Theorem 1

For all integers, ${\displaystyle n>1}$,

${\displaystyle \int _{[0,1]^{n}}{\frac {1}{1-\prod _{i=1}^{n}x_{i}}}\prod _{i=1}^{n}\mathrm {d} x_{i}=\zeta (n)}$

Proof

Using the power series of ${\displaystyle (1-x)^{-1}}$,

${\displaystyle \int _{[0,1]^{n}}{\frac {1}{1-\prod _{i=1}^{n}x_{i}}}\prod _{i=1}^{n}\mathrm {d} x_{i}=\int _{[0,1]^{n}}\sum _{j=0}^{\infty }\left(\prod _{i=1}^{n}x_{i}\right)^{j}\prod _{i=1}^{n}\mathrm {d} x_{i}}$

Evaluating,

${\displaystyle =\int _{[0,1]^{n}}\sum _{j=0}^{\infty }\prod _{i=1}^{n}x_{i}^{j}\mathrm {d} x_{i}}$

${\displaystyle =\sum _{j=0}^{\infty }\prod _{i=1}^{n}\int _{0}^{1}x_{i}^{j}\mathrm {d} x_{i}}$

Evaluating the integral,

${\displaystyle =\sum _{j=0}^{\infty }\prod _{i=1}^{n}{\frac {1}{j+1}}}$

Evaluating the product,

${\displaystyle =\sum _{j=1}^{\infty }\prod _{i=1}^{n}{\frac {1}{j}}}$

${\displaystyle =\sum _{j=1}^{\infty }{\frac {1}{j^{n}}}}$

Using the definition of the zeta function that holds only for ${\displaystyle \Re n>1}$,

${\displaystyle \int _{[0,1]^{n}}{\frac {1}{1-\prod _{i=1}^{n}x_{i}}}\prod _{i=1}^{n}\mathrm {d} x_{i}=\sum _{j=1}^{\infty }{\frac {1}{j^{n}}}=\zeta (n)}$ for all integers ${\displaystyle n>1}$ ${\displaystyle \blacksquare }$

Note

It can be noted that,

${\displaystyle \int _{0}^{1}{\frac {1}{1-x}}\mathrm {d} x}$

fails to converge, as ${\displaystyle -\lim _{x\to 1}\log(1-x)=\infty }$.