How do I calculate the limit of this integral? Using an appropriate probability distribution or otherwise show that
$$\lim_{n\to\infty} \int_0^n e^{-x}{x^{n-1}\over(n-1)!}dx
=0.5$$
 A: Let $\{X_n\}$ a sequence of independent identically distributed random variable of exponential law of mean $1$, that is, a density of $X_1$ is 
$$f(x):=e^{-x}\chi_{\{x\geq 0\}}.$$
We want to know a density $f_n$ of $S_n:=\sum_{k=1}^nX_k$. We can use induction: $f_n(x)=e^{-x}\frac{x^{n-1}}{(n-1)!}\chi_{\{x\geq 0\}}$. It's true for $n=1$ and if it's true for a $n$, we use convolution:
\begin{align}
f_{n+1}(x)&=\int_{\Bbb R}f_n(t)f_1(x-t)dt\\
&=\int_{\Bbb R}e^{-t}\frac{t^{n-1}}{(n-1)!}\color{green}{\chi_{(0,+\infty)}(t)}e^{-(x-t)}\color{red}{\chi_{(0,+\infty)}(x-t)}dt\\
&=e^{-x}\int_{\color{green}0}^{\color{red}x}\frac{t^{n-1}}{(n-1)!}dt\\
&=e^{—x}\frac{x^n}{n!}.
\end{align}
Denote $I_n:=\int_0^ne^{—x}\frac{x^{n-1}}{(n-1)!}dx$. We have, since $X_n\geq 0$ and the integrand is a density of $S_n$, that
\begin{align}
I_n&=P\left(\sum_{j=1}^nX_j\leq \color{red}n\right)\\
&=P\left(\sum_{j=1}^n(X_j\color{red}{-1})\leq 0\right)\\
&=P\left(\frac{\sum_{j=1}^nX_j\color{red}{-E[X_j]}}{\sqrt n}\leq 0\right),
\end{align}
the expectation of $X_1$ being $1$.
Since the set $(\infty,0]$ has a boundary of measure $0$ and $\frac{\sum_{j=1}^nX_j-E[X_j]}{\sqrt n}$ converges in law to a normal law of mean $0$ and variance $1$, say $N$ (it's given by the central limit theorem), we have by portmanteau theorem,
$$\lim_{n\to +\infty}I_n=P(N\leq 0)=1/2,$$
$N$ being symmetric. 
A: How about:
$\mathcal{L}( \int_0^n e^{-x}{x^{n-1}\over(n-1)!}dx) = \frac{1}{s}\mathcal{L}(e^{-x}{x^{n-1}\over(n-1)!}) = \frac{1}{s(s + 1)^n}$
$\mathcal{L}^{-1}(\frac{1}{s(s + 1)^n}) = 1 - \frac{\Gamma(n,n)}{\Gamma(n)}$ 
$\lim_{n\to\infty} 1 - \frac{\Gamma(n,n)}{\Gamma(n)} = 1 - \frac{1}{2} = \frac{1}{2}$
Edit:
The above hack could rightly be criticized as sort of "begging the question"; it's certainly not clear that evaluating the limit involving the upper incomplete gamma function is any easier than the original problem involving the lower! So, let's try this:
Define:
$P(n) = \frac{1}{\Gamma(n)}\int_0^n e^{-x}{x^{n-1}}dx$
$Q(n) = \frac{1}{\Gamma(n)}\int_n^\infty e^{-x}{x^{n-1}}dx$
So it should be fairly obvious that for all real n:
$P(n) + Q(n) = 1$
Then:
$\mathcal{L}(P(n) + Q(n)) = \mathcal{L}(1) \implies \frac{1}{s(s + 1)^n} + \mathcal{L}Q(n) = \frac{1}{s}$
$\implies \mathcal{L}Q(n) = \frac{1}{s} - \frac{1}{s(s + 1)^n}$
$\lim_{x\to\infty} P(n) = \lim_{s\to 0}s\mathcal{L}P(n) = 1$
$\lim_{x\to 0} Q(n) = \lim_{s\to\infty}s\mathcal{L}Q(n) = 1$
So, 
$\lim_{n\to\infty} P(n) + Q(n) = 1$
$\lim_{n\to\infty}P(n) - Q(n) = 0$,
from which the respective limits follow.
