Greens Functions in Birrell and Davies "QFT in Curved Space" book I am studying QFT in curved spacetimes from Birrell's and Davies' book. I am trying to derive the expressions for the Green's functions for a scalar field in flat space. My attempt, according to the insightful answer to a question I have made previously (Green Function expressed in terms of Hankel function (of the second kind)), has taken me to a final expression for the Greens function for a massive scalar following the Feynman prescription,
$$G_F(x,x')=-
\frac{i\pi}{(4\pi i)^{n/2}}\bigg(\frac{2m^2}{-\sigma+i\epsilon}\bigg)^{(n-2)/4}
\int_{-\infty}^0\frac{dt}{t^{n/2}}\ e^{\frac{1}{2}[2m^2(\sigma-i\epsilon)]^{1/2}(t-1/t)}$$
from which I suppose that the integral can be identified as the integral representation of Hankel of the second kind. This completes the steps in deriving the expression, but it seems to me that it requires a leap of faith from my part, since $t$ is supposed to be a variable whose possible values are restricted on the negative real axis! On the other hand, the contour in the integral representation of the Hankel function of the second kind starts at $-\infty$, passes from $-i$ and then goes to zero! So, if anyone could somehow explain why I am allowed to identify the integral as the Hankel function, despite the fact that
$$H_{\frac{1}{2}n-1}\{[2m^2(\sigma-i\epsilon)]^{1/2}\}
=\frac{1}{\pi i}\int_{C}\frac{dt}{t^{n/2}}\ 
e^{\frac{1}{2}[2m^2(\sigma-i\epsilon)]^{1/2}(t-1/t)}$$
would be nice. In the equation above $C$ is the contour I have described earlier.
This is a tiny little detail I couldn't find in any book on mathematical methods I searched for. However, my real problem lies in attempting to derive the expression for the massless propagator in $n=4$ dimensions. My attempt is as follows:
$$D_F(x,x')=\int\frac{d^4k}{(2\pi)^4}\frac{e^{i\vec{k}\cdot(\vec{x}-\vec{x}')-ik^0(t-t')}}
{(k^0)^2-|\vec{k}|^2+i\epsilon}=
-i\int_0^{\infty}dte^{i\epsilon t}\int\frac{d^4k_E}{(2\pi)^4}
e^{-t[k_E^2-ik_E\cdot(x_E-x_E')/t]}$$
and upon performing the momentum integral, we conclude that
$$D_F(x,x')=-\frac{i}{(4\pi)^2}
\int_0^{\infty}\frac{dt}{t^2}e^{i\epsilon t+\sigma/2t}.$$
My questions are the following:

*

*How does this expression yield the desired result $$D_F(x,x')=(i/8\pi^2\sigma)-(1/8\pi)\delta(\sigma)~?\tag{2.78}$$ I have tried neglecting $i\epsilon$ prescription (just out of curiosity) and I obtain the first contribution + diverging corrections from the lower bound of the $dt$ integral of the form
$$\lim_{\delta\to0}\frac{1}{\sigma}e^{\sigma/2\delta}$$
which is not exactly a $\delta$ function in any (known to me) way...

*Is there a way of setting $m^2=0$ and $n=4$ in the expression for the Greens function in the massive case, and recover the desired result?

*If we chose to work with performing the countour integrals, how would the second part occur? Is there some trick I am missing (I will redo the calculations on that part, so feel free to skip this third question if it is not something quick)?

Any help will be appreciated.
P.S.: I realise that some of the info I am asking for can be found in "mathematical methods in physics books". If there are any suggestions, please let me know.
 A: Well, let's see.

*

*There is evidently an implicitly written Cauchy principal value in eq. (2.78). After trivially shifting $x^{\prime}=0$ and using the Sokhotski-Plemelj theorem we can rewrite the 3+1D Feynman propagator (2.78) as
$$ iD_F(\vec{r},t_M)~=~\frac{1}{8\pi^2}\frac{1}{-\sigma+i\epsilon}
~=~\frac{1}{4\pi^2}\frac{1}{r^2-t_M^2+i\epsilon}, \qquad 2\sigma~:=~t_M^2-r^2,\tag{2.78'}$$
where $\epsilon>0$ is an infinitesimally small regulator.


*The Euclidean Greens functions can e.g. be found via Fourier transformation. The Euclidean 4D massless Greens function
$$ D_E(\vec{r},t_E) ~=~ \frac{1}{4\pi^2}\frac{1}{t_E^2+r^2 } \tag{E}$$
is a limit of the Euclidean massive Greens function (which is a modified Bessel function of the second kind).


*Wick rotation $$t_E~=~e^{i(\frac{\pi}{2}-\epsilon)} t_M~=~i(1-i\epsilon)t_M~=~ (i+\epsilon)t_M\tag{W}$$ yields
$$ D_E(\vec{r}, (i+\epsilon)t_E)~\stackrel{(E)+(W)}{=}~\frac{1}{4\pi^2}\frac{1}{r^2-t_M^2+ i\epsilon~ {\rm sgn}(t_M)}, \tag{EW}$$
which matches the Feynman propagator (2.78') for $t_M>0$.
A: Minor complementary answer to the well informative answer posted by @Qmechanic. The only minor detail that was not so clear to me after I read the answer posted by @Qmechanic was how to derive the expression for the Feynman propagator of a massless scalar in $n=4$ dimensions. I have found that and hence I post an overview of the steps for completeness sake...

*

*Start with the expression all the Green functions obey
$$\mathcal{G}(x,x')=\int\frac{d^nk}{(2\pi)^n}\frac{e^{i\vec{k}\cdot(\vec{x}-\vec{x}')-ik^0(t-t')}}{(k^0)^2-|\vec{k}|^2-m^2}$$
and set $m=0$ and $n=4$.


*Use the Feynman prescription s.t. the Feynman propagator emerges (i.e. add $+i\epsilon$ to the denominator)
$$D_F(x,x')=
\int\frac{d^4k}{(2\pi)^4}\frac{e^{i\vec{k}\cdot(\vec{x}-\vec{x}')-ik^0(t-t')}}{(k^0)^2-|\vec{k}|^2+i\epsilon}$$


*Use the identity
$$\frac{i}{k^2+i\epsilon}=\int_0^{\infty}dte^{it(k^2+i\epsilon)}$$
and then change the perform the momentum integral first by completing the square.


*The result should be
$$D_F(x,x')=\frac{-1}{(4\pi)^2}\int_0^{\infty}\frac{dt}{t^2}
e^{-\epsilon t}e^{-i\sigma/2t}$$
where $\sigma=\frac{1}{2}(x-x')^2$. The result can be re-written (by performing the substitution $u=1/t$) as
$$D_F(x,x')=\frac{-1}{(4\pi)^2}\int_0^{\infty}du e^{iu(-\sigma/2+i\epsilon)}$$
if we substitute $\epsilon/u^2$ with $\epsilon$.


*Performing the integral and using the Sokhotski-Plemelj theorem yields the correct result
$$D_F(x,x')=\frac{i}{8\pi^2\sigma}-\frac{1}{8\pi}\delta(\sigma)$$
Thank you @Qmechanic for all the useful tips!
