D'Alembert operator Green function in arbitrary dimension I am interested to learn about the Green function for the D'Alembert operator in arbitrary dimensions. While searching through the web I came across the following document:
https://math.dartmouth.edu/~ahb/notes/waveequation.pdf
There the fundamental solution for the general Green function is given in eq. (14). Special cases for 3, 2 and 1 dimensions are given further below. Now, I was trying to start with eq. (14) and reduce it to the 3 dimensional result eq. (15) by setting $d=3$ and taking the appropriate limit - but I do not get the same result. In fact, restricting to the imaginary part I actually get zero in the limit $\epsilon\rightarrow 0^+$. It would be great if someone could demonstrate how to do this. Thanks for any suggestion!
 A: The general expression given in the paper as eq. (14) for the wave propagator $G_n(x)$ in $n$ spacetime dimensions is, denoting $x=(x^0,\mathbf x)$ and $x^2=(x^0)^2-|\mathbf x|^2$,
$$\begin{align}
G_n(x) &= \frac{\Gamma(n/2)}{(n-2)\pi^{n/2}}\lim_{\epsilon\to0^+} \mathrm{Im}[-(x^0-i\epsilon)^2+|\mathbf x|^2]^{1-n/2}\\
&=\frac{\Gamma(n/2)}{(n-2)\pi^{n/2}}\lim_{\epsilon\to0^+} \mathrm{Im}(-x^2+i\epsilon \,\mathrm{sign}(x^0))^{1-n/2}
\end{align}$$
For n=4 (i.e. three space dimensions),
$$
G_4(x)=\frac{1}{2\pi^2}\lim_{\epsilon\to 0^+}\mathrm{Im}\frac{1}{-x^2+i\epsilon \,\mathrm{sign}(x^0)}\,.
$$
Now, recall the distributional identity
$$
\lim_{\epsilon\to 0^+}\frac{1}{t\pm i\epsilon} =\mathrm{PV}\,\frac{1}{t}\mp i\pi\delta(t)\,,
$$
where $\mathrm{PV}$ is the Cauchy principal value. Thus,
$$
G_4(x)=-\frac{\delta(x^2)}{2\pi}\mathrm{sign}(x^0)\,.
$$
Using also 
$$
\delta(f(t))=\frac{\delta(t-f^{-1}(0))}{|f'(t)|},
$$
we arrive at
$$
G_4(x)=-\frac{\delta(x^0-|\mathbf x|)}{4\pi |\mathbf x|}+\frac{\delta(x^0+|\mathbf x|)}{4\pi |\mathbf x|}\,.
$$
This is, up to a sign, the next to last equation eq. (15).
