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Consider the field of $p$-adic numbers $\mathbb Q_p$. Define the character $\chi(u p^n) = e(p^n)$ for all $n \in \mathbb Z$ and all unit $u$. In particular it is trivial on integers. This allows to define a Fourier transform and I am interested in computing the one of $\mathbb{1}_{\mathcal O_p}$ (characteristic function of the $p$-adic integers $\mathcal O_p$.

I tried to cut by valuations, without success (the measure is normalized so that $\mathcal O_p$ gets measure one). Let $y = p^{-n} v$ for $n \in \mathbb Z$ and $v$ a unit: \begin{align} \int_{\mathbb Q_p} e_p(-x y) \mathbb{1}_{\mathcal O_p}(x) dx & = \int_{\mathcal O_p} e_p(-x y) dx \\ & = \sum_{k \geq n} \int_{\mathcal p^k O_p^\times} e_p(-x y) dx + \sum_{0\leq k < n} \int_{\mathcal p^k O_p^\times} e_p(-x y) dx \end{align}

Now I can write $x = p^k u \in p^k \mathcal{O}_p^\times$, and get \begin{align} & \sum_{k \geq n} \int_{\mathcal O_p^\times} e_p(-p^{k-n} uv) p^{-k} du + \sum_{0 \leq k < n} \int_{\mathcal p^k O_p^\times} e_p(-p^{k-n} uv) p^{-k} du \\ & = \sum_{k \geq n} p^{-k} \mathrm{vol}(\mathcal O_p^\times) + \sum_{0 \leq k < n} e(p^{k-n}) p^{-k} \mathrm{vol}(\mathcal O_p^\times) \end{align}

When $n \leq 0$, i.e. $y$ is integer, I indeed get 1 as expected. However when it is not the case I don't know how to compute this sum (and, expectedly, get 0).

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    $\begingroup$ What’s $e(p^n)$ or $e_p$ ? $\endgroup$ – Lubin Jan 17 at 3:48
  • $\begingroup$ @Lubin $e=e_p$ is defined by $e(x) = \exp(2i\pi \{x\}_p)$ where $\{x\}_p$ is the fractional $p$-adic part of $x$. So essentially if $x=p^n$, then $e(x) = \exp(2i\pi p^n)$. $\endgroup$ – Gory Jan 17 at 6:24
  • $\begingroup$ And then $e(p^n)=1$? $\endgroup$ – Lubin Jan 17 at 21:34
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For $a,b\in \Bbb{Z}_p$ and $k\ge 0$ an integer $$\exp(-2i\pi ab/p^k)=\exp(-2i\pi (ab\bmod p^k)/p^k)$$ is well-defined. Then

$$\int_{\mathbb Z_p} \exp(-2i\pi x y)dx=\lim_{n\to \infty}\frac1{p^n} \sum_{a=0}^{p^n -1} \exp(-2i\pi a y)= 1_{y\in \Bbb{Z}_p}$$

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