$$I=\int_0^1\int_0^1\int_0^1\int_0^1\frac{\mathrm dx~\mathrm dy~\mathrm dz~\mathrm dw}{(wxy-1)(zwx-1)(yzw-1)(xyz-1)}=?$$
I think $I$ diverges because both $$\int_0^1\int_0^1\frac{\mathrm dx~\mathrm dy}{(x-1)(y-1)}$$ and $$\int_0^1\int_0^1\int_0^1\frac{\mathrm dx~\mathrm dy~\mathrm dz}{(xy-1)(zx-1)(yz-1)}$$ diverge as well (by Mathematica). However, Mathematica cannot do $\iiiint$..
Any ideas ?
By expanding each term of the $\frac{1}{1-efg}$ kind as a geometric series, then performing termwise integration, we get that the quadruple integral equals
$$ \int_{(0,1)^4}\sum_{a,b,c,d\geq 0}x^{b+c+d}y^{a+c+d}z^{a+b+d}w^{b+c+d}\,d\mu$$
$$=\sum_{a,b,c,d\geq 0}\frac{1}{(1+b+c+d)(1+a+c+d)(1+a+b+d)(1+b+c+d)} $$ but by the AM-GM inequality the last series is greater than
$$ \sum_{a,b,c,d\geq 0}\frac{1}{\left(1+\frac{3}{4}(a+b+c+d)\right)^4}=\sum_{m\geq 0}\frac{\binom{m+3}{3}}{\left(1+\frac{3}{4}m\right)^4} $$ which is divergent by comparison with the harmonic series.
The same argument works in arbitrary dimension.