The integral $\int_0^1 dx_1 \cdots dx_n \frac{1}{[(x_1+ \cdots +x_n)(1-x_1- \cdots -x_n)]^\alpha},$ Does anyone see a way to compute the integral 
$$\int_0^1 dx_1 \cdots dx_n \frac{1}{[(x_1+ \cdots +x_n)(1-x_1- \cdots -x_n)]^\alpha},$$
where $n$ is a positive integer, and $\alpha$ is real/ The limits to all variables are from $0$ to $1$. 
I have tried writing $x_1+A$ (grouping $A=x_2 + \cdots x_n$) and performing the $x_1$ integral, but I can't seem to find a closed form solution. 
EDIT: 
I have tried utilizing Beta functions. For instance, for $n=1$ the integral is equal to 
$$B(1-\alpha,1-\alpha).$$
One can go on, but I get terms like $B(n,\alpha_1,\alpha_2)$ (this is the incomplete Beta function). I have also asked this question, about whether I can relate $B(n,\alpha_1,\alpha_2)$ to $$B(1,\alpha_1,\alpha_2)$$.

I'm not completely sure what the integration domain is. If it is really supposed that all variables are ranging from $0$ to $1$, then $1-x_1-\ldots-x_n$ gets negative when all of $x_1,\ldots,x_n$ are close to $1$ (provided $n>1$), and complex numbers enter the picture. I suspect that the integration domain is really the simplex $$0<x_1,\ldots,x_n,x_1+\ldots+x_n<1.$$
If that suspicion is right, then our integral $$I=\int\limits_{\substack{0<x_1,\ldots,x_n\\x_1+\ldots+x_n<1}}\frac{dx_1\cdots dx_n}{[(x_1+\ldots+x_n)(1-x_1-\ldots-x_n)]^\alpha}$$ can be evaluated using the substitution $y_k=x_1+\ldots+x_k$ ($1\leqslant k\leqslant n$); the Jacobian is $1$, thus $$I=\int\limits_{0<y_1<\ldots<y_n<1}\frac{dy_1\cdots dy_n}{[y_n(1-y_n)]^\alpha}=\frac{1}{(n-1)!}\int_0^1\frac{y^{n-1}\,dy}{[y(1-y)]^\alpha}=\frac{\mathrm{B}(n-\alpha,1-\alpha)}{(n-1)!}$$ after "inner" integration over $y_1,\ldots,y_{n-1}$ under $0<y_1<\ldots<y_{n-1}<y_n:=y$.
