TITLE: Find $\underbrace{\int_0^1 \cdots \int_0^1}_{a+b \text{ times}} \frac{dx_1 \,dx_2 \cdots dx_{a+b}}{(1+x_1 x_2 \cdots x_a)(1+x_1 x_2 \cdots x_{a+b})}$
QUESTION [2 upvotes]: Can we find a closed form or reduce or even evaluate the below weird integrals?
$$\underbrace{\int_0^1 \cdots \int_0^1}_{a+b \text{ times}} \dfrac{dx_1 \, dx_2 \cdots dx_{a+b}}{(1+x_1 x_2 \cdots x_a)(1+x_1 x_2 \cdots x_{a+b})}$$
I have tried to evaluate the above expressions for some very small values of $a$ and $b$, and I got it in form of logarithms inside a double integral and stuff. However I have no idea what it would be in general. Your insight would be very useful, Thanks.

REPLY [6 votes]: Reduction to double integrals with logarithms is based on $$\idotsint\limits_{[0,1]^n}f(x_1\cdots x_n)\,dx_1\cdots dx_n=\frac1{(n-1)!}\int_0^1(-\log x)^{n-1}f(x)\,dx$$ for any $n>0$ and "good enough" $f$. Then, say, for $a,b>0$
$$
\idotsint\limits_{[0,1]^{a+b}}\frac{dx_1\cdots dx_{a+b}}{\big(1-\prod_{k=1}^a x_k\big)\big(1-\prod_{k=1}^{a+b}x_k\big)}
\\=\frac1{(a-1)!(b-1)!}\int_0^1\int_0^1\frac{(-\log x)^{a-1}(-\log y)^{b-1}}{(1-x)(1-xy)}\,dx\,dy
\\=\frac1{(a-1)!(b-1)!}\sum_{n,m\geqslant 0}\int_0^1\int_0^1 x^n(xy)^m(-\log x)^{a-1}(-\log y)^{b-1}\,dx\,dy
\\=\sum_{n,m\geqslant 0}(n+m+1)^{-a}(m+1)^{-b}=\zeta(a+b)+\zeta(a,b)
$$
expresses in terms of multiple zeta function. Similarly, the $\pm$ variants use alternating zetas.
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