Integrating an additive character over a local field Let $F$ be a non-Archimedean local field, $\psi$ a non-trivial additive character of $F$. Let $\mathfrak{o}$ be the ring of integers of $F$, and $\mathfrak{p}$ be the maximal ideal of $F$. Endow $F$ with a Haar measure (possibly normalized so that $\mu(\mathfrak{o}) = 1$).
I want to show that
$$ \int_{\mathfrak{p}^{-1}-\mathfrak{o}} \psi(-x) \ dx = -1 $$
and
$$ \int_{\mathfrak{p}^{-(k+1)}-\mathfrak{p}^{-k}} \psi(-x) \ dx = 0 $$
for all $k \geq 1$.
This feels like it's in the style of when you sum over all the non-trivial roots of unity, you get $1$. But I can't quite seem to translate this intuition into a proof. Any help will be greatly appreciated.
This integral shows up in Bump's Automorphic forms and representations, Ch. 4.6 p. 499.
Edit 1: I should mention that $\psi$ is trivial on $\mathfrak{o}$, and the codomain of $\psi$ seems to be $\mathbb{C}^\times$, the multiplicative group.

Since $\psi$ is trivial on $\mathfrak{o}$, we have
$$ \int_{\mathfrak{o}}\psi(-x)\ dx = \mu(\mathfrak{o}) = 1. $$
Since $\psi$ is non-trivial on the additive group $\mathfrak{p}^{-1}$, by the orthogonality of characters, we have that the pairing of $\psi(-x)$ with the trivial character on $\mathfrak{p}^{-1}$ is 0, and hence
$$ 0 = \int_{\mathfrak{p}^{-1}}\psi(-x)\ dx = \int_{\mathfrak{p}^{-1}-\mathfrak{o}}\psi(-x)\ dx + \int_{\mathfrak{o}}\psi(-x)\ dx = \int_{\mathfrak{p}^{-1}-\mathfrak{o}}\psi(-x)\ dx + 1. $$
A similar argument holds by induction to the additive groups $\mathfrak{p}^{-k}$.
