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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...

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3 Answers 3

up vote 3 down vote accepted

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, $$ u_n\sim\frac1{(n+2)!}\left(\frac\pi2\right)^{n+2}. $$

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Thank you very much for this nice answer did! I think the difficult point is still the same: how did you find out this suitable substitution? –  Chon Oct 28 '12 at 11:58
    
As always, the idea comes from the Laplace method. –  Did Oct 28 '12 at 12:06
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This is not a complete answer, but an improved inequality. From $$ \arcsin x\le \frac{\pi}{2}\,x $$ you get $$ u_n\le\frac{1}{(n+1)!}\Bigr(\frac{\pi}{2}\Bigl)^n. $$

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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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$\cos(t) \sim_{t \rightarrow \pi/2^{-}} \frac{\pi}{2}-t$. Integrating the equivalence relation, I get: $ \int_x^{\pi/2} t^n\cos(t) \mathrm dt \sim_{x \rightarrow \pi/2^{-}} \int_x^{\pi/2} t^n \left(\frac{\pi}{2}-t \right) \mathrm dt$... Could you explain how you get: $ u_{n} \sim_{n \rightarrow \infty} \frac{1}{n!}\int_0^{\pi/2} t^n (\frac{\pi}{2} - t) \mathrm dt$ from $\cos(t) \sim_{t \rightarrow \pi/2^{-}} \frac{\pi}{2}-t$ please ? –  Chon Oct 28 '12 at 13:16
    
@Chon I've made it a bit more explicit. –  WimC Oct 28 '12 at 14:11
    
OK thanks! My last question is: How do you prove: $ \sum_{m=0}^{\infty} f(m,n) \sim_{n\rightarrow \infty}f(0,n)$ with $ f(m,n)=\frac{(-1)^m}{(n+2m+2)!}\left(\frac{\pi}{2}\right)^{n + 2m + 2}$? The condition $\forall m\geq1: f(m,n)=o(f(0,n))$ is not sufficient so what is a correct proof of this equivalence relation? –  Chon Oct 28 '12 at 15:00
    
@chon It is an alternating series, so it suffices to consider only the first two partial sums. These will give upper and lower bounds for $u_n$ that are tight enough. –  WimC Oct 28 '12 at 15:03
    
I just needed to use the alternating series test, thanks! –  Chon Oct 28 '12 at 20:23
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