This question expands on [this one from MSE](https://math.stackexchange.com/questions/1327543/does-the-euler-product-for-the-dirichlet-beta-function-converge-for-all-re).


In the literature about Dirichlet $L$-series, I found that their Euler products:


$$L(s, \chi) =\prod\_p \bigg(\frac {1}{1-\frac{\chi(p)}{p^s}} \bigg)$$


are typically considered to be only converging for $\Re(s)>1$.


However, there seems to be an exception to this rule since Euler proved that:


$$L(1, \chi\_4) =\prod\_p \bigg(\frac {p}{p-\chi\_4(p)} \bigg)=\prod\_p \bigg(\frac {p}{p-\sin\left(\frac{p \,\pi}{2}\right)} \bigg)=\frac{3}{4}\cdot\frac{5}{4}\cdot\frac{7}{8}\cdot\frac{11}{12}\cdot\frac{13}{12}\dots=\beta(1)=\frac{\pi}{4}$$


does converge (albeit slowly).


I then decided to explore values for $\Re(s) \lt 1$ and numerical evidence suggests that the Euler-product:


$$\prod\_p \bigg(\frac {p^s}{p^s-\sin\left(\frac{p \,\pi}{2}\right)} \bigg)$$


also (slowly) converges in the domain $\frac12 < \Re(s) \le 1$.


**Questions:**


1) Is the Euler product for $L(s,\chi\_4)$ the only one known to converge for $s=1$?


2) Does this particular Euler product indeed converge in the right half of the strip?


Thanks.



Does the Euler product for $L(s,\chi_4)$ also converge in the right half of the critical strip?
By taking the logarithm (or log derivative) of $L(s)$, you get a Dirichlet series whose convergence relates directly to the zeros. Your question about convergence in the right half of the critical strip is equivalent to the Riemann hypothesis for $\chi$. Similarly, (conditional) convergence of the Euler product (again as the "Euler sum" after taking the logs) on the 1-line is equivalent to the lack of zeros on this line.


