Let $F$ be a local field and $\omega : F^{\times} \to \mathbb {S}^1$ be a unitary character. The local zeta integral is defined to be $$ z(s,\omega,f) = \int_{F^{\times}} f(x)\omega(x)\omega_s(x)d^{\times}x $$ where $f \in S(F)$ be a Schwartz function and $\omega_s(x) = |x|^{s}$. This function is absolutely convergent when Re(s)>0 for all $f \in S(F)$.
I'm reading Kudla's paper Tate's thesis in the book Introduction to langlands program. And I'm confused with the sentence that the poles of local zeta integral is caused by $f(0)$ (in non-archimedean case) and the whole Taylor series of $f$ at $0$ (in archimedean case).
You can find the sentence at the beginning of page P121 of the paper.
Thankyou for your help!
I have found the answer. The key point is that if $f(0) =0$，then there is a good decomposition of Supp($f$).
Let the conductor of $\omega$ be $c$, which means $\omega$ is trivial on $U_0=\mathcal{O}^{\times}$ if $c=0$, otherwise it is trivial on $U_c = 1+\mathfrak{p}^c$.
Let's see what will happen if $f(0)=0$. Since $f$ is local constant, we can choose some $m > c$ such that near $1$, $f$ is constant on $1 +\mathfrak{p}^m $. Then we can decompose the compact set Supp($f$), say, Supp($f$) = $\cup_{i=1}^N a_i U_m$.
Notice that this can be done only in $f(0)=0$!!! If $f(0)\ne 0$, such a decomposition will never contain $0$.
Then let's calculate local zeta integral in this case. $$z(s,f,\omega) = \sum^N_{i=1} \int_{a_i U_m} f(x) \omega(x)|x|^{s-1}dx =\sum^N_{i=1}f(a_i)\omega(a_i)|a_i|^{s-1} \int_{U_m}dx $$ which is holomorphic on the whole complex plane.
But assume $f(0)\ne 0$, following the standard calculation of local zeta integral(see the book of Ramakrishnan & Valenze, P246 Theorem 7-2), we can only get the local zeta integral convergence for Re(s)>0.
I think that is why Kudla say $f(0)$ accounts for the poles of local zeta integral.