Feynman diagrams, can't Wick-rotate due to poles in first and third $p_0$ quadrants? I have a confusion about relating general diagrams (involving multiple propagators) in Minkowski vs Euclidean signature, which presumably should be identical (up to terms which are explicitly involved in Wick-rotation). I'm confident the resolution to my issue is simple, as it's on quite a fundamental/elementary topic. Please, I would appreciate any help on this.

In Minkowski signature $(-,+++)$, the scalar Feynman (causal) propagator is given by:
$$\Delta_M(x-y)=\int \frac{d^4 k}{(2\pi)^4}e^{i k\cdot (x-y)}\Delta_M(k) \tag{1}$$
$$\Delta_M(k)=\frac{i}{k^2+m^2-i\epsilon}=\frac{i}{-k_0^2+|\vec k |^2+m^2-i\epsilon} \tag{2}$$
Note that $\Delta(k)$ has two poles in the $k_0$ complex plane, at $k_0=\omega (\vec k )-i\epsilon$ and $k_0=-\omega (\vec k )+i\epsilon$. The poles reside in the second and fourth quadrants of the $k_0$ plane. The $k_0$-integral in the propagator is along the real-axis, from $-\infty$ to $+\infty$.
In Euclidean signature (++++), the scalar propagator is given by:
$$\Delta_E(x-y)=\int \frac{d^4 k}{(2\pi)^4}e^{i k\cdot (x-y)}\Delta_E(k) \tag{3}$$
$$\Delta_E(k)=\frac{-1}{k^2+m^2}=\frac{-1}{k_4^2 +|\vec k |^2 + m^2} \tag{4}$$
In the Euclidean propagator, the poles appear at $k_4=\pm i\omega (\vec p )$, and the $k_4$-integral is over the real-axis.
It's easy to see that when $\Delta_{M,E}(k)$ are integrated over the $k_{0,4}$ variables, they are numerically equal, since one can continuously deform the contour in the Minkowski integral to obtain the Euclidean integral. I've even heard the Feynman diagram be called the "Euclidean" propagator, presumably for this very reason. This idea is illustrated in the picture below, which shows the $k_{0,4}$ integrals happening in their respective complex planes.

Now consider two momentum-space propagators multiplying each other, as one might naturally get in a loop diagram. In Minkowski signature this would look like
$$\int \frac{d^4 k}{(2\pi)^4}\frac{f(k)}{\left(k^2+m^2-i\epsilon \right)\left((k-p)^2+m^2-i\epsilon \right)} \tag{5}$$
where $f(k)$ is some regular function of $k$ (i.e. no singularities). Note that in the $k_0$-integral we will have 4-poles, 2 for each propagator. The poles coming from the second propagator will be shifted by $p$.
Now consider the equivalent "diagram" in Euclidean signature. Let's define $p^\mu = (p_1,p_2,p_3 , ip_4 )$ where $p_{1,2,3,4}$ are all real. This is the Wick-rotated $p^\mu$. We will get
$$\int \frac{d^4 k}{(2\pi)^4}\frac{f(k)}{\left(k^2+m^2 \right)\left((k-p)^2+m^2\right)} \tag{6}$$
The integrand again has 4 poles, but this time, because of $p$, the second propagator need not have poles symmetrically positioned on opposite sides of the imaginary axis. This prevents us from immediately Wick-rotating the Minkowski diagram into the Euclidean one. See the following picture.

So it seems the Euclidean answer will be wildly different than the Minkowski answer, but for a seemingly superficial reason. The fact that they will be wildly different can be seen by the fact that, assuming the $k_{0,4}$ integrals can be done by closing the contour in a half-plane and using residue theorem, pole-closing will give us different poles, when something tells me they should be the same. It seems that this is boiling down to the existence of poles in the first and third quadrants of in the $k_0$ Minkowski plane.
I am confident about the expressions I've given for Minkowski signature, (1) (2) and (5), as they are totally standard in ordinary QFT classes. Therefore I believe my error lies in Wick-rotation, in the Euclidean expressions. For example, we don't normally Wick-rotate (5) until we have already combined all the propagators via Feynman parameters. This makes sense, since after combining propagators we reduce our original $2N$ poles in the $k_0$ plane into $2$ poles each of order $N$. We can now continuously deform the contour to be parallel with the imaginary axis.
So please, help me. What is going on here? What's wrong with my Euclidean expressions, (3) (4) and (6)?
 A: *

*Well, according to physics lore, the Wick rotation [i.e. the analytic continuation between the Minkowski (M) and Euclidean (E) formulations] should work, so if one encounters poles or branch cuts during the deformation of integration contour, they should be taken into account.


*That being said, it's possible to rewrite OP's one-loop diagram (5) $^1$
$$\begin{align}
I_M(p_M)&\cr
~:=~~~~~~&\int\! \frac{d^d k^{\bullet}_M}{(2\pi)^d}\frac{1}{\left(k_M^2+m^2-i\epsilon \right)\left((p_M\!-\!k_M)^2+m^2-i\epsilon \right)}  \cr 
~=~~~~~~&\int\! \frac{d^d k^{\bullet}_M}{(2\pi)^d}\int_0^1\!dx \frac{1}{\left[x\left((p_M\!-\!k_M)^2+m^2-i\epsilon \right)+(1\!-\!x)\left(k_M^2+m^2-i\epsilon \right) \right]^2}   \cr
~=~~~~~~&\int\! \frac{d^d k^{\bullet}_M}{(2\pi)^d}\int_0^1\!dx \frac{1}{\left[k_M^2-2xk_M\cdot p_M+xp_M^2 +m^2-i\epsilon \right]^2}   \cr
~\stackrel{\ell^{\mu}_M=k^{\mu}_M-x p^{\mu}_M}{=}&\int_0^1\!dx\int\! \frac{d^d \ell_M}{(2\pi)^d} \frac{1}{\left[\ell_M^2+x(1\!-\!x)p_M^2 +m^2-i\epsilon \right]^2}\cr
~\stackrel{\ell^0_M=i\ell^0_E}{=}~~&i\int_0^1\!dx\int\! \frac{d^d \ell_E}{(2\pi)^d} \frac{1}{\left[\ell_E^2+x(1\!-\!x)p_M^2 +m^2-i\epsilon \right]^2}\cr 
~\stackrel{d=4-\varepsilon}{=}~~~&\frac{i\Gamma(\frac{\varepsilon}{2})}{(4\pi)^{2-\varepsilon/2}}\int_0^1\!dx\left(x(1\!-\!x)p_M^2 +m^2-i\epsilon\right)^{-\varepsilon/2}\cr 
~=~~~~~~&\frac{i}{(4\pi)^2}\left[\frac{2}{\varepsilon} -\int_0^1\!dx\ln\left\{\frac{e^{\gamma}}{4\pi}\left(x(1\!-\!x)p_M^2 +m^2-i\epsilon\right)\right\} +{\cal O}(\varepsilon)\right] 
\end{align}\tag{A} $$
with the help of the Feynman parametrization, so that instead of 2 different propagators with 4 poles, the same propagator appears twice. After an appropriate shift of the loop momentum integration variable $k^{\mu}_M\to \ell^{\mu}_M$, there are only 2 poles in the quadrants II & IV,
$$ \ell^0_{M}~=~\left\{\begin{array}{lcl} 
\pm\left(\sqrt{\omega^2} -i\epsilon\right)&{\rm for}& \omega^2~>~0, \cr
\pm\left(i\sqrt{-\omega^2} -\epsilon\right)&{\rm for}& \omega^2~<~0,
\end{array}\right. \tag{B} $$
where
$$ \omega^2~~:=~\vec{\ell}^2 +x(1-x)p_M^2 +m^2. \tag{C}  $$
To perform the Wick rotation in eq. (A), assume that the external momentum $p^{\mu}_M$ is near the mass-shell $p_M^2\approx -m^2$, so that the $x$-integration doesn't cross the branch cut of the complex $\ln$ function.


*For comparison, the corresponding Euclidean one-loop diagram is
$$\begin{align}
I_E(p_E)&\cr
~:=~~~~~~&\int_{\mathbb{R}^d}\! \frac{d^d k_E}{(2\pi)^d}\frac{1}{\left(k_E^2+m^2 \right)\left((p_E\!-\!k_E)^2+m^2 \right)}\cr
~=~~~~~~&\ldots\cr
~=~~~~~~&\int_{\mathbb{R}^d}\! \frac{d^d k_E}{(2\pi)^d}\int_0^1\!dx \frac{1}{\left[k_E^2-2xk_E\cdot p_E+xp_E^2 +m^2 \right]^2}   \cr
~\stackrel{k^{\mu}_E=\ell^{\mu}_E+x p^{\mu}_E}{=}~&\int_0^1\!dx\int_{\mathbb{R}^d}\!\frac{d^d \ell^{\bullet}_E}{(2\pi)^d} \frac{1}{\left[\ell_E^2+x(1\!-\!x)p_E^2 +m^2 \right]^2}\cr 
~\stackrel{d=4-\varepsilon}{=}~~~&\frac{\Gamma(\frac{\varepsilon}{2})}{(4\pi)^{2-\varepsilon/2}}\int_0^1\!dx\left(x(1\!-\!x)p_E^2 +m^2\right)^{-\varepsilon/2}\cr 
~=~~~~~~&\frac{1}{(4\pi)^2}\left[\frac{2}{\varepsilon} -\int_0^1\!dx\ln\left\{\frac{e^{\gamma}}{4\pi}\left(x(1\!-\!x)p_E^2 +m^2\right)\right\} +{\cal O}(\varepsilon)\right]  
\end{align} \tag{D}$$
Note that the RHS of eq. (A) is the imaginary unit $i$ times the RHS of eq. (D) if we identify the external momentum $$p^0_M~=~ip^0_E.\tag{E}$$


*As OP points out in this accompanying Math.SE post,

*

*it is important in the above calculation (A) that the external momentum $p^0_M$ is real when we shift the loop momentum variable $k^{\mu}_M\to \ell^{\mu}_M$. If $p^0_M$ is imaginary, we would shift the integration contour away from the real axis. So when we then shift the integration contour back, we may pick up residues.


*Similarly for the Euclidean calculation (D), but now it is $p^0_E$ that should be real.
In light of eq. (E), there are unaccounted residues in at least 1 of the calculations (A) and (D).
--
$^1$ The bullet $\bullet$ in the integration measure indicates the position the spacetime index. The Minkowski sign convention is $(-,+,+,+)$. In this answer, we will implicitly assume that UV divergences (for large $k$) can and have been properly regularized, e.g. via dimensional regularization.
