# Find $\underbrace{\int_0^1 \cdots \int_0^1}_{a+b \text{ times}} \frac{dx_1 \,dx_2 \cdots dx_{a+b}}{(1+x_1 x_2 \cdots x_a)(1+x_1 x_2 \cdots x_{a+b})}$

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.

• Please don't delete questions after someone has gone to the effort of answering.
– robjohn
Commented Apr 27, 2021 at 7:01

## 1 Answer

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.

• What about the first integral? first equation of my question Commented Apr 27, 2021 at 6:11
• @BooleanWick: See the last sentence. (I can't say much more.) Commented Apr 27, 2021 at 6:13
• I didn't get your third step, could you please elaborate? Commented Apr 27, 2021 at 6:15
• @BooleanWick: Geometric series $(1-A)^{-1}(1-B)^{-1}=\sum_{n,m\geqslant 0}A^n B^m$? Commented Apr 27, 2021 at 6:18