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I want to find the Feynman Green's function of the D'Alembertian operator but I get stuck at one point.

$G(x-y)$ satisfies

$$\square_xG(x-y)=\delta^{(4)}(x-y),\tag{1}$$ where $\square_x=\eta_{\mu\nu}\frac{\partial}{\partial x^\mu}\frac{\partial}{\partial x^\nu}$ and $\eta_{\mu\nu}=\text{diag}(1,-1,-1,-1)$.

Fourier-transforming the equation, using that

$$G(x-y)=\int\frac{d^4k}{(2\pi)^4}\tilde{G}(k)e^{ik_\mu(x^\mu-y^\mu)}\tag{2}$$

I get

$$-\int\frac{d^4k}{(2\pi)^4}k_\mu k^\mu\tilde{G}(k)e^{ik_\mu(x^\mu-y^\mu)}=\int\frac{d^4k}{(2\pi)^4}e^{ik_\mu(x^\mu-y^\mu)},\tag{3}$$

so (calling $k^0\equiv\omega$ and $|\vec{k}|\equiv k$)

$$\tilde{G}(k)=\frac{-1}{k_\mu k^\mu}=\frac{-1}{\omega^2-k^2}=\lim_{\epsilon\to0}\frac{-1}{[\omega-(k-i\epsilon)][\omega+(k-i\epsilon)]}.\tag{4}$$

Now I replace this in $(2)$

$$G(x-y)=\lim_{\epsilon\to0}\frac{-1}{(2\pi)^4}\int d^3\vec{k}~e^{-i\vec{k}\cdot(\vec{x}-\vec{y})}\int\limits_{-\infty}^\infty d\omega\frac{e^{i\omega(x^0-y^0)}}{[\omega-(k-i\epsilon)][\omega+(k-i\epsilon)]}\tag{5}.$$

Doing the integral in $\omega$ using residues and setting $\epsilon\to0$ I get

$$\begin{align} G(x-y)=&\frac{i}{(2\pi)^3}\int d^3\vec{k} \frac{e^{-ik|\vec{x}-\vec{y}|\cos\theta}}{2k}e^{ik|x^0-y^0|}=...= \\\\ =&\frac{-1}{4\pi|\vec{x}-\vec{y}|}\int\limits_0^\infty \frac{dk}{2\pi}\Big(e^{-ik|\vec{x}-\vec{y}|}-e^{ik|\vec{x}-\vec{y}|}\Big)e^{ik|x^0-y^0|}\tag{6} \end{align}$$

I know the answer involves Dirac deltas, but I can't see how to extend this integral from $-\infty$ to $\infty$, maybe I'm doing something wrong. Any help would be appreciated.

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  • $\begingroup$ Which Green's function? Which boundary conditions? The Feynman Green's function? $\endgroup$
    – Qmechanic
    Oct 20, 2021 at 19:13
  • $\begingroup$ yes, Feynman Green's function $\endgroup$
    – AFG
    Apr 15, 2023 at 16:25
  • $\begingroup$ Related: physics.stackexchange.com/q/614754/2451 $\endgroup$
    – Qmechanic
    Apr 17, 2023 at 10:13

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