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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!

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    $\begingroup$ $|f(x)|$ decreases faster than every (negative) power of $|x|$ so that $\int_{|x| > 1} f(x) |x|^{\sigma-1}d^\times x$ converges absolutely for every $\sigma$. The basic case is $F = \mathbb{R}, f(x) = e^{-\pi x^2}, \chi(x) = 1$ or $f(x) = x e^{-\pi x^2}, \chi(x) = \text{sign}(x)$ letting us obtain $\Gamma(s/2)$ or $\Gamma((s+1)/2)$ for the local factor at $\infty$ of Dirichlet L-functions $\endgroup$ – reuns Aug 29 '17 at 9:41
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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.

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