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This is Problem (6.1) from Schwartz's QFT and the Standard Model. I am trying to directly calculate, by performing the integral over momenta, the explicit position-space expression of the Feynman propagator. That is, performing the integral

\begin{equation} D_{F} = \int \, \frac{d^4 k}{(2 \pi)^4} \, \frac{i}{k^2 - m^2 + i \varepsilon} e^{i k (x_{1} - x_{2})}. \end{equation} I know that one can quote the solution that this equals some function including a modified Bessel function of the second kind as

\begin{equation} D_{F} = \frac{m}{4 \pi^2 r + i \varepsilon} K_{1} (mr), \end{equation} for spacelike separations, where $r = \vec{x}_{1} - \vec{x}_{2}$.

You are asked, after working this expression out, to take the massless limit, which evaluates to

\begin{equation} \lim_{m \rightarrow 0} D_{F} = \frac{1}{4 \pi^2 r^2 + i \varepsilon}. \end{equation} I understand the logic in these posts (The poles of Feynman propagator in position space, How to obtain the explicit form of Green's function of the Klein-Gordon equation?) in how to manipulate the integral to get it into an integral representation of $K_{1}$, but Question 1 would be how is it that this limit arrived at? I have looked at some of the different representation of $K_{1}$ and am still unsure how this limit is calculated. My shaky first thoughts is that it approaches $\frac{1}{mr}$, and that you just say $mr(i \varepsilon) = i \varepsilon$ as this $\varepsilon$ is only used for keeping track of contours, etc., and will be taken to zero. This is what seems to be occurring in the top answer in this post: Propagator of a scalar in position space. $K_{1}$ does seem to visually appear like a $1/r$ type function when plotted.

My other question (I will be happy to move this to separate post if inappropriate) involves an apparently different derivation of the above, some of the steps of which I would appreciate being explained, or hinted at. For completeness, I want to derive the dependence on $K_{1}$, not just state it and take the limit:

enter image description here

Question 2: Specifically, I don't quite understand how the integral over $d k^{0}$ is being performed from the second the third line (have we transformed to a frame in which $\vec{k} = 0$?). The question mentions use of Schwinger parameters, which I think is perhaps treated in the linked posts above, but I'm not sure as they don't seem to be in the form stated in Schwartz.

I'm a bit unsure on the last couple of steps but I assume that I'll find some integral trig representation of $K_{1}$ somewhere that'll match up.

Question 3: Why can we just invoke Lorentz symmetry here to reinstate the rest of the coordinate dependence? Is this just because we transformed frames to ease calculation?

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    $\begingroup$ As far as your first question is concerned, there are a lot of sources online about limiting behavior of special functions. Indeed $K_1 (z) \sim 1/z$ for small $z$ and therefore $mK_1(mr)\rightarrow 1/r$ for $m\rightarrow 0$. I think that recalling that the $+i\varepsilon$ prescription is just a regulator will also clear up any confusion on why is the denominator $4\pi^2 r^2 +i\varepsilon$ and not $4\pi^2 r^2 +i\varepsilon r$. $\endgroup$
    – mb28025
    Commented Dec 7, 2023 at 20:20
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    $\begingroup$ As far as your second question goes, try this. It really goes through all the steps and explicitly states many details that are usually assumed the reader is aware of. $\endgroup$
    – mb28025
    Commented Dec 7, 2023 at 20:52
  • $\begingroup$ A lot of mathematical facts are simply noticing of relations. For example, the integral representation of special functions might simply be taken as their definitions, and then shown to solve the differential equations, etc. Or some other equally tenuous path. There seems to be a sign mistake in the quoted solutions when promoting to Lorentz invariance. As for why promotion works, you have to convince yourself when you study tensors that specific functional form dependences is a thing. $\endgroup$
    – naturallyInconsistent
    Commented Dec 8, 2023 at 5:58
  • $\begingroup$ Thank you @mb28025 for the resources; they look very helpful. $\endgroup$
    – agfdsa
    Commented Dec 11, 2023 at 16:06
  • $\begingroup$ @naturallyInconsistent Would you be able to elaborate at all on your last sentence? Do you refer to the equal treatment of spacetime coordinates, sort of like space- and time-like coordinates are treated equivalently in the KG equation (at least in terms of order of derivatives) in contrast to the Schrodinger equation? So then you enforce identical dependence in the time-coordinate when now taking $\Delta t \neq 0$ as you have just derived in the spatial coordinates? $\endgroup$
    – agfdsa
    Commented Dec 11, 2023 at 16:13

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