TITLE: A multiple integral question
QUESTION [7 upvotes]: Proving that
$$\lim_{n\to\infty}\underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{n \text{ times}}\frac{1}{(x_1\cdot x_2\cdots x_n)^2+1} \mathrm{d}x_1\cdot\mathrm{d}x_2\cdots\mathrm{d}x_n=1$$

REPLY [12 votes]: Since
$$
\int\limits_{0}^{1}\ldots\int\limits_{0}^{1}dx_n\ldots dx_1=1
$$
it is enough to show that
$$
\lim\limits_{n\to\infty}\int\limits_{0}^{1}\ldots\int\limits_{0}^{1}\left(1-\frac{1}{(x_1\cdot\ldots\cdot x_n)^2+1}\right)dx_n\ldots dx_1=0
$$
which is equivalent to
$$
\lim\limits_{n\to\infty}\int\limits_{0}^{1}\ldots\int\limits_{0}^{1}\frac{(x_1\cdot\ldots\cdot x_n)^2}{(x_1\cdot\ldots\cdot x_n)^2+1}dx_n\ldots dx_1=0
$$
Now note that
$$
\begin{align}
0&\leq \lim\limits_{n\to\infty}\int\limits_{0}^{1}\ldots\int\limits_{0}^{1}\frac{(x_1\cdot\ldots\cdot x_n)^2}{(x_1\cdot\ldots\cdot x_n)^2+1}dx_n\ldots dx_1\\
&\leq \lim\limits_{n\to\infty}\int\limits_{0}^{1}\ldots\int\limits_{0}^{1}(x_1\cdot\ldots\cdot x_n)^2dx_n\ldots dx_1\\
&=\lim\limits_{n\to\infty}\left(\frac{1}{3}\right)^n=0
\end{align}
$$
And the result follows.