

Nevertheless, I would like to know if there are any pure math methods for solving this (that don't involve anything statistical). The integral itself, from Wolfram|Alpha, is $$\int_0^n\frac{x^n\text{e}^{-x}}{n!}\text{d}x=\frac{\Gamma(n+1)-\Gamma(n+1,n)}{n!}$$ which doesn't seem to be very useful (it's just rewriting this integral in a Gamma function form). Any help would be appreciated.

  • The rewriting of the integral in terms of the incomplete Gamma function does have one interesting implication: Since $\Gamma(n+1)=n!$ for nonnegative integer $n$, the desired limit amounts to showing $$1-\frac{\Gamma(n+1,n)}{\Gamma(n+1)}\to \frac12,$$ or $\Gamma(n+1,n)/\Gamma(n+1)\to 1/2$ as $n\to\infty$. – Semiclassical Jan 25 '21 at 06:18

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Not sure it is as pure math as you'd wish but here goes. Let: $$ I_{m,n} = \int_0^m \frac{x^ne^{-x}}{n!}\text{d}x $$ Using a simple integration by parts yields, for $n>0$: $$ I_{m,n} -I_{m,n-1} = -\frac{m^ne^{-m}}{n!} $$ Furthermore: $$ I_{m,0} = \int_0^m e^{-x}\text{d}x = 1 - e^{-m} $$ Then: $$ I_{m,n} = I_{m,0} + \sum_{k=1}^n (I_{m,k} - I_{m,k-1}) $$

$$ I_{m,n} = 1 - e^{-m} - \sum_{k=1}^n \frac{m^ke^{-m}}{k!} $$

Which, when setting $m$ to $n$:

$$ I_{n,n} = \int_0^n \frac{x^ne^{-x}}{n!}\text{d}x = 1 - e^{-n} \sum_{k=0}^n \frac{n^k}{k!} $$

That was the rather easy part. You can then conclude using the result of this question (where most of the answers essentially go back the other way, so really you're just better off just picking up from where your integral appears).

Bill O'Haran
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