Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|cite|improve this question
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

2 Answers 2

up vote 9 down vote accepted

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 }}.$

share|cite|improve this answer
I need to check this, but this seems elegant. Thanks. –  B Armbruster Dec 19 '10 at 21:37

Or just type Series[Integrate[(1 - x)^N Exp[x N], {x, 0, 1}], {N, Infinity, 1}] into Mathematica, and preserve those valuable neurons.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.