Let $\nu \leq \infty$ and $\mathbb{Q}_\nu$ denote a local field. Suppose $s \in \mathbb{C}$ with $\Re(s) > 0$. Let $\Phi: \mathbb{Q}_\nu \to \mathbb{C}$ be a Schwartz-Bruhat function. Let $\omega : \mathbb{Q}_\nu^\times \to \mathbb{C}^\times$ be a local unitary character. Defining $$Z_\nu(s, \Phi, \omega) := \int_{\mathbb{Q}_\nu^\times} \Phi(x) \omega(x) |x|_\nu^s \, d^\times x$$ as the local integral associated to $\omega$ and $\Phi$.
Question 1: Is the following proof for the convergence of following integral correct? Suppose $\Re(s) >0$ as $\Phi$ has compact support then there exists $0 < K,M < \infty$ so that \begin{align*} \left|\int_{\substack{x \in \mathbb{Q}_\nu^\times \\ |x|_\nu >1}} \Phi(x) \omega(x) |x|_\nu^s \, d^\times x\right| &=\left| \int_{\substack{x \in \mathbb{Q}_\nu^\times \\ 1<|x|_\nu\leq K }}\Phi(x) \omega(x) |x|_\nu^s \, d^\times x \right|\\ &\leq \int_{\substack{x \in \mathbb{Q}_\nu^\times \\ 1<|x|_\nu\leq K }} M K^{\Re(s)} \, d^\times x\\ &\leq \mu\{1 < |x|_\nu \leq K\} M K^{\Re(s)}\\ &<\infty \end{align*} where $\mu$ denotes the Haar measure on the multiplicative group on $\mathbb{Q}_\nu^\times$.
Question 2 (Page 59): Suppose $\nu = p$, a finite prime $p$ and $\omega$ is ramified with conductor $p^r$. Defining $\Phi^\circ(x) := e^{- 2 \pi i\{x\}}$ if $x \in p^{-r} \mathbb{Z}_p$ and $0$ otherwise. Where $$\{x\} := \begin{cases} \sum_{i=-k}^{-1} a_i p^i &\ \text{if}\ x = \sum_{i=-k}^\infty a_i p^i \in \mathbb{Q}_p\\ 0 & \ \text{otherwise} \end{cases}.$$ Then, \begin{align} Z_p(s, \Phi^\circ, \omega) &= \frac{p}{p-1} \int_{p^{-r} \mathbb{Z}_p \setminus \{0\}} e^{- 2\pi i\{x\}} \omega(x) |x|_p^s \frac{dx}{|x|_p}\\ &= \frac{p}{p-1} \sum_{l=1}^r \sum_{\substack{j=1\\ (j,p) = 1}}^{p^r} \int_{p^{-l}(j + p^r \mathbb{Z}_p)} e^{- 2\pi i\{x\}} \omega(x) |x|_p^s \frac{dx}{|x|_p}\\ &= p^{-r} \frac{p}{p-1} \sum_{l=1}^r p^{ls} \omega(p)^{-l} \sum_{\substack{j=1\\ (j,p) =1}}^{p^r} e^{- \frac{2 \pi i j}{p^l}} \omega(j)\\ &= \frac{p^{-r + 1}}{p-1} p^{rs} \omega(p)^{-r} \sum_{\substack{j=1\\ (j,p) =1}}^{p^r} e^{- \frac{2 \pi i j}{p^l}} \omega(j). \end{align}
- How do you obtain the decomposition of $p^{-r}\mathbb{Z}_p \setminus \{0\}$ to obtain the Gauss sum?
- Why does $\sum_{\substack{j=1\\ (j,p) =1}}^{p^r} e^{- \frac{2 \pi i j}{p^l}} \omega(j) =0$ whenever $l < r$?
Question 3: Assuming that $$\widehat{\Phi^\circ}(x) := \begin{cases} p^r& \text{if}\ x \in 1 + p^r \mathbb{Z}_p\\ 0& \text{otherwise}\end{cases}.$$ Then, \begin{align} Z_p(1-s, \widehat{\Phi^\circ}, \overline{\omega}) &= p^r \frac{p}{p-1} \int_{1 + p^r \mathbb{Z}_p} \overline{\omega(x)} |x|_p^{1-s} \frac{dx}{|x|_p}\\ &= p^r \frac{p}{p-1}\int_{1 + p^r \mathbb{Z}_p} \frac{dx}{|x|_p}\\ &= \frac{p}{p-1}. \end{align}
- Is there is mistake in this computation? As I believe it should be compute as follows $$Z_p(1-s, \widehat{\Phi^\circ}, \overline{\omega}) = p^r \frac{p}{p-1} \int_{1 + p^r \mathbb{Z}_p} \overline{\omega(x)} |x|_p^{1-s} \frac{dx}{|x|_p} = p^r \frac{p}{p-1} \int_{1 + p^r \mathbb{Z}_p} \frac{dx}{|x|_p^s} = p^r \cdot \frac{p}{p-1} p^{rs} p^{-r} = \frac{p}{p-1} p^{rs}.$$
Source: Goldfeld-Hundley, automorphic representations and $L$-functions for the general linear group (Pg 55-60).