Calculation of Tate epsilon factor in the ramified case Let $F$ be a nonarchimedean local field, $\chi$ a ramified character of $F^{\ast}$, $\psi$ a nontrivial character of $F$, and $dx$ a Haar measure on $F$ with respect to which the Fourier transform is defined.  The local Tate epsilon factor $\epsilon(\chi,\psi,dx)$ is defined via the local functional equation
$$\epsilon(\chi,\psi,dx) \int\limits_{F^{\ast}} f(x)\chi(x) d^{\ast}x = \int\limits_{F^{\ast}} \hat{f}(x) \chi^{-1}(x)|x| d^{\ast}x$$
for all suitable functions $f \in C_c(F)$.  Taking $f$ to be a characteristic function of a suitable open compact subgroup of $F$, one can realize $\epsilon(\chi,\psi,dx)$ as a "principal value integral"
$$\epsilon(\chi,\psi,dx) = \int\limits_{F} \chi^{-1}(x) \overline{\psi(x)} d^{\ast}x \tag{1}$$
in the sense that the right hand side becomes constant and equal to the left hand side when evaluated over $\mathfrak p_F^{-k}$ for sufficiently large $k$.
Let $f$ and $d$ be the conductors of $\chi$ and $\psi$.  Tate's article in Corvallis II describes $\epsilon(\chi,\psi,dx)$ more specifically as the right hand side of (1) where the integral is only taken over the annulus of $x \in F^{\ast}$ with $\operatorname{ord}(x) = -d-f$.
Here $\operatorname{ord}(c) = d +f$.

My question is, what can be said about the summands
$$\int\limits_{\pi^n \mathcal O^{\ast}} \chi^{-1}(x) \psi(x)dx$$
Are they all equal to zero except for $n = -d-f$?  In Tate's thesis, he shows they are zero for $n > -d-f$.  I was wondering whether the ones for $n < -d-f$ are individually zero, or whether they just cancel each other out in the long term.
 A: Let me for completeness repeat the argument from the proof of Lemma 1.1.1 in "Some remarks on local newforms for $(2)$" by Ralf Schmidt.
Define $$I(n)=\int_{\pi^n \mathcal{O}^{*}} \chi^{-1}(x)\psi(x)dx.$$ Since The case $n>-d-f$ seems to be known we focus on $n<-d-f$. (Since we are only interested in showing that these integrals vanish the precise normalisation of the measure is not so important. However, for convenience say dx is the additive Haar measure on $F$ normalised such that it gives measure $1$ to $\mathcal{O}$.) We decompose the integral as follows: $$I(n) = \vert \pi^n\vert \chi^{-1}(\pi^n)\sum_{y\in \mathcal{O}^*/(1+\pi^d \mathcal{O})}\chi^{-1}(y)\int_{1+\pi^d\mathcal{O}}\chi^{-1}(x)\psi(xy\pi^n)dx.$$ By the definition of the conductor we find that $\chi^{-1}$ restricted to $1+\pi^d\mathcal{O}^*$ is trivial. Thus we have \begin{align}\int_{1+\pi^d\mathcal{O}}\chi^{-1}(x)\psi(xy\pi^n)dx &= \int_{1+\pi^d\mathcal{O}}\psi(xy\pi^n)dx = \int_{1+\pi^d\mathcal{O}}\chi^{-1}(x)\psi(xy\pi^n)dx \\ &= \vert \pi^d \vert \psi(y\pi^n)\int_{\mathcal{O}}\psi(xy\pi^{n+d})dx.\end{align} Finally, by assumption on $n$ we have $n+d<-f$ such that, according to the definition of $f$, we have $$ \int_{\mathcal{O}}\psi(xy\pi^{n+d})dx=0.$$ This is enough to conclude that $I(n)=0$ for $n<-d-f$.
