Let's consider the following Euler product ($s=\sigma+it)$:

$$ P(s)=\prod_{p \; \text{prime}} \frac{1}{1-p^{-s}} \; e^{-p^{-s}}$$

So for $\sigma>1$, it is clear the product converges and we have:

$$ P(s)=\zeta(s) \prod_{p \; \text{prime}} e^{-p^{-s}} = \zeta(s) B(s)$$

But for $\frac{1}{2} <\sigma<1$, the product $P(s)$ is also converging (taking the log of $P(s)$ we see it is absolutely convergent), so on this interval (if $B(s)$ can be analytically continued), then can we write that we have also following relation ?:

$$ P(s)=\prod_{p \; \text{prime}} \frac{1}{1-p^{-s}} \; e^{-p^{-s}} = \zeta(s) B(s)$$

(even if on this interval the Euler product does not converge for $P(s)$ or $\zeta(s)$. We have $P(s)$ and $\zeta(s) B(s)$ which coincide for $\sigma >1$).

I would say no, as the zeros of $\zeta$ seems to disappear in the product... but why exactly the equality does not hold ?

Note: maybe in my example the function $B(s)$ is not optimum because I do not know if it can be prolonged analytically for $\frac{1}{2} <\sigma<1$, but imagine a function that makes the Euler product to converge (same principle as above) and can be prolonged analytically for $\frac{1}{2} <\sigma<1$, then can we say that equality will hold also for $\frac{1}{2} <\sigma<1$.

Let's consider the following Euler product ($s=\sigma+it)$:

$$ P(s)=\prod_{p \; \text{prime}} \frac{1}{1-p^{-s}} \; e^{-p^{-s}}$$

So for $\sigma>1$, it is clear the product converges and we have:

$$ P(s)=\zeta(s) \prod_{p \; \text{prime}} e^{-p^{-s}} = \zeta(s) B(s)$$

But for $\frac{1}{2} <\sigma \leq 1$, the product $P(s)$ is also converging (taking the log of $P(s)$ we see it is absolutely convergent), so on this interval (if $B(s)$ can be analytically continued), then can we write that we have also following relation ?:

$$ P(s)=\prod_{p \; \text{prime}} \frac{1}{1-p^{-s}} \; e^{-p^{-s}} = \zeta(s) B(s)$$

(even if on this interval the Euler product does not converge for $P(s)$ or $\zeta(s)$. We have $P(s)$ and $\zeta(s) B(s)$ which coincide for $\sigma >1$).

I would say no, as the zeros of $\zeta$ seems to disappear in the product... but why exactly the equality does not hold ?

Note: maybe in my example the function $B(s)$ is not optimum because I do not know if it can be prolonged analytically for $\frac{1}{2} <\sigma \leq 1$, but imagine a function that makes the Euler product to converge (same principle as above) and can be prolonged analytically for $\frac{1}{2} <\sigma \leq 1$, then can we say that equality will hold also for $\frac{1}{2} <\sigma \leq 1$.

Let's consider the following Euler product ($s=\sigma+it)$:

$$ P(s)=\prod_{p \; prime} \frac{1}{1-p^{-s}} \; e^{-p^{-s}}$$

So for $\sigma>1$, it is clear the product converges and we have:

$$ P(s)=\zeta(s) \prod_{p \; prime} e^{-p^{-s}} = \zeta(s) B(s)$$

But for $\frac{1}{2} <\sigma<1$, the product $P(s)$ is also converging (taking the log of $P(s)$ we see it is absolutely convergent), so on this interval (if $B(s)$ can be analytically continued), then can we write that we have also following relation ?:

$$ P(s)=\prod_{p \; prime} \frac{1}{1-p^{-s}} \; e^{-p^{-s}} = \zeta(s) B(s)$$

(even if on this interval the Euler product does not converge for $P(s)$ or $\zeta(s)$. We have $P(s)$ and $\zeta(s) B(s)$ which coincide for $\sigma >1$).

I would say no, as the zeros of Zeta seems to disappear in the product... but why exactly the equality does not hold ?

Note: maybe in my example the function $B(s)$ is not optimum because I do not know if it can be prolonged analytically for $\frac{1}{2} <\sigma<1$, but imagine a function that makes the Euler product to converge (same principle as above) and can be prolonged analytically for $\frac{1}{2} <\sigma<1$, then can we say that equaility will hold also for $\frac{1}{2} <\sigma<1$.

Let's consider the following Euler product ($s=\sigma+it)$:

$$ P(s)=\prod_{p \; \text{prime}} \frac{1}{1-p^{-s}} \; e^{-p^{-s}}$$

So for $\sigma>1$, it is clear the product converges and we have:

$$ P(s)=\zeta(s) \prod_{p \; \text{prime}} e^{-p^{-s}} = \zeta(s) B(s)$$

But for $\frac{1}{2} <\sigma<1$, the product $P(s)$ is also converging (taking the log of $P(s)$ we see it is absolutely convergent), so on this interval (if $B(s)$ can be analytically continued), then can we write that we have also following relation ?:

$$ P(s)=\prod_{p \; \text{prime}} \frac{1}{1-p^{-s}} \; e^{-p^{-s}} = \zeta(s) B(s)$$

(even if on this interval the Euler product does not converge for $P(s)$ or $\zeta(s)$. We have $P(s)$ and $\zeta(s) B(s)$ which coincide for $\sigma >1$).

I would say no, as the zeros of $\zeta$ seems to disappear in the product... but why exactly the equality does not hold ?

Note: maybe in my example the function $B(s)$ is not optimum because I do not know if it can be prolonged analytically for $\frac{1}{2} <\sigma<1$, but imagine a function that makes the Euler product to converge (same principle as above) and can be prolonged analytically for $\frac{1}{2} <\sigma<1$, then can we say that equality will hold also for $\frac{1}{2} <\sigma<1$.