Evaluate $\int_0^1···\int_0^1\frac{dx_n...x_1}{(1+x_1+x_2+ · · ·+x_n)^n}$ using Fubini's Theorem 
Evaluate using Fubini's Theorem
$\displaystyle\int_{[0,1]^n}\frac{1}{(1+x_1+x_2+ · · ·+x_n)^n}d(x_1,...,x_n)$

My idea:
Fubini’s Theorem $n$ times, to write it as $n$ interated integrals over $[0, 1]$:
$\displaystyle\int_0^1···\int_0^1\frac{1}{(1+x_1+x_2+ · · ·+x_n)^n}dx_n...x_1$
$\displaystyle\int_0^1\frac{1}{1+x_1}\int_0^1\frac{1}{(1+x_1+x_2)^2}\int_0^1···\int_0^1\frac{1}{(1+x_1+x_2+ · · ·+x_n)^n}dx_n...x_1$
And then I'm not sure where to go from there or if I'm on the right lines in the first place. Any help?
 A: If $X_1,X_2,\ldots,X_n$ are independent random variables, uniformly distributed over $[0,1]$, and $S=X_1+X_2+\ldots+X_n$, your integral equals the average value of $\frac{1}{(1+S)^n}$. By the law of large numbers it is expected to be pretty small, close to $\frac{1}{\left(1+\frac{n}{2}\right)^n}$.
Its explicit value can be simply computed by integrating with respect to $x_1$ first, then $x_2$ and so on. For instance, if $n=3$
$$\begin{eqnarray*} \int_{[0,1]^3}\frac{d\mu}{(1+x_1+x_2+x_3)^3} &=& \frac{1}{2}\int_{[0,1]^2}\left[\frac{1}{(1+x_2+x_3)^2}-\frac{1}{(2+x_2+x_3)^2}\right]\,d\mu\\&=&\int_{0}^{1}\frac{dx_3}{(1+x_3)(2+x_3)(3+x_3)} \end{eqnarray*}$$
and the general case boils down to integrating
$$ \int_{0}^{1}\frac{dx_n}{(1+x_n)(2+x_n)\cdots(n+x_n)} $$
By partial fraction decomposition we have
$$ \frac{1}{(1+x_n)(2+x_n)\cdots(n+x_n)} = \frac{1}{(n-1)!}\sum_{k=0}^{n-1}\frac{(-1)^k \binom{n-1}{k}}{x_n+k+1} $$
hence the explicit value of the integral is

$$ \int_{[0,1]^n}\frac{d\mu}{(1+x_1+\ldots+x_n)^n}=\frac{1}{(n-1)!}\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}\log\left(\frac{k+2}{k+1}\right).$$

By exploiting Frullani's integral (leading to an integral representation for $\log\frac{k+2}{k+1}$) and the binomial theorem we also get the identity

$$\begin{eqnarray*} \int_{[0,1]^n}\frac{d\mu}{(1+x_1+\ldots+x_n)^n}&=&\frac{1}{(n-1)!}\int_{0}^{+\infty}\frac{(1-e^{-x})^n}{xe^x}\,dx\\ &=& \frac{1}{(n-1)!}\int_{0}^{1}\frac{t^n}{-\log(1-t)}\,dt.\end{eqnarray*}$$

from which it follows that
$$ \int_{[0,1]^n}\frac{d\mu}{(1+x_1+\ldots+x_n)^n} \sim \frac{1}{n!\log(n)}. $$
