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

Beginner2022-06-22Added 21 answers

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.

$\begin{array}{rl}\sqrt{n}{\int}_{\pi /2-\epsilon}^{\pi /2}{\mathrm{sin}}^{n}x\phantom{\rule{thinmathspace}{0ex}}dx& \approx \sqrt{n}{\int}_{\pi /2-\epsilon}^{\pi /2}{\left(1-\frac{1}{2}{\left(\frac{\pi}{2}-x\right)}^{2}\right)}^{n}dx=\sqrt{n}{\int}_{0}^{\epsilon}{\left(1-\frac{1}{2}{t}^{2}\right)}^{n}dt\\ & ={\int}_{0}^{\sqrt{n}\epsilon}{\left(1-\frac{{s}^{2}}{2n}\right)}^{n}ds\to {\int}_{0}^{+\mathrm{\infty}}{e}^{-{s}^{2}/2}ds=\sqrt{\frac{\pi}{2}}.\end{array}$

I let you make this argument rigorous. You may study the Laplace method.