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Euler for a third day, or: the second Euler product for zeta

(For those who missed them, the first day was about π, and the the second day was about ζ(2); what will the third day reveal… read on!).

According to Hadamard’s factorization for the “completed” zeta function

\xi(s)=s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)

(which is an entire function of s, and is invariant under the replacement of s by 1-s), we can write

\xi(s)=e^{a+bs}\prod_{\rho}{\Bigl(1-\frac{s}{\rho}\Bigr)e^{s/\rho}},

where the product runs over all non-trivial zeros of the Riemann zeta function (those in the critical strip), and a and b are some constants.

Now, as in Euler’s original definition of the Gamma function, it is tempting to replace the exponential terms by

(1-1/\rho)^{-s},

which leads to

\xi(s)=\Delta^s e^{a+bs}\prod_{\rho}{\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{1}{\rho}\Bigr)^{-s}},

with

\Delta=\prod_{\rho}{e^{1/\rho}(1-1/\rho)}.

(All these products converge absolutely for all s since the series with general term |ρ|2 converges). Now, since

\xi(0)=\xi(1)=1,

(easily remembered because the zeta function has a simple pole with residue 1 at s=1), we can plug in the value s=0 and s=1 to get

1=e^a,\quad\quad 1=\Delta e^b,

which leads to what may be anachronistically called Euler’s second product for the Riemann zeta function:

\xi(s)=\prod_{\rho}{\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{1}{\rho}\Bigr)^{-s}}

(I have never seen this formula before, but of course it is very unlikely to be new!)

One may even mix this with Euler’s gamma formula

\frac{1}{\Gamma(1+z)}=\prod_{k\geq 1}{\Bigl(1+\frac{z}{k}\Bigr)\Bigl(1+\frac{1}{k}\Bigr)^{-z}}

and the well-known value Γ(1/2)=π1/2, which together give the nice expression

\pi^{-s/2}\Gamma(s/2)=\frac{2^{1-s}}{s}\prod_{k\geq 1}{\Bigl(1+\frac{s}{2k}\Bigr)^{-1}\Bigl(1+\frac{1}{2k}\Bigr)^{-s}}

from which we can incorporate the trivial zeros of the zeta function at -2, -4, -6, etc, in the product, and deduce

\zeta(s)=\frac{2^{1-s}}{s-1}\prod_{\rho}{\Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{1}{\rho}\Bigr)^{-s}}

where the product runs now over trivial and non-trivial zeros!

Now, it is more than tempting to specialize the value of s; the first formula, for s=2, leads to

\xi(2)=2\pi^{-1}\Gamma(1)\zeta(2)=\prod_{\rho}{\Bigl(1-\frac{2}{\rho}\Bigr)\Bigl(1-\frac{1}{\rho}\Bigr)^{-2}},

or in other words to the relation

\frac{\pi}{3}=\prod_{\rho}{\Bigl(1-\frac{1}{(\rho-1)^2}\Bigr)}

where the product is — once more — over non-trivial zeros. Again, I would bet this has already appeared somewhere, but I had never seen it.

Using the table of the first 10000 positive ordinates of zeros of zeta (found here), one gets (putting in the complex conjugates, of course) the values

1.02737,\quad 1.04070,\quad 1.04569,\quad 1.04692

with the first 10, 100, 1000 and 10000 zeros, respectively, compared with π/3=1.04720.

One Response to “Euler for a third day, or: the second Euler product for zeta”

  1. moonface wrote:

    Very cute!

    Reply

    Saturday, December 12, 2009 at 3:21 | Permalink

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