# limit of definite integral as $N \to \infty$

I'm interested in $\theta(N):=\int_0^1 (1-x)^{N-1} e^{xN} dx$. I'd like to show that $\theta(N)\sim c/\sqrt{N}$ as $N\to\infty$ and determine $c$. Any ideas?

• This is homework. Presumably an application of the principle of stationary phase. – Denis Serre Dec 19 '10 at 20:50
• no, research. I'll check out the stationary phase approximation page on wikipedia... Thanks. – B Armbruster Dec 19 '10 at 21:07
• The function doesn't oscillate though. So I'm not sure stationary phase approximation is appropriate. Mathematica gives $e^N N^{-N} (\Gamma(N)-\Gamma(N,N))$. However, now I'm stuck with finding the asymptotics of $\Gamma(N,N)$ which doesn't seem easier. – B Armbruster Dec 19 '10 at 21:12
• To the OP, a hint: re-write $(1-x) = \exp \ln (1-x)$ and apply Laplace's method. – Willie Wong Dec 19 '10 at 21:16

Denote by $I$ your integral. Then, $I = e^N \int_0^1 {x^{N - 1} e^{ - xN} \,{\rm d}x} = \frac{{\Gamma (N)e^N }}{{N^N }}\int_0^N {\frac{{x^{N - 1} e^{ - x} }}{{\Gamma (N)}}\,{\rm d}x}.$ Now, if $X_1,\ldots,X_N$ are independent and identically distributed exponential(1) random variables, then their sum $X_1 + \cdots + X_N$ has gamma density $x^{N-1}e^{-x}/\Gamma(N)$, $x>0$. Thus, the last integral above is equal to ${\rm P}(X_1 + \cdots + X_N \leq N)$, or equivalently to ${\rm P}(X_1 + \cdots + X_N - N \leq 0)$. By the central limit theorem, ${\rm P}(X_1 + \cdots + X_N - N \leq 0) \to 1/2$ as $N \to \infty$ (since the $X_i$ have expectation equal to $1$). So, using $N^N \sim N!e^N /\sqrt {2\pi N}$, we get $I \sim \sqrt {\frac{\pi }{2}} \frac{1}{{\sqrt N }}.$