Minimum and maximum of a partial Euler product? 
Question: If if $n\in\mathbb{N}$ and $s\in \mathbb{C},$ say $s=\sigma+t\sqrt{-1},$ then Dirichlet Beta function is defined to be
  $$
\beta(s)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s};
$$
  which  for Re(s)>1, has the Euler product representation over prime numbers $p:$
$$
L(s)=\prod_{p>2}{ 1\above 1.5pt 1-(-1)^{(p-1)/2}p^{-s}};
$$
  equiv.,
  $$
\prod_{p \equiv 1 \pmod 4}\frac{1}{1-p^{-s}} \prod_{p \equiv 3 \pmod 4}\frac{1}{1+p^{-s}}.
$$
  If $x$ is a positive number I write $L_x(s)$ where for the above product is restricted to those odd prime numbers that are less than or equal to $x.$ I set
  $$
m(s)=\operatorname*{min}_{x} L_x(s)
$$
  and
  $$
M(s)=\operatorname*{max}_{x} L_x(s)
$$
  Can we show that $m(s)$ and $M(s)$ exist and compute (estimate) there values ? 


Since the product converges I am certain $L_x(s)$ is bounded and so has a maximum ? For example I think I have correctly that 
$$
\lim_{x \to \infty}L_x(1)=\frac{\pi}{4}
$$
This is just the Euler product representation of $\frac{\pi}{4}$ and so $L_{x}(1)$ is the partial Euler product up to some magnitude $x.$ I would like to make the claim that over all $x$ that $\operatorname*{min}_{x} L_x(1)=\frac{3}{4}
$ and that $\operatorname*{max}_{x} L_x(1)=\frac{15}{16}.$ That the minimum might be equal to $\frac{3}{4}$ might straightforward but even there I am not certain. 

Similar but not quite the same question:


*

*Partial Euler Products

*how to prove that the partial euler product of primes less than or equal x is bounded below by log x
 A: With $$\mathfrak{M}(s) = \sup_x\ \Re(\sum_{p \le x} (-1)^{(p-1)/2} p^{-s}), \qquad \Re(s) > 1$$
the $\Bbb{Q}$-linear independence of the $\log p$ implies $\lim_{k \to \infty} \mathfrak{M}(\sigma+it_k) = \sum_p p^{-\sigma}$ for a sequence $t_k\to \infty$ where each $t_k$ is the imaginary part of a zero of $\beta(s)-\zeta_2(\sigma)$. Moreover $\beta(s+it_k)$ converges uniformly to $\zeta(s)$ on $\Re(s) \ge 1+\delta$ and the same holds for the partial sums of their Euler product :
For $\epsilon> 0$ small enough, if $\mathfrak{M}(\sigma+it)$ is $\epsilon$-close to $ \sum_p p^{-\sigma}$ then $\beta(\sigma+it)$ is $\epsilon$-close to $\zeta_2(\sigma)=(1-2^{-\sigma})\zeta(\sigma)$ and $\beta'(\sigma+it)$ is $\epsilon$-close to $\zeta_2'(\sigma)$, and since $\beta''(s)$ is bounded it means there is a zero of $\beta(s)-\zeta_2(\sigma)$ at $s = \sigma+it+\frac{ \zeta_2(\sigma )- \beta(\sigma+it)}{\zeta_2'(\sigma)} + O(\epsilon^2) = \sigma+it + O(\epsilon)$. 
In other words the rate of convergence of $\sup_{t \le T} \mathfrak{M}(\sigma+it) \to \sum_p p^{-\sigma}$ depends on the density of zeros of $\beta(s)-\zeta_2(\sigma)$ around $\Re(s)=\sigma$. 
The latter is research level, there are some methods and results about it in the literature.
