MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 0 down vote accepted

No to the stronger statement. Here is the first simple example. If you take the zeta integral of a function f and a non trivial Dirichlet character at s, then take absolute values inside the integral you obtain the zeta integral of |f| with s`=Re s. The former has no pole at zero but the later. So the order of divergence might change. Tates thesis explains this pretty well.

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.

share|cite|improve this answer
Dear Palm, I am really thankful for your kind reply and making a good counterexample. I was also benefited from you before. I hope your good luck! – Jude Apr 15 '13 at 9:01
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
I understood! Thanks again. – Jude Apr 15 '13 at 11:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.