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This is from a practice prelim exam and I know I should be able to get this one. $$ \lim_{n\to\infty} n^{1/2}\int_0^\infty \left( \frac{2x}{1+x^2} \right)^n $$

I have tried many different $u-$substitions but to no avail. I have tried $$ u = \log(1+x^2) $$ $$ du = \frac{2x}{1+x^2}dx $$ but did not get anywhere

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How do you take into account the $2x$ raised to the $n$. –  Eager Student May 7 '13 at 0:53
    
My answer avoids doing a substitution for the full problem, but generally the trick is trigonometric jobs when you have this sort of expression, with a complicated denominator with a large power. –  Sharkos May 7 '13 at 1:22
    
What exam is this? –  Mhenni Benghorbal May 7 '13 at 5:10
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4 Answers

up vote 2 down vote accepted

A related problem. Use the change of variables $ \frac{1}{1+x^2}=t $, then we have

$$ n^{1/2}\int_0^\infty \left( \frac{2x}{1+x^2} \right)^ndx = \sqrt{n}2^{n-1} \int_0^1 t^{\frac{n}{2}-\frac{3}{2}}(1-t)^{\frac{n}{2}-\frac{1}{2}}dt$$

$$=\sqrt{n}2^{n-1}\beta\left(\frac{n}{2}-\frac{1}{2},\frac{n}{2}+\frac{1}{2}\right)=\sqrt{n} 2^{n-1}\frac{\Gamma(\frac{n}{2}-\frac{1}{2})\Gamma(\frac{n}{2}+\frac{1}{2})}{\Gamma(n)}=I(n), $$

where $\beta$ is the $\beta$ function. Taking the limit of $I(n)$, we have

$$ \lim_{n\to \infty} I(n)=\sqrt{2\pi}. $$

You can use the Stirling' approximation $n!=\Gamma(n+1) \sim \left(\frac{n}{e}\right)^n\sqrt{2 \pi n} $ of the gamma function to evaluate the limit

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This was the trick I was looking for! Thanks a lot! –  Eager Student May 7 '13 at 3:24
    
@EagerStudent: You are very welcome. –  Mhenni Benghorbal May 7 '13 at 4:22
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Let $x = \tan(t)$. We then get that \begin{align} I(n) & = \int_0^{\infty} \left(\dfrac{2x}{1+x^2} \right)^n dx = \int_0^{\pi/2} \sin^n(2t) \sec^2(t) dt = 2^n \int_0^{\pi/2} \sin^n(t) \cos^{n-2}(t) dt\\ & = 4\int_0^{\pi/2}\sin^2(t) \sin^{n-2}(2t)dt \tag{$\star$} \end{align} Replacing $t$ by $\pi/2-t$, we get $$I(n) = 4 \int_0^{\pi/2} \cos^2(t) \sin^{n-2}(2t) dt \tag{$\perp$}$$ Adding $(\star)$ and $(\perp)$, we get that $$2I(n) = 4 \int_0^{\pi/2} \sin^{n-2}(2t)dt \implies I(n) = 2 \int_0^{\pi/2} \sin^{n-2}(2t)dt$$ I trust you can take it from here, using this post, which evaluates $\displaystyle \int_0^{\pi} \sin^{k}(t) dt$ and using Stirling (or) Wallis formula.

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How do you go from $cos^2(t)\sin^{n-2}(2t)$ to $\sin^{n-2}(2t)$? –  TCL May 7 '13 at 1:17
    
@TCL Updated the answer. Add $(\star)$ and $(\perp)$ to get it. –  user17762 May 7 '13 at 1:19
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Probably not how you want this done, but a useful and elegant method nonetheless:

Note that $$f(x)=\frac{2x}{1+x^2}$$ has the property that $f(x)<1$ except at $f(1)=1$. Since exponentiation to the power of $n\to\infty$ kills off anything less than $1$, we expect that the dominant contribution is near $1$.

$f^n \equiv \exp(n \log f)$ and $n\log(f(1+u))=-n\frac{1}{2}u^2+n\mathcal O(u^3)$. Then letting $v=\sqrt{n} u$, $$\int_0^\infty n^{1/2}\mathrm d x \exp(-\frac{1}{2}v^2+\mathcal O(n^{-1/2})) \approx \int_{-\infty}^\infty \mathrm d v\, e^{-v^2/2} = \sqrt{2\pi}$$

Verification: WolframAlpha.


Edit: This is called Laplace's method amongst other things, and is a core part of the general theory of asymptotics of integrals. (It's also how Stirling's approximation is most easily derived! Here one applies the method to the $\Gamma$ function.) It can be tightened up and made rigorous straightforwardly by estimating errors. You can also derive an asymptotic expansion by considering the correction terms.

One useful insight it offers is why the $\sqrt{n}$ is really there: it's because the peak contributing the dominant term has a width decaying as $1/\sqrt{n}$.

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Very nice. +1.$ $ –  Potato May 7 '13 at 1:21
    
Thanks! It's too straightforward not to share, really - most other methods are secretly using this in the background (in an appeal to the asymptotics of some non-trivial function like $\Gamma$ or $\beta$) so not understanding why it all works is a shame. –  Sharkos May 7 '13 at 1:32
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$$\lim_{n\to\infty} n^{1/2}\int_0^\infty \left( \frac{2x}{1+x^2} \right)^n dx = {1\over\sqrt{n}}\lim_{n\to \infty} \int_0^{\pi/2} \left(2\tan(x)\over \sec^2(x)\right)^n{ \sec^2(x)\,dx}$$ Can you use this?

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