Let $F$ be a nonarchimedean local field with ring of integers $\mathcal O_F$, maximal ideal $\mathfrak p$, and uniformizer $\varpi$. Let $\chi$ be a nontrivial character of $\mathcal O_F^{\ast}$, and $\psi$ a nontrivial character of $F$. Let $f$ and $d$ be the conductors of $\chi$ and $\psi$. This means that $f \geq 0$ and $d$ are the unique integers satisfying:
$\chi$ is trivial on $1+ \mathfrak p^{f}$, but not on $1+\mathfrak p^{f-1}$ (if $\chi$ is trivial on $1+\mathfrak p$, we understand $1+\mathfrak p^{f-1} = \mathcal O_F^{\ast}$).
$\psi$ is trivial on $\mathfrak p^{-d}$, but not on $\mathfrak p^{-d-1}$.
The local epsilon factor as defined by Tate can be calculated by a "Gauss sum" obtained by integrating $\chi$ against $\psi$. To calculate this Gauss sum, I am trying to show that for $0 \neq a \in F$
$$S = \int\limits_{\mathcal O_F^{\ast}} \chi(t)\psi(at)d^{\ast} t = 0$$ whenever $\operatorname{ord}(a) \neq -d -f$. Two out of the three cases are proved in Tate's thesis, and I've written their proofs below. The third case I haven't been able to prove yet.
Case 1: $\operatorname{ord}(a) \geq -d$.
In this case, $at \in \mathfrak p^{-d}$ for all $t \in \mathcal O_F^{\ast}$, so $\psi(at) = 1$ and $S = \int\limits_{\mathcal O_F^{\ast}} \chi(t) d^{\ast}t = 0$, since we are assuming $\chi$ is nontrivial.
Case 2: $-d -f < \operatorname{ord}(a) < -d $.
This case only occurs when $f \geq 2$. Let $a = u \varpi^n$, where $n = \operatorname{ord}(a)$. Here we use the fact that $-n-d$ is a positive integer, and $\mathcal O_F^{\ast}$ is a union of sets of the form $z + \mathfrak p^{-n-d}$, where $0 \neq z \in \mathcal O_F$ with $\operatorname{ord}(z) < -n - d$. On such sets, the integral is
$$\int\limits_{z + \mathfrak p^{-n-d}} \chi(t)\psi(at)d^{\ast}t$$
Write $z = v \varpi^k$, with $0 \leq k < - n - d$. We have for $x \in \mathfrak p^{-n-d}$,
$$a(z+x) = az+ax $$ with $\psi$ taking the constant value $\psi(az)$, since $\psi(az+ax) = \psi(az)\psi(ax)$ with $$\operatorname{ord}(ax) = n + \operatorname{ord}(x) \geq n + (-n-d) = -d$$ so that $\psi(ax) = 1$ for all $x \in \mathfrak p^{-d-n}$.
Thus
$$\int\limits_{z+\mathfrak p^{-n-d}} \chi(t)\psi(at) d^{\ast}t = \psi(az)\int\limits_{z+\mathfrak p^{-n-d}} \chi(t) d^{\ast}t = \psi(az) \chi(z)\int\limits_{1 + \mathfrak p^{-n-d-k}} \chi(1+t) d^{\ast}t $$ with $$-n-d-k < (d+f) - d - k = f - k \leq f$$ Since $\chi$ is therefore nontrivial on $1+\mathfrak p^{-d-n-f}$, the given integral must be zero.
Case 3: $\operatorname{ord}(a) < -d -f$
I was thinking this case might not actually be true. The epsilon factor is given by a principal value integral which stabilizes over a large set. Maybe integrals for $|a|$ this large might not be relevant for the calculation of local factors.