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Zaza
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On modified Euler product

I know this forum is for professional Mathematicians. I didn't know if the following question is fit or not fit for the site. If not, feel free to delete it.

Consider the modified Euler product as follows:

$$F(s) = \prod_{p} \left( 1 - \frac{c}{p^s} \right)^{-\ln(p)}$$

Here $c$ is a constant

My questions are

  1. Is there a compact representation for this product?

  2. What are some non-trivial properties of this product?

  3. Value of the regularized sum:

$$-\sum_{p}\ln\left( 1 - \frac{1}{(ep)^{1/2}} \right){\ln(p)}$$

Help me understand the analytic continuation of this function.

Or consider

$$f(s) = \sum_{p}\left( 1 - \frac{1}{(ep)^{1/2}} \right)^{s\ln(p)}$$

$p$ belongs to set of primes

We can find f'(x)'s analytic continuation for some region containing s=0.

(This is the main motivation behind the question)

Edit: Now by using the answer by R.Furman how can we calculate the product's value 'uniquely' at 1/2?

I think this is equal to zero under Riemann hypothesis. Also I think if its equal to zero then RH is true . So, I need more insight into this. Any comments regarding this are welcome.

Edit:

We can also have a generalized conjecture:

$$ \prod_{p} \left( 1 - \frac{1}{e^{\alpha}p^{\beta}} \right)^{-\ln(p)}=\left|\frac{(\zeta({\alpha}+{\beta})({\alpha}+{\beta}-1)}{{\alpha}-{\beta}+1}\right|$$

Here, $ {\alpha} >0$ ; ${\beta} \in [1/2,1)$ s.t.

${\alpha}+{\beta} \neq1$

Any ideas about attacking this are welcome.

References:

[1] Germund Dahlquist ; "On the analytic continuation of Eulerian products"

[2] https://mathworld.wolfram.com/PrimeProducts.html ( and all the references therein)

[3] https://en.m.wikipedia.org/wiki/Euler_product

[4] Kimoto , Wakayama "Remarks on Zeta Regularized Products"

Progress post:

On infinite sum containing logarithmic derivative of Zeta function and Möbius function:

Recently I found a similar post while surfing through the site:

Does this product have analytic continuation?

If anyone could give answer in the context of the above post it will be very nice.

Zaza
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