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I have to compute the limit $\lim_{n\to +\infty}I_n$, where: $$\qquad I_n=\int_{[0,1]^n}\sqrt{\frac{1}{n}\sum_{i=1}^n x_i^2}\,d\mu.$$ I believe that its value is just $\frac{1}{\sqrt{3}}$, since the mean value of $x_i^2$ over the unit hypercube is $\frac{1}{3}$. Numerical experiments agree with this conjecture.

Moreover, since the square root is a concave function, Jensen's inequality gives $$ I_n \leq \frac{1}{\sqrt{3}}$$ and $\{I_n\}_{n\in\mathbb{N}}$ looks to be a monotonic sequence, but an extra insight is needed to prove the conjecture, maybe a sort of converse of Jensen's inequality or a clever application of Fubini's theorem.

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There's one tricky way to calculate such a limit: use Strong Law of Large Numbers

Let $X_1, X_2, . . . , X_n$ be independent random variables, each having a uniform distribution over $[0,1]$. You have:

$$\qquad I_n=\int_{[0,1]^n}\sqrt{\frac{1}{n}\sum_{i=1}^n x_i^2}\,d\mu=E\left[\sqrt{\frac{1}{n}\sum_{i=1}^{n}}X_{i}^2\right]$$

From Strong Law of Large Numbers:

$$\sqrt{\frac{1}{n}\sum_{i=1}^{n}X_{i}^2} \to \sqrt{E[X_1^{2}]}=\frac{1}{\sqrt{3}}$$

Now because $0\leq \sqrt{\frac{1}{n}\sum_{i=1}^n x_i^2}\leq 1$ from Dominated convergence theorem you have that the limit is $\frac{1}{\sqrt{3}}$.

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Clever solution, but I would like a more elementary one, with some explicit bound on $\frac{1}{\sqrt{3}}-I_n$. Nonetheless, I will accept this answer if nothing different come out. – Jack D'Aurizio Jul 6 '14 at 21:28
@JackD'Aurizio +1 Ha ha, I was going to say the opposite. This is not a tricky solution, it is the obvious approach, at least to a probabilist. – Byron Schmuland Jul 6 '14 at 21:31
@math110 : I bet you were interested, too. – Jack D'Aurizio Jul 6 '14 at 22:14
@ByronSchmuland: I think that even Mathematics is a matter of taste, sometimes :) – Jack D'Aurizio Jul 6 '14 at 22:34
up vote 3 down vote accepted gives a way to bound the difference between $\frac{1}{\sqrt{3}}$ and the given integral. If we take $X_1,\ldots,X_n$ as indipendent random variables with a uniform distribution over $[0,1]$, then $X_1^2,\ldots,X_n^2$ are bounded indipendent random variables with density function $$ f(x)=\frac{1}{2\sqrt{x}}\cdot\mathbb{1}_{(0,1]}$$ and mean value $\frac{1}{3}$. If we set $\bar{X}=\frac{1}{n}\sum_{j=1}^{n}X_j^2$, the Hoeffding's inequality gives: $$ \mathbb{P}\left[\left|\bar{X}-\mathbb{E}[\bar{X}]\right|\geq t\right]\leq 2\exp\left(-2t^2 n\right),$$ hence by choosing $t=\sqrt{\frac{\log n}{4n}}$ we have: $$ \mu\left(\left\{\bar{x}\in[0,1]^n : \left|-\frac{1}{3}+\frac{1}{n}\sum_{i=1}^{n}x_i^2\right|\geq\sqrt{\frac{\log n}{4n}}\right\}\right)\leq\frac{2}{\sqrt{n}},$$ so: $$\int_{[0,1]^n}\sqrt{\frac{1}{n}\sum_{i=1}^{n}x_i^2}d\mu\geq\left(1-\frac{2}{\sqrt{n}}\right)\sqrt{\frac{1}{3}-\sqrt{\frac{\log n}{4n}}}$$ and $$\int_{[0,1]^n}\sqrt{\frac{1}{n}\sum_{i=1}^{n}x_i^2}d\mu\geq\frac{1}{\sqrt{3}}\left(1-\sqrt{\frac{\log n}{n}}\right)$$ for any $n\geq e^{64}.$

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So we have a probabilistic argument (a concentration inequality) through which is possible to estimate the convergence rate, too. – Jack D'Aurizio Jul 9 '14 at 7:36

The evaluation of limit is essentially same as here.

We have, $\displaystyle \lim\limits_{n\to\infty}\int_{0}^{1}\cdots\int_{0}^{1} \left(\dfrac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{n}\right)dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{3}$

We can show $\displaystyle \lim\limits_{n\to\infty}\int_{0}^{1}\cdots\int_{0}^{1} \left(\dfrac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{n}\right)^k \,dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{3^k}$, in the same way as the linked answer.

Thus, $\displaystyle \lim\limits_{n\to\infty}\int_{0}^{1}\cdots\int_{0}^{1} \left(\dfrac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{n}\right)^{1/2}dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{\sqrt{3}}$ follows from Weierstrass-Approximation of the function $f(x) = \sqrt{x}$, $x \in [0,1]$.

I'm not aware of the Asymptotics of these multiple-integral limits.

Edit: Set $E_n(\delta) := \left\{ (x_1,x_2,\cdots,x_n) \in [0,1]^n : \left| \dfrac{\sum_{i=1}^n x_i^2}{n} - \dfrac{1}{3} \right| > \delta \right\}$

Then, $\displaystyle \frac{1}{\sqrt{3}}-I_n \le \int_{[0,1]^n} \left|\sqrt{\frac{\sum\limits_{i=1}^n x_{i}^2}{n}} - \frac{1}{\sqrt{3}} \right|\,dx < \sqrt{3}\int_{[0,1]^n} \left|\frac{\sum\limits_{i=1}^n x_{i}^2}{n} - \frac{1}{3} \right|\,dx $

$$ < \sqrt{3}\left(\int_{E_n(\delta)}\,dx + \int_{[0,1]^n\setminus E_n(\delta)}\delta\,dx \right)$$

$$ < \sqrt{3}\delta + \sqrt{3}\int_{E_n(\delta)}\,dx$$

Since, by Chebychev's Inequality: $\displaystyle \int_{E_n(\delta)}\,dx \le \frac{1}{n^2\delta^2} \int_{[0,1]^n}\left(\sum\limits_{i=1}^n \left(x_i^2-\frac{1}{3}\right)\right)^2\,dx = \frac{1}{n^2\delta^2} \int_{[0,1]^n}\sum\limits_{i=1}^n \left(x_i^2-\frac{1}{3}\right)^2\,dx = O(1/n)$

$\implies \displaystyle \frac{1}{\sqrt{3}} - I_n < \sqrt{3}\delta + O(1/n) $

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You gave me a great idea, anyway. If we define $P_n(x)$ as the rescaled Legendre polynomial $\frac{1}{n!}\frac{d^n}{dx^x}\left((x(1-x))^n\right)$ we have: $$\sqrt{x}=\frac{2}{3}-\sum_{k=1}^{+\infty}\frac{2\,P_k(x)}{(2k-1)(2k+3)}$$ uniformly over $[0,1]$ with a very fast convergence, hence we can bring all the calculations through this approximation. – Jack D'Aurizio Jul 8 '14 at 19:42

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