[math-fun] euler's infinite product, primes of the form 4n+1, 4n+3... ??

1 Sep 2011

      Hello,

  I am working on a simple problem which is the
following,

  consider the Euler product with primes,

    infinity
    --------'
   '  |  |       1
      |  |    -------- = Zeta(2)
      |  |         1
      |  |    1 - ----
     p = 2          2
                   p

this is well known. Now if we take
only the primes of the form 4n+1 and 4n+3
and split (avoiding the number 2), then
the product is

  p1*p3 = Pi^2/8 ,

  p1 is the Euler product with primes of
the form 4n+1 and p3 = with the primes of
the form 4n+3.

  Now these 2 values are approximately :

p1 = 1.05618212168678504596905 and
p3 = 1.16807558536624241169267

  The product of p1*p3 being as expected Pi^2/8.

  the good question is : What is p1 made of actualy ?
and p3 ?   I tried all the tricks I know and came
with NO answer at all. My first guess was a simple
rational combination : this is wrong, my second guess
was that p1 is a algebraic multiple of Pi : wrong too.
And then log(p1) is related to some log(Gamma(a/b)) values ?
There are no numerical evidence of that, then I tried
all the tables, all the algorithms against that number
and my quest was incorrect.

  Does anybody knows what are these numbers ? Is this
a trivial question related to Dirichlet series, I don't
get it.

  The values are correct to about 8 digits.

  I also tried with the exponent being, 3,4 and
8 : nothing too, a complete mystery.

    Any clue would help, thanks in advance.

  My gut feeling is that p1 should be simple
(as well as p3 ).

  Simon Plouffe