# Riemann Hypothesis/Appendix

Theorem 1

For all integers, $n>1$ ,

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

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

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

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

$=\sum _{j=0}^{\infty }\prod _{i=1}^{n}{\frac {1}{j+1}}$ Evaluating the product,

$=\sum _{j=1}^{\infty }\prod _{i=1}^{n}{\frac {1}{j}}$ $=\sum _{j=1}^{\infty }{\frac {1}{j^{n}}}$ Using the definition of the zeta function that holds only for $\Re n>1$ ,

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

It can be noted that,

$\int _{0}^{1}{\frac {1}{1-x}}\mathrm {d} x$ fails to converge, as $-\lim _{x\to 1}\log(1-x)=\infty$ .