LOG#079. Zeta multiple integral.

My second post this day is a beautiful relationship between the Riemann zeta function, the unit hypercube and certain multiple integral involving a “logarithmic and weighted geometric mean”. I discovered it in my rival blog, here:

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

First of all, we begin with the Riemann zeta function:

\displaystyle{\zeta (s)=\sum_{n=1}^\infty n^{-s}=\sum_{n=1}^\infty \dfrac{1}{n^{s}}}

Obviously, \zeta (1) diverges (it has a pole there), but the zeta value in s=2 and s=3 can take the following multiple integral “disguise”:

\displaystyle{\zeta (2) =-\int_0^1 \dfrac{\ln (x)}{1-x}dx=-\left(-\dfrac{\pi^2}{6}\right)=\dfrac{\pi^2}{6}}

\displaystyle{\zeta (3)=-\dfrac{1}{2}\int_0^1\int_0^1\dfrac{\ln (xy)}{1-xy}dxdy}

Moreover, we can even check that

\displaystyle{\int_0^1\int_0^1\int_0^1\dfrac{\ln (xyz) }{1-xyz}=-\dfrac{\pi^4}{30}=-3\zeta (4)}

In fact, you can generalize the above multiple integral over the unit hypercube

H_n(1)=\left[0,1\right]^n=\left[0,1\right]\times \underbrace{\cdots}_{n}\times \left[0,1\right]

and it reads

(1) \boxed{\displaystyle{-n\zeta (n+1)=\int_0^1\cdots \int_0^1 \dfrac{\ln (x_1 x_2\cdots x_n)}{1-x_1x_2\cdots x_n}dx_1dx_2\cdots dx_n}}

or equivalently

(2) \boxed{\displaystyle{\zeta (n+1)=-\dfrac{1}{n}\int_0^1\cdots \int_0^1\dfrac{\displaystyle{\ln \prod_{i=1}^n x_i \prod_{i=1}^n dx_i}}{\displaystyle{1-\prod_{i=1}^n x_i}}}}

I consulted several big books with integrals (specially some russian “Big Book” of integrals, series and products or the CRC handbook) but I could not find this integral in any place. If you are a mathematician reading my blog, it would be nice if you know this result. Of course, there is a classical result that says:

\displaystyle{\zeta (n)=\left(\int_0^1\right)^n\dfrac{\displaystyle{\prod_{i=1}^n dx_i}}{\displaystyle{1-\prod_{i=1}^n x_i}}}

but the last boxed equation was completely unknown for me. I knew the integral represeantations of \zeta (2) and \zeta (3) but not that general form of zeta in terms of a multidimensional integral. I like it!

In fact, it is interesting (but I don’t know if it is meaningful at all) that the last boxed integral (2) can be rewritten as follows

(3) \boxed{\displaystyle{\zeta\left(n+1\right)=\int_0^1\cdots\int_0^1\left(\dfrac{1}{\displaystyle{1-\prod_{i=1}^n x_i}}\right)\ln\left(\dfrac{1}{\displaystyle{\sqrt[n]{\prod_{i=1}^n x_i}}}\right)\left(\prod_{i=1}^n dx_i\right)}}

or equivalently

(4) \boxed{\displaystyle{\zeta \left(n+1\right)=-\int_0^1\cdots \int_0^1 \omega (x_i) \ln \left(\overline{X}_{GM}\right) d^nX}}

where I have defined the weight function

\displaystyle{\omega (x_i)=\dfrac{1}{\displaystyle{1-\prod_{i=1}^n x_i}}}

and the geometric mean is

\displaystyle{\overline{X}_{GM}=\sqrt[n]{\prod_{i=1}^n x_i}}

and the volume element reads

d^nX=dx_1dx_2\cdots dx_n

I love calculus (derivatives and integrals) and I love the Riemann zeta function. Therefore, I love the Zeta Multiple Integrals (1)-(2)-(3)-(4). And you?

PS: Contact the author of the original multidimensional zeta integral ( his blog is linked above) and contact me too if you know some paper or book where those integrals appear explicitly. I believe they can be derived with the use of polylogarithms and multiple zeta values somehow, but I am not an expert (yet) with those functions.

PS(II): In math.stackexchange we found the “proof”:

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

\displaystyle{I=\dfrac{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}dx_i}

Then
\displaystyle{I=\dfrac{1}{n-1}\iiint_{0 < u_i < \infty}\frac{u}{1-e^{-u}}e^{-u}\prod_{i=1}^{n-1}du_i}
\displaystyle{I=\frac{1}{n-1} \int_0^{\infty} \dfrac{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\}}
\displaystyle{I=\dfrac{1}{n-1} \int_0^{\infty} \frac{u du }{e^u - 1 } \dfrac{u^{n-2}}{(n-2)!}=\frac{1}{\Gamma(n)} \int_0^{\infty} \frac{u^{n-1}}{e^u - 1} du=\dfrac{1}{\Gamma(n)} \Gamma(n)\zeta(n)=\zeta(n)}

Advertisements