Convergence of local zeta function I'm working with local zeta functions. For an Archimedean local field $F$ we say that $f:F\longrightarrow\mathbb C$ is a Schwartz-Bruhat function (I will be working in the Archimedean case only, i.e. $F=\mathbb R$ or $F=\mathbb C$) if $f\in\mathcal C^{\infty}(F)$ and if $f(z)p(z)\to 0$ when $z\to\infty$ for all polynomials $p(z)$. We define for a quasi-character $\chi$ the local zeta function by the formula
$$Z(f,\chi)=\int_{F^{\times}}f(x)\chi(x)\,d^{\times}x,$$
where $d^{\times}x=\frac{dx}{|x|}$ and $dx$ is the usual Lebesgue measure. I'm having problems with showing the absolute convegence of this integral. In the course of the problem, we arrive at my concrete question:

Given a Schwartz-Bruhat function $f$ and $\sigma>0$, why is the integral $$\int_{F^{\times}}|f(x)||x|^{\sigma-1}\ dx$$ finite?

One attempt is to take a compact neighborhood $K$ of $0$, and to bound $f$ by a certain positive constant $C$, so that the integral becomes
$$\left(\int_{F-K}+\int_{K-\{0\}} \right)|f(x)||x|^{\sigma-1}dx\leq\int_{F-K}|f(x)||x|^{\sigma-1}dx+C\int_{K-\{0\}}|x|^{\sigma-1}dx,$$
and we know that the second term in the last sum is finite for $\sigma>0$. I've tried to bound the first term using the properties of $f$ but I have not been able to reach somethig interesting. Any help is appreciated very much!
 A: For the sake of completeness, here are some details for anyone coming across this page with the same question. This isn't the cleanest way to prove it, but I think all the nit-picky details are here.
Let's call $K=\{x\in F:|x|_F\leq 1\}$. Again, write 
$$\begin{align*}
\int_{F^{\times}}|f(x)||x|^{\sigma-1}\ dx &= \left(\int_{F-K}+\int_{K-\{0\}} \right)|f(x)||x|^{\sigma-1}dx\\
&\leq\int_{F-K}|f(x)||x|^{\sigma-1}dx+C\int_{K-\{0\}}|x|^{\sigma-1}dx.
\end{align*}$$
So, we need only show the convergence of the first term of the final line, as the latter converges for $\sigma>0$ by basic calculus.
In the case $F = \mathbb R$, you can prove convergence by comparison to, for example $1/|x|^2$ since
$$\begin{align*}
\lim_{x\to\infty}\frac{|f(x)||x|^{\sigma-1}}{1/|x|^2} &= \lim_{x\to\infty}|f(x)||x|^{\sigma + 1}=0,
\end{align*}$$
implying there is some $M>0$ such that $|f(x)||x|^{\sigma-1} \leq 1/|x|^2$ for all $x > M$.  
In the case $F = \mathbb C$, consider $g(r) = \int_0^{2\pi}|f(re^{i\theta})|d\theta$.
Note that for any $d > 0$, 
$$\begin{align*}
\lim_{r\to\infty} |g(r)||r|^d
&= \lim_{r\to\infty} |r|^d\cdot\int_0^{2\pi}|f(re^{i\theta})|d\theta\\
&= \lim_{r\to\infty} \int_0^{2\pi}|re^{i\theta}|^d|f(re^{i\theta})|d\theta\\
&= \int_0^{2\pi}\lim_{r\to\infty} |re^{i\theta}|^d|f(re^{i\theta})|d\theta = 0,
\end{align*}$$
where moving the limit inside is justified by, say, Dominated convergence since for large enough $r$ the functions are bounded by a constant, and the since $f$ is Schwartz-Bruhat, the limit is zero.
Hence, if we rewrite
$$
\begin{align*}
\int_{\mathbb C-K}|f(x)||x|^{\sigma-1}dx 
&= \int_1^\infty\int_0^{2\pi}|f(re^{i\theta})||re^{i\theta}|^{\sigma-1}d\theta dr\\
&= \int_1^\infty\int_0^{2\pi}|f(re^{i\theta})|d\theta |r|^{\sigma-1}dr\\
&= \int_1^\infty|g(r)||r|^{\sigma-1}dr\end{align*},$$
we reduce to the real case, which we've already proven.
