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Let \(\Delta^\text{ret}(x-y), \Delta^\text{adv}(x-y), \Delta_F(x-y)\) and be the retarded, advanced Green functions and Feynman propagator respectively. Using Fourier transform, the three Green functions can be written as contour integral \[\frac{1}{(2\pi)^4}\oint_C \frac{d^4k e^{ik(x-y)}}{(k^2-m^2)}\] The \(k_0\) integral is the first integral to be evaluated by the method of contour integration in complex \(k_0\) plane. The point marked on the real line are the poles of the integrand at \(k_0=\pm\sqrt{\vec{k}^2+m^2}\). Four contours are shown in the figure below. These contours are to be suitably closed to form the contour \(C\) for evaluation of the Green function. Identify the contours corresponding to each of the three the advanced, retarded and Feynman propagators. \\ \FigBelow{10,10}{120}{30}{FourContours}{} Give brief reasons in support of your answer. In each case draw the closed contour \(C\).

A Quick Answer is given separately. See scan of hand written sheets. Detailed explanation of essential points of the solution is given below.

Main Steps in Solution :: Some details
In this problem \(k_0\) integration is to be carried out using the method of
contour integration in complex plane. I will explain and write the solution in
several steps.
Write out the expression in full glory.

The integral to be evaluated, apart from an overall constant is
\begin{equation}
\int d^3k \int_{-\infty}^\infty dk_0 \frac{e^{ik_0x^0-
\vec{k}\cdot\vec{x}}}{k_0^2- \omega^2}, \qquad \omega\equiv +\sqrt{\
\vec{k}^2-m^2}.
\end{equation}

Stare at the expression

First of all it is an improper integral because the lower limit is
\(-\infty\) and upper limit is \(+\infty\). An integral where one, or both, the
limits go to infinity,is an {\it improper integral}.

An integral is also an improper integral the function becomes
infinite at one or more points in the range of integration.
The improper integrals are defined through a limiting procedure.
Sometimes an improper integrals can be given well defined, unique, meaning
though a limiting procedure. In such a case we say that the integral exists as
an improper integral. For example \(\int_0^1\frac{dx}{\sqrt{x}}\) is a well
defined improper integral and has a value \(\sqrt{2}\). In general
\(\int_0^1 \frac{dx}{x^c}\) exists when \(-1<c<0\). For positive \(c\) there is
the integral exists and
no problem any way.
An integral which does not exist as a improper integral, is a singular
integral.
The given integral is a singular integral. The integrand blows up at
\(k_0=\pm\omega\) and these points fall inside the range of integration.
As a result the integral is does not exist. Also that the answer
depends on how we handle the integral. }
In general there will be several ways of defining a singular integral by
prescribing a method of avoiding the singularity.

Remember any such method will
be one definition and different methods will give different answers.

One such method is use contour integration in complex plane.
Given integral as contour integral in complex

\(k_0\) plane.}
Any real integral \(\int_a^b f(x) dx\) can be trivially written as a an
integral in complex plane
\begin{equation}
\int_a^b f(x) dx = \int_\gamma f(z) dz
\end{equation}
where \(\gamma\) is real line interval from \(a\) to \(b\).
So we write our given integral as
\begin{equation}
\int_\gamma dk_0 \frac{e^{ik_0x^0 - \vec{k}\cdot\vec{x}}}{k_0^2-\omega^2}
\end{equation}
where the contour \(\gamma\), in complex \(k_0\)plane, is to be chosen so as
to avoid the singularity.

Statement of the problem

In the given problem some choices are shown, there
are many more possibilities.
Also do remember that infinite contours are to be handled by
suitable limiting procedure.
If we choose one of the four contours and try to evaluate the integral, we
shall get well defined answer.{\tt of course different answers for different
cases.

Choose any one of the four contour

Suppose we have chosen one of four contours, we can compute the \(k_0\) 

integral. This is done by making use of Cauchy residue theorem. That requires
us to close the contour first before we apply residue theorem.
In this problem we first truncate the given contour from \(-R \) to \(+R\) and
take the limit \(R\to\infty\) at the end.

Next we add a semicircle of radius \(R\) with centre at the origin. We can do
this in two ways. Take the semi-circle in upper half plane or lower half plane.
Having closed the contour by choosing semicircle one way or other, we then
apply the residue theorem.

Which semi-circle?

In this class of problems with exponentials, the choice of contour is dictated
by the argument of the exponential which in our case is \(ik_0x^0\) and
and \(k_0\) is complex. So let us write it as a sum of real and
imaginary parts \(k_0=k_1+i k_2\).

Case I :: Assume \(x_0>0\)\\}
Then we get
\begin{equation}
\exp(ik_0 x^0)= \exp( ik_1x_0 - k_2x_0).
\end{equation}
{As \(R\to\infty\), \(k_2\to \infty\) in the upper half plane and the exponential will go to zero, because \(k_2\) is positive. }
So will the integral along the semicircle go to zero as \(R\to \infty\)
Case II :: Assume \(x_0<0\)\\}
Then we get
\begin{equation}
\exp(ik_0 x^0)= \exp( ik_1x_0 - k_2x_0)=\exp( ik_1x_0 + k_2|x_0|) .
\end{equation}

As \(R\to\infty\), \(k_2\to \infty\) in the lower half plane
\(k_2\) negative and large. Therefore the exponential will go to zero. 
So will the integral along the semicircle go to zero as \(R\to
\infty\)