# On the absolute convergence of the local-zeta integral.

Though I am in a situation considering only local-zeta integral, to explain my question briefly, let me ask it in quite general form.

Let $f(s,g)$ be a two variable smooth good (in a suitable sense) function and let $F(s)=\int_{G}f(s,g)dg$. Assume $F(s)$ is absolutely convergent for $\Re(s)>0$ and has meromorphic continuation to all $s \in C$(complex number). Also we know that $$\lim_{s\to 0} s^{m}\cdot F(s)$$ exist for some positive integer $m>0$. Then can we say that $$\lim_{s\to 0} s^{m}\cdot \int_{G}|f(s,g)|dg$$ exist? Or more weakly, can we ensure that $$\lim_{s\to 0} s^{m+1}\cdot \int_{G}|f(s,g)|dg=0?$$

If it does not holds, would you suggest some mild assumption that compels this to hold? In my case, I am considering only when $F(s)$ is given as a local-zeta integral from some tempered representation and some degenerate principal series representation. Then the local-zeta integral has the above properties. If it does not hold in general, then does it hold for local-zeta integral?

Any help or comments will be greatly appreciated!

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Also, no for the second statement. Take $G$ being a couple of copies of the multiplicative group and argue as above. In this manner, you obtain something with arbitrary high multiplicity at zero, although the limit of the original integrals is convergent.
I shouldn't have said Dirichlet character in a local setting though, but I am sure that one can guess what I meant: a character which is non trivial when restricted to $o^\times_v$. – Marc Palm Apr 15 '13 at 9:11