证明

\[
\lim_{n\rightarrow+\infty}\int_0^1\cdots\int_0^1\frac{n}{x_1+x_2+\cdots+x_n}dx_1 dx_2\cdots dx_n=2.
\]

[分析]

$n=1$ 时, 

\[
\int_0^1 \frac{1}{x_1}dx_1=\ln x\biggr|_{0}^{1}=+\infty.
\]

$n=2$ 时,

\[
\begin{split}
\int_0^1\int_0^1\frac{2}{x_1+x_2}dx_1 dx_2&=\int_0^1 \int_0^1\frac{2}{t+x}dt dx\\
&=\int_0^1\biggl[2\ln(t+x)\biggr|_{t=0}^{t=1}\biggr]dx\\
&=2\int_0^1[\ln(1+x)-\ln x]dx\\
&=2\int_0^1\ln(1+\frac{1}{x})dx\\
&=2\biggl[x\ln(1+\frac{1}{x})\biggr|_0^1-\int_0^1 xd\ln(1+\frac{1}{x})\biggr]\\
&=2\biggl[\ln 2-\int_0^1 x\cdot\frac{1}{1+\frac{1}{x}}\cdot\frac{-1}{x^2}dx\biggr]\\
&=2\biggl[\ln 2+\int_0^1\frac{1}{x+1}dx\biggr]\\
&=4\ln 2.
\end{split}
\]