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I am looking for the solution of this integral:

$$ \lim_{n \rightarrow \infty} \int_0^1\cdots\int_0^1 \sin(\sqrt[n]{x_1\cdots x_n}) \, dx_1\cdots dx_n $$

I know that $X_1,\ldots,X_n \sim \mathcal{U}[0,1]$. I tried to use law of large numbers and compute the expected value but it failed because I can't apply this law to product of random variables. Does anyone have any idea?

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  • $\begingroup$ What is the $x \to n$ doing? There is no $x$ on the right side, which might approach $n.$ $\endgroup$
    – coffeemath
    Aug 25, 2016 at 17:01
  • $\begingroup$ There have to be multiple typos here. This question makes no sense. $\endgroup$
    – Paul
    Aug 25, 2016 at 17:02
  • $\begingroup$ Sorry, its of course $n \rightarrow \infty.$ My mistake. $\endgroup$ Aug 25, 2016 at 17:17

1 Answer 1

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Note if $X_i\sim U(0,1)$ then $-\log X_i\sim Exp(1)$ iid thus $\dfrac{1}{n}\sum_{i=1}^n -\log(X_i)\to 1$ by SLLN.

Hence $\dfrac{1}{n}\sum_{i=1}^n \log(X_i)\to -1$ from which you can see that $(\prod_{i=1}^nX_i)^{1/n}\to e^{-1}$ a.s.

Now you want to find $\lim_{n\to\infty} E(\sin(\prod_{i=1}^n X_i)^{1/n})$ actually. Since $|\sin(t)|\leq 1$ for any $t$ you can use Bounded Convergence Theorem as follows: Since $\prod_{i=1}^n X_i^{1/n}\to e^{-1}$ a.s. we have $\sin(\prod_{i=1}^n X_i^{1/n})\to \sin (e^{-1})$ a.s. and therefore, $E(\sin(\prod_{i=1}^n X_i)^{1/n})\to E(\sin(e^{-1}))=\sin(e^{-1})$.

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