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I would like to find an equivalent of:

$$ u_n = \int_0^1 (\arctan x)^n \mathrm dx$$

which might be: $$ u_n \sim \frac{\pi}{2n} \left(\frac{\pi}{4} \right)^n$$

$$ 0\le u_n\le \left( \frac{\pi}{4} \right)^n$$

So $$ u_n \rightarrow 0$$

In order to get rid of $\arctan x$ I used the substitution $$x=\tan \left(\frac{\pi t}{4n} \right) $$

which gives:

$$ u_n= \left(\frac{\pi}{4n} \right)^{n+1} \int_0^n t^n\left(1+\tan\left(\frac{\pi t}{4n} \right)^2 \right) \mathrm dt$$

But this integral is not easier to study!

Or: $$ t=(\arctan x)^n $$

$$ u_n = \frac{1}{n} \int_0^{(\pi/4)^n} t^{1/n}(1+\tan( t^{1/n})^2 ) \mathrm dt $$

How could I deal with this one?

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4 Answers 4

up vote 2 down vote accepted

$$ \begin{align} &\int_0^1\arctan^n(x)\,\mathrm{d}x\\ &=\int_0^{\pi/4}x^n\sec^2(x)\,\mathrm{d}x\\ &=\int_0^{\pi/4}\left(\frac\pi4-x\right)^n\sec^2\left(\frac\pi4-x\right)\,\mathrm{d}x\\ &=\frac{(\pi/4)^{n+1}}{n}\int_0^ne^{-x}\color{#C00000}{e^x\left(1-\frac{x}{n}\right)^n}\color{#00A000}{\sec^2\left(\frac\pi4-\frac{\pi x}{4n}\right)}\,\mathrm{d}x\\ &=\frac{(\pi/4)^{n+1}}{n}\int_0^ne^{-x}\color{#C00000}{\small\left(1-\frac{x^2}{2n}-\frac{8x^3-3x^4}{24n^2}+O\left(\frac1{n^3}\right)\right)}\color{#00A000}{\small\left(2-\frac{\pi x}{n}+\frac{\pi^2x^2}{2n^2}+O\left(\frac1{n^3}\right)\right)}\,\mathrm{d}x\\ &=\frac{(\pi/4)^{n+1}}{n}{\small\left(2-\frac{\pi+2}{n}+\frac{2+3\pi+\pi^2}{n^2}+O\left(\frac1{n^3}\right)\right)} \end{align} $$

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Thank you very much robjohn for this much more precise asymptotic expansion! It took me some time to get your result! –  Chon Jan 18 '13 at 17:10
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To get a rough asymptotic answer, you can substitute $$x=1-\frac{y}{n}.$$ Then $$ \frac{u_n}{(1/n) (\pi/4)^n}=\int_0^n \left(\frac{\arctan (1-\frac{y}{n})}{\pi/4}\right)^n \, dy. $$ For fixed $y$, as $n\to\infty$,
$$ \frac{\arctan (1-\frac{y}{n})}{\pi/4}=1-\frac{2}{\pi n} y + O(\frac{1}{n^2})= \exp(-\frac{2y}{\pi n} + O(\frac{1}{n^2})), $$ so, pointwise, the integrand converges to $e^{-2y/\pi}$. Since we have the bound $$ \arctan (1-\frac{y}{n})\le \frac{\pi}{4} e^{-2y/\pi n}, \qquad 0\le y\le n, \ \ \ (*) $$ we can apply the dominated convergence theorem to conclude that $$ u_n \sim \frac{1}{n}(\frac{\pi}{4})^n \int_0^\infty e^{-2y/\pi} \, dy=\frac{\pi}{2n} (\frac{\pi}{4})^n. $$ Also, from the integral and (*), $$ 0\le u_n\le \frac{\pi}{2n}(\frac{\pi}{4})^n, \qquad \rm for\ all\ \it n. $$

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How did you get: $ \arctan (1-\frac{y}{n})\le \frac{\pi}{4} e^{-2y/\pi n} $? The RHS $e^{-2y/\pi n} $ is integrable but should not depend on $n$. –  Chon Jan 18 '13 at 18:15
    
The left- and right-hand sides are both $\pi/4$ at $y/n=0$. Then, take derivatives with respect to $y/n$. The derivative of the LHS is bounded above by $-1/2$, and the derivative of the RHS is bounded below by $-1/2$. –  David Moews Jan 18 '13 at 21:55
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From your simplification, $$ u_n = \left(\frac{\pi}{4n}\right)^{n+1} \int_0^n t^n \left(1 + \tan^2\left(\frac{\pi t}{4n}\right) \right) dt $$ note that the integrand is non-negative and in fact on $[0,n]$, $0 \leq \tan(\pi t/4n) \leq 1$. Applying this we get $$ \left(\frac{\pi}{4n}\right)^{n+1} \int_0^n t^n dt \leq u_n \leq \left(\frac{\pi}{4n}\right)^{n+1} \int_0^n 2t^n dt $$ which gives $$ \left(\frac{\pi}{4n}\right)^n \frac{\pi}{4(n+1)} \leq u_n \leq \left(\frac{\pi}{4n}\right)^n \frac{\pi}{2(n+1)} $$ Hope that helps.

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$\int_0^nt^n\,\mathrm{d}t=\frac{n^{n+1}}{n+1}$ –  robjohn Jan 18 '13 at 7:23
    
$ \left(\frac{\pi}{4}\right)^n \frac{\pi}{4(n+1)} \leq u_n \leq \left(\frac{\pi}{4}\right)^n \frac{\pi}{2(n+1)} $. $ \left(\frac{\pi}{4}\right)^n \frac{\pi}{2(n+1)} \sim \frac{\pi}{2n} \left(\frac{\pi}{4} \right)^n $ but $ \left(\frac{\pi}{4}\right)^n \frac{\pi}{4(n+1)} \sim \frac{\pi}{4n} \left(\frac{\pi}{4} \right)^n $ so we don't get the result. –  Chon Jan 18 '13 at 18:22
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Another (simpler) approach is to substitute $x = \tan{y}$ and get

$$u_n = \int_0^{\frac{\pi}{4}} dy \: y^n \, \sec^2{y}$$

Now we perform an analysis not unlike Laplace's Method: as $n \rightarrow \infty$, the contribution to the integral is dominated by the value of the integrand at $y=\pi/4$. We may then say that

$$u_n \sim \sec^2{\frac{\pi}{4}} \int_0^{\frac{\pi}{4}} dy \: y^n = \frac{2}{n+1} \left ( \frac{\pi}{4} \right )^{n+1} (n \rightarrow \infty) $$

The stated result follows.

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