Multivariable integral limitation proof Please prove the following formula.
$$
\lim_{n\to\infty}\int_0^1\cdots\int_0^1 \frac{n}{\sum_{i=1}^n x_i}dx_1\cdots dx_n = 2
$$
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$\ds{\lim_{n\to\infty}\int_{0}^{1}\cdots\int_{0}^{1}
     {n \over \sum_{i = 1}^{n}x_{i}}\dd x_{1}\ldots\dd x_{n} = 2}$

\begin{align}&\int_{0}^{1}\cdots\int_{0}^{1}
{n \over \sum_{i = 1}^{n}x_{i}}\dd x_{1}\ldots\dd x_{n}
=n\int_{0}^{1}\cdots\int_{0}^{1}\int_{0}^{\infty}\exp\pars{-t\sum_{i = 1}^{n}x_{i}}
\,\dd t\,\dd x_{1}\ldots\dd x_{n}
\\[3mm]&=n\int_{0}^{\infty}\pars{\int_{0}^{1}\expo{-tx}\,\dd x}^{n}\,\dd t
=n\int_{0}^{\infty}\pars{1 - \expo{-t} \over t}^{n}\,\dd t\end{align}

Can you take it from here ?.
It seems that the integrand behaves as $\ds{\expo{-nt/2}}$ when $\ds{n \gg 1}$ since $\ds{{1 - \expo{-t} \over t} \sim\pars{1 - {t \over2}}}$ when $\ds{t \sim 0}$.  The main contribution comes from $\ds{t \gtrsim 0}$. That integral and limit was posted recently.
A: Hint: Assuming that, as n grows larger and larger, the sequence $x_i$ becomes more equidistributed, then the sum in the denominator will also draw closer and closer to a certain well-known formula.
