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Take the following 'tweaked' Euler product ($i = \text{the imaginary unit})$:

$$Eul_i(s) := \prod_{p\in\mathbb{P}} = \dfrac{1}{1-\dfrac{1}{(ip)^s}}$$

It is not difficult to see that for $s=2k+1, k=0,1,2...$ the following must hold:

$$|Eul_i(s)| = \sqrt{\dfrac{\zeta(4s)}{\zeta(2s)}}$$

I am particularly interested whether the complex value for $s=1$, i.e.:

$$Eul_i(1) := \prod_{p\in\mathbb{P}} = \dfrac{1}{1-\dfrac{1}{(ip)}}$$

converges or not. Have managed to test it up to primes $\le n=4.260.000.000$ (~32 bit limit of Pari) and found it converging towards $-0.79991... + 0.13456... i$ (but not sure what happens for larger primes).

Grateful for any thoughts.

Thanks.

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1 Answer 1

up vote 2 down vote accepted

The logarithm of the denominator is

$$ \log\left(1-(\mathrm ip)^{-1}\right)=\log\left(1+\mathrm ip^{-1}\right)=\log\sqrt{1+p^{-2}}+\mathrm i\arctan p^{-1}\approx\frac12p^{-2}+\mathrm ip^{-1}\;. $$

The sum over the real parts converges, but the sum over the imaginary parts diverges because the sum over $p^{-1}$ diverges. You can't see it diverging because it only goes with $\log\log n$ if you sum up to $p\lt n$.

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Vielen Dank, Joriki. That's exactly what I was looking for! –  Agno Feb 11 '13 at 16:22

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