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:


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}

\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)}


3 Comments on “LOG#079. Zeta multiple integral.”

  1. You could also try a different way. From (2) onwards you could just split the logarithm into a sum of n terms over xi multiplied with your weight function. If you also change variables as xi –> exp(ui) to change product to sum and absorb the minus sign inside, the whole thing starts looking suspiciously like the weighted ratio of the arithmetic mean of exponentials (-1/n in the denominator) but not inside the hypercube any more. But then, for n going to Infinity you would just get the mean of inf. integrals of exp(ui)!

  2. Ask it in some forum of mathematics, or in some usenet specialized group or google.groups.
    This young bloggers don’t remember the old basic internet resources! 🙂

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