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I would like to find a simple equivalent of:

$ u_{n}=\frac{1}{n!}\int_0^1 (\arcsin x)^n \mathrm dx $

We have:

$ 0\leq u_{n}\leq \frac{1}{n!}\left(\frac{\pi}{2}\right)^n \rightarrow0$

So $ u_{n} \rightarrow 0$

Clearly:

$ u_{n} \sim \frac{1}{n!} \int_{\sin(1)}^1 (\arcsin x)^n \mathrm dx $

But is there a simpler equivalent for $u_{n}$?

Using integration by part:

$ \int_0^1 (\arcsin x)^n \mathrm dx = \left(\frac{\pi}{2}\right)^n - n\int_0^1 \frac{x(\arcsin x)^{n-1}}{\sqrt{1-x^2}} \mathrm dx$

But the relation

$ u_{n} \sim \frac{1}{n!} \left(\frac{\pi}{2}\right)^n$

seems to be wrong...

3 Answers 3

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The change of variable $x=\cos\left(\frac{\pi s}{2n}\right)$ yields $ u_n=\frac1{n!}\left(\frac\pi2\right)^{n+2}\frac1{n^2}v_n, $ with $ v_n=\int_0^n\left(1-\frac{s}n\right)^n\,\frac{2n}\pi \sin\left(\frac{\pi s}{2n}\right)\,\mathrm ds. $ When $n\to\infty$, $\left(1-\frac{s}n\right)^n\mathbf 1_{0\leqslant s\leqslant n}\to\mathrm e^{-s}$ and $\frac{2n}\pi \sin\left(\frac{\pi s}{2n}\right)\mathbf 1_{0\leqslant s\leqslant n}\to s$. Both convergences are monotonic hence $v_n\to\int\limits_0^\infty\mathrm e^{-s}\,s\,\mathrm ds=1$. Finally, $ u_n\sim\frac1{(n+2)!}\left(\frac\pi2\right)^{n+2}. $

  • 0
    As always, the idea comes from the [Laplace method](http://en.wikipedia.org/wiki/Laplace%27s_method#The_idea_of_Laplace.27s_method).2012-10-28
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This is not a complete answer, but an improved inequality. From $ \arcsin x\le \frac{\pi}{2}\,x $ you get $ u_n\le\frac{1}{(n+1)!}\Bigr(\frac{\pi}{2}\Bigl)^n. $

1

Substitute $x = \sin(t)$ to get

$ u_n = \frac{1}{n!}\int_0^{\pi/2} t^n \cos(t) dt. $

The following equalities hold for all integers $m \geq 0$ (which can be checked by partial integration):

$ \frac{1}{n!}\int_0^{\pi/2} t^n \left(\tfrac{\pi}{2} - t \right)^m dt = \frac{m!}{(n + m + 1)!}\left(\frac{\pi}{2}\right)^{n + m + 1}. $

Then it follows from the Taylor expansion of $\cos(t)$ around $t = \pi/2$

$ \cos(t) = \sin \left(\tfrac{\pi}{2} - t \right) = \sum_{m=0}^{\infty}\frac{(-1)^m}{(2m + 1)!}\left(\tfrac{\pi}{2} - t\right)^{2m+1} $ that

$ u_n = \sum_{m=0}^{\infty} \frac{(-1)^m}{(n+2m+2)!}\left(\frac{\pi}{2}\right)^{n + 2m + 2}. $ The growth in $n$ is determined by the first term ($m=0$) so $ u_n \sim \frac{1}{(n+2)!}\left(\frac{\pi}{2}\right)^{n + 2}. $

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    I just needed to use the alternating series test, thanks!2012-10-28