TITLE: Iterated integral question
QUESTION [31 upvotes]: Show $$\lim_{n \to\infty} \int_0^1 \cdots  \int_0^1  \int_0^1 \frac{ x_1^2 + \cdots +  x_n^2}{x_1 + \cdots + x_n} \, dx_1 \cdots dx_n = \frac 2 3.$$
Not sure how to start off this iterated integral question, any help would be appreciated.

REPLY [2 votes]: We follow the idea of @peter, and also prepare for an integration by parts. We try, however, to keep the argument as elementary as possible.
$$\begin{aligned}
I_n & =\int_{[0,1]^n}\frac{x^2+x_2^2+\cdots+x_n^2}{x+x_2+\cdots+x_n}\,dx\,dx_2\,\ldots\,dx_n\\
& = n\int_0^{+\infty}\int_0^{1}x^2e^{-\lambda x}\,dx \biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^{n-1}\,d\lambda\\
& = n\int_0^{+\infty}\biggl(\frac{2}{\lambda^3}-\frac{e^{-\lambda}(2+2\lambda+\lambda^2)}{\lambda^3}\biggr)\biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^{n-1}\,d\lambda\\
& = \int_0^{+\infty}\biggl(\frac{2}{\lambda^3}-\frac{e^{-\lambda}(2+2\lambda+\lambda^2)}{\lambda^3}\biggr)
\biggl(\frac{e^{-\lambda}}{\lambda}-\frac{1-e^{-\lambda}}{\lambda^2}\biggr)^{-1}
\frac{d}{d\lambda}\biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^n\,d\lambda\\
\end{aligned}
$$
It happens that
$$
\biggl(\frac{2}{\lambda^3}-\frac{e^{-\lambda}(2+2\lambda+\lambda^2)}{\lambda^3}\biggr)
\biggl(\frac{e^{-\lambda}}{\lambda}-\frac{1-e^{-\lambda}}{\lambda^2}\biggr)^{-1}
=\frac{\lambda}{e^\lambda-1-\lambda}-\frac{2}{\lambda}.
$$
Thus, integrating by parts,
$$
\begin{aligned}
I_n & =\biggl[\biggl(\frac{\lambda}{e^\lambda-1-\lambda}-\frac{2}{\lambda}\biggr)
\biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^n\biggr]_0^{+\infty}\\
& \quad-
\int_0^{+\infty}\biggl(\frac{2}{\lambda^2}-\frac{\lambda^2}{(e^\lambda-1-\lambda)^2}-\frac{\lambda-1}{e^{\lambda}-1-\lambda}\biggr)
\biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^n
\,d\lambda\\
& =\frac{2}{3}-
\int_0^{+\infty}\biggl(\frac{2}{\lambda^2}-\frac{\lambda^2}{(e^\lambda-1-\lambda)^2}-\frac{\lambda-1}{e^{\lambda}-1-\lambda}\biggr)
\biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^n
\,d\lambda.
\end{aligned}
$$
We need to show that the last integral tends to $0$ as $n\to+\infty$. 
The absolute value of the first factor is bounded by some constant $C$.
Indeed, close to $0$ a Maclaurin expansion shows that it is of size 
$1/18+O(\lambda)$. Moreover, it is continuous for $0<\lambda<+\infty$ and tends to $0$ 
as $\lambda\to +\infty$. For the other factor, one can use that
$$
0<\frac{1-e^{-\lambda}}{\lambda}\leq \frac{2}{2+\lambda}\quad\text{for $\lambda>0$.}
$$
Thus,
$$
\begin{aligned}
0 & < \biggl|\int_0^{+\infty}\biggl(\frac{2}{\lambda^2} -\frac{\lambda^2}{(e^\lambda-1-\lambda)^2}-\frac{\lambda-1}{e^{\lambda}-1-\lambda}\biggr)
\biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^n\,d\lambda\biggr|
\\
& \leq \int_0^{+\infty} M\biggl(\frac{2}{2+\lambda}\biggr)^n\,d\lambda =\frac{2M}{n-1}\to 0\quad\text{as $n\to+\infty$.}
\end{aligned}
$$
We conclude that $\lim_{n\to+\infty}I_n=2/3$. Of course, Peter will receive the bounty.