Clarifying differences between Feynman prescriptions (i.e., pole shifts) and contour deformations

Question

Summary and question

I am working to clarify my understanding of differences (if any) between the so-called Feynman prescription and contour deformations to evaluate improper integrals. For concreteness, consider the integral $$ J=\int_{-\infty}^{\infty}\phi(x)dx =\int_{-\infty}^\infty \frac{xe^{iAx}}{(x+k)(x-k)}dx, $$ with $A>0$. For context, I encountered integrals of this form when deriving the Green's function for the Helmholtz equation in three dimensions. In such a derivation, it is common to use contour integration in conjunction with the Feymnan prescription to bump the poles off of the real axis. A seemingly equivalent approach would be to use contour integration in conjunction with contour deformations to detour out and around the poles on the real axis. The top-rated answer posted to a related question even states these are equivalent so long as the deformations and pole shifting occur in corresponding directions (link). However, this is seemingly not the case as discussed in another related post (link) and as I'll summarize below.

My question is two-fold:

1. Do I have it correct that the contour deformation method will evaluate the principal value $p.v.J$ (regardless the choice of deformation orientations), while the Feynman prescription method (without symmetric averaging) will evaluate $p.v.J$ up to some residue terms as in the Sokhotski-Plemelj Theorem? If so, this seems to contradict various statements in literature about the equivalence of "contour detours" and "pole shifting".
2. In the context of Green's functions and causality, what intuition would lead us to choose Feynman prescriptions over contour deformations during derivations? Or is it simply hind-sight? I ask this because only the Feynman prescription method allows for recovery of the (causal) Helmhotz Green's function $G(r)=\frac{1}{4\pi r}e^{ikr}$.

Contour deformations

Consider the contour integral $$ \int_{C}\frac{ze^{iAz}}{(z+k)(z-k)}dz, $$ where $C=\cup_{i=1}^6C_i$ with $$ \begin{aligned} C_1&=\{z\in\mathbb{C}: z=x+0i, x\in(-R,-k-\epsilon)\},\\ C_2&=\{z\in\mathbb{C}: z=-k+\epsilon e^{i\theta}, \theta\in[\pi, 0]\},\\ C_3&=\{z\in\mathbb{C}:z=x+0i, x\in(-k+\epsilon, k-\epsilon)\},\\ C_4&=\{z\in\mathbb{C}: z=k+\epsilon e^{i\theta}, \theta\in[-\pi, 0]\},\\ C_5&=\{z\in\mathbb{C}: z=x+0i x\in(k+\epsilon, R)\},\\ C_6&=\{z\in\mathbb{C}: z=Re^{i\theta}, \theta\in[0,\pi]\}. \end{aligned} $$ This contour closes in the upper half-plane and contains the pole at $z=k$. Taking $R\to\infty$ and $\epsilon\to 0$, invoking appropriate theorems (Jordan's lemma, dominated convergence, residue) and evaluating each contour segment we obtain: $$ \begin{aligned} 2\pi i\text{Res}\left(\phi(z), z=k\right) &= \lim_{\substack{R\to\infty\\\epsilon\to 0}}\int_{C}\phi(z)dz\\ &=\lim_{\substack{R\to\infty\\\epsilon\to 0}}\int_{C_1} + \int_{C_2} + \int_{C_3} + \int_{C_4} + \int_{C_5} + \int_{C_6}\\ &\Updownarrow\\ \pi i e^{iAk} &= p.v.J -\pi i\frac{e^{-iAk}}{2} + \pi i\frac{e^{iAk}}{2} + 0. \end{aligned} $$ Rearranging terms and simplifying we get $$ p.v.J = \pi i \cos(Rk). $$ Note, this result should be (we've checked a few) independent of the orientation of the detour contours $C_2, C_4$ as the residue term on the LHS will compensate for the different choices.

Feynman prescription (pole shifts)

Now consider the contour integral $$ \int_{C}\phi_\epsilon(z)dz := \int_{C}\frac{ze^{iAz}}{(z-(-k-i\epsilon))(z-(k+i\epsilon))}dz, $$ where the contour $C=C_1\cup C_2$ with $$ \begin{aligned} C_1 &= \{z\in\mathbb{C}: z=x+0i, x\in(-R,R)\},\\ C_2 &= \{z\in\mathbb{C}: z=Re^{i\theta},\theta\in[0,\pi]\}. \end{aligned} $$ This contour closes in the upper half-plane and due to the shift of the poles contains only the pole at $z=k+i\epsilon$. Taking $R\to \infty$ and invoking appropriate theorems (Jordan's lemma, residue) we evaluate $$ \begin{aligned} 2\pi i\text{Res}(\phi_\epsilon(z),z=k+i\epsilon) &= \int_{C}\phi_\epsilon(z)dz\\ &=\int_{C_1} + \int_{C_2}\\ &\Updownarrow\\ \pi i e^{iA(k+\epsilon i)} = \int_{-\infty}^{\infty}\phi_\epsilon(x)dx. \end{aligned} $$ Taking the limit $\epsilon\to 0$ (and blindly assuming dominated convergence holds) we obtain $$ J = \pi i e^{iAk}. $$

Note, this approach is essentially the idea of the Sokhotski-Plemelj theorem. As such, taking a symmetric pole shift ($k\to k-\epsilon i$ and $ -k\to -k + \epsilon i$) and averaging resulting values would recover $p.v.J$.

Answer 1

I've come across an article that resolves my question: Schmalz et. al, On the derivation of the Green function for the Helmholtz equation using generalized functions. To fully see how this is applicable to my question, I should provide a little more context.

I was originally deriving the 3D Helmholtz Green's function and (erroneously) trying to indicate how the choice of contour would give varying boundary conditions similar to selecting different pole-shifts in the Feynman prescription. I'm now convinced that this is not the case, i.e., pole-shifting and contour-detours are not equivalent. Regardless the choice of contour detours (enclosing or excluding poles) and correctly accounting for for all contour contributions and residue(s), the result will be the same---the principal value $p.v.J$. This would be the same as symmetrically averaging different pole shifts (or Feynman prescriptions) and thus not equivalent to a single pole-shift.

Derivation of $G$ using homogeneous and particular solutions

The above linked article performs exactly the rigorous derivation I was attempting. The primary difference between their derivation and mine, is their use of the homogeneous and particular solutions to the PDE whose sum satisfies both the boundary conditions and inhomogeneous RHS simultaneously.

The particular solution $G_p$

In brief, the authors derive the particular solution to the PDE $$ \nabla^2G_p(\mathbf{r},\mathbf{r}') + k^2G_p(\mathbf{r},\mathbf{r}') = \delta(\mathbf{r}-\mathbf{r}'), $$ in exactly the same manner as I was, i.e., taking the spatial Fourier transform of the equation, solving for $\widehat{G}$ and then taking the inverse Fourier transform via spherical coordinates and ultimately contour integration. Their result: $$ G_p = \frac{1}{4\pi}\cos(k(\mathbf{r}-\mathbf{r}')), $$ which is the principal value of the associated contour integral (there is an extra component of the integrand from what I have posted in the original question, but the concept is the same).

The homogeneous solution $G_h$

The authors then prove and derive the form of the homogeneous solution satisfying $$ \begin{aligned} &\nabla^2 G_h(\mathbf{r},\mathbf{r}') + k^2G_h(\mathbf{r},\mathbf{r}') = 0.\\ \end{aligned} $$ Without diving into the details and instead appealing only to intuition, one can take the Fourier transform of the homogeneous equation above and observe that $(-|\boldsymbol{\xi}|^2 + k^2)\hat{G}_h = 0$, and thus $$ \widehat{G}_h = C(\boldsymbol{\xi})\delta(k^2-|\boldsymbol{\xi}|^2), $$ where $\boldsymbol{\xi}$ denotes the spatial Fourier variable, and $C(\boldsymbol{\xi})$ possibly depends on orientation on the spherical shell $k^2 = |\boldsymbol{\xi}|^2.$ Furthermore, the authors are able to show $C(\boldsymbol{\xi})\equiv C$ for some constant $C$ and thus recovering $G$ from $\widehat{G}$ is straightforward.

Summing the results

Finally, adding together the homogeneous and particular solutions they obtain $$ G(\mathbf{r},\mathbf{r}') = \frac{1}{4\pi}\cos(k(\mathbf{r}-\mathbf{r}')) + C\frac{1}{4\pi}\sin(k(\mathbf{r}-\mathbf{r}')), $$ which gives an added degree of freedom (namely, $C$) to satisfy appropriate boundary conditions (e.g., Sommerfeld radiation for causality).

My key takeaways

Observing the authors derivation, it becomes clear that a single regularization to the improper integral is not apriori enough to satisfy both the inhomogeneous RHS of the PDE as well as the Sommerfeld radiation condition (causality). In the particular regularization given by the Feynman prescription of shifting the positive pole into the positive half-plane and the negative pole to the ngative half-plane, it turns out to be sufficient and the constant $C$ associated with the homogeneous solution would then be $C=0$ (the authors discuss this point in Section IV of their paper Schmalz et. al). Other choices of regularizing this integral can be made, but we may then require a non-zero contribution from the homogeneous solution to satisfy the given boundary conditions.

Finally, to answer my questions:

1. Contour detours and Feynman prescriptions (i.e., pole-shifting) are two different methods of regularizing an improper integral but are not equivalent.
2. I believe a lot of literature regarding Green's functions derivations overlooks the difference between these two regularization principals, erroneously assuming they are equivalent when minor comments are made about the contour detours and pole shifts. The Feynman prescription approach is objectively easier, recovers the correct solution so long as the pole-shifts are selected correctly, and avoids the complexity of the homogeneous solution contribution ($\delta$-shells and all). Presently I'm not convinced there is any physical intuition in choosing this approach aside from hind-sight.