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How can I find an approximation of $$ n^a \int_{0}^{\pi/n}\sin^b(t)dt $$ when $ n\rightarrow \infty$, $(a,b>0)$ ?

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Thank you very much! – Chon Oct 25 '11 at 12:59

1 Answer 1

up vote 2 down vote accepted

Since $\sin(t)\sim t$ when $t\to0$, the integral is equivalent to $$ \left[(b+1)^{-1}t^{b+1}\right]_0^{\pi/n}=(b+1)^{-1}(\pi/n)^{b+1}, $$ and the equivalent you are looking for is $\pi^{b+1}(b+1)^{-1}n^{a-b-1}$.

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