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This question expands on this one from MSE.

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.

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    $\begingroup$ The Euler product for every Dirichlet series (for a nonprincipal character) converges at $s=1$, but that's more-or-less equivalent with the Prime Number Theorem in arithmetic progressions, and in particular Euler's heuristic argument for $\chi_4$ does not readily yield a rigorous proof for that $L$-function. Convergence for real part $> 1/2$ is equivalent to the Riemann hypothesis for the same $L$-function. (While I was editing this ABCDveve wrote similarly in his/her answer.) $\endgroup$ Commented Jun 17, 2015 at 21:40
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    $\begingroup$ To see the equivalence that Noam mentions, see Theorem 3.3 (and the two lemmas preceding it) in math.uconn.edu/~kconrad/articles/eulerprod.pdf, setting $d = 1$ and $\alpha_{p,1} = \chi(p)$. Note that for infinite products like an Euler product, the term "converges" means "converges and is not $0$." Since nobody has ever proved the $L$-function of a nontrivial Dirichlet character has no zeros in any vertical strip $1-\varepsilon < {\rm Re}(s) < 1$, it is basically hopeless at present to expect anyone to prove the Euler product converges at any $s$ with ${\rm Re}(s) < 1$. $\endgroup$
    – KConrad
    Commented Jun 17, 2015 at 23:14
  • $\begingroup$ Related question here: mathoverflow.net/questions/63714/… $\endgroup$
    – GH from MO
    Commented Jun 17, 2015 at 23:57
  • $\begingroup$ @NoamD.Elkies Why is the trivial character corresponding to zeta exception to convergence? $\endgroup$
    – joro
    Commented Jun 18, 2015 at 13:41
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    $\begingroup$ Good question. The answer is basically the pole at $s=1$. The connection with the $L$-function zeros is via the logarithmic derivative $L'/L$, which is singular when $L$ has either a zero or a pole. Dirichlet $L$-functions other than the Riemann zeta function have no poles, so only their zeros affect the analysis; but when you try to do the same for $\zeta$ you run into the $s=1$ pole before any zero shows up. $\endgroup$ Commented Jun 18, 2015 at 15:08

1 Answer 1

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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.

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  • $\begingroup$ If $\chi$ is kronecker character it is nonnegative. Let $t=0$. Every term is real greater than one. I don't see how this will converge to the $L$ function (computing a single value above the correct is counterexample). $\endgroup$
    – joro
    Commented Jun 18, 2015 at 6:36
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    $\begingroup$ @joro: The only character modulo $n$ that is nonnegative is the trivial character modulo $n$. $\endgroup$
    – GH from MO
    Commented Jun 18, 2015 at 6:37
  • $\begingroup$ @GHfromMO Thank you. My mistake. So why the answer doesn't apply to the trivial character which is zeta AFAICT? $\endgroup$
    – joro
    Commented Jun 18, 2015 at 7:03
  • $\begingroup$ @joro: Roughly speaking, the reason is the oscillation of $\chi$ (when $\chi$ is nontrivial). More precisely, while $\sum p^{-s}$ over the primes clearly diverges unless $\Re s>1$, the oscillating sum $\sum\chi(p)p^{-s}$ has a chance to converge for smaller values of $\Re s$ assuming the primes are sufficiently well-distributed mod $n$ (where $n$ is the conductor of $\chi$). And, indeed, GRH implies that the latter sum converges for $\Re s>1/2$. $\endgroup$
    – GH from MO
    Commented Jun 18, 2015 at 7:41
  • $\begingroup$ @joro: Two related posts are mathoverflow.net/questions/63714/… and mathoverflow.net/questions/181241/… $\endgroup$
    – GH from MO
    Commented Jun 18, 2015 at 7:43

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