Contradiction in simple complex integral: should this pole contribute

calculuscomplex integrationcomplex-analysismathematical physicsphysics

This is a follow-up question to a post I made on Physics SE. TLDR; Essentially I have derived an apparent contradiction, starting from a simple integral. It feels like I'm missing something quite elementary and trivial, but I cannot seem to resolve it.

Consider the following integral, whose approximate form is commonly finds in physics.

$$f(x)=\int_{-\infty}^{+\infty}\frac{dz}{\left(-z^2 + m^2 – i\epsilon\right )^2 \left(-(z-x)^2+ M^2 -i \epsilon\right)} \tag{1}$$

$m,M>0$ are positive real numbers, $x$ is a positive real number (which we may allow to be complex), and $\epsilon>0$ is an infinitesimal small number which shifts the poles away from the real axis. In the complex $z$ plane, the integral looks as follows:

enter image description here

The blue dots correspond to the double-poles at $z=\pm (m – i\epsilon)$, and red dots correspond to the single-poles at $z=x\pm (M-i\epsilon )$. The green line represents the integration contour. Notice that the pole at $z=x-M+i\epsilon$ resides in the $1^{\textrm{st}}$ quadrant.

Now let's apply the technique of Feynman parameterization, which combines both the rational terms in the integrand into a single rational function.

$$\begin{align}
f(x)&=\int_{-\infty}^{+\infty}dz \int_0^1 du_1 \int_0^{1-u_1} du_2 \frac{2!}{\left[
(1-u_1-u_2)\left( -(z-x)^2+ M^2 -i \epsilon \right)+u_2 \left( -z^2 + m^2 – i\epsilon\right) +u_3 \left( -z^2 + m^2 – i\epsilon \right) \right]^3} \\
&= \int_0^1 du_1 \int_0^{1-u_1} du_2 \int_{-\infty}^{+\infty}dz \frac{2}{\left[-(z-(1-u_1-u_2)x)^2-(u_1+u_2)(1-u_1-u_2)x^2+(1-u_1-u_2)M^2+(u_1+u_2)m^2-i\epsilon\right]^3} \\
&= \int_0^1 du_1 \int_0^{1-u_1} du_2 \int_{-\infty}^{+\infty}dz \frac{2}{\left[-z^2+g(u_1,u_2,x)^2-i\epsilon\right]^3} \tag{2}
\end{align}$$

In the final expression above, note that no matter if $g^2$ is positive or negative, the triple-poles will always lie in the $2^{\textrm{nd}}$ and $4^{\textrm{th}}$ quadrants.

enter image description here

Since no poles $1^{\textrm{st}}$ and $3^{\textrm{rd}}$ quadrants, we may rotate the contour counter-clockwise by $90^\circ$, also known as Wick-Rotation. We now have the following expression:

$$f(x)=\int_0^1 du_1 \int_0^{1-u_1} du_2 \int_{-\infty}^{+\infty} i dz \, \frac{1}{\left[z^2+g(u_1,u_2,x)^2 -i\epsilon\right]^3} \tag{3}$$

Let's now reverse all the previous algebraic steps, but using the Wick-rotated expression. This will finally give us (compare with (1)):

$$f(x)=\int_{-\infty}^{+\infty}\frac{d(iz)}{\left(z^2 + m^2 -i\epsilon\right )^2 \left(-(iz-x)^2+ M^2 -i\epsilon\right)} \tag{4}$$

But this is none other than the Wick-rotated version of the original integral! This is explicitly telling us that:

$$\int_{\Gamma_1} I(z) dz= \int_{\Gamma_2} I(z) dz \tag{5}$$

where $\Gamma_1$ is the contour along the real axis, $\Gamma_2$ is the contour along the imaginary axis, and $I(z)$ is the integrand in (1). Note that this is expression was derived by @QMechanic in the PSE post mentioned previously. The essence of this derivation is found in most introductory textbooks on Quantum Field Theory.

On the other hand, if we tried to smoothly rotate the original contour by $90^\circ$, we would unavoidably pick up the residue of the pole in the $1^{\textrm{st}}$ contour.

$$\int_{\Gamma_1} I(z) dz= 2\pi i \,\textrm{Res} \left(I(z);z=x-M-i\epsilon \right)+ \int_{\Gamma_2} I(z) dz \tag{6}$$

Below you can see an illustration of this intuitive equation.

enter image description here

Combining (5) and (6), we see that the residue must be zero. But we can calculate this residue and show it isn't zero, a contradiction!

$$\textrm{Res} \left(I(z);z=x-M-i\epsilon \right)=\frac{1}{2M\left(-(x-M)^2+m^2\right)}\overset{?}{\neq} 0 \tag{7}$$

So what's going on here? How can I resolve this contradiction?

Edit 1: A possible resolution.

I've just realized an issue in the algebraic steps. In (2), as per @Maxim's comment, I shifted $z\rightarrow z+(1-u_1-u_2)x$ in order to remove the term linear in $z$. This was fine, since $x$ was real-valued.

However when reversing these steps later, going from (3) to (4), the equivalent shift would be $z\rightarrow z+i(1-u_2-u_2)x$. This is not valid since for arbitrary $x$ the shift may hop over one of the poles! The resulting shifted-contour would not be homotopic to the original one! For it to be valid we would need $x$ to be imaginary.

If we have analytic continuation in mind, we could simply assume at this step that $x$ is actually purely imaginary. But then our original shift $z\rightarrow z+(1-u_1-u_2)x$ would not be valid for the same reason!

Best Answer

A quick numerical experiment in python gives no reason to believe that the integrals (1) and (4) should give different results.

from scipy.integrate import quad, nquad
import cmath

def integrand( z, m, M, eps, x, which ):
    ie = complex(0,1)*eps
    tmp1 = (-z**2 + m**2 - ie)**2
    tmp2 = (-(z-x)**2 + M**2 - ie)
    if which == "re":
       return( (1/tmp1/tmp2).real )
    else:
       return( (1/tmp1/tmp2).imag )

def integrand_wick( z, m, M, eps, x, which ):
    ie = complex(0,1)*eps
    iz = complex(0,1)*z
    i = complex(0,1)
    tmp1 = (z**2 + m**2 - ie)**2
    tmp2 = (-(iz-x)**2 + M**2 - ie)
    if which == "re":
       return( (i/tmp1/tmp2).real )
    else:
       return( (i/tmp1/tmp2).imag )


 m = 1.0
 M = 1.0
 eps = 1.0
 x = 1.0

 print( quad( integrand, -20, 20, args = ( m, M, eps, x, "re" ) ) )
 print( quad( integrand, -20, 20, args = ( m, M, eps, x, "im" ) ) )
 print( quad( integrand_wick, -20, 20, args = ( m, M, eps, x, "re" ) ) )
 print( quad( integrand_wick, -20, 20, args = ( m, M, eps, x, "im" ) ) )

The output is

(-0.49033163694725157, 5.153607661012333e-09)
(-0.35979795020946703, 3.859702046164447e-09)
(-0.49033176263804457, 4.841029106680306e-11)
(-0.35979807319732326, 4.2856183711173664e-11)

The very small numbers are the accuracies achieved by quad.

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