Test vector for local zeta integral with ramified character Suppose $\pi$ is an unramified principal series representation of ${\rm GL}_2(F)$, where $F$ is a non-archimedean local field with integers $\mathfrak{o}$. Let $W$ be a function in its Whittaker model. If $\chi$ is a quasi-character of $F^\times$ then we define its local zeta integral as 
$$ Z(W, s, \chi, g) = \int_{F^\times} W((\begin{smallmatrix} y & 0 \\ 0 & 1 \end{smallmatrix})g) \chi(y) |y|^{s-1/2}\, d^\times y,$$ see e.g. (6.28) of Gelbart's book. Then for $W= W_0$ the unique $K = {\rm GL}_2(\mathfrak{o})$-invariant Whittaker function with $W_0(1)=1$, we have  $$Z(W_0,s,\chi,1) = L(s, \pi \otimes \chi),$$ see e.g. prop 6.17b of Gelbart.
Note that if $\chi$ is ramified, then we have simply $$L(s,\pi \otimes \chi)=1.$$ Let's write $y \in F^\times$ as $p^n x,$ where $x \in \mathfrak{o}^\times$, and $p$ is a uniformizer, and $\chi$ as $\chi(p^nx) = |p^n|^{s'} \chi^*(x),$ where $ \chi^*$ is a character of $\mathfrak{o}^\times$. Then we have
$$Z(W,s, \chi, 1) = \sum_{n \in \mathbb{Z}} W((\begin{smallmatrix} p^n & 0 \\ 0 & 1 \end{smallmatrix}))|p^n|^{s-1/2+s'} \int_{x \in \mathfrak{o}^\times} \chi^*(x)\,dx,$$
which is $0$ if $\chi$ is ramified. Contradiction.
Am I correct to interpret the above to mean that the $W_0$ that I have chosen above is not the correct choice of test vector for the zeta integral when $\chi$ is ramified? If so, what is the correct test vector?
 A: Yes.  The issue is that $Z(-, s, \chi, 1)$ is a linear functional which transforms by $\chi$ on $\mathfrak o^\times$ (say in the Kirillov model).  So if you plug in a function which is $\mathfrak o^\times$-invariant and $\chi$ is ramified, you must get 0. 
A: [Expanding my comment to an answer]
The new vector of $\pi$ is not the correct test vector to use when $\chi$ is ramified; and if Gelbart really claims this holds for all $\chi$, then he is wrong.
You can decompose the Kirillov model of $\pi$ as a (countably infinite) direct sum of eigenspaces for the action of $\mathfrak{o}^\times$. It's easy to check that any function that is in an eigenspace other than the $\chi^{-1}$-eigenspace will be sent to 0 by $Z(-, s, \chi, 1)$. Conversely, a Schwartz function on $F$ which lies in the $\chi^{-1}$-eigenspace for $\mathfrak{o}^\times$ had better vanish at 0 and hence have compact support in $F^\times$; so this eigenspace is precisely the linear combinations of the functions $\chi^{-1} \cdot 1_{\varpi^n \mathfrak{o}^\times}$, for $n \in \mathbf{Z}$ (exercise). So the space of functions $\{ Z(W, s, \chi, 1) \}$ for varying $W$ is exactly the space of polynomials in $q^{-s}$, and the function with Kirillov model $\chi^{-1} 1_{\varpi^n \mathfrak{o}^\times}$ will be a test vector for any $n$, although $n = 0$ is surely the most obvious choice. (Note that the vector with Kirillov function $\chi^{-1} 1_{\mathfrak{o}^\times}$ maps to the new vector of $\pi \otimes \chi$.)
