Tristian Velazquez

2022-06-21

Compute ${I}_{n}=\underset{n\to \mathrm{\infty }}{lim}\sqrt{n}{\int }_{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{n}xdx$

trajeronls

You may show that the main contribution comes from the neighbourhood of $\frac{\pi }{2}$. Then, by dominated convergence,
$\begin{array}{rl}\sqrt{n}{\int }_{\pi /2-\epsilon }^{\pi /2}{\mathrm{sin}}^{n}x\phantom{\rule{thinmathspace}{0ex}}dx& \approx \sqrt{n}{\int }_{\pi /2-\epsilon }^{\pi /2}{\left(1-\frac{1}{2}{\left(\frac{\pi }{2}-x\right)}^{2}\right)}^{n}dx=\sqrt{n}{\int }_{0}^{\epsilon }{\left(1-\frac{1}{2}{t}^{2}\right)}^{n}dt\\ & ={\int }_{0}^{\sqrt{n}\epsilon }{\left(1-\frac{{s}^{2}}{2n}\right)}^{n}ds\to {\int }_{0}^{+\mathrm{\infty }}{e}^{-{s}^{2}/2}ds=\sqrt{\frac{\pi }{2}}.\end{array}$
I let you make this argument rigorous. You may study the Laplace method.

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