TITLE: How to prove $\lim_{n \to +\infty} \sqrt{n}\int_0^\pi{\cos(\frac{t}{2})^n}dt>0$
QUESTION [6 upvotes]: I want to prove 
$$\lim_{n \to +\infty}\sqrt{n}\int_0^\pi{\cos\left(\frac{t}{2}\right)^n}dt>0.$$
First, I consider $$\lim_{n \to +\infty}\sqrt{n}\int_0^\pi{\cos\left(\frac{t}{2}\right)^n}\sin\left(\frac{t}{2}\right)dt,$$ which is smaller than what I want, but the second integral leads to $$\lim_{n \to +\infty}\frac{\sqrt{n}}{2(n+1)}.$$ So it does not work.

REPLY [3 votes]: This integral can be analyzed using Laplace's Method.  Write 
$$\cos{\left(\frac{t}{2}\right)}^n = e^{n \log{\cos{(t/2)}}}$$
Note that $\log{\cos{(t/2)}}$ has a maximum at $t=0$; as $n \to \infty$ there is an increasingly narrow and high peak about $t=0$.  Thus we may Taylor expand about $t=0$ and ignore contributions away from a small neighborhood of $t=0$:
$$\int_0^{\pi} dt \, \cos{\left(\frac{t}{2}\right)}^n \sim \int_0^{\epsilon} dt \, e^{n \log{(1-t^2/8)}} $$
where $\epsilon = O\left(n^{-1/2}\right)$.  Because contributions away from $[0,\epsilon]$ are exponentially small, we may extend the integral out to infinity within the error of the leading-order approximation.  Taylor expanding the log term, we find
$$\int_0^{\pi} dt \, \cos{\left(\frac{t}{2}\right)}^n \sim \int_0^{\infty} dt \, e^{-n t^2/8}  = \frac12 \sqrt{\frac{8 \pi}{n}}$$
Therefore 
$$\lim_{n \to \infty} \sqrt{n} \int_0^{\pi} dt \, \cos{\left(\frac{t}{2}\right)}^n =\sqrt{2 \pi}$$