$\lim_{n\rightarrow\infty} n \int_{0}^{1} \frac{{x}^{n-2}}{{x}^{2n}+x^n+1} \mbox {d}x$

Evaluating

$$L = \lim_{n\rightarrow\infty} n \int_{0}^{1} \frac{{x}^{n-2}}{{x}^{2n}+x^n+1} \mbox {d}x$$

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A general strategy here is to find some simpler expression which gives a lower bound and some simpler expression which gives an upper bound and hope that in the limit they become equal. – Qiaochu Yuan Jun 7 '12 at 20:39
I think you should add "super-duper mega awsome" to the title. – Michael Greinecker Jun 7 '12 at 21:04

Marvis showed that $$I_n = \int_0^1 \dfrac{nx^{n-2}}{x^{2n} + x^n + 1} dx = \int_0^1 \dfrac{dt}{t^{1/n}(t^2 + t + 1)}.$$ At this point one can use monotone convergence as Davide commented, but since Chris is interested in a more direct approach, here is one.

We have $$I_n \geq \int_0^1 \dfrac{dt}{t^2 + t + 1}=\frac{\pi}{3\sqrt3},$$ and for $n>1$ and for any small $\delta>0$ $$I_n\leq \int_0^\delta \dfrac{dt}{t^{1/n}} + \int_\delta^1 \dfrac{dt}{\delta^{1/n}(t^2 + t + 1)}=\frac{\delta^{1-1/n}}{1-1/n}+\frac1{\delta^{1/n}}\Big(\frac{2\pi}{3\sqrt3}-\frac{2\arctan(\frac{2\delta+1}{\sqrt3})}{\sqrt3}\Big).$$ An inspection of the limit as $n\to\infty$ of the right hand side shows that $$I_n\leq \delta+\frac{2\pi}{3\sqrt3}-\frac{2\arctan(\frac{2\delta+1}{\sqrt3})}{\sqrt3}+E(\delta,n),$$ where $E(\delta,n)\to0$ as $n\to\infty$ for any fixed $\delta>0$. Since $\arctan(z)$ is continuous at $z=\frac1{\sqrt3}$ with $\arctan(\frac1{\sqrt3})=\frac\pi6$, for any given $\varepsilon>0$, we can choose $\delta>0$ so small that $$\delta+\frac{2\pi}{3\sqrt3}-\frac{2\arctan(\frac{2\delta+1}{\sqrt3})}{\sqrt3}<\frac{\pi}{3\sqrt3}+\frac\varepsilon2.$$ For sufficiently large $n$, we can also ensure $E(\delta,n)<\frac\varepsilon2$, implying that there is a threshold value $N$ such that $$\frac{\pi}{3\sqrt3}\leq I_n<\frac{\pi}{3\sqrt3}+\varepsilon,$$ for all $n>N$.

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nice proof. Thanks! – Chris's sis the artist Jun 7 '12 at 21:59

Let $x^n = t$. Then $n x^{n-1} dx = dt \implies nx^{n-2} dx = \dfrac{dt}x = \dfrac{dt}{t^{1/n}}$.

$$I_n = \int_0^1 \dfrac{nx^{n-2}}{x^{2n} + x^n + 1} dx = \int_0^1 \dfrac{dt}{t^{1/n}(t^2 + t + 1)}$$ Hence, $$\lim_{n \rightarrow \infty} I_n = \int_0^1 \dfrac{dt}{t^2 + t + 1} = \dfrac{\pi}{3 \sqrt{3}}$$

But can I change the limit and integral?...

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@Chris But I still need to argue why I can change the limit and integral and at the moment I don't have an argument. – user17762 Jun 7 '12 at 20:42
@ZevChonoles I might be missing something trivial, but why is the denominator ($t^{1/n}(t^2 + t + 1)$) in the integral $\geq 1$? For $t$ sufficiently close to $0$, the value is less than $1$. – user17762 Jun 7 '12 at 20:47
@Marvis: I was originally writing my comment about the integral $$\int_0^1\frac{nx^{n-2}}{x^{2n}+x^n+1}dx$$ until I realized that's not what we're taking the limit of. The DCT would work for the above, because the numerator is always $\leq n$ and the denominator is always $\geq 1$. Unfortunately I don't think there's any way of taking the limit before the change of variables (?) – Zev Chonoles Jun 7 '12 at 20:49
Isn't there monotone convergence? – Davide Giraudo Jun 7 '12 at 20:51
I don't see how this is very simple, – Gigili Jun 7 '12 at 21:42