On modified Euler product 
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

*

*Is there a compact representation for this product?


*What are some non-trivial  properties of this product?


*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.
So, I  need more insight into this. Any comments regarding this are welcome.
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.
 A: I'll expand on Ralph's answer to describe how to evaluate $F(1/2)$.
Ralph's main point is that $\log F(s)$ is well-approximated by a sum of logarithmic derivatives of $\zeta(s)$. Writing it out explicitly, we have that
$$ \begin{align}
  \log(F(s)) &= -c \frac{\zeta'(s)}{\zeta(s)} - \big(\tfrac{c^2}{2} - c\big) \frac{\zeta'(2s)}{\zeta(2s)} \\
&\qquad + \sum_p \log p \Big( \frac{c^3}{3} p^{-3s} + \frac{c^4}{4} p^{-4s} + \cdots\Big) \\
&\qquad + c\sum_p \log p \big(p^{-3s} + p^{-4s} + \cdots) \\
&\qquad + (\tfrac{c^2}{2} - c)\sum_p \log p \big( p^{-4s} + p^{-6s} + p^{-8s} + \cdots \big)
\end{align}.$$
The first line comes from matching the first two terms in the expansion of $\log F(s)$, as in Ralph's answer. It would be possible to carry this on for mor terms if desired. The second line is the remaining terms of the original expansion of $\log F(s)$. The third line are the remaining terms from $c \zeta'(s)/\zeta(s)$. And the fourth line are the remaining terms from $(c^2/2 - c) \zeta'(2s)/\zeta(2s)$.
For evaluation, one can use typical continuations for $\zeta$ and $\zeta'$ to evaluate the first line, and the remaining three lines are all absolutely convergent as long as $\lvert c p^{-s} \rvert < 1$. For evaluation, it's easier to recollect the various infinite expansions and to have something instead like
$$ \begin{align}
  \log(F(s)) &= -c \frac{\zeta'(s)}{\zeta(s)} - \big(\tfrac{c^2}{2} - c\big) \frac{\zeta'(2s)}{\zeta(2s)} \\
&\qquad + \sum_p \log p \Big(
  - \log(1 - cp^{-s}) - \big( cp^{-s} + c^2 p^{-2s}/2 \big)
\Big) \\
&\qquad + c\sum_p \log p \Big(\frac{p^{-3s}}{1 - p^{-s}}\Big)\\
&\qquad + (\tfrac{c^2}{2} - c)\sum_p \log p \Big(\frac{p^{-4s}}{1 - p^{-2s}}\Big)
\end{align}.$$
With this, using the first primes up to $10^5$ in sage, I estimate $\log(F(0.5)) \approx -0.3312$ when $c = e^{-1/2}$. The (very simple, hastily written) code for this is below.
c = e**-0.5

tot1 = 0
for p in primes(100000):
    tot1 += RR(log(p) * (- log(1 - c * p**-0.5) - (c * p**-0.5 + c*c * p**-1 / 2)))

tot2 = 0
for p in primes(100000):
    tot2 += RR(log(p) * p**-1.5/(1 - p**-0.5))

tot3 = 0
for p in primes(100000):
    tot3 += RR(log(p) * p**-2 / (1 - p**-1))

actual_total = tot1 + c * tot2 + (c*c/2 - c) * tot3 - c * zetaderiv(1, 0.5)/zeta(0.5)

# Note the other zeta term divides by zeta(1), and thus vanishes

print(actual_total)
```

A: Expanding as a power series in $x$
$$\log(1-x) = -x - \frac{x^2}2 -\frac{x^3}3 - \cdots $$
Thus
\begin{eqnarray*}
\log F(s) &=& \sum_p -\log(p)\log (1-c p^{-s}) \\
&=& \sum_p \log(p) \left(c p^{-s} + \frac{c^2}2p^{-2s} + \cdots\right)
\end{eqnarray*}
Similarly, applying to $\zeta(s)$ and differentiating:
\begin{eqnarray*}
\log \zeta(s) &=& \sum_p -\log (1-p^{-s}) \\
&=& \sum_p \left(p^{-s} + p^{-2s}/2 + \cdots\right) \\
\frac{\zeta'(s)}{\zeta(s)} &=& -\sum_p \log(p) \left( p^{-s} + p^{-2s} + \cdots \right)
\end{eqnarray*}
Comparing the two gives
$$\log F(s) = -c\frac{\zeta'(s)}{\zeta(s)} + (c-c^2/2)\frac{\zeta'(2s)}{\zeta(2s)} + \cdots $$
The remarkable thing about the series on the right hand side is that the $k$-term is meromorphic and to the right of $1/k$ it is further holomorphic with an absolutely convergent sum representation.  This allows you to understand the analytic continuation of $\log F(s)$ to the right of any $\epsilon>0$ using only a finite number of sums.  This analysis should also show that there is a dense set of poles near the imaginary axis, leading to the function not continuing past there.  Note though that the poles of on the half-line (or to the right of it if RH is false) lead to $F(s)$ having essential singularities.
