D'Alembertian Green's Function and Wick Rotation

Consider the wave equation:

$$\square A(t,x^i) = S(t,x^i) ,$$

where $\square = -\partial_\mu \partial^\mu = \partial_t ^2 - \nabla^2$, $S(t,x^i)$ is the source term and $A(t,x^i)$ is the field of interest.

There are various ways to find the Green's function for the wave equation.

$$\square G(t,x^i) = \delta^{(4)}(t,x^i) \left( = \delta(t) \delta^{(3)}(x^i) \right)$$

For example, Landau starts by setting $G(t,x^i) = \chi(t) \cdot \frac{1}{4\pi r^2}$ and deduce that $\chi(t) = \delta(t-r)$ ($r:= (x_i x^i )^{1/2}$). On the other hand, the most popular and systematic way is to apply Fourier transformation:

$$\ast \ \ast \ \ast$$

$$\ast \ \ast \ \ast$$

After dealing with some complex integration with an appropriate contour, one gets $G_{\rm{ret}}(t,x^i) = \delta(t-r) \frac{1}{4\pi r^2}$. However, what I found fascinating was the following approach, using Wick Rotation.

First, one performs the Wick rotation $t = x^0 := x^4 / i$ . Then,

$$\square = \eta^{\mu\nu} \partial_\mu \partial_\nu = \delta^{AB}\partial_A \partial_B \ \ \ (\mu, \nu = 0,1,2,3, \ A, B = 1,2,3,4)$$

so,

$$\frac{1}{\square} \delta^{(4)}(x) = \frac{1}{4\pi^2 } \cdot \frac{1}{\delta_{AB} x^A x^B},$$

in analogy with the three-dimensional Euclidean result

$$\frac{1}{- \nabla^2} \delta^{(3)}(x) = \frac{1}{4\pi } \cdot \frac{1}{r^1}$$

( $4 \pi$ and $4\pi^2$ is the solid angle in $\mathbb{E}^3$ and $\mathbb{E}^4$ , respectively).

Therefore,

$$A(t,x^i) = \frac{1}{\square} S(t,x^i) = \frac{1}{\square} \iiint d^3 x' \int dx'^4 \ S(x'^4/i, x'^i) \delta ^{(4)} (x)$$

$$= \iiint d^3 x' \int dx'^4 \ S(x'^4/i, x'^i) \frac{1}{4 \pi^2} \frac{1}{-(t-t')^2 + R^2 }$$

$$= \frac{i}{4 \pi^2} \iiint d^3 x' \int^{-i \infty}_{i \infty} dt' \ S(t', x'^i) \frac{1}{-(t-t')^2 + R^2 }$$

$$= \frac{i}{4 \pi^2 } \iiint d^3 x' \int^{i \infty}_{-i \infty} dt' \frac{S(t', x'^i)}{ (t' -t +R)(t'-t - R ) }$$

where $R:= \left( (x_i - x'_i)(x^i - x'^i) \right)^{1/2}$. This approach is adapted from Lightman et al.'s "Problem book in relativity and gravitation, Princeton University Press."

At the last step, the author performs complex integration with a following path,

and gets the correct result (the retarded Green's function):

$$= \frac{i}{4 \pi^2 } \iiint d^3 x' \left[ (2\pi i)\cdot \frac{ S(t-R , x'^i)}{-2R} \right]$$

$$= \frac{1}{4 \pi } \iiint d^3 x' \left[ \frac{S(t-R , x'^i)}{R} \right]$$

However, I am not sure that why should the contour pass between two poles $t' = t-R$ and $t' = t+R$. I understand that the orientation of folding the original contour (the imaginary line $- i \infty$ to $i \infty$) corresponds to different boundary conditions (retarded and advanced), but what if $0<t-R$ or $t+R < 0$ so that there are no residue contribution from poles and the $t'$-integral goes to zero? (Note that the original contour passes over $0$.)