Compute I n = lim n → ∞ n ∫ 0 π 2 sin n ⁡ x d x

Question

Compute       I    n    =      lim          n      →      ∞            n        ∫          0                      π        2                  sin          n        ⁡  x  d  x

trajeronls · Accepted Answer

You may show that the main contribution comes from the neighbourhood of

\frac{π}{2}

. Then, by dominated convergence,

\begin{aligned} \sqrt{n} \int_{π / 2 - ε}^{π / 2} \sin^{n} x d x & \approx \sqrt{n} \int_{π / 2 - ε}^{π / 2} {(1 - \frac{1}{2} {(\frac{π}{2} - x)}^{2})}^{n} d x = \sqrt{n} \int_{0}^{ε} {(1 - \frac{1}{2} t^{2})}^{n} d t \\ = \int_{0}^{\sqrt{n} ε} {(1 - \frac{s^{2}}{2 n})}^{n} d s \to \int_{0}^{+ \infty} e^{- s^{2} / 2} d s = \sqrt{\frac{π}{2}} . \end{aligned}

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

Compute I n = lim n → ∞ n ∫ 0 π 2 sin n...

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