Explicit form of scalar propagator

[parton]

Hi!

I have encountered a little problem. I want to show
that the explicit form of the Feynman propagator for massless scalar fields is given by:

$$\begin{align}<br /> G_F(x) & = - \lim_{\epsilon \to +0} \int \dfrac{\mathrm{d}^{4}k}{(2 \pi)^{4}} \dfrac{1}{k^{2} + i \epsilon} \mathrm{e}^{- i k x} <br /> \\<br /> & = - \lim_{\epsilon \to +0} \dfrac{1}{4 \pi^{2}} \dfrac{1}{x^{2} -i \epsilon} <br /> \end{align}$$

And I would like to do that directly, i.e., without starting from the massive case and considering the limit $m \to 0$.

I found a script where one can find a derivation of that result,

http://mo.pa.msu.edu/phy853/lectures/lectures.pdf

on pages 58-59.

There, the integration is split up into the imaginary and real part.

But for the imaginary part, the author finds:

$$\begin{align}<br /> G_F,i(x) & = \dfrac{1}{4 \pi^{2} r} \int_{0}^{\infty} \mathrm{d}k \cos(k x_{0}) \sin(k r)<br /> \\<br /> & = \dfrac{1}{16 \pi^{2} r i} \int_{0}^{\infty} \mathrm{d}k \left[ \mathrm{e}^{ik(x_{0}+r)} - \mathrm{e}^{-ik(x_{0}+r)} + \mathrm{e}^{-ik(x_{0}-r)} - \mathrm{e}^{ik(x_{0}-r)} \right]<br /> \\<br /> & = - \frac{1}{8 \pi^{2} r} \left[ \dfrac{1}{x_{0} + r} - \dfrac{1}{x_{0} - r} \right]<br /> \\<br /> & = - \frac{1}{4 \pi^{2} x^{2}}<br /> \end{align}$$

Here, I don't understand how the integration is performed. I think if you integrate every exponential term it will diverge at $\infty$, but somehow the author ends up with a finite term.

Does anyone understand what is going on here?

Furthermore, I think we should end up with Cauchy's principal value and not just 1/x^2, right?

Maybe, someone has a more "elegant" way of deriving the massless propagator?

 

[dauto]

Make a change of variables k'=ik or k'=-ik as needed

 

[parton]

Make a change of variables k'=ik or k'=-ik as needed

Thank you for your hint, but somehow I don't see how that could help.
If I make the substitution $k' = i k$ I find:

$$\begin{align}<br /> \int_{0}^{\infty} \mathrm{d}k & \left[ \mathrm{e}^{ik(x_{0}+r)} - \mathrm{e}^{-ik(x_{0}+r)} + \mathrm{e}^{-ik(x_{0}-r)} - \mathrm{e}^{ik(x_{0}-r)} \right] =<br /> \\<br /> & = -i \int_{0}^{i \infty} \mathrm{d}k \left[ \mathrm{e}^{k(x_{0}+r)} - \mathrm{e}^{-k(x_{0}+r)} + \mathrm{e}^{-k(x_{0}-r)} - \mathrm{e}^{k(x_{0}-r)} \right]<br /> \\<br /> & = -i \left[ \dfrac{1}{x_{0} + r} \left( \mathrm{e}^{k(x_{0}+r)} + \mathrm{e}^{-k(x_{0}+r)} \right) - \dfrac{1}{x_{0} - r} \left( \mathrm{e}^{k(x_{0}-r)} + \mathrm{e}^{-k(x_{0}-r)} \right) \right] \Bigg|_{0}^{i \infty}<br /> \\<br /> & = -2 i \left[ \dfrac{1}{x_{0} + r} \mathrm{cosh}(k (x_{0} + r)) - \dfrac{1}{x_{0} - r} \mathrm{cosh}(k(x_{0}-r)) \right] \Bigg|_{0}^{i \infty}<br /> \end{align}$$

And this will diverge for $k \to i \infty$.

Do you have an idea to resolve this problem?

 

[Ravi Mohan]

Multiply integrand by $\exp(-\alpha k)$, integrate it and then take limit of $\alpha$ as zero.
$\alpha$ is real and positive.

 

[parton]

Multiply integrand by $\exp(-\alpha k)$, integrate it and then take limit of $\alpha$ as zero.
$\alpha$ is real and positive.

OK, thanks a lot . This seems to work.

But what I still not undestand is: Why can the result $- \dfrac{1}{4 \pi^{2} x^{2}}$ be undestood as Cauchy's principal value?