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Opalg · #3 November 25th, 2008, 04:53 AM

Quote:

Originally Posted by alexthepenguin

prove that

$\lim_{n\rightarrow \infty} \sqrt{n} \int_0^{\pi/2} \cos^n x d\mu=\int_0^\infty e^{-x^2/2}$

this is based on the assumption that $\ln(\cos x) \le \frac{-x^2}{2}$ for all x in 0..pi/2 ( i have done this part)

Substitute y = x√n to get $\sqrt n\int_0^{\pi/2}\cos^nx\,dx = \int_0^{\pi\sqrt n/2}\cos^n(y/\sqrt n)\,dy$ .

Use good ol' l'Hôpital to check that $\lim_{x\to0}\frac{\ln(\cos x)}{x^2} = \frac12$ . Put $x=y/\sqrt n$ in that, to see that $\lim_{n\to\infty}n\ln\bigl(\cos(y/\sqrt n)\bigr) = -\frac{y^2}2$ . Taking exponentials, $\lim_{n\to\infty}\cos^n(y/\sqrt n) = e^{-y^2/2}$ . Also, the inequality $\ln(\cos x)\leqslant-x^2/2$ shows that $\cos^n(y/\sqrt n) \leqslant e^{-y^2/2}$ . You can then apply the Dominated Convergence Theorem to get $\lim_{n\to\infty} \int_0^{\pi\sqrt n/2}\cos^n(y/\sqrt n)\,dy = \int_0^\infty e^{-y^2/2}dy$ .