Theorem?: Let $n$ be a positive interger, $n>1$, then Riemann zeta function can be expressed in terms of a multiple integral which exhibits the following form:

$$ \displaystyle \zeta(n)=-\frac{1}{n-1}\ \underset{n-1}{\underline{\ \int_0^1...\int_0^1}} \dfrac{ \ln(\prod_{i=1}^{n-1}x_i) \ \prod_{i=1}^{n-1}dx_i}{1-\prod_{i=1}^{n-1}x_i} $$

for $1<n<9$ we get these first cases:

$$ \displaystyle \begin{align} \displaystyle \int_0^1 \frac{\ln x }{1-x } \, dx =-\frac{\pi ^2}{6} = - \zeta(2) \\ \\ \\ \int _0^1\int _0^1\frac{\ln(x y)}{1-x y}dxdy =-2\zeta(3) \\ \\ \\ \int _0^1\int _0^1\int _0^1\frac{\ln(x y z)}{1-x y z }dxdy dz =-\frac{\pi ^4}{30} = - 3\zeta(4) \\ \\ \\ \int _0^1\int _0^1\int _0^1\int _0^1\frac{\ln(x y z w)}{1-x y z w}dxdy dz dw =-4 \zeta(5) \\ \\ \\ \int _0^1\int _0^1\int _0^1\int _0^1\int _0^1\frac{\ln(x y z w t)}{1-x y z w t}dxdy dz dw dt =-\frac{\pi ^6}{189}= - 5\zeta(6) \\ \\ \\ \int _0^1\int _0^1\int _0^1\int _0^1\int _0^1\int _0^1\frac{\ln(x y z w t r)}{1-x y z w t r}dxdy dz dw dt dr =-6\zeta(7) \\ \\ \\ \int _0^1\int _0^1\int _0^1\int _0^1\int _0^1\int _0^1\int _0^1\frac{\ln(x y z w t r s)}{1-x y z w t r s}dxdy dz dw dt dr ds =-\frac{7\pi ^8}{9450}= - 7\zeta(8) \end{align} $$

But, obviously we are looking for a general proof or a counter example.

Regards http://tardigrados.wordpress.com/2013/01/08/la-funcion-zeta-de-riemann-definida-en-terminos-de-integrales-multiples/


Just change variables from $x_i$ to $u_i = -\log x_i$ and let $u = \sum_{i=1}^{n-1} u_i$. For $n \ge 2$, we have:

$$\begin{align}&\frac{1}{n-1}\iiint_{0 < x_i < 1} \frac{-\log(\prod_{i=1}^{n-1} x_i)}{1 - \prod_{i=1}^{n-1} x_i} \prod_{i=1}^{n-1} d x_i\\ =&\frac{1}{n-1} \iiint_{0 < u_i < \infty} \frac{u}{1 - e^{-u}} e^{-u} \prod_{i=1}^{n-1} du_i\\ =&\frac{1}{n-1} \int_0^{\infty} \frac{u du }{e^u - 1 }\left\{\iint_{\stackrel{u_2,\ldots,u_{n-1} > 0}{u_2+\cdots+u_{n-1} < u}}\prod_{i=2}^{n-1} du_i \right\}\\ =&\frac{1}{n-1} \int_0^{\infty} \frac{u du }{e^u - 1 } \frac{u^{n-2}}{(n-2)!}\\ =&\frac{1}{\Gamma(n)} \int_0^{\infty} \frac{u^{n-1}}{e^u - 1} du\\ =&\frac{1}{\Gamma(n)} \Gamma(n)\zeta(n)\\ =&\zeta(n) \end{align}$$

  • $\begingroup$ At first glance, your derivation seems to be ok, congratulations and thank you very much. $\endgroup$ – user72430 Apr 14 '13 at 13:45

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