What is the limit $$ \lambda =\lim\limits_{n \to \infty}{n\int_0^{\frac{\pi}{2}}(\sin x)^{2n} dx}$$ I would like to find it without Wallis' integral formula: I mean without evaluating the closed form. Is it possible? Thanks.
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$\begingroup$ I didn't look too deep, did you try the induction on $n$? I mean take $\sin^{2n} x = \sin x \sin^{2n-1}x$, integrate by parts, etc. $\endgroup$– AlexAug 27, 2014 at 13:54
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$\begingroup$ @Alex I thought about it but got stuck. Thanks. $\endgroup$– eventAug 27, 2014 at 13:59
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$\begingroup$ how did you do it? Define $I_n = n \int_{0}^{\frac{\pi}{2}} \sin^{2n}x dx$, do the IBP. You should see the pattern. $\endgroup$– AlexAug 27, 2014 at 14:02
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$\begingroup$ @Alex but that's reinventing the Wallis wheel, and I don't really understand why te OP wants to do without it, since it's so simple. $\endgroup$– Jean-Claude ArbautAug 27, 2014 at 14:02
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$\begingroup$ OK I admit I never heard of Wallis integral before. Induction+IBS seem like a logical choice for this problem. $\endgroup$– AlexAug 27, 2014 at 14:03
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Note that $$ \int_{0}^{\pi/2}(\sin x)^{2n}dx=\int_{0}^{\pi/2}(\cos x)^{2n}dx=\frac{1}{2}\int_{-\pi/2}^{\pi/2}e^{-n f(x)}dx, $$ where $f(x)=-2\log\cos x$. As $n\rightarrow \infty$, the integral is dominated by the region around $x=0$, since that is the unique minimum of $f(x)$. But for small $x$ we have $$ f(x)=-2\log\cos x \approx -2\log(1-x^2/2)\approx x^2. $$ So $$ \int_{0}^{\pi/2}(\sin x)^{2n}dx \;\sim\; \frac{1}{2}\int_{-\infty}^{\infty}e^{-nx^2}dx=\frac{\sqrt{\pi}}{2\sqrt{n}} $$ as $n\rightarrow \infty$.
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$\begingroup$ How can I not upvote the Laplace method? $\endgroup$ Aug 27, 2014 at 15:22
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