Elementary question about endpoint singularities In George Sterman's book "An Introduction to Quantum Field Theory", on pages 413-414, there is a description of the endpoint singularity. One begins with the function
$$ I(w) ~=~ \int_{\zeta_a}^{\zeta_b}\! d\zeta\,F(\zeta, w)\tag{13.2} $$
of a rational function $F(\zeta, w)$ of $\zeta$ which has at worst, isolated poles in $\zeta$ with positions $\zeta = \xi_i(w)$ for some $w$.
Suppose one of the poles migrates to $\zeta_a$. In the language of the book, there exists $w = w_0$ such that $\zeta_a = \xi_i(w_0)$. Consider the analytic continuation of $I(w)$ from $w_1$ to $w_2$ along a path $R_j$, which avoids $w_0$.
According to the book,

Any such continuation corresponds to a path $\rho_j$ of pole $\xi_i(w)$ in the $\zeta$ plane from $\xi_i(w_1)$ to $\xi_i(w_2)$. The latter paths are of two types: paths like $\rho_1$ go around the contour, while those like $\rho_2$ cross the contour, which then encloses the extra pole at $\xi_i(w_2)$.
The two analytic continuations thus differ by $2\pi i Z(w_2)$ where $Z_i(w_2)$ is the residue of the pole of $F(\zeta, w)$ at $\zeta = \xi_i(w_2)$.

My question is: why is there the extra contour around the pole at $\xi_i(w_2)$ in the second case at all?
This is probably a very trivial question.
EDIT: Here is a Google books link to the pages concerning this issue.
 A: Use fig 13.2 of [2] as reference.
Taking the example Qmechanic uses, the idea is that 
$I(\omega) = \int_{\zeta_a}^{\zeta_b}\frac{Z(\omega)}{\zeta - \xi(\omega)} d\zeta$
$I(\omega)$ needs to be analytically continued from $\omega_1 \rightarrow \omega_2$.
The pole of the integrand  travels from $\zeta = \xi(\omega_1) \rightarrow \zeta = \xi(\omega_2)$ in the complex plane of $\zeta$.
At some point $\omega' \in (\omega_1, \omega_2)$, the expression for $I(\omega')$ is no longer valid, since $\xi(\omega')$ falls on the contour defined in $I(\omega)$, seen in the 3rd diagram in Fig 13.2 a [2].
But, we could avoid this whole situation by analytically continuing from $\omega_1 \rightarrow \omega_2$ along a path which avoids the contour of $I(\omega)$ altogether, which is what happens in 2nd diagram in Fig 13.2 a (second line left)[2]. Analytical continuation along this path in $\omega$-plane gives the same function $I(\omega)$ written above.
For the problematic path (3rd diagram in Fig 13.2 a [2]), analytical continuation would require that we bypass the pole somehow such that the new expression $I'(\omega)$ is a analytical function for all points $\omega \in (\omega_1, \omega_2)$.
A simple way to do this is to define
$$I'(\omega) = \int_\mathcal{C}\frac{Z(\omega)}{\zeta - \xi(\omega)} d\zeta$$
Where the contour is now

Why does this work?
The integrand is defined throughout  $\omega \in (\omega_1, \omega_2)$ and it matches with the expression $I(\omega)$ when $\xi(\omega)$ lies in the region below the contour of $I(\omega)$, where it was supposed to be an analytic function to begin with.
Now, it can be seen easily that $I'(\omega)-I(\omega) = 2\pi\iota Res(\omega_2)$
[2]: G. Sterman, Intro to QFT, 1993; Fig 13.2 p. 414
A: I) Let there be given a meromorphic function $\zeta \mapsto F_{w}(\zeta)$
in the $\zeta$-plane with a single (not necessarily simple) pole at the position $\zeta=\xi(w)$, where $\xi$ is a holomorphic function, and $w\in \mathbb{C} $ is an external parameter.
Ref. 1 is considering the contour integral 
$$\tag{A}I_{\Gamma,w} ~=~ \int_{\Gamma} \! d\zeta ~F_w(\zeta), $$
along an open oriented curve $\Gamma:[a,b]\to \mathbb{C}$ with boundary conditions
$$\tag{B} \Gamma(a)~=~\zeta_a\quad\text{and}\quad \Gamma(b)~=~\zeta_b. $$
II) The integral (A) is well-defined if the pole $\xi(w)$ does not lie on the integration contour $\Gamma([a,b])$.
On the other hand, the integral (A) does not change if we deform the
integration contour from $\Gamma_0$ to $\Gamma_1$ via a homotopy
$$ H: [0,1] \times  [a, b]\to \mathbb{C},\qquad H( 0,\cdot)~=~ \Gamma_0,\qquad H(1, \cdot)~=~ \Gamma_1, $$
$$\tag{C} H(\cdot,a)~=~\zeta_a,\qquad  H(\cdot,b)~=~\zeta_b, $$
without crossing the pole $\xi(w)$, cf. Cauchy's integral theorem.
III) Now Ref. 1 is studying the monodromy of an end-point singularity, say at the upper end-point $\zeta_b=\xi(w_b)$. This means effectively considering a closed oriented curve $\gamma:[0,1]\to \mathbb{C}$ in the $w$-plane 
$$\tag{D}  w~=~\gamma(t), \qquad t\in [0, 1], $$
that encircles $w_b$. Let's call the common start and end-point for $w_0\equiv w_1$.
This, in turn, induces a closed curve $\xi\circ \gamma$ in the $\zeta$-plane 
$$\tag{E}  \zeta~=~\xi\circ \gamma(t), \qquad t\in [0, 1], $$
that encircles $\zeta_b$, say, in the positive direction/counter-clockwise.
When we go along the curve (E), we must deform/adjust the integration contour $\Gamma_t\equiv H(t,\cdot)$ correspondingly in order to avoid that the singularity touches the integration contour. The curve (E) pushes the the integration contour $\Gamma_t\equiv H(t,\cdot)$ in front of it, so to speak, $t\in [0, 1]$. 
After a full circle, the change in the integral (A) becomes
$$\tag{F} I_{\Gamma_1,w_1}-I_{\Gamma_0,w_0}
~=~ \oint_{C(\xi(w_0))} \! d\zeta ~F_w(\zeta)
~=~ 2\pi i~Z_{w_0}, $$
where
$$\tag{G}  Z_w~:=~{\rm Res}\left(F_{w},\zeta=\xi(w)\right), $$
cf. the Residue theorem. 
Here $C(\xi(w_0))$ denotes a simple closed contour in the $\zeta$-plane oriented in positive direction around the common start and end-point $\xi(w_0)$ for the closed curve (E).
We stress that the residue (F) is taken at the common start and end-point $\xi(w_0)$ for the closed monodromy curve (E), rather than the end-point $\zeta_b$ of the open integration contour $\Gamma_0$, as the name end-point singularity may perhaps naively suggest.
References:


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*G. Sterman, Intro to QFT, 1993; p. 413-414.

