Tate gamma factor as a principal value integral Let $F$ be a local field, $\chi$ a multiplicative character of $F^{\ast}$, and $\psi$ an additive character of $F$.  The gamma factor $\gamma(s,\chi,\psi)$ is defined by means of the local functional equation
$$Z(\hat{f},\chi^{-1},1-s) = \gamma(s,\chi,\psi)Z(f,\chi,s)$$
where $f$ is any locally constant and compactly supported complex valued function, the Fourier transform $\hat{f}$ is defined by $\hat{f}(x) = \int\limits_F f(x)\psi(xy) \space dy$, the Haar measure $dy$ is taken to be self-dual with respect to $\psi$, and the zeta integral $Z(f,\chi,s)$ is defined by
$$\int\limits_{F^{\ast}} f(x)\chi(x)|x|^s d^{\ast}x. \tag{$d^{\ast}x = \frac{dx}{|x|}$}$$
I have read that the gamma factor can be calculated informally as an "principal value integral''
$$\gamma(s,\chi,\psi)^{-1} = \lim\limits_{N \to \infty} \int\limits_{|x| \leq N} \chi(x)|x|^s \psi(x) \space \frac{dx}{|x|}$$
without reference to a choice of test function $f$.  Why is this?  Also, when $\chi$ is ramified, why do all of the "annuli''
$$\int\limits_{\varpi^k \mathcal O_F^{\ast} } \chi(x) |x|^s \space \psi(x) \frac{dx}{|x|}$$
vanish except for one?
 A: For the second question, let $\varpi$ be a uniformizer, and set $\mathfrak p = \varpi \mathcal O_F$.  Let's define the conductor of $\chi$ to be $f$ if $\chi$ is trivial on $1+\mathfrak p^f$, but not on $1+\mathfrak p^{f-1}$ (here we set $1+\mathfrak p^0 = \mathcal O_F^{\ast}$).  Let's also define the conductor of $\psi$ to be $d$ if $\psi$ is trivial on $\mathfrak p^d$, but not on $\mathfrak p^{d-1}$.  Thus saying $\chi$ or $\psi$ is unramified means its conductor is $0$.
Lemma 1: Assume $\chi$ is ramified (that is, $f > 0$).  If $f \neq d$, then
$$\int\limits_{\mathcal O_F^{\ast}} \chi(x) \psi(x) d^{\ast}x = \int\limits_{\mathcal O_F^{\ast}} \chi(x) \psi(x) dx = 0$$
Proof: The first equality is always true because on $\mathcal O_F^{\ast}$, we have $d^{\ast}x=dx$.
Assume $d > f$, so that $\psi$ is nontrivial on $\mathfrak p^f$.    Then
$$\int\limits_{\mathcal O_F^{\ast}} \chi(x) \psi(x) dx = \int\limits_{\mathcal O_F^{\ast}/(1+\mathfrak p^f)} \space \space \int\limits_{1+\mathfrak p^f} \chi(xy)\psi(xy) \space dy \space dx.$$
The inner integral is (noting that $\chi(y) = 1$ for all $y \in 1+\mathfrak p^f$
$$ \chi(x) \int\limits_{1+\mathfrak p^f}  \psi(xy)dy = \chi(x)\int\limits_{\mathfrak p^f} \psi(x(1+y))dy = \chi(x) \psi(x) \int\limits_{\mathfrak p^f} \psi(y)dy = 0. $$
On the other hand, if $f > d$, then $\chi$ is nontrivial on $1 + \mathfrak p^{f-1}$, but $\psi$ is trivial on $\mathfrak p^{f-1}$.  Thus
$$\int\limits_{\mathcal O_F^{\ast}} \chi(x) \psi(x) dx = \int\limits_{\mathcal O_F^{\ast}/(1+\mathfrak p^{f-1})} \space \space \int\limits_{1+\mathfrak p^{f-1}} \chi(xy) \psi(xy) \space dy \space dx$$
where the inner integral is
$$ \chi(x)\psi(x) \int\limits_{1+\mathfrak p^{f-1}} \chi(y) \psi(x(y-1)) dy = \chi(x) \psi(x) \int\limits_{1+\mathfrak p^{f-1}} \chi(y) dy = 1. \space \space \blacksquare$$
Corollary: Whenever $k \neq d - f$, we have
$$ \int\limits_{ \varpi^k\mathcal O_F^{\ast}} \chi(x) \psi(x) d^{\ast}x = 0$$
Proof: This is because
$$\int\limits_{ \varpi^k\mathcal O_F^{\ast}} \chi(x) \psi(x) d^{\ast}x = \chi(\varpi^k) \int\limits_{\mathcal O_F^{\ast}} \chi(x) \psi(\varpi^kx) d^{\ast}x$$
and the conductor of the character $x \mapsto \psi(\varpi^kx)$ is $d-k$.  $\space \space$ $\blacksquare$
It is also worth mentioning what happens in the unramified case.
Proposition: Assume $\chi$ is unramified.  Then
$$\int\limits_{\varpi^k\mathcal O_F^{\ast}} \chi(x) \psi(x) d^{\ast}x=0$$
whenever $d-k \geq 2$.
Proof: In this case, if we let $\psi'(x) = \psi(\varpi^kx)$, the integral is equal to
$$ \chi(\varpi^k)\int\limits_{\mathcal O_F^{\ast}} \psi'(x) dx = \chi(\varpi^k) \int\limits_{\mathcal O_F^{\ast}/(1+\mathfrak p^{d-k-1})} \space \space \int\limits_{1+\mathfrak p^{d-k-1}} \psi'(xy) \space dy \space dx.$$
The conductor of $\psi'$ is $d-k \geq 1$, so $\psi'$ is nontrivial on $\mathfrak p^{d-k-1}$, with $d-k-1 \geq 1$.  So the inner integral is
$$\int\limits_{\mathfrak p^{d-k-1}} \psi'(x(1+y)) \space dy = \psi(x) \int\limits_{\mathfrak p^{d-k-1}}  \psi'(xy) \space dy = 0.$$
