Is Riemann zeta function essentially the only L-function with a pole?
问题内容
I mean, if a function $F(s)$ is in Selberg class, and $F(s)$ has a pole of order m at $s=1$, is it true that there exists a function $G(s)$ in Selberg class such that $F(s)=\zeta^m(s)G(s)$, and $G$ is entire?
回答 (1)
In the full Selberg class this is not known unconditionally. It is a standard consequence of Selberg's Conjecture B. More precisely, Lemma 5.6 in Dixit's survey says that if $F\in\mathcal S$ has a pole of order $m$ at $s=1$, then Conjecture B implies $$ F(s)=\zeta(s)^m L(s), \qquad L\in\mathcal S. $$ Since elements of $\mathcal S$ are holomorphic except possibly for a pole at $s=1$, and since the pole of $F$ has already been fully accounted for by $\zeta(s)^m$, the remaining factor $L$ is entire.
Thus the answer is: expected yes; known conditional on Selberg's Conjecture B; not known in general.